Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

SOLUTIONS MANUAL

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Taking another time-derivative: 𝑥̈ = 𝑥̈ cos 𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 cos 𝑥̈ sin 𝑥̈ − 𝑥̈2 cos 𝑥̈ sin 𝑥̈ + 𝑥̈𝑥̈ cos 𝑥̈ cos 𝑥̈+ — 𝑥̈𝑥̈ sin 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈ cos 𝑥̈ cos 𝑥̈ − 2𝑥̈𝑥̈ sin 𝑥̈ sin 𝑥̈ − 2𝑥̈𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ 𝑥̈ = 𝑥̈ sin 𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 sin 𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 sin 𝑥̈ sin 𝑥̈ + 𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈+ + 𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈𝑥̈ cos 𝑥̈ cos 𝑥̈ 𝑥̈ = 𝑥̈ cos 𝑥̈ − 2𝑥̈𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 cos 𝑥̈ Equating each 𝑥̈, 𝑥̈, and 𝑥̈ acceleration expressed in spherical coordinates with its respective acceleration terms gives us the equations of motion for the two-body problem in terms of spherical coordinates 𝑥̈, 𝑥̈, and 𝑥̈ 𝑥̈ cos 𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 cos 𝑥̈ sin 𝑥̈ − 𝑥̈2 cos 𝑥̈ sin 𝑥̈ + 𝑥̈𝑥̈ cos 𝑥̈ cos 𝑥̈+ — 𝑥̈𝑥̈ sin 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈ cos 𝑥̈ cos 𝑥̈ − 2𝑥̈𝑥̈ sin 𝑥̈ sin 𝑥̈ − 2𝑥̈𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈+ 𝑥̈ sin 𝑥̈ cos 𝑥̈ + =0 𝑥̈2 𝑥̈ sin 𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 sin 𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 sin 𝑥̈ sin 𝑥̈ + 𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈+ + 𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈ cos 𝑥̈ sin 𝑥̈ + 2𝑥̈𝑥̈𝑥̈ cos 𝑥̈ cos 𝑥̈ 𝑥̈ sin 𝑥̈ sin 𝑥̈ + =0 𝑥̈2 𝑥̈ cos 𝑥̈ 𝑥̈ cos 𝑥̈ − 2𝑥̈𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈ sin 𝑥̈ − 𝑥̈𝑥̈2 cos 𝑥̈ + =0 2 𝑥̈ √ where we used the fact that 𝑥̈ = 𝑥̈ = 𝑥̈2 + 𝑥̈2 + 𝑥̈2.

2

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 2 Prove that for the unperturbed two-body problem, orbital energy is constant.

Solution 2 Start with the vis-viva equation, Equation (2.50): 𝑥̈ = 𝑥̈ − 𝑥̈ 2

𝑥̈

To prove that energy is constant, we need to take its time derivative and show that it is equal to zero: 𝑥̈𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ = − 𝑥̈𝑥̈ 𝑥̈𝑥̈ ( 2 ) 𝑥̈𝑥̈ [(𝑥̈ 𝑥̈ 𝑥̈)1/2 ]

=

𝑥̈ 𝑥̈ 𝑥̈ + 𝑥̈ 𝑥̈ 𝑥̈ (

2

)

— 𝑥̈

1 — 𝑥̈−3 (2𝑥̈ 𝑥̈ 𝑥̈) [ 2

)

Recall that 𝑥̈ = 𝑥̈ = −𝑥̈𝑥̈ and 𝑥̈ = 𝑥̈, so 𝑟3 𝑥̈𝑥̈ 𝑥̈𝑥̈

= 𝑥̈ 𝑥̈ (

−𝑥̈ 𝑥̈ 𝑥̈ 3 )

𝑥̈𝑥̈ +

𝑥̈ 3

𝑥̈ 𝑥̈

𝑥̈𝑥̈ 𝑥̈𝑥̈ = −𝑥̈ 𝑥̈ ( 3 ) + 𝑥̈ 𝑥̈ 3 = 0 𝑥̈ 𝑥̈ Thus, 𝑥̈𝑥̈ = 0 which means that orbital energy is constant. 𝑑𝑡

3

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 3 Prove that the angular momentum vector and eccentricity vector are orthogonal to each other.

Solution In order to prove that two vectors are orthogonal, or perpendicular, to each other, one must show that their dot product is zero.

Starting with the definitions of eccentricity, Equation (2.27), 𝑥̈ =

𝑥̈ × 𝑥̈ 𝑥̈

—

𝑥̈ 𝑥̈

and angular momentum, Equation (2.41), 𝑥̈ = 𝑥̈ × 𝑥̈ we take the dot product between angular momentum and eccentricity, 𝑥̈ × 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ = 𝑥̈ 𝑥̈ 𝑥̈ − 𝑥̈ ) ( 1 1 = 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) (𝑥̈ × 𝑥̈) 𝑥̈ 𝑥̈ − 𝑥̈ 𝑥̈ where we used the definition of angular momentum for the second term. We then use the scalar triple product on the above equation, which, for three generic vectors 𝑥̈, 𝑥̈, and 𝑥̈ is 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) = 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) = 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) This helps us simplify the first term as 1 1 1 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) 𝑥̈ 𝑥̈ 𝑥̈ = 0 = = 𝑥̈ 𝑥̈ 𝑥̈ and the second term as 1 1 1 — 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) = − 𝑥̈ 𝑥̈ (𝑥̈ × 𝑥̈) = 𝑥̈ 𝑥̈ 𝑥̈ = 0 − 𝑥̈ 𝑥̈ 𝑥̈ which proves that 𝑥̈ 𝑥̈ 𝑥̈ = 0 and thus 𝑥̈ 𝑥̈ 𝑥̈ = 0.

4

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 4 Prove that for the unperturbed two-body problem, orbital motion is confined to a plane (this is the orbital plane).

Solution Planar motion means that the angular momentum (𝑥̈) is conserved. Thus, showing that the time derivative of 𝑥̈ is zero is sufficient to show that the motion is constrained to a plane. This derivation is shown in Section 2.3, and summarized here.

Taking the cross product of 𝑥̈ with Equation (2.9) gives 𝑥̈𝑥̈ 𝑥̈ × 𝑥̈ + 3 = ( 𝑥̈ ) 0 𝑥̈𝑥̈ 𝑥̈ × 𝑥̈ + 𝑥̈ × 3 = 0 𝑥̈ The second term is zero (cross product of a vector with itself), simplifying the above equation to 𝐫×𝐫 =0 which is equivalent to 𝑑 (𝐫 × 𝐫) = 𝑑 (𝐫 × 𝐯) = 𝑑𝐡 =0 𝑑𝑡 𝑑𝑡 𝑑𝑡 which proves that angular momentum is constant, and thus orbital motion is planar for the unperturbed two-body problem.

5

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 5 Show that the speed of a spacecraft in a circular orbit is √ 𝑥̈ 𝑥̈𝑥̈ = 𝑥̈ Solution Starting with the vis-viva equation (Equation (2.50)), 𝑥̈2 𝑥̈

=

−𝑥̈ 2𝑥̈

−

2 𝑥̈

we let 𝑥̈ = 𝑥̈ since a circular orbit has a constant radius, which is equal to its semimajor axis. This gives 𝑥̈2 𝑥̈ 2

=

−𝑥̈ 2𝑥̈

−

𝑥̈ 𝑥̈2 𝑥̈ 1 𝑥̈ = − 2 𝑥̈ 2 𝑥̈ Solving for 𝑥̈ gives √ 𝑥̈𝑥̈ =

6

𝑥̈ 𝑥̈

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 6 A spacecraft in Earth orbit passes through its apogee (apoapsis with respect to Earth) at a radius of 𝑥̈𝑥̈ = 9, 500 km, and with speed 𝑥̈𝑥̈ = 5.95 km/s. Determine how much time (in hours) elapses before the spacecraft returns to apogee.

Solution Computing this time means computing orbital period. From Equation (2.57), √ 𝑥̈3 𝑥̈ = 2𝑥̈ 𝑥̈ which means that we need to first compute the semimajor axis of the orbit. Using the vis-viva equation, Equation (2.50), 𝑥̈2 𝑥̈ −𝑥̈ — = 2 𝑥̈ 2𝑥̈ we solve for 𝑥̈ to get 𝑥̈ =

− 𝑥̈ 𝑥̈2 − 2𝑥̈ 𝑟

which is equivalent to Equation (2.61).

We can use the radius and speed that we are given, 𝑥̈𝑥̈ and 𝑥̈𝑥̈, to compute semimajor axis using the above equation, resulting in 𝑥̈ = 8, 216.31 km, which corresponds to an orbital energy of 𝑥̈ = −24.2566 𝑥̈𝑥̈2/𝑥̈2.

Using the equation for orbital period gives √ 3 𝑥̈ = 2𝑥̈ 𝑥̈ = 7, 411.8 𝑥̈ 𝑥̈ which corresponds to 8.05884 hours.

7

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 7 Starting with any known relationships, show that the orbital eccentricity can be expressed as √ 2𝑥̈𝑥̈ 𝑥̈ = 1 + 𝑥̈ where 𝑥̈ is semi-latus rectum, 𝑥̈ is orbital energy, and 𝑥̈ is the gravitational parameter.

Solution Starting with Equation (2.35) 𝑥̈ = 𝑥̈(1 − 𝑥̈2) and plugging Equation (2.50)

−𝑥̈

𝑥̈ = into it, results in Simplifying, 𝑥̈ =

2𝑥̈

−𝑥̈ 𝑥̈(1 − 𝑥̈2) 2𝑥̈

2𝑥̈𝑥̈ 𝑥̈ 𝑥̈2 =

= 𝑥̈2 − 1 2𝑥̈𝑥̈ 𝑥̈

+1

√ 2𝑥̈𝑥̈ 𝑥̈ = 1 + 𝑥̈ where we took only the positive root since 𝑥̈ > 0.

8

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 8 Radar measurements give the following data for a spacecraft in Earth orbit at some time 𝑥̈1: • 𝑥̈ = 8, 000 km • 𝑥̈ = −0.8201 km/s • 𝑥̈ = 5.635 × 10−2 deg/s Calculate: a Orbital energy, in km2/s2 b Semimajor axis, in km c Angular momentum, in km2/s d Semi-latus rectum, in km e Eccentricity f Radii of periapsis and apoapsis, in km g Orbital period, in hours h Flight-path angle at time 𝑥̈1 in rad Solution Note that the 𝑥̈ measurement must be converted into rad/s, which gives 𝑥̈ = 9.835 × 10−4 rad/s.

Computing the requested quantities gives: a Orbital energy: using Equation (2.46), 𝑥̈ = −18.5364 km2/s2 b Semimajor axis: using Equation (2.50), 𝑥̈ = 10, 751.8 km c Angular momentum: using Equation (2.47), 𝑥̈ = 62, 943.6 km2/s d Semi-latus rectum: using Equation (2.30), 𝑥̈ = 9, 939.52 km e Eccentricity: using Equation (2.35), 𝑥̈ = 0.27486 f Radii of periapsis and apoapsis: using Equation (2.37), 𝑥̈𝑥̈ = 7, 796.55 km and 𝑥̈𝑥̈ = 13, 707 km g Orbital period: using Equation (2.57), 𝑥̈ = 3.08198 hours h Flight-path angle (FPA) at time 𝑥̈1: FPA is the current spacecraft heading, i.e. the angle between its velocity and the local horizon, which can be computed as 𝑥̈ = ± arccos 𝑥̈ ( 𝑟𝑣 ) = −0.103858 rad. In order to resolve the ± in this equation, refer to Equation (2.74)

9

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 9 At time 𝑥̈ = 0, a spacecraft on an Earth orbit with 𝑥̈ = 50, 612 km and 𝑥̈ = 0.63174 is located at 𝑥̈ = 42.319𝑥̈. At what time did the satellite previously pass through periapsis?

Solution Here, we are asked to solve for time of periapse passage, 𝑥̈𝑥̈. In order to do so, we use Kepler’s Time Equation, or Equations (2.96) and (2,97). However, we must first convert true anomaly into eccentric anomaly using Equation (2.91) keeping in mind that all calculations must be done using radians, not degrees, √ 1 − 𝑥̈ 𝑥̈ 𝑥̈ = 2 arctan tan = 0.363695 𝑥̈𝑥̈𝑥̈ ( 2 )] [ 1 + 𝑥̈ where no quadrant check is needed since 𝑥̈ and 𝑥̈ always reside in the same quadrant.

2

2

From Equations (2.96) and (2.97), √ 𝑥̈ − 𝑥̈ sin 𝑥̈ =

𝑥̈ (𝑥̈ − 𝑥̈𝑥̈) 𝑥̈3

Solving for 𝑥̈𝑥̈ with the fact that 𝑥̈ = 0 gives √ 𝑥̈ 𝑥̈𝑥̈ = − 3 (𝑥̈ − 𝑥̈ sin 𝑥̈) = −2, 506.23 𝑥̈ 𝑥̈ which is equal to approximately -41.77 minutes. In other words, the spacecraft passed through periapsis 41.77 min prior to 𝑥̈ = 0.

10

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 10 A satellite on an Earth orbit with 𝑥̈ = 12, 587 km and 𝑥̈ = 0.42 passed through periapsis at 10:04:06 (time here is given in the form in 24-hour format). Calculate the time, in 24-hour format, at which the satellite will reach 𝑥̈ = 297𝑥̈.

Solution Here, we are asked to solve for elapsed time since periapse passage, (𝑥̈ − 𝑥̈𝑥̈). In order to do so, we use Kepler’s Time Equation, or Equations (2.96) and (2,97). However, we must first convert true anomaly into eccentric anomaly using Equation (2.91) keeping in mind that all calculations must be done using radians, not degrees, √ 1 − 𝑥̈ 𝑥̈ 𝑥̈ = 2 arctan tan = −0.74628 𝑥̈𝑥̈𝑥̈ ( 2 )] [ 1 + 𝑥̈ where no quadrant check is needed since 𝑥̈ and 𝑥̈ always reside in the same quadrant. However, 2

2

since we are asked to find when the spacecraft will arrive at the given true anomaly, we must add 2𝑥̈ to 𝑥̈ to compute the future spacecraft passage, resulting in 𝑥̈ = 5.537 rad. From Equations (2.96) and (2.97), 𝑥̈ − 𝑥̈ sin 𝑥̈ = Solving for (𝑥̈ − 𝑥̈𝑥̈) gives

√

√

𝑥̈ (𝑥̈ − 𝑥̈𝑥̈) 𝑥̈3

𝑥̈

𝑥̈ − 𝑥̈𝑥̈ =

(𝑥̈ − 𝑥̈ sin 𝑥̈) = 13, 022 𝑥̈ 𝑥̈3 So, adding 13,022 s to 10:04:06 means adding 3 hours (10,800 s), 37 minutes (2,220 s) and 2 s, resulting in a time of 13:41:08.

11

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 11 A satellite in Earth orbit passes through perigee at 𝑥̈0 = 978 s on an orbit with 𝑥̈ = 0.7 and 𝑥̈ = 30, 500 km. Using the Newton-Raphson iteration method and an absolute tolerance of Δ𝑥̈ = 0.001, calculate the true anomaly 𝑥̈ at 𝑥̈ = 35, 000 s. Do not implement a relative tolerance check. This problem must be done by hand and calculator (i.e., no computer programs or use of calculator solver routines). Show each step in the iteration – 𝑥̈𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈𝑥̈, and tolerance check.

Solution We must use Kepler’s Time Equation (Equations (2.96) and (2.97)) to solve for eccentric anomaly 𝑥̈ at the given time √ 𝑥̈ 𝑥̈ 𝑥̈ = (𝑥̈ − 𝑥̈𝑥̈) = 𝑥̈ − 𝑥̈ sin 𝑥̈ 𝑥̈3 √ 𝑥̈ where 𝑥̈ = (𝑥̈ − 𝑥̈ ) = 4.03254 rad. 𝑥̈3

0

Following the Newton-Raphson method discussed in Section 2.6.1, we set up 𝑥̈ (𝑥̈) and 𝑥̈ ′(𝑥̈) as follows 𝑥̈ (𝑥̈) = 𝑥̈ − 𝑥̈ sin(𝑥̈) − 𝑥̈ 𝑥̈ ′(𝑥̈) = 1 − 𝑥̈ cos(𝑥̈) where only 𝑥̈ is unknown. For Newton-Raphson, the update step for each iteration is 𝑥̈

𝑥̈𝑥̈𝑥̈

= 𝑥̈ − 𝑥̈ (𝑥̈𝑥̈𝑥̈𝑥̈) 𝑥̈𝑥̈𝑥̈ 𝑥̈ ′(𝑥̈𝑥̈𝑥̈𝑥̈)

We then use 𝑥̈𝑥̈𝑥̈𝑥̈ = 𝑥̈ as the initial guess. To check the absolute tolerance for every step, we check if |𝑥̈𝑥̈𝑥̈𝑥̈ − 𝑥̈𝑥̈𝑥̈𝑥̈| < Δ𝑥̈. Iterations are carried on until the absolute tolerance is met. This results in: Iteration 𝑥̈𝑥̈𝑥̈𝑥̈ (rad) 𝑥̈𝑥̈𝑥̈𝑥̈ (rad) |𝑥̈𝑥̈𝑥̈𝑥̈ − 𝑥̈𝑥̈𝑥̈𝑥̈| (rad) 1 4.03254 3.65453 0.3780 2 3.65453 3.67606 0.0215 3 3.67606 3.67601 <0.001 Thus, 𝑥̈ = 3.67601 rad from which 𝑥̈ can be computed using Equation (2.91) as √ 𝑥̈ + 1 𝑥̈ 𝑥̈ = 2 arctan tan = 3.3706 𝑥̈𝑥̈𝑥̈ ( 2 )] [ 𝑥̈ − 1 which is also equivalent to 𝑥̈ = 193.12𝑥̈.

12

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 12 Write a computer program that a takes as input • A position vector (in km) and a velocity vector (in km/s) in ECI (Earth-Centered Inertial) coordinates.

• An initial time, 𝑥̈1 • A later time, 𝑥̈2 • Tolerances 𝑥̈𝑥̈ (relative tolerance) and 𝑥̈𝑥̈ (absolute tolerance) for the iterative procedure used to solve Kepler’s time equation. b Outputs the input data, with labels and units, and calculates the following: Classical orbital elements 𝑥̈, 𝑥̈, 𝑥̈, 𝑥̈, and Ω (Note that these are 5 of the 6 COEs) c The DCM 𝑥̈𝑥̈𝑥̈ that converts from ECI to perifocal coordinates. d Position (in km) and velocity (in km/s) vectors in perifocal coordinates at time 𝑥̈1. e True and eccentric anomalies 𝑥̈1 and 𝑥̈1 at time 𝑥̈1. f Time of periapsis passage, 𝑥̈𝑥̈. g True and eccentric anomalies 𝑥̈2 and 𝑥̈2 at time 𝑥̈2. h Position (in km) and velocity (in km/s) vectors in perifocal coordinates at time 𝑥̈2. i Position (in km) and velocity (in km/s) vectors in ECI coordinates at time 𝑥̈2. j Outputs the values in items (b) – (i) with labels and units.

Run the program with the input values: 𝑥̈𝑥̈ = 5, 214.1𝑥̈ + 6, 322.9𝑥̈ − 3, 001.7𝑥̈ 𝑥̈𝑥̈ 𝑥̈𝑥̈ = −4.1549𝑥̈ + 3.1666𝑥̈ − 4.5044𝑥̈ 𝑥̈𝑥̈/𝑥̈ 𝑥̈1 = 0 𝑥̈ 𝑥̈2 = 12, 500 𝑥̈ Δ𝑥̈ = 10−6 Δ𝑥̈ = 10−6 Solution The resulting quantities for each computation are summarized in the table below

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Part Quantity

b

c

d e f g h i

Value and Unit

Equation(s) Used

𝑥̈

9,108.95 km

(2.61)

𝑥̈

0.201529

(2.62)

𝑥̈

0.759947 rad

(2.64)

𝑥̈

2.09983 rad

(2.71), (2.72)

Ω

3.62717 rad

(2.68), (2.69)

𝑥̈𝑥̈𝑥̈

𝑥̈ 0.7384 −0.3179 𝑥̈ 𝑥̈ 0.5928 0.7265 𝑥̈ 𝑥̈−0.3215 0.6093 𝑥̈

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

55.10𝑥̈ + 8, 727.72𝑥̈ + 0𝑥̈ km

(2.78), (2.79); or (2.152)

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

−6.7535𝑥̈ + 1.40369𝑥̈ + 0𝑥̈ km/s

(2.152)

𝑥̈1

1.56448 rad

(2.73)

𝑥̈1

1.3617 rad

(2.91)

𝑥̈𝑥̈

-1,603.6 s or 7,048.3 s

(2.96), (2.97)

𝑥̈2

3.71099 rad

(2.73) (compute after 𝑥̈2)

𝑥̈2

3.83091 rad

(2.96), (2.97) (must iterate)

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

−8, 864.9𝑥̈ − 5, 678.5𝑥̈ + 0𝑥̈ km

(2.78), (2.79); or (2.152)

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

3.64103𝑥̈ − 4.32705𝑥̈ + 0𝑥̈ km/s

(2.152)

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

−9, 909.7𝑥̈ − 1, 304.3𝑥̈ − 3, 299.1𝑥̈ km

(𝑥̈𝑥̈𝑥̈ )𝑥̈ 𝑥̈𝑥̈𝑥̈𝑥̈

𝑥̈𝑥̈𝑥̈𝑥̈ 𝑥̈

0.5947 𝑥̈ 𝑥̈ −0.3477𝑥̈ 𝑥̈ 0.7249 𝑥̈ 𝑥̈

0.123732𝑥̈ − 4.30091𝑥̈ + 3.66979𝑥̈ km/s

𝑥̈3(𝑥̈)𝑥̈1(𝑥̈)𝑥̈3(Ω)*

𝑥̈𝑥̈ 𝑥̈

𝑥̈ 𝑥̈𝑥̈𝑥̈

(𝑥̈ ) 𝑥̈

𝑥̈

*the notation 𝑥̈𝑥̈ (𝑥̈) indicates a direction cosine matrix (DCM) about the j-axis (j = 1, 2, 3) through an angle 𝑥̈. DCMs should not be confused with rotation matrices. A rotation matrix is the inverse of a DCM.

14

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 13 Consider a spacecraft in orbit around the Earth with the following Initial Conditions (ICs; position and velocity vectors) in an Earth-Centered Inertial (ECI) frame at some time 𝑥̈0: 𝑥̈𝑥̈ = 666.780𝑥̈ + 7, 621.32𝑥̈ + 2, 338.97𝑥̈ 𝑥̈𝑥̈ 𝑥̈𝑥̈ = −7.22541𝑥̈ + 0.266320𝑥̈ + 1.52177𝑥̈ 𝑥̈𝑥̈/𝑥̈ Consider the Cartesian formulation of the J2-perturbed Equations of motion for the two-body problem: { } 3 𝑥̈ 𝑥̈ 𝑥̈𝑒2 𝑥̈ = − 𝑥̈ 1 − 2 —1 𝑥̈ 2 ] 2 (5 𝑥̈ 𝑥̈ 2 )[ 𝑥̈ 2 3 { } 3 𝑥̈ 𝑥̈ 𝑥̈𝑒2 𝑥̈ = − 𝑥̈ 1 − 2 —1 ] 2 ( 𝑥̈ 2 } 𝑥̈ 5 𝑥̈ 2 𝑥̈ { )[ 2 3 3 𝑥̈ 𝑥̈2 𝑥̈ 2 𝑥̈ 5 𝑥̈ 𝑥̈ = − —3 1− ] 𝑥̈ 2 ( 𝑥̈ 2 )[ 𝑥̈ 3 2 𝑥̈2 where: 𝑥̈ = 398, 600 km3/s2 is the gravitational parameter of the Earth, 𝑥̈𝑥̈ = 6, 378.14 km is the e√quatorial radius of the Earth, 𝑥̈2 = 1.08263 × 10−3 the coefficient of Earth’s oblateness, and 𝑥̈ = 𝑥̈2 + 𝑥̈2 + 𝑥̈2 is the magnitude of the position of the spacecraft with respect to Earth’s center.

Set up a state 𝑥̈ = {𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈} and rewrite the above EOMs in state-space form using these given states. Then, using a computer program such as MATLAB, numerically integrate the ICs of this spacecraft using the above EOMs for 30 days. Provide two figures: 1. Figure 1: a 3D plot (x vs. y vs. z) of the spacecraft orbit (in red) with respect to Earth’s center in ECI coordinates. Also, for reference plot the Earth as a blue sphere having radius equal to 𝑥̈𝑥̈.

2. Figure 2: subplots showing: (a) Semimajor Axis (SMA) in km vs. time (b) Eccentricity vs. time (c) Inclination in degrees vs. time (d) Orbital period in seconds vs. time (e) Perigee in km vs. time (f) Apogee in km vs. time If using MATLAB, compare the difference between using different integration schemes, such as ode45 vs. ode113. What do you notice? Which one is faster for a given accuracy?

Solution 15

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Following the numerical integration methods discussed in Section 2.8, we obtain the graphs shown in Figures1 and 2. Numerical integration using both MATLAB’s ode45 and ode113 shows that ode113 is faster than ode45.

16

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 1: Spacecraft Orbit in ECI

Figure 2: Deviation of Orbital Parameters 17

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Interplanetary Astrodynamics Chapter 3 Problem Solutions

Problem 1 Compute the period, time unit (in seconds), and dimension unit (in kilometers) of the systems listed in Table 3.1. You may find it useful to consult the appendix for various constants you will need, such as semi-major axes and gravitational parameters.

Solution The value of dimension (or length) unit (LU) is simply the value of semimajor axis 𝑥̈, while the period is computed as given by Equation (3.10) √ 𝑥̈3 𝑥̈ = 2𝑥̈ 𝑥̈ (𝑥̈1 + 𝑥̈2) where 𝑥̈ = 6.6743 × 10−20 km3kg-1s-2 is the universal gravitational constant. Then, the time unit (TU) can be computed as given by Equation (3.11) 1 𝑥̈ 𝑥̈ =

𝑥̈ 2𝑥̈

Compiling the results for all systems listed in Table 3.1 gives the following numerical results

1

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

System

Mass Ratio, 𝑥̈

1 LU (km)

Period (s)

1 TU (s)

Earth-Moon

1.215060e-02

3.844000e+05 2.356987e+06 3.751261e+05

Sun-Mercury

1.660148e-07

5.790931e+07 7.599252e+06 1.209459e+06

Sun-Venus

2.447835e-06

1.082041e+08 1.940945e+07 3.089110e+06

Sun-Earth

3.003460e-06

1.495978e+08 3.155262e+07 5.021756e+06

Sun-Mars

3.227137e-07

2.279422e+08 5.934516e+07 9.445075e+06

Sun-Jupiter

9.536922e-04

7.783274e+08 3.742697e+08 5.956687e+07

Sun-Saturn

2.857260e-04

1.426983e+09 9.294258e+08 1.479227e+08

Mars-Phobos

1.660513e-08

9.376000e+03 2.755916e+04 4.386176e+03

Mars-Deimos

2.299699e-09

2.345800e+04 1.090626e+05 1.735786e+04

Jupiter-Ganymede

7.803794e-05

1.070400e+06 6.180722e+05 9.836925e+04

Saturn-Titan

2.365667e-04

1.221870e+06 1.377519e+06 2.192389e+05

Pluto-Charon

1.085112e-01

1.959140e+04 5.516461e+05 8.779721e+04

Note about the pronunciation of Pluto’s moon, Charon: the pronunciation "Sharon" is commonly accepted among many English speakers. However, in Greek mythology, the name originates from Charon or Kharon (𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈) and was the name of the ferryman who would carry the dead across the river Styx. The Greek letter X (capitalized) or 𝑥̈ (lower case) in the name 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ has a sound similar to the Spanish j in j𝑥̈mon. All of Pluto’s moons (Charon, Hydra, Kerberos, Nix, and Styx) are named after Greek mythology.

2

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 2 Compute the x- and y-coordinates of the collinear Lagrange points for the following systems: (a) Earth-Moon (b) Sun-Mercury (c) Sun-Venus (d) Sun-Earth (e) Sun-Mars (f) Sun-Jupiter (g) Sun-Saturn (h) Mars-Phobos (i) Mars-Deimos (j) Jupiter-Ganymede (k) Saturn-Titan (l) Pluto-Charon Solution The y-coordinate of all collinear points is zero, so it is only necessary to solve for the x-coordinate. In order to do so, we use Equation (3.39) to set up a function of the form 𝑥̈ (𝑥̈) =

(1 − 𝑥̈)(𝑥̈ − 𝑥̈) 𝑥̈(𝑥̈ + 1 − 𝑥̈) + − 𝑥̈ |𝑥̈ − 𝑥̈|3 |𝑥̈ + 1 − 𝑥̈|3

Solving for 𝑥̈ (𝑥̈) = 0 gives three real-valued solutions which correspond to the x-coordinates of 𝑥̈1, 𝑥̈2, and 𝑥̈3, which we here denote as 𝑥̈𝑥̈1 , 𝑥̈𝑥̈2 , and 𝑥̈𝑥̈3 , respectively. Solving for the zeroes of this algebraic function can be cumbersome due to the presence of the absolute values. MATLAB’s fzero command was used here to obtain the results shown in the following table.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

System

𝑥̈

𝑥̈𝑥̈1 (ND)

𝑥̈𝑥̈2 (ND)

𝑥̈𝑥̈3 (ND)

Earth-Moon

1.215060e-02

-8.369151e-01

-1.155682e+00

1.005063e+00

Sun-Mercury

1.660148e-07

-9.961939e-01

-1.003815e+00

1.000000069e+00

Sun-Venus

2.447835e-06

-9.906823e-01

-1.009371e+00

1.0000010199e+00

Sun-Earth

3.003460e-06

-9.900266e-01

-1.010034e+00

1.000001251e+00

Sun-Mars

3.227137e-07

-9.952513e-01

-1.004763e+00

1.000000e+00

Sun-Jupiter

9.536922e-04

-9.323699e-01

-1.068826e+00

1.000397e+00

Sun-Saturn

2.857260e-04

-9.547492e-01

-1.046069e+00

1.000000134e+00

Mars-Phobos

1.660513e-08

-9.982321e-01

-1.001770e+00

1.00000000691e+00

Mars-Deimos

2.299699e-09

-9.990851e-01

-1.000915e+00

1.00000000095e+00

Jupiter-Ganymede

7.803794e-05

-9.705872e-01

-1.029842e+00

1.000033e+00

Saturn-Titan

2.365667e-04

-9.575005e-01

-1.043252e+00

1.000099e+00

Pluto-Charon

1.085112e-01

-5.930430e-01

-1.262516e+00

1.045139e+00

In this table, we used additional significant figures for some 𝑥̈𝑥̈3 values that are extremely small.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 3 How do the values you computed in Problem 2 compare with the approximations for 𝑥̈1, 𝑥̈2, and 𝑥̈3 given by Equation (3.55)?

Solution Equation (3.55) gives an approximation for the x-coordinates of the 𝑥̈1, 𝑥̈2, and 𝑥̈3, which we here denote as 𝑥̈𝑥̈1 , 𝑥̈𝑥̈2 , and 𝑥̈𝑥̈3 , respectively, so that 𝑥̈ 1/3 𝑥̈𝑥̈1 ≈ ( ) — 1 3 𝑥̈ 1/3 𝑥̈𝑥̈2 ≈ − ( ) − 1 3 5𝑥̈ 𝑥̈𝑥̈3 ≈ 1 + ( 12 ) The following table summarizes the results of computing approximate values for 𝑥̈𝑥̈1 , 𝑥̈𝑥̈2 , and 𝑥̈𝑥̈3 .

System

𝑥̈

𝑥̈𝑥̈1 (ND)

𝑥̈𝑥̈2 (ND)

𝑥̈𝑥̈3 (ND)

Earth-Moon

1.215060e-02

-8.405986e-01

-1.159401e+00 1.005063e+00

Sun-Mercury

1.660148e-07

-9.961893e-01

-1.003811e+00 1.000000e+00

Sun-Venus

2.447835e-06

-9.906556e-01

-1.009344e+00 1.000001e+00

Sun-Earth

3.003460e-06

-9.899962e-01

-1.010004e+00 1.000001e+00

Sun-Mars

3.227137e-07

-9.952441e-01

-1.004756e+00 1.000000e+00

Sun-Jupiter

9.536922e-04

-9.317511e-01

-1.068249e+00 1.000397e+00

Sun-Saturn

2.857260e-04

-9.543323e-01

-1.045668e+00 1.000119e+00

Mars-Phobos

1.660513e-08

-9.982311e-01

-1.001769e+00 1.000000e+00

Mars-Deimos

2.299699e-09

-9.990848e-01

-1.000915e+00 1.000000e+00

Jupiter-Ganymede

7.803794e-05

-9.703702e-01

-1.029630e+00 1.000033e+00

Saturn-Titan

2.365667e-04

-9.571178e-01

-1.042882e+00 1.000099e+00

Pluto-Charon

1.085112e-01

-6.692871e-01

-1.330713e+00 1.045213e+00

Computing the percentage difference between the numerical solutions (Problem 2) and the approximate solutions using Equation (3.55) can be done as follows |𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ − 𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈. | | × 100% Percentage Difference = || | 𝑥̈ 𝑥̈ , 𝑥̈ | | 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

with 𝑥̈ = 1, 2, 3. Doing so results in the following table.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

System

𝑥̈

% Difference 𝑥̈𝑥̈1

% Difference 𝑥̈𝑥̈2

% Difference 𝑥̈𝑥̈3

Earth-Moon

1.215060e-02

0.438203905350070

0.320785222575412 0.000009769945384

Sun-Mercury

1.660148e-07

0.000469844938868

0.000465080575677 0.000000000000022

Sun-Venus

2.447835e-06

0.002699586399057

0.002632645018871 0.000000000000022

Sun-Earth

3.003460e-06

0.003076736262845

0.002995083495722

< 10−14

Sun-Mars

3.227137e-07

0.000726291929042

0.000717105059948

< 10−14

Sun-Jupiter

9.536922e-04

0.066416675305139

0.054036055092968 0.000000004715131

Sun-Saturn

2.857260e-04

0.043688710896332

0.038405000317756 0.000000000126795

Mars-Phobos

1.660513e-08

0.000102884310336

0.000102399485460

< 10−14

Mars-Deimos

2.299699e-09

0.000027723644489

0.000027656019694

< 10−14

Jupiter-Ganymede

7.803794e-05

0.022356697113660

0.020608597303759 0.000000000002576

Saturn-Titan

2.365667e-04

0.039989328222425

0.035454070329668 0.000000000071958

Pluto-Charon

1.085112e-01

11.391846098281569 5.124834377092897 0.007098841803256

From the above table, one can see how the approximation introduces errors on the order of << 1%, except for the Pluto-Charon system, which has a relatively large 𝑥̈ value. The second-worst approximation is for the Earth-Moon system, where errors are on the order of 10−1%.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 4 The nonlinear equations of motion in the restricted three-body problem are: (1 − 𝑥̈)(𝑥̈ − 𝑥̈) 𝑥̈(𝑥̈ + 1 − 𝑥̈) 𝑥̈ − 2𝑥̈ − 𝑥̈ = − − 𝑥̈23 𝑥̈1 3 𝑥̈𝑥̈ (1 − 𝑥̈ + 2𝑥̈ − 𝑥̈ = − 𝑥̈)𝑥̈ — 3 𝑥̈2 3 𝑥̈𝑥̈ 𝑥̈1 𝑥̈ = − (1 − — 3 𝑥̈2 𝑥̈)𝑥̈ where 1 𝑥̈3 √ 𝑥̈1 = (𝑥̈ − 𝑥̈)2 + 𝑥̈2 + 𝑥̈2 𝑥̈2 = √ (𝑥̈ + 1 − 𝑥̈)2 + 𝑥̈2 + 𝑥̈2 For the restricted three-body problem, the complete set of equations of motion for the linear solutions about the Lagrange points are found as: { } (𝑥̈𝑥̈ − 1 1 (𝑥̈𝑥̈ + 1 − 𝑥̈)2 —3 + 𝑥̈𝑥̈ − 2𝑥̈𝑥̈ − 𝑥̈𝑥̈ = 𝑥̈)2 ] + 𝑥̈ (1 − 𝑥̈) 3 5 ] [𝑟 1 — 3 𝑟 2𝑥̈ 2𝑒 −𝑥̈𝑥̈ 5 𝑥̈ 𝑟1𝑥̈ [𝑥̈3 (𝑥̈𝑥̈ − 𝑥̈)𝑥̈𝑥̈ (𝑥̈𝑥̈ + 1 − 𝑥̈)𝑥̈𝑥̈ + 𝑥̈𝑥̈ 3(1 − 𝑥̈) + 3𝑥̈ 5 ] [ 𝑟1𝑥̈5 𝑟2𝑥̈ (𝑥̈𝑥̈ − 𝑥̈)𝑥̈𝑥̈ (𝑥̈𝑥̈ + 1 − 𝑥̈)𝑥̈𝑥̈ 3(1 − 𝑥̈) + 3𝑥̈ + 5 5 ] [ 𝑟 𝑟2𝑥̈ } { 1 1 𝑥̈ 𝑥̈2 𝑥̈2 1 — 3 5𝑥̈ — 𝑥̈𝑥̈ (1 − 𝑥̈) 3 — 3 5𝑥̈ + [𝑟1𝑥̈ [𝑟2𝑥̈3 𝑟2𝑥̈ ] 𝑟2𝑥̈] 𝑥̈

𝑥̈𝑥̈ + 2𝑥̈𝑥̈ − 𝑥̈𝑥̈ = 𝑥̈𝑥̈

Using these equations, assess the stability of 𝑥̈1 for the systems listed in Problem 2.

Solution We can simplify the equations given in the problem statement since the y- and z-coordinates for 𝑥̈1 are zero, so that √ 𝑥̈1𝑥̈ = (𝑥̈𝑥̈ — 𝑥̈)2 = 𝑥̈1𝑥̈ − 𝑥̈ 1

𝑥̈2𝑥̈ = √ + 1 − 𝑥̈)2 = 𝑥̈1𝑥̈ + 1 − 𝑥̈ (𝑥̈𝑥̈ 1

where the subscript ‘𝑥̈’ here denotes the fact that we are considering equilibrium points of the system. Here, we will show a detailed solution for the Earth-Moon system. For all other systems, the solution method is identical, and while the numerical values are slightly different, the end result (that 𝑥̈1 is unstable) is the same.

7

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

For the Earth-Moon system, 𝑥̈1𝑥̈ = −0.8369 (refer to the solution for Problem 2), so that the perturbed equations provided in the problem statement simplify to 𝑥̈𝑥̈ − 2𝑥̈𝑥̈ − 𝑥̈𝑥̈ = −𝑥̈𝑥̈ (−10.2935) 𝑥̈𝑥̈ + 2𝑥̈𝑥̈ − 𝑥̈𝑥̈ = −𝑥̈𝑥̈ (5.1468)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

In matrix form

{ } { } { } { } 1 0 𝑥̈𝑥̈ 0 −2 −11.2935 0 𝑥̈𝑥̈ 𝑥̈𝑥̈ 0 + + = 0 4.1468] 𝑥̈𝑥̈ [0 1] 𝑥̈𝑥̈ [2 0 ] 𝑥̈𝑥̈ [ 0

We assume that this linear system of ODEs has a solution of the form { } 𝑥̈𝑥̈ = 𝑥̈ 𝑥̈𝑥̈𝑥̈ 𝑥̈𝑥̈ so that

{ } { } 𝑥̈𝑥̈ = 𝑥̈𝑥̈ 𝑥̈ 𝑥̈𝑥̈ 𝑥̈𝑥̈ { } { } 𝑥̈𝑥̈ 𝑥̈𝑥̈ = 𝑥̈𝑥̈ 𝑥̈𝑥̈ 𝑥̈2

Substituting these into the matrix equation for 𝑥̈𝑥̈ and 𝑥̈𝑥̈ we get { } { } 𝑥̈𝑥̈ 0 𝑥̈2 − 11.2935 −2𝑥̈ = 0 2𝑥̈ 𝑥̈2 + 4.1468] 𝑥̈𝑥̈ [ Here, the solutions are the trivial solution (which gives no information on the system) or the matrix in the above equation is not invertible, i.e. its determinant is zero, so det

𝑥̈2 − 11.2935 −2𝑥̈ =0 2𝑥̈ 𝑥̈2 + 4.1468] [

This gives us the following characteristic equation (𝑥̈2 − 11.2935)(𝑥̈2 + 4.1468) + 4𝑥̈2 = 0 𝑥̈4 − 3.1467𝑥̈2 − 46.832 = 0 which is a quadratic equation in 𝑥̈2. Solving for 𝑥̈2 using the quadratic equation and then taking further square roots results in the following four values of 𝑥̈ 𝑥̈1 = 2.932 𝑥̈2 = −2.932 𝑥̈3 = 2.334 i 𝑥̈4 = −2.334 i In order for a dynamical system to be stable, all roots of the characteristic equation must have non-positive real parts. However, since 𝑥̈1 is a positive real number, the system is unstable at the computed equilibrium point, which means that the Lagrange point 𝑥̈1 is unstable.

9

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 5 Compute the value of the Jacobi integral at each Lagrange point computed in Problem 2 assuming zero-velocity components in the x-, y-, and z-directions.

Solution From Equation (3.63), the Jacobi constant can be computed as follows 𝑥̈ 1 1− 𝑥̈ = − (𝑥̈2 + 𝑥̈2) − — 𝑥̈ 2 𝑥̈2 𝑥̈1 where we assumed that all velocity components are zero, as stated by the problem. Furthermore, 𝑥̈1 and 𝑥̈2 can be simplified since the y- and z-components of the collinear Lagrange points are zero, so 𝑥̈1 = 𝑥̈𝑥̈𝑥̈ − 𝑥̈ 𝑥̈2 = 𝑥̈𝑥̈𝑥̈ + 1 − 𝑥̈ where 𝑥̈ = 1, 2, 3 for 𝑥̈1, 𝑥̈2, and 𝑥̈3, respectively. Thus, the Jacobi constant simplifies to 1 2 1 − 𝑥̈ 𝑥̈ — 𝑥̈ = − 𝑥̈𝑥̈𝑥̈ − 2 |𝑥̈𝑥̈ — 𝑥̈| + 1 − 𝑥̈| 𝑥̈ |𝑥̈𝑥̈ 𝑥̈

where we need to use absolute values for the denominators since 𝑥̈1 > 0 and 𝑥̈2 > 0. Computing the Jacobi constant using the x-coordinates of each collinear point (refer to Problem 2) results in the following table System

𝑥̈

𝑥̈ (𝑥̈1)

𝑥̈ (𝑥̈2)

𝑥̈ (𝑥̈3)

Earth-Moon

1.215060e-02

-1.594170625224589 -1.586080287273612 -1.506073582531158

Sun-Mercury

1.660148e-07

-1.500065070997697 -1.500064960320992 -1.500000083007400

Sun-Venus

2.447835e-06

-1.500388858029825 -1.500387226125325 -1.500001223917438

Sun-Earth

3.003460e-06

-1.500445344888560 -1.500443342561633 -1.500001501729906

Sun-Mars

3.227137e-07

-1.500101245712571 -1.500101030569587 -1.500000161356849

Sun-Jupiter

9.536922e-04

-1.519378038491350 -1.518742117111655 -1.500476836539595

Sun-Saturn

2.857260e-04

-1.508910362635931 -1.508719852663865 -1.500142862147275

Mars-Phobos

1.660513e-08

-1.500014053103714 -1.500014042033623 -1.500000008302565

Mars-Deimos

2.299699e-09

-1.500003765337075 -1.500003763803943 -1.500000001149850

Jupiter-Ganymede

7.803794e-05

-1.503820911405483 -1.503768882392033 -1.500039018906516

Saturn-Titan

2.365667e-04

-1.507882004358728 -1.507724273455416 -1.500118282765730

Pluto-Charon

1.085112e-01

-1.810171662165219 -1.739669120616945 -1.553995665696784

As discussed in Chapter 3, various correct definitions of the Jacobi constant exist. Using a definition that 10

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

is different from what we used here will result in different Jacobi constant values.

11

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 6 Using the nonlinear equations of motion in the restricted three-body problem: (a) if the state vector is: 𝑥̈ = (𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈)𝑥̈ , and using the values of 𝑥̈ = 0.16 and the following initial conditions: 𝑥̈(0) = (−1.024114 0 0.528576 0 0.381137 0)𝑥̈ , numerically integrate the equations of motion for one period (you will have to experimentally determine what the period is (hint: think of the symmetry of the problem and where does the orbit cross the plane of symmetry again?)). Show plots of x vs. y, x vs. z, y vs. z and a three-dimensional plot of x vs. y vs. z. You should get a periodic, halo orbit. Identify the period of the orbit. (b) Confirm the invariance of these equations of motion under the transformation 𝑥̈ → −𝑥̈ and 𝑥̈ → −𝑥̈. (c) For another case, we don’t have the exact initial conditions for 𝑥̈0 and 𝑥̈0. If 𝑥̈ = 0.2 and the initial conditions are 𝑥̈(0) = (−1.01 0 0.58 0 0.413250 0)𝑥̈ , use a differential correction process to correct the initial conditions to obtain a periodic halo orbit. Show three iterations of your differential correction results with the correct-ed initial conditions and period (or half period) after each iteration.

Solution (a) Through trial and error, it is possible to determine that the period of the orbit is 2.52 TU. Alternatively, one can set an event flag to stop the integration once the state returns to the initial state (within some tolerance). The plots we obtain from this numerical integration are shown in Figure 1. The resulting orbit is an 𝑥̈2 Northern halo orbit. Notice that the value of 𝑥̈ used here (0.16) is considered to be rather large – compare it with 𝑥̈ for Earth-Moon and Pluto-Charon.

Although not requested by the problem, one can compute the Lagrange point locations for 𝑥̈1, 𝑥̈2, and 𝑥̈3 (refer to Problem 2) resulting in the following x-coordinates 𝑥̈𝑥̈1 = −0.502942319513751 𝑥̈𝑥̈2 = −1.271094441718751 𝑥̈𝑥̈3 = 1.066420639638059 (b) Integrating the equations of motion backwards in time using the ICs in part (a) results in the same orbit. However, a more rigorous proof is done by replacing 𝑥̈ with −𝑥̈ and 𝑥̈ with −𝑥̈ in the CR3BP equations of motion: 𝑥̈ − 2𝑥̈ − 𝑥̈ = Ω𝑥̈ 𝑥̈ + 2𝑥̈ − 𝑥̈ = Ω𝑥̈ 𝑥̈ = Ω𝑥̈ Looking at the above equations term-by-term, and performing the required substitutions (𝑥̈ → −𝑥̈

12

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 1: Orbital Plot for Part (a) and 𝑥̈ → −𝑥̈), gives 𝑥̈ → 𝑥̈ −𝑥̈𝑥̈ 𝑥̈ 𝑥̈ → 𝑥̈ = 𝑥̈ −𝑥̈ = 𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈𝑥̈ → Ω𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ → − = −𝑥̈ 𝑥̈ 𝑥̈ 𝑥̈ → −𝑥̈ Ω𝑥̈ → −Ω𝑥̈ 𝑥̈ → 𝑥̈ Ω𝑥̈ → Ω𝑥̈

13

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Thus, the new equations of motion become 𝑥̈ − 2𝑥̈ − 𝑥̈ = Ω𝑥̈ −𝑥̈ − 2𝑥̈ + 𝑥̈ = −Ω𝑥̈ 𝑥̈ = Ω𝑥̈ The x- and z-equations are identical to the original equations of motion, while the y-equation can be multiplied by −1 in order to get the original y-equation of motion. This proves that the CR3BP equations of motion are invariant (do not change) under the transformation 𝑥̈ → −𝑥̈ and 𝑥̈ → −𝑥̈. (c) Using a differential corrector (Section 3.7.1) to correct for 𝑥̈0 and 𝑥̈0 for the given fixed value of 𝑥̈0 means using Equations (3.112) and (3.119). After three iterations, the new 𝑥̈0 and 𝑥̈0 are within ~ 10−5 accuracy of the desired values, and these numerical results are shown in the table below.

Iterating for more steps resulting in 𝑥̈0 and 𝑥̈0 with even higher accuracy, if desired.

Element

Iteration 1

Iteration 2

Iteration 3

𝑥̈ 2

1.3121

1.3466

1.3465

𝑥̈0

-1.0100

-1.0141

-1.0145

𝑥̈0

0

0

0

𝑥̈0

0.5800

0.5879

0.5876

𝑥̈0

0

0

0

𝑥̈0

0.4133

0.4133

0.4133

𝑥̈0

0

0

0

𝑥̈(𝑥̈ /2)

-0.8115

-0.8143

-0.8144

𝑥̈(𝑥̈ /2)

-2.6553×10−6

-7.1126×10−6

1.5844×10−6

𝑥̈(𝑥̈ /2)

-0.0851

-0.0912

-0.0912

𝑥̈(𝑥̈ /2)

-0.0150

-0.0033

-7.6783×10−7

𝑥̈(𝑥̈ /2)

-2.0501

-1.9706

-1.9707

𝑥̈(𝑥̈ /2)

0.0206

-3.3031×10−4

-1.0398×10−4

𝑥̈𝑥̈0

0.0041

3.6758×10−4

-1.4878×10−5

𝑥̈𝑥̈0

-0.0079

2.3535×10−4

3.9651×10−5

14

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 7 Numerically integrate the equations of motion of the circular restricted three-body problem using the mass ratio of the Earth-Moon system (𝑥̈ = 0.01215060) for the following sets of initial conditions: (a) 𝑥̈(0) = (−1.180732845 0 0.013007284 0 0.156830543 0)𝑥̈ (b) 𝑥̈(0) = (−1.178004826 0 0.052029136 0 0.169784129 0)𝑥̈ (c) 𝑥̈(0) = (−1.125032004 0 0.182101977 0 0.225430661 0)𝑥̈ (d) 𝑥̈(0) = (−1.014568158 0 0 0 − 0.9385765507 0)𝑥̈ (e) 𝑥̈(0) = (−0.750000000 0 0.6204002141 0 − 0.3228789168 0)𝑥̈ (f) 𝑥̈(0) = (−1.156957 0 0 0 0.48000000 0)𝑥̈ You will need to determine the period of these orbits to correctly plot them (hint: think of the symmetry of the problem). What orbit types do you get? Match and verify that the orbits you obtained are the same as the orbits plotted on Figure 3.55.

Figure 3.56: Various Periodic Orbits in the CR3BP Solution Performing the required numerical integration and plotting each orbit reveals that the given initial conditions (ICs) correspond to the plotted orbits as summarized in the following table 15

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

ICs

Orbit

(a)

Small halo (blue)

(b)

Medium halo (red)

(c)

Large halo (orange)

(d)

Planar Lyapunov (purple)

(e)

Vertical Lyapunov (green)

(f)

Distant retrograde orbit (light blue)

16

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 8 In the planar restricted three-body problem, the equations for the zero-velocity curve can be written as: 𝑥̈ (1 − 𝑥̈) 𝑥̈ 2 + 𝑥̈(1 − 𝑥̈) — 𝑥̈ 2 + 𝑥̈ = − + 2 2 2 and

(1

𝑥̈1 )

2(2

𝑥̈2 )

2

1 1 − 𝑥̈ 𝑥̈ 𝑥̈ = − (𝑥̈2 + 𝑥̈2) − − 2 𝑥̈1 𝑥̈2

(a) Show that these two expressions are equivalent. (b) Using the first equation above, show that the zero-velocity curves disappear where 𝑥̈ =

1 2

[−3 + 𝑥̈(1 − 𝑥̈)]

(c) For a system with 𝑥̈ = 0.25, choose the value of the Jacobi constant half-way between 𝑥̈(𝑥̈1) and 𝑥̈(𝑥̈2), i.e. the arithmetic average between 𝑥̈(𝑥̈1) and 𝑥̈(𝑥̈2) (be sure to state this value of the Jacobi constant) and develop a computer program to plot the zero-velocity curve for this value. Be sure to annotate the location of the collinear libration points. Test your code at one more location between 𝑥̈(𝑥̈2) and 𝑥̈(𝑥̈3) and identify the Jacobi constant value you use.

Solution Note that 𝑥̈ and 𝑥̈ are used interchangeably in the literature to both denote the Jacobi constant. (a) We start by setting the 𝑥̈ equations equal to each other 𝑥̈ (1 − 𝑥̈) 𝑥̈ 2 + 𝑥̈(1 − 𝑥̈) 1 1 − 𝑥̈ 𝑥̈ — 𝑥̈ 2 + — + = − (𝑥̈2 + 𝑥̈2) − − 2 2 2

(1

𝑥̈1 )

2(2

𝑥̈2 )

2

2

𝑥̈1

𝑥̈2

and we expand terms and simplify —

𝑥̈/ ) (1 − 𝑥̈) 𝑥̈ 2 + −(1 −/ / /

2

1

/

𝑥̈ 1

𝑥̈ 𝑥̈/ 𝑥̈(1 − − 𝑥̈2− 𝑥̈) + 2 2// 𝑥̈2 2

1 / 𝑥̈/ ) 𝑥̈ = − (𝑥̈2 + 𝑥̈2) + −(1 −/ — / / 2

(1 − 𝑥̈) 2 𝑥̈ 2 𝑥̈(1 − 𝑥̈) 1 − 𝑥̈1 − 𝑥̈2 + = − (𝑥̈2 + 𝑥̈2) 2 2 2 2 Recall that the position vectors are defined as 𝑥̈𝑥̈ = (𝑥̈ − 𝑥̈)𝑥̈ + 𝑥̈𝑥̈ 𝑥̈𝑥̈ = [𝑥̈ + (1 − 𝑥̈)]𝑥̈ + 𝑥̈𝑥̈ so 17

/

𝑥̈1

𝑥̈2 //

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

𝑥̈ 2 = 𝑥̈2 − 2𝑥̈𝑥̈ + 𝑥̈2 + 𝑥̈2 𝑥̈12 2 = 𝑥̈2 + 2𝑥̈(1 − 𝑥̈) + (1 − 𝑥̈)2 + 𝑥̈2

18

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Substituting the equations for 𝑥̈2 and 𝑥̈2 into the prior equations gives 1

1 − 𝑥̈ —

2

2

( — 2𝑥̈𝑥̈ + 𝑥̈ 𝑥̈ 2

2

𝑥̈

2

2

+ 𝑥̈ ) − [𝑥̈ 2

2

+ 2𝑥̈(1 − 𝑥̈) + (1 − 𝑥̈)

1

𝑥̈(1 −

2

𝑥̈) + 𝑥̈ ] +

2

2

2

= − (𝑥̈ + 𝑥̈ ) 2

Multiplying all by −2 and simplifying (1 − 𝑥̈)(𝑥̈2 − 2𝑥̈𝑥̈ + 𝑥̈2 + 𝑥̈2) + 𝑥̈[𝑥̈2 + 2𝑥̈(1 − 𝑥̈) + (1 − 𝑥̈)2 + 𝑥̈2] − 𝑥̈(1 − 𝑥̈) = (𝑥̈2 + 𝑥̈2 Considering the left-hand side of the equation, we can expand it and simply as 𝑥̈ 2−cc 2𝑥̈c 𝑥̈ + 𝑥̈/2 + 𝑥̈2 − 𝑥̈ 𝑥̈/2 + 2𝑥̈ 𝑥̈2 − 𝑥̈/3 − 𝑥̈ 𝑥̈/2 + 𝑥̈ 𝑥̈/2 + 𝑥̈𝑥̈/2 + 2𝑥̈ 𝑥̈ − 2𝑥̈ 𝑥̈2 + 𝑥̈ − 2𝑥̈/2 + /

/

+ 𝑥̈/3 + 𝑥̈ 2/𝑥̈ − 𝑥̈ + 𝑥̈/2 = 𝑥̈2 + 𝑥̈2 /

/

/

/

/

/

/

/

/

/

which is equal to the right-hand side and thus proves that the two given expressions for the Jacobi constant are equivalent. (b) We can show that zero-velocity curves disappear at 𝑥̈4 and 𝑥̈5, where 𝑥̈1 = 𝑥̈2 = 1. We start with 𝑥̈ (1 − 𝑥̈) 𝑥̈ 2 + 𝑥̈(1 − 𝑥̈) — 𝑥̈ 2 + 𝑥̈ = − + 2 2 2

(1

𝑥̈1 )

2(2

𝑥̈2 )

2

and substitute 𝑥̈1 = 𝑥̈2 = 1 into it to get 𝑥̈ 𝑥̈(1 − (1 − 𝑥̈) (1 + 2) − (1 + 2) + 𝑥̈) 2 2 2 3 3 𝑥̈(1 − 𝑥̈) = − (1 − 𝑥̈) − 𝑥̈ + 2 2 2 3 3 / 3 / 𝑥̈(1 − 𝑥̈) = − + /𝑥̈ − /𝑥̈ + 2 /2 /2 2 1 = [−3 + 𝑥̈(1 − 𝑥̈)] 2

𝑥̈ = −

which is equal to a constant and proves the fact that zero-velocity curves disappear at 𝑥̈4 and 𝑥̈5. (c) For 𝑥̈ = 0.25 the x-coordinates of the Lagrange points can be computed as 𝑥̈𝑥̈1 = −0.360743428367017 𝑥̈𝑥̈2 = −1.265858102510350 𝑥̈𝑥̈3 = 1.103166848822924 Refer to the solution for Problem 2 for computations related to these values. Furthermore, 𝑥̈1 and 𝑥̈2 can be simplified since the y- and z-components of the collinear Lagrange points are zero, so 𝑥̈1 = 𝑥̈𝑥̈𝑥̈ − 𝑥̈ 𝑥̈2 = 𝑥̈𝑥̈𝑥̈ + 1 − 𝑥̈ where 𝑥̈ = 1, 2, 3 for 𝑥̈1, 𝑥̈2, and 𝑥̈3, respectively. Thus, the Jacobi constant simplifies to 1 2 1 − 𝑥̈ 𝑥̈ 19

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

— 𝑥̈(𝑥̈𝑥̈) = − 𝑥̈𝑥̈𝑥̈ − 2 |𝑥̈𝑥̈ — 𝑥̈| 𝑥̈

𝑥̈

20

𝑥̈ + 1 − 𝑥̈|

| 𝑥̈

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

where we need to use absolute values for the denominators since 𝑥̈1 > 0 and 𝑥̈2 > 0. The corresponding Jacobi constant values to each Lagrange point are 𝑥̈(𝑥̈1) = −1.935329401439718 𝑥̈(𝑥̈2) = −1.780597028114743 𝑥̈(𝑥̈3) = −1.622470510138496 Note that in the literature 𝑥̈ and 𝑥̈ are used interchangeably to denote the Jacobi constant. The halfway values of the Jacobi constant between 𝑥̈1 and 𝑥̈2 and between 𝑥̈2 and 𝑥̈3 are therefore 1 𝑥̈𝑥̈1/𝑥̈2 = [𝑥̈(𝑥̈1) + 𝑥̈(𝑥̈2)] = −1.857963214777230 2 1 𝑥̈𝑥̈2/𝑥̈3 = [𝑥̈(𝑥̈2) + 𝑥̈(𝑥̈3)] = −1.701533769126619 2 Figures 2 and 3 show zero-velocity surfaces where the perimeter of the surfaces represent the Jacobi constants computed above, 𝑥̈𝑥̈1/𝑥̈2 and 𝑥̈𝑥̈2/𝑥̈3 , respectively. The physical sizes of 𝑥̈1 and 𝑥̈2 in the figures are arbitrary, and are used only to visualize the position of the primaries. The Lagrange points are also plotted for reference.

21

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 2: Zero-velocity surface plot for 𝑥̈ values up to 𝑥̈𝑥̈1/𝑥̈2

22

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 3: Zero-velocity surface plot for 𝑥̈ values up to 𝑥̈𝑥̈2/𝑥̈3

23

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Interplanetary Astrodynamics Chapter 4 Problem Solutions

Problem 1 Using Equation (4.8), plot a graph of Earth’s obliquity vs. time between January 1, 2000 and January 1, 3000. How do the initial and final values of obliquity compare? What is their difference?

Create another plot using the same approximation, but this time from January 1, 2000 for 10,000 years. When does the approximation start diverging significantly from the prediction given in Figure 4.4?

Solution Plotting Earth’s using Equation (4.8) from January 1, 2000 to January 1, 3000 (10 centuries) gives Figure 1. The difference between final and initial values is -0.12952○, i.e. the Earth’s obliquity on January 1, 3000 is estimated to be 0.12952○ less than that on January 1, 2000. Repeating this process for 100 centuries gives Figure 2. The final value of obliquity estimated with this approximation is 22.637279○ which is visually very similar to that reported in Figure 4.4. The computed approximation seems to start diverging from Figure 4 around 90 centuries, where the minimum value of obliquity happens if using Equation (4.8), but it occurs later (around 100 centuries) in Figure 4.4.

1

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 1: Earth’s Obliquity for 10 centuries since J2000

Figure 2: Earth’s Obliquity for 100 centuries since J2000 2

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 2 Consider an interplanetary Hohmann transfer from Earth to Mars (refer to Chapter 5 if needed).

Using the resultant Mars arrival 𝑥̈8, compute the impact parameter and eccentricity of the arrival trajectory assuming that your targeted periapsis with respect to Mars has an altitude of 300 km.

Solution As provided in Appendix A, the symbol for Mars is D, and the orbital and physical parameters of interest for this problem are also taken from Appendix A 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈D = 43, 050 km3/s2 𝑥̈D = 3, 397 km 𝑥̈D{@ = 227, 942, 167.9 km 𝑥̈C{@ = 149, 597, 800 km Thus, the targeted orbit at Mars arrival has periapsis radius 𝑥̈𝑥̈{D = 𝑥̈D . 300km = 3,697 km.

In order to compute the 𝑥̈8{D, we need to solve part of this interplanetary Hohmann transfer (as discussed in Chapter 5). We compute the semimajor axis and energy of the Hohmann transfer orbit (subscript ‘H’) as 1 ˘ 𝑥̈𝑥̈ = C{@ . D{@ = 188, 769, 983.9 km ` 𝑥̈ 𝑥̈ 2

𝑥̈𝑥̈ = — = —351.486 km2/s2 𝑥̈@ 2𝑥̈𝑥̈ Then, we compute the heliocentric speeds that the spacecraft would have at apoapsis (i.e. right before encountering Mars) and that Mars has as

d

.

𝑥̈𝑥̈{@ = 2𝑥̈𝑥̈ 2𝑥̈@ = 21.47926 km/s 𝑥̈D{ c @

𝑥̈D{@ =

𝑥̈ 𝑥̈

@

= 24.12810 km/s

D{@

Thus, 𝑥̈8{D is just the difference of these two speeds 𝑥̈8{D = 𝑥̈D{@ — 𝑥̈𝑥̈{@ = 2.64884 km/s To compute the impact parameter, we use Equation (4.15)

d

Δ = 𝑥̈𝑝{D 1 .

2𝑥̈D = 7, 683.42 km 2 𝑟𝑝{D 𝑣8{D

To compute the eccentricity of the Mars arrival hyperbolic orbit, we use Equations (2.50) and (2.37), so that the hyperbolic semimajor axis and eccentricity become 3

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

𝑥̈ =

—𝑥̈D

= —6, 135.66 km 2 𝑣8{ D 𝑥̈𝑥̈{ 𝑥̈ = 1 — = 1.60254 D

𝑥̈

4

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 3 Compute the Julian Date (JD) and Modified Julian Date (MJD) for a given time (e.g., when your homework is due). Make sure to use enough decimal places so that your accuracy is within 1 ms (10´3 s). Then, compute the Greenwich location angle corresponding to this time.

Solution For this problem, we will use the date: January 1, 2025 00:00:00 UTC. To compute JD, using Equation (4.45) gives 2,460,676.5000000.

To compute MJD, we use the computed JD and Equation (4.46), giving 60,676.0000000.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 4 Create a ground track similar to that of Figure 4.7 for an Earth orbiting spacecraft that has classical orbital elements with respect to ECI as follows: (a) semimajor axis: 6671 km (b) eccentricity: 0.3 (c) inclination: 45○ (d) RAAN: 20○ (e) argument of periapsis: 95○ Assume that the spacecraft starts at periapsis (true anomaly = 0 ○) on January 1, 2030, and propagate its orbit using the unperturbed two-body model for 5 full orbits.

Solution Following the steps given in Sections 4.3 and 4.3.1 gives the ground track shown in Figure 3, where tick marks are each separated by 60 seconds. For reference, 𝑥̈ 𝑥̈ = 2462502.50000 and 𝑥̈𝑥̈ = 1.75069204 rad.

Figure 3: Ground Track

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 5 Compute the coordinates of your school/institution using the topocentric equatorial coordinates (IJK) discussed in Section 4.3.2 for a given time (e.g., when your homework is due). Compute these coordinates assuming that the Earth is a perfect sphere (flattening parameter = 0) and using a flattening parameter of 0.003353. How do the results differ?

Solution For this example, we will use Embry-Riddle Aeronautical University – Prescott Campus as the institution of choice. This location has the following altitude 𝑥̈, latitude 𝑥̈, and longitude 𝑥̈ 𝑥̈ = 1.572 km 𝑥̈ = —112.450644○ 𝑥̈ = 34.617177○ Additionally, from Appendix A we know that Earth’s equatorial radius is 𝑥̈C = 6, 378.140 km and we are given Earth’s flattening as 𝑥̈ = 0.003353.

Using Equation (4.23) with 𝑥̈ = 0 gives

˜

¸

𝑥̈C

𝑥̈ = a . 𝑥̈ cos 𝑥̈ cos 𝑥̈ = —2, 005.0185 km 2) sin2 𝜙 1 — (2𝑓 — 𝑓 ˜ ¸ 𝑥̈C 𝑥̈ = a . 𝑥̈ cos 𝑥̈ sin 𝑥̈ = —4, 852.3614 km 1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈ 𝑥̈ (1 — 𝑥̈ )2

˜ 𝑥̈ =

a

¸

C

1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈

sin 𝑥̈ = 3, 624.2537 km

. 𝑥̈

while using 𝑥̈ = 0.003353 gives

˜

𝑥̈ = a

𝑥̈C

𝑥̈ = a

𝑥̈C

˜ 1 — (2𝑓 — 𝑓 2) sin2 𝜙

. 𝑥̈ cos 𝑥̈ cos 𝑥̈ = —2, 007.1875 km ¸ . 𝑥̈ cos 𝑥̈ sin 𝑥̈ = —4, 857.6105 km

1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈ 𝑥̈ (1 — 𝑥̈ )2

˜ 𝑥̈ =

¸

a

C

1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈

¸

. 𝑥̈

sin 𝑥̈ = 3, 603.8906 km

The percent differences in the x-, y-, and z- components are 0.1081%, 0.1081%, and 0.5650%, respectively.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 6 Compute the coordinates of the NASA Perseverance landing location using the topocentric equatorial coordinates (IJK) discussed in Section 4.3.2 for a given time (e.g., when your homework is due). Compute these coordinates assuming that Mars is a perfect sphere (flattening parameter = 0) and using a flattening parameter of 0.00648. How do the results differ?

Solution For this problem, we will use the landing date of Perseverance, i.e. February 18, 2021, where the rover was located at altitude 𝑥̈, latitude 𝑥̈, and longitude 𝑥̈ 𝑥̈ = 2.478 km 𝑥̈ = 77.451○ 𝑥̈ = 18.445○ This data can be obtained directly from JPL Horizons. Additionally, from Appendix A we know that Mars’s equatorial radius is 𝑥̈D = 3, 389.5 km and we are given Earth’s flattening as 𝑥̈ = 0.00648.

Using Equation (4.23) with 𝑥̈ = 0 gives

˜

¸

𝑥̈D 𝑥̈ = a . 𝑥̈ cos 𝑥̈ cos 𝑥̈ = 699.1294 km ˜ 1 — (2𝑓 — 𝑓 2) sin2 𝜙 ¸ 𝑥̈D 𝑥̈ = a . 𝑥̈ cos 𝑥̈ sin 𝑥̈ = 3, 140.8550 km 1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈ 𝑥̈ (1 — 𝑥̈ )2

˜ 𝑥̈ =

a

¸

D

1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈

sin 𝑥̈ = 1, 073.2021 km

. 𝑥̈

while using 𝑥̈ = 0.00648 gives

˜

𝑥̈ = a

𝑥̈D

𝑥̈ = a

𝑥̈D

˜ 1 — (2𝑓 — 𝑓 2) sin2 𝜙

. 𝑥̈ cos 𝑥̈ cos 𝑥̈ = 699.5816 km ¸ . 𝑥̈ cos 𝑥̈ sin 𝑥̈ = 3, 142.8863 km

1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈ 𝑥̈ (1 — 𝑥̈ )2

˜ 𝑥̈ =

¸

a

D

1 — (2𝑥̈ — 𝑥̈ 2) sin2 𝑥̈

¸

. 𝑥̈

sin 𝑥̈ = 1, 060.0337 km

The percent differences in the x-, y-, and z- components are 0.0646%, 0.0646%, and 1.2423%, respectively.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 7 Compute the JD and MJD for your next birthday. If you know it, include the hours and minutes correspond to your time of birth, otherwise, use midnight UTC. Solution For this problem, we will use the date: January 22, 2025 15:05:00 UTC. To compute JD, using Equation (4.45) gives 2,460,698.12847.

To compute MJD, we use the computed JD and Equation (4.46), giving 60,697.62847.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 8 Using Equation (4.42), compute the date on which TDB and TCB will start differing by more than a minute.

Solution From Equation (4.42), we find the difference between TCB and TDB as 𝑥̈ 𝑥̈𝑥̈ — 𝑥̈ 𝑥̈𝑥̈ = 1.550519768 × 10´8Δ𝑥̈ . 6.55 × 10´5 where Δ𝑥̈ is 𝑥̈ 𝑥̈𝑥̈ 𝑥̈𝑥̈ — 𝑥̈0, i.e. the time elapse from Jan 1, 1977 00:00:00 UTC in seconds (we multiplied the 86,400 to turn this quantity into seconds). Setting 𝑥̈ 𝑥̈𝑥̈ — 𝑥̈ 𝑥̈𝑥̈ = 60 s and solving for Δ𝑥̈ gives Δ𝑥̈ =

60 — 6.55 × 10´5 1.550519768 × 10´8

„ 44, 788 days or 122.7 years

Adding these days to Jan 1, 1977 00:00:00 UTC leads to Aug 17, 2099, which is the day upon which TCB and TDB will be 60 seconds apart from each other.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 9 Compute the ephemeris of the Moon using Simpson’s approximation (discussed in Section 4.5) for a given time, e.g. from when your homework is due for the following 10 years. Plot the orbital elements of the Moon with respect to ECI for this time frame.

Solution For this problem, we will use the initial and final dates Jan 1, 2025 and Jan, 1, 2035, respectively.

Computing the lunar ephemeris for these dates using Simpson’s approximation results in the classical orbital elements shown in Figures 4 and 5.

Figure 4: Classical Orbital Elements of the Moon with respect to ECI (1 of 2)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 5: Classical Orbital Elements of the Moon with respect to ECI (2 of 2)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 10 Using JPL Horizons, compute the position of Earth and Mars for a given time (e.g., when your homework is due). Compute the OWLT for this time. Then, obtain the positions of Earth and Mars from JPL Horizons for one synodic period and plot the OWLT for this time.

Solution For this problem, we will use the initial date of Jan 1, 2025. As provided in Appendix A, the synodic period of Earth and Mars is 2.1351 years (or 780 days). This results in a final date of Feb 20, 2027.

The positions of Earth and Mars in the J2000 ecliptic reference frame at the initial date are listed in the following tables. 𝑥̈C (AU)

𝑥̈C (AU)

𝑥̈C (AU)

-0.184413 0.962073 0.000129 𝑥̈D (AU)

𝑥̈D (AU)

𝑥̈D (AU)

-0.527416 1.520324 0.044935 The OWLT between Earth and Mars for this synodic period is plotted on Figure 6. The maximum and minimum OWLT computed are 20.158436 minutes and 5.341873 minutes, respectively.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 6: OWLT between Earth and Mars for one Synodic Period (High-Fidelity Ephemeris)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 11 Repeat Problem 10, but use the low-fidelity ephemeris model discussed in this chapter (Standish and Williams’ method).

Solution For this problem, we will use the initial date of Jan 1, 2025. As provided in Appendix A, the synodic period of Earth and Mars is 2.1351 years (or 780 days). This results in a final date of Feb 20, 2027.

The positions of Earth and Mars in the J2000 ecliptic reference frame at the initial date are listed in the following tables. 𝑥̈C (AU)

𝑥̈C (AU)

𝑥̈C (AU)

-0.178693 0.966957 -0.000055 𝑥̈D (AU)

𝑥̈D (AU)

𝑥̈D (AU)

-0.521810 1.525229 0.044761 The OWLT between Earth and Mars for this synodic period is plotted on Figure 7. The maximum and minimum OWLT computed are 20.160159 minutes and 5.341875 minutes, respectively. Note that OWLT is almost identical to what was computed in the previous problem using a high-fidelity ephemeris model.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 7: OWLT between Earth and Mars for one Synodic Period (Low-Fidelity Ephemeris)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 12 Using the method explained in Section 4.6, compute the position of the Sun with respect to the Earth for a given time (e.g., when your homework is due).

Solution For this problem, we will use Jan 1, 2025 as the given date. The following tables shows the parameters, values, and referenced equations used to compute solar ephemeris with respect to ECI for the given time, and the resulting Sun’s coordinates at the given time, respectively.

Parameter

Value

Reference

𝑥̈𝑥̈@

281.39198554○ Equation (4.52)

𝑥̈@

358.03080941○ Equation (4.53)

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

281.32482029○ Equation (4.54)

𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

0○

Equation (4.54)

𝑥̈@

0.98330568 AU

Equation (4.55)

𝑥̈C

23.43602619○

Equation (4.8)

𝑥̈@ (AU)

𝑥̈@ (AU)

𝑥̈@ (AU)

0.19311875 -0.88474131 -0.38352241 Although not requested by the problem statement, computing the Sun’s position with respect to the Earth for 10 years from the given date results in the components and magnitudes shown in Figure 8.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 8: Sun’s Position with respect to the Earth

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 13 Using JPL Horizons and the low-fidelity model discussed in Section 4.7.1, generate the ephemeris data for Venus from a given time (e.g., when your homework is due) for 10 years. How do the two methods compare? Plot the difference in distance of the planet with respect to the Sun and the planet’s orbital speed.

Solution In order to use Standish and William’s approximation, we need to refer to Appendix C, and use the appropriate values corresponding to the celestial body of interest in combination with Equations (4.57) - (4.59). After obtaining the orbital elements of the body of interest, we can use the methods and equations in Chapter 2 to obtain position and velocity of any given body with respect to the gravitational source (here, the Sun).

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Venus, this results in the plots shown in Figure 9 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 10.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 9: Venus’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 10: Venus’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 14 Repeat Problem 13 for Mercury.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Mercury, this results in the plots shown in Figure 11 where the Earth is also shown for comparison. Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 12.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 11: Mercury’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 12: Mercury’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 15 Repeat Problem 13 for Jupiter.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Jupiter, this results in the plots shown in Figure 13 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 14.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 13: Jupiter’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 14: Jupiter’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 16 Repeat Problem 13 for Saturn.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Saturn, this results in the plots shown in Figure 15 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 16.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 15: Saturn’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 16: Saturn’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 17 Repeat Problem 13 for Uranus.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Uranus, this results in the plots shown in Figure 17 where the Earth is also shown for comparison. Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 18.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 17: Uranus’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 18: Uranus’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 18 Repeat Problem 13 for Neptune.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Neptune, this results in the plots shown in Figure 19 where the Earth is also shown for comparison. Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 20.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 19: Neptune’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 20: Neptune’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 19 Repeat Problem 13 for Pluto.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Pluto, this results in the plots shown in Figure 21 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 22.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 21: Pluto’s Distance and Speed (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 22: Pluto’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 20 Use Standish and Williams’ method to estimate the date on which the next conjunction between Earth and Mars occurs, i.e. when Earth and Mars are on the opposite side of the Sun. Solution For this problem, we will use Jan 1, 2025 as the starting date, i.e. we want to estimate when the next Earth-Mars conjunction happens from this given date. To do this, we will look at dates that go from this given date for a full Earth-Mars synodic period (780 days), i.e. until Feb 20, 2027.

Two possible ways to estimate this are to 1. find the maximum distance between Earth and Mars during the given synodic period; or 2. find when the angle between the positions of Earth and Mars is close to 180○; or 3. plot the orbits of Earth and Mars and visually estimate when the planets are in conjunction (we won’t show this method) For the first method, we compute the ephemeris data and search for the maximum distance, i.e. ` ˘ 𝑥̈𝑥̈𝑥̈𝑥̈ = max |rD{@ — rC{@| for the given dates. This also means that the one-way light time (OWLT) is maximized. Referring to Figure 7, we can estimate this date to be around Dec 1, 2025.

For the second method, we compute the ephemeris data and search for the value of the angle between the Earth and Mars position vectors (Δ𝑥̈) that is closest to 180○, i.e. we compute Δ𝑥̈ using the definition of the dot product ˆ ˙ rD{@ ¨ rC{@ Δ𝜃 = arccos |rD{@||rC{@| Plotting Δ𝑥̈ as a function of time for the given dates results in Figure 23, from which we can estimate the conjunction date to be around Jan 9, 2026.

While these are estimates, if the orbits of the planets were exactly circular and coplanar, these methods would provide the same (exact) answer.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 23: Angle between Earth’s and Mars’s position vectors

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Interplanetary Astrodynamics Chapter 5 Problem Solutions

Problem 1 A spacecraft is launched into a circular orbit around the Earth at an altitude of 450 km. The upper stage of the launch vehicle initiates a Hohmann transfer to GEO through a GTO. How much Δ𝑥̈ is the launch vehicle expected to perform (Δ𝑥̈1) and how much Δ𝑥̈ is the spacecraft’s propulsion system expected to deliver in order to perform the adequate orbit insertion at GEO (Δ𝑥̈2)? How long does the transfer take?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Earth are 𝑥̈C = 398, 600 km3/s2 𝑥̈C = 6, 371 km Therefore, the spacecraft starts in a LEO having radius and velocity 𝑥̈1 = 𝑥̈C . 450 km = 6, 821 km c 𝑥̈ 𝑥̈1 = = 7.64442 km/s C 𝑥̈ 1

The spacecraft needs to transfer to GEO, i.e. an orbit which has period equal to the length of a sidereal day, or 23 hours 56 minutes and 4.091 seconds, which gives 𝑥̈𝑥̈𝑥̈𝑥̈ = 23 × 3600 . 56 × 60 . 4.091 = 86, 164.091 s Thus, the semimajor axis corresponding to this orbital period is

ˆ

𝑎𝐺𝐸𝑂 = 𝜇1{3 C

˙

𝑥̈𝐺𝐸𝑂 2{3 = 42, 164.15 km 2𝜋

Using Equations (5.4) – (5.10) with 𝑥̈1 = 6, 821 km and 𝑥̈2 = 𝑥̈𝑥̈𝑥̈𝑥̈ gives Δ𝑥̈1 = 2.38553 km/s Δ𝑥̈2 = 1.452085 km/s Δ𝑥̈𝑥̈𝑥̈𝑥̈ = 3.83762 km/s And the time of flight is

d 𝑥̈ 𝑥̈𝑥̈ = 𝑥̈

(𝑥̈1 . 𝑥̈2)3 8𝑥̈C1

= 5.29820 hours

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 2 A space vehicle in a circular orbit at an altitude of 300 km above Mars’s surface executes a Hohmann transfer. The total Δ𝑥̈ available is 0.35 km/s. This means that this Δ𝑥̈ needs to be split between Δ𝑥̈1 (to initiate the transfer) and Δ𝑥̈2 (to complete the transfer). What is the maximum altitude of the final orbit that this spacecraft can reach with the allotted Δ𝑥̈? How long does this transfer take?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Mars are 𝑥̈D = 43, 050 km3/s2 𝑥̈D = 3, 389.5 km so that the spacecraft starts at a radius and orbital speed 𝑥̈1 = 𝑥̈D . 300 km = 3, 689.5 km c 𝑥̈ 𝑥̈1 = = 3.40706 km/s D 𝑥̈ 1

We then use Equation (5.10)

ˆ

Δ𝑥̈𝑥̈𝑥̈𝑥̈ 1 = 1— 𝑣𝑐1 𝑅 with Δ𝑥̈𝑥̈𝑥̈𝑥̈

˙c

2𝑥̈ 𝑅. 1

.,

1 𝑅

—1

= 0.102728 to numerically solve for 𝑥̈, which leads to 𝑥̈ = 1.24293. This means that the 𝑥̈𝑥̈1

final altitude 𝑥̈2 that the spacecraft reaches is 𝑥̈2 = 𝑥̈2 — 𝑥̈D = 𝑥̈𝑥̈1 — 𝑥̈D = 1, 196.27 km

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 3 A spacecraft completes an interplanetary Hohmann transfer from Earth to Venus. The crew wants to come back to Earth. (a) How long do they have to wait before Earth and Venus align themselves to allow for a Hohmann transfer from Venus to Earth? (b) What is the angle between Earth and Venus at departure? (c) What is the angle between Earth and Venus at arrival?

Solution Using Appendix A, we can find the gravitational parameter of the Sun, the Earth’s and Venus’s distances from the Sun 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈C{@ = 149, 597, 800 km 𝑥̈B{@ = 108, 204, 088.7 km We then compute the orbital periods of Earth and Venus, along with their mean motions

d 𝑇C = 2𝜋

𝑥̈ 3C{@

d

𝑥̈B = 2𝑥̈

𝜇@ = 365.25 days

𝑟 3 {@ B

𝑥̈@

= 224.65 days

360˝ = 0.985626 ○ /day 𝑥̈C 360˝ 𝑥̈B = = 1.602493 ○ /day 𝑥̈B

𝑥̈C =

Computing the semimajor axis and time of flight of an Earth-Venus Hohmann transfer gives 1 𝑥̈𝑥̈ = (𝑥̈C{@ . 𝑥̈B{@) = 128, 900, 944 km 2

d

𝑇 𝑂𝐹𝐻 = 𝜋

𝑥̈𝐻3 = 146 days 𝜇@

Figure1 shows the positions of Earth and Venus at departure from Earth and arrival at Venus for the outbound (Earth-Venus) trajectory, while Figure 2 shows the positions of Earth and Venus at departure from Venus and arrival at Earth for the inbound, or return, (Venus-Earth) trajectory.

For the outbound transfer (Earth-Venus) we arbitrarily set Earth’s angle at 𝑥̈C,𝑥̈𝑥̈𝑥̈ = 0˝ at departure (or launch) from Earth. Therefore, for the outbound trajectory we can compute the Earth and Venus angles when the spacecraft leaves Earth at departure ‘dep’ and at Venus upon arrival ‘arr’

3

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

as follows ○

𝑥̈C,𝑥̈𝑥̈𝑥̈ = 0

○

𝑥̈B,𝑥̈𝑥̈𝑥̈ = 180 ○ 𝑥̈C,𝑥̈𝑥̈𝑥̈ = 𝑥̈C𝑥̈ 𝑥̈𝑥̈𝑥̈ = 143.9 ˝ ˝ ˝ 𝑥̈B,𝑥̈𝑥̈𝑥̈ = 𝑥̈B,𝑥̈𝑥̈𝑥̈ — 𝑥̈B𝑥̈ 𝑥̈𝑥̈𝑥̈ = 234 — 180 = —54 This means that at departure Venus is 54˝ behind Earth and at arrival Venus is 𝑥̈B,𝑥̈𝑥̈𝑥̈ — 𝑥̈C,𝑥̈𝑥̈𝑥̈ =36.1˝ ahead of Earth.

For the return trip from Venus, we must wait until Earth and Venus are once again aligned appropriately to perform a Hohmann transfer – this time from Venus to Earth. Mathematically, this means that the angle between Earth and Venus must grow by 360˝ — 2(𝑥̈B,𝑥̈𝑥̈𝑥̈ — 𝑥̈C,𝑥̈𝑥̈𝑥̈ ) = 287.8˝ which corresponds to a wait time of Δ𝑥̈ = 𝑥̈B — 𝑥̈C = 0.616866 ˝/day 287.8˝ 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ = = 466.55 days Δ𝑥̈ To summarize a the crew must wait approximately 467 days b For the outbound trajectory, at departure Venus is 54˝ behind Earth, while at arrival Earth is 36.1˝ ahead of Venus c For the inbound trajectory, at departure Earth is 36.1˝ behind Venus, while at arrival Venus is 54˝ ahead of Earth

4

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 1: Earth-Venus Hohmann Transfer (outbound)

5

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 2: Earth-Venus Hohmann Transfer (outbound)

6

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 4 For a Hohmann transfer from Earth to Venus, plot C3 vs. time for a window that begins 15 days before the optimal Hohmann geometry and ends 15 days after the Hohmann alignment. Increment launch date by 1 day, and use the time of flight of 146 days. Keep the date of arrival constant.

Solution In order to compute 𝑥̈3, we need to first compute the solution to Lambert’s problem (refer to Sections 5.3.1 or 5.3.2) using r1 = 𝑥̈C{@ [cos 𝑥̈ sin 𝑥̈ 0] r2 = 𝑥̈B{@ [—1 0 0] v C{@ =

c

𝑥̈@ 𝑟C{@ [— sin 𝑥̈ cos 𝑥̈ 0]

such that 𝑥̈ changes following Earth’s mean motion, or 𝑥̈C =

360˝ 𝑥̈C

which gives 𝑥̈ = 𝑥̈CΔ𝑥̈ with Δ𝑥̈ < [—15, 15] days. Then, solving Lambert’s problem with 𝑥̈ 𝑥̈𝑥̈= 146 days gives the required initial velocity for the transfer to take place, or v1. We can therefore compute the velocity of the spacecraft at Earth’s sphere of influence as 𝑥̈8{C = |v1 — vC{@| From which we compute 𝑥̈3 2

𝑥̈3 = 𝑥̈8{C This results in the plot shown in Figure 3.

7

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 3: Earth-Venus 𝑥̈3 as a function of time

8

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 5 Starting with Equations (5.12) – (5.15) and the definitions of R and S, prove Equation (5.16). Hint: an equation that you will find useful in this derivation is that of angular momentum 𝑥̈ written as b

,

a function of apoapse radius ra and periapse radius rp only, i.e. 𝑥̈ = 2𝑥̈ Solution

𝑥̈𝑥̈𝑥̈ 𝑟𝑥̈ 𝑎`𝑟𝑝

For this derivation, we will use the given definitions of 𝑥̈ and 𝑥̈ such that 𝑥̈2 𝑥̈1 𝑥̈𝑥̈ 𝑥̈ = 𝑥̈2 𝑥̈𝑥̈ 𝑥̈𝑥̈ = 𝑥̈1 𝑥̈ =

Referring to Figure 5.2 in Chapter 5, our goal is to write the various velocities shown based on the initial circular speed and the above definitions. We therefore start by writing the initial and final circular speeds 𝑥̈𝑥̈1 = 𝑥̈𝑥̈2 =

c 𝑥̈ 𝑥̈1 c 𝑥̈ 𝑥̈2

For the bi-elliptic transfer, two transfer trajectories are used having periapse and apoapse radii as given in the following table Transfer Periapsis Apoapsis 1

𝑟𝑝1 = 𝑟1

𝑟𝑎1 = 𝑟𝑖

2

𝑟𝑝2 = 𝑟2

𝑟𝑎2 = 𝑟𝑖

b 𝑥̈ 𝑥̈

,

Using the suggested equation, 𝑥̈ = 2𝑥̈

momenta of these transfer trajectories as

𝑥̈ 𝑥̈

𝑟𝑎`𝑟𝑝

and definitions of 𝑥̈ and 𝑥̈, we compute the angular

c a2𝑥̈

𝑥̈𝑥̈ , 𝑥̈1 1 = 𝑥̈𝑥̈ . c 1 a 𝑥̈ , = 𝑥̈ 2𝑥̈ 𝑥̈2 2 𝑥̈ .

1 𝑥̈

9

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

and we compute the velocities on these trajectories at periapsis and apoapsis

c

𝑥̈1 a 𝑥̈𝑝1 = 2𝑥̈ 𝑟1 =

c

𝑥̈𝑥̈

1 2𝑥̈𝑥̈ , 1 = 𝑅𝑆 . 1𝑥̈𝑐1 c 𝑅𝑆 . 1 𝑟 c a 𝑥̈1 𝑥̈𝑥̈ 1 1 2𝑥̈𝑥̈ 1 𝑥̈𝑎1 = = 𝑥̈𝑐1 = 2𝑥̈ , 𝑟𝑖 𝑅𝑆 .

1 𝑅𝑆 𝑅𝑆 .

1 𝑅𝑆 𝑟 1 c c 𝑥̈2 a 𝑥̈ 1 2𝑥̈ 𝑣𝑐 𝑣𝑝2 = = 2𝜇 , = 𝑟2 𝑆. 1 2 c 𝑆 . 1 𝑟12 c 𝑥̈2 a 𝑥̈ 1 1 2𝑥̈ 1 𝑥̈𝑎2 = 2𝑥̈ = 𝑥̈𝑐2 = , 𝑟𝑖 𝑆. 1𝑆 𝑆 . 1 𝑆 𝑟2

Thus, each Δ𝑥̈ maneuver is the difference of the required vs. current velocities, such that

¸

˜c

2𝑥̈𝑥̈ Δ𝑥̈1 = 𝑥̈ — 𝑥̈ = — 1 𝑥̈1 𝑥̈𝑥̈ .

1 1 𝑥̈ 𝑥̈ 𝑥̈ ˜c c 1 ¸ 2𝑥̈ 1 1 2𝑥̈𝑥̈ 1 , Δ𝑥̈2 = 𝑥̈𝑥̈2 — 𝑥̈𝑥̈1 = 𝑥̈𝑥̈1 𝑥̈ . 1 𝑥̈ 𝑥̈ 𝑥̈𝑥̈ . 1 𝑥̈𝑥̈

—

˜c Δ𝑥̈3 = 𝑥̈𝑥̈2 — 𝑥̈𝑥̈2 =

2𝑥̈

,

1

𝑥̈ . 1 𝑥̈

—,

1

¸

𝑥̈

𝑥̈𝑥̈1

where we used the fact that 𝑥̈2 = 𝑥̈𝑥̈1. Combining all Δ𝑥̈ maneuvers together and dividing by 𝑥̈𝑥̈1 gives Δ𝑥̈𝑥̈𝑥̈𝑥̈ = Δ𝑥̈1 . Δ𝑥̈2 . Δ𝑥̈3 Δ𝑥̈1 Δ𝑥̈2 Δ𝑥̈3 Δ𝑥̈𝑥̈𝑥̈ 𝑥̈

𝑥̈𝑥̈1

=

.

𝑥̈𝑥̈1

𝑥̈𝑥̈1

.

𝑥̈𝑥̈1

which, combining all the previous Δ𝑥̈ expressions in terms of 𝑥̈ and 𝑥̈ can be simplified to Δ𝑥̈𝑥̈𝑥̈𝑥̈ = 𝑣𝑐1

d

2 (𝑅𝑆 . 1)𝑅𝑆

c (𝑅𝑆 — 1) .

which is equivalent to Equation (5.16), thus completing the proof.

10

ˆ

2(1 . 𝑥̈) 1 — 1. , 𝑅𝑆 𝑅

˙

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 6 Compute the Δ𝑥̈ and TOF for a bi-elliptic heliocentric transfer between Earth (1 AU) and Uranus (19.19 AU) using S = 2. How do Δ𝑥̈ and TOF for this transfer compare with those of a Hohmann transfer?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Earth are 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈1 = 𝑥̈C{@ = 149, 597, 800 km 𝑥̈2 = 𝑥̈G{@ = 2, 870, 991, 219 km The ratio of 𝑥̈2 to 𝑥̈1 is 𝑥̈ = 𝑟𝑥̈1 2 = 19.1914, which is larger than the critical R value discussed in Chapter 5, so we expect the Δ𝑥̈ for a bi-elliptic transfer to be lower than a Hohmann transfer. Also, since 𝑥̈ = 2, then 𝑥̈𝑥̈ = 𝑥̈𝑥̈2 = 5, 741, 982, 438 km.

We start in a heliocentric orbit having radius 𝑥̈1 so the orbital speed is c 𝑥̈ = 29.7833 km/s 𝑥̈𝑥̈1 = @ 𝑥̈ C{@

Computing the total Δ𝑥̈ for the bi-elliptic transfer using Equation (5.16) and the Hohmann transfer using Equation (5.10) gives ˙ Δ𝑥̈𝑥̈𝑥̈𝑥̈ = 0.535175 𝑥̈𝑥̈1 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ Δ𝑥̈𝑥̈𝑥̈𝑥̈ = 0.526874 ˙ 𝐵𝑖´𝑒𝑙𝑙𝑖𝑝𝑡𝑖𝑐

𝑣𝑐1 so that the total Δ𝑥̈ for each transfer type is Δ𝑥̈𝑥̈𝑥̈𝑥̈)𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ = 15.9393 km/s Δ𝑥̈𝑥̈𝑥̈𝑥̈)𝑥̈𝑥̈´𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ = 15.6920 km/s This means that this type of bi-elliptic transfer (with 𝑥̈ =2) has a Δ𝑥̈ saving of 0.247259 km/s, or approximately 1.55% when compared to a Hohmann transfer. However, the difference in time of flight is much more significant ˘ 1` 𝑥̈ = 𝑥̈ . 𝑥̈ = 1, 510, 294, 510 km 𝑥̈ G{@ 2 C{@

d

𝑇 𝑂𝐹𝐻𝑜ℎ𝑚𝑎𝑛𝑛 = 𝜋 𝑥̈ 𝑥̈𝑥̈𝑥̈𝑥̈´𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈

= 𝑥̈

𝑥̈𝐻3 = 16.0403 years 𝜇@ ˜d (𝑥̈𝑥̈ . (𝑥̈1 . . 𝑥̈2)3 𝑥̈𝑥̈)3 8𝑥̈@ 8𝑥̈@ 11

= 120.927 years

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

d

¸

which means that this bi-elliptic transfer takes approximately 7.53897 times longer than a Hohmann transfer.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 7 A resupply of potatoes for Mark Watney is currently orbiting Mars in a circular orbit with altitude of 2,000 km (green orbit in Figure 5.71). The resupply vehicle performs a retrograde Δ𝑥̈ that puts it on an elliptical orbit (blue orbit) intersecting Mars at an angle 𝑥̈ 30˝ = from the line of apsides (LOA) of the new orbit, as shown in Figure 5.71. Calculate the Δ𝑥̈ (in km/s) required to perform this orbital maneuver.

Figure 5.71: Mark Watney, Visibly Angry Solution Let’s define point A as the starting spacecraft location at apoareion, and point B as the point where the spacecraft impacts Mars. Then, in polar coordinates, the(𝑥̈, 𝑥̈) coordinates of points A and B are( 𝑥̈𝑥̈= 𝑥̈D .2, 000 km, 190˝) and (𝑥̈D, — 30˝ ,) respectively, where 𝑥̈𝑥̈=5, 389.5 km. Let’s denote with subscripts ‘1’ and ‘2’ the initial and final orbits of the spacecraft, respectively. Then, the orbital speed and angular momentum of the spacecraft at point A on orbit 1 are: 𝑥̈𝑥̈)1

c 𝑥̈ = 2.81896 km/s D = 𝑥̈ 𝑥̈

𝑥̈1 = 𝑥̈𝑥̈𝑥̈𝑥̈)1 = 15, 192.8 km2/s Evaluating the orbit equation at points A and B on orbit 2 gives us the following system of

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

equations in terms of the angular momentum and eccentricity of orbit 2, i.e. 𝑥̈2 and 𝑥̈2 𝑥̈𝑥̈ =

𝑥̈2/𝑥̈D 2

s c

c˝ ´1 1 . 𝑥̈2c coc s (c180 ) 2 𝑥̈2 / 𝑥̈D s̋ ?2 c 𝑥̈D = 1 . 𝑥̈ coc s(c —c 3 2c 30 )

which simplifies to 𝑥̈22 = 𝑥̈𝑥̈(1 — 𝑥̈ 𝑥̈2) D

ˆ , ˙ ℎ22 3 = 𝑥̈D 1 . 𝑥̈2 𝑥̈D 2 Equating these two expressions and solving first for 𝑥̈2 and then for 𝑥̈2 gives 𝑥̈2 = 0.240243 𝑥̈2 = 13, 242.7 km2/s Thus, the orbital speed of the spacecraft on orbit 2 at point A can be computed 𝑥̈) = 𝑥̈ 2

𝑥̈2 𝑥̈𝐴

= 2.45713 km/s

So the required Δ𝑥̈ to transfer from orbit 1 to orbit 2 is Δ𝑥̈ = 𝑥̈𝑥̈)1 — 𝑥̈𝑥̈)2 = 0.361831 km/s

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 8 A spacecraft is orbiting Earth with semimajor axis 𝑥̈1 = 15,000 km, and eccentricity 𝑥̈1 = 0.45. (a) Calculate the flight-path angle 𝑥̈1 when the spacecraft reaches a true anomaly of 𝑥̈1 = 258˝. (b) Calculate the Δ𝑥̈ needed to circularize the orbit at 𝑥̈1 = 258˝ where the radius is 𝑥̈1. (c) How much has the line of apsides rotated from the initial elliptical orbit to the final circular orbit?

Solution As provided by Appendix A, the gravitational parameter of the Earth is 𝑥̈C= 398, 600 km3 /s2. Let ‘1’ denote the initial orbit and ‘2’ denote the final orbit. Then, 𝑥̈1 = 15, 000 km 𝑥̈1 = 0.45 (a) We are given that 𝑥̈1 = 258˝. We start by computing the semilatus rectum, energy, radius, speed, and angular momentum of the spacecraft on orbit 1 using the information that we are given 𝑥̈1 = 𝑥̈1(1 — 𝑥̈1 2) = 11, 362.5 km 𝑥̈C 𝑥̈1 = 2𝑥̈ = —13.2867 km 2/s 2 1 𝑥̈1 𝑥̈ = = 13, 197.2 km 1 1 . 𝑥̈1 cos 𝑥̈1

d ˆ

𝑥̈1 = 2 𝑥̈1 .

𝑥̈C

˙

= 5.81664 km/s

𝑟1

,

𝑥̈1 = 𝑥̈C𝑥̈1 = 69, 052.5 km2/s Thus, the flight path angle 𝑥̈ can be computed as

ˆ

𝑥̈1 = ± arccos

𝑥̈1

˙

= ±0.452061 rad

𝑟1𝑣1

To solve the sign ambiguity, we refer to where on the orbit the spacecraft is located, which can be determined by the value of 𝑥̈1. Since 180˝ c 𝑥̈1 c 360˝, then 𝑥̈1 must be negative, i.e. 𝑥̈1 = —0.452061 rad or —25.9012˝. (b) To circularize an orbit it means to transfer the spacecraft from orbit 1 to a circular orbit (orbit 2). Thus, we must have that 𝑥̈2 = 0 𝑥̈2 = 𝑥̈𝑥̈

c =

𝑥̈

= 5.49575 km/s

C

𝑥̈

1

Thus, the change in flight path angle is Δ𝑥̈ = |𝑥̈2 — 𝑥̈1| = 0.452061 rad 15

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Using the law of cosines, we can compute the Δ𝑥̈ needed to perform this circularization maneuver

a

Δ𝑥̈ = 𝑥̈21. 𝑥̈2 — 2 2𝑥̈1𝑥̈2 cos Δ𝑥̈ = 2.55444 km/s (c) Since the line of apsides (LOA) for a circular orbit is undefined, we cannot determine the LOA rotation from orbit 1 to orbit 2.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 9 Calculate the Δ𝑥̈ required to change the inclination of an Earth orbit at apoapsis, using the following data: 𝑥̈ = 13,952 km, 𝑥̈ = 0.47,= 𝑥̈1 28.5˝, 𝑥̈=2 63.4˝, 𝑥̈=0˝. Note that these orbital elements define an orbit with periapsis at the ascending node.

Solution As provided by Appendix A, the gravitational parameter of the Earth is 𝑥̈= C 398, 600 km3/s2. We can summarize the given orbital parameters as follows 𝑥̈ = 13, 952 km 𝑥̈ = 0.47 𝑥̈1 = 28.5˝ 𝑥̈2 = 63.4˝ 𝑥̈ = 0˝ The inclination change maneuver needs to perform an inclination change of Δ𝑥̈ 2 𝑥̈= 1 34.9˝. = 𝑥̈— ˝ Since 𝑥̈ = 0 , the inclination change maneuver must be performed at the ascending or descending node (here, periapsis or apoapsis) in order to be a pure inclination change, i.e. the other orbital parameters are left unchanged by the maneuver. Since Δ𝑥̈ for an inclination change is directly proportional to the transverse speed of the spacecraft, we would perform this maneuver at apoapsis, where the spacecraft is going slower. Let’s compute the apoapse speed of the spacecraft 𝑥̈C = —14.2847 km 2/s 2 2 𝑥̈ 𝑥̈𝑥̈ = 𝑥̈(1 . 𝑥̈) = 20, 509.4 km 𝑥̈ =

c

𝑥̈𝑥̈ = 2(𝑥̈ .

𝑥̈ C ) = 3.20944 km/s 𝑥̈ 𝑥̈

Thus, the Δ𝑥̈ required to perform the inclination change is

ˆ

˙

Δ𝑥̈ Δ𝑥̈ = 2𝑥̈𝑥̈ sin 2 ˆ ˙ Δ𝑥̈ = 1.92485 km/s = 2𝑥̈𝑥̈ sin 2

17

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 10 Consider a spacecraft in GEO. It is desired to shift its longitude by 12˝ westward using three revolutions of its phasing orbit. Calculate the orbital period of the phasing orbit, the total TOF and Δ𝑥̈ for such phasing maneuver.

Solution As provided by Appendix A, the gravitational parameter of the Earth is 𝑥̈C=398, 600 km3/s2.

Furthermore, from the solution of Problem 1, we know that 𝑥̈𝑥̈𝑥̈𝑥̈ = 86, 164 s 𝑥̈𝑥̈𝑥̈𝑥̈ = 𝑥̈𝑥̈𝑥̈𝑥̈ = 42, 164.15 km 𝑥̈𝑥̈𝑥̈𝑥̈ = 3.074659 km/s from which we can compute the mean motion of a spacecraft in GEO 2𝑥̈ = 7.29212 × 10 ´5 rad/s 𝑥̈𝑥̈𝑥̈𝑥̈ = 𝑥̈𝑥̈𝑥̈𝑥̈ Shifting spacecraft position means performing a phasing maneuver in 𝑥̈ = 3 revolutions. We start by computing the orbital period of the phasing orbit 𝑥̈𝑥̈𝑥̈ for the required phasing angle 𝑥̈ = 12˝ (note that for westward phasing 𝑥̈ > 0, while for eastward phasing 𝑥̈ c 0)

`

˘

𝑥̈𝑥̈𝑥̈𝑥̈ 𝑥̈ 𝑥̈𝑥̈𝑥̈ = 2𝑥̈𝑥̈ . 𝑥̈ from which 𝑥̈𝑥̈𝑥̈ =

2𝑥̈𝑥̈ . 𝑥̈ 𝑥̈𝑥̈𝑥̈𝑥̈ 𝑥̈ 𝑥̈𝑥̈𝑥̈𝑥̈ ’ 2𝑥̈ / 𝑥̈

= / . 𝑥̈𝑥̈𝑥̈𝑥̈ 𝑛 𝑥̈𝑥̈𝑥̈ / 𝑥̈ = 86, 164 . 957.4 = 87, 121.4 s Thus, the time of flight for this maneuver to occur is 𝑥̈ 𝑥̈𝑥̈ = 𝑥̈ 𝑥̈𝑥̈𝑥̈ = 261, 364 s or approximately 3.02505 days. We then compute the semimajor axis, apoapse radius, angular momentum, and periapse velocity associated with the phasing orbit in order to quantify Δ𝑥̈

ˆ

˙

a

c

𝑥̈𝑥̈𝑥̈ 2{3 𝜇1{3 𝑎𝑝ℎ = C = 42, 475.9 km 2𝜋 𝑥̈𝑥̈,𝑥̈𝑥̈ = 2𝑥̈𝑥̈𝑥̈ — 𝑥̈𝑥̈𝑥̈𝑥̈ = 42, 787.6 km 𝑥̈𝑝ℎ = 2𝑥̈C

=

𝑥̈𝑥̈,𝑥̈𝑥̈

𝑥̈𝑥̈

𝑥̈

𝑥̈𝑥̈,𝑥̈𝑥̈ 𝑥̈𝑥̈𝑥̈𝑥̈ 𝑟𝑎,𝑝ℎ . 𝑟𝐺𝐸𝑂

𝑥̈

18

= 130, 115 km 2/s

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

=

𝑥̈𝑥̈𝑥̈

3.08592 km/s

19

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Therefore, the required Δ𝑥̈ to initiate the phasing maneuver is Δ𝑥̈1 = 𝑥̈𝑥̈,𝑥̈𝑥̈ — 𝑥̈𝑥̈𝑥̈𝑥̈ = 11.2626 m/s and the Δ𝑥̈2 required to complete the maneuver is equal to Δ𝑥̈1, so that Δ𝑥̈2 = Δ𝑥̈1 = 11.2626 m/s Δ𝑥̈𝑥̈𝑥̈𝑥̈ = Δ𝑥̈1 . Δ𝑥̈2 = 22.5252 m/s

20

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 11 In this problem use canonical units such that 𝑥̈ =4𝑥̈2 LU3/TU2, where 1 LU = 1 AU and 1 TU = 1 year. Planet Express has been tasked with delivering a shipment of buggalo milk from Mars to a colony of Amphibiosans orbiting Jupiter. Assume all transfers in this problem are prograde and that Mars and Jupiter are in circular orbits around the Sun. Also, ignore the gravity of Mars and Jupiter. (a) Determine the type of orbit (i.e., conic section) for the heliocentric cruise required to transfer from Mars to Jupiter provided that the transfer angle Δ𝑥̈ is 192.558˝ and the time of flight (TOF) is 1.4 years. (b) Treating the transfer angle Δ𝑥̈ as a variable, determine the range of values for the speed 𝑥̈1 (in LU/TU) that the Planet Express Ship (“Old Bessie”) would require for a minimum-energy transfer ellipse as it enters the transfer orbit at Mars.

Solution We use the constants given in the problem statement so that 𝑥̈@ = 4𝑥̈ LU3/TU2 𝑥̈1 = 𝑥̈D{@ = 1.523679 LU 𝑥̈2 = 𝑥̈E{@ = 5.202887 LU where here 1 LU is equivalent to 1 AU and 1 TU is equivalent to 1 year. We are also given the transfer angle and time of flight Δ𝑥̈ = 192.558˝ = 3.3608 rad 𝑥̈ 𝑥̈𝑥̈ = 1.4 TU (a) In order to determine the orbit type for this transfer, we start by computing the chord and semiperimeter

a

𝑥̈ = 𝑥̈21. 𝑥̈2 2— 2𝑥̈1𝑥̈2 cos Δ𝑥̈ = 6.6983 LU 1 𝑥̈ = (𝑥̈1 . 𝑥̈2 . 𝑥̈) = 6.7124 LU 2 Then, since 180˝ c Δ𝑥̈ c 360˝ we compute the parabolic time of flight as follows

,

‰ 2 “ 3{2 3{2 𝑥̈ 𝑥̈ . (𝑥̈ — 𝑥̈) = 1.304897 TU 𝑝 = , 3 𝜇@ Since 𝑥̈ 𝑥̈𝑥̈ = 1.4 TU > 𝑥̈𝑥̈, the orbit must be elliptical. (b) Since we want to compute the range of 𝑥̈1 for the minimum energy case, we start by computing the minimum energy semimajor axis as a function of Δ𝑥̈, which is the only unspecified parameter 𝑥̈ 1 1a 2 2 𝑥̈ 𝑚𝑖𝑛 = = (𝑥̈1 . 𝑥̈2 ) . 4 𝑥̈1 . 𝑥̈2 — 2𝑥̈1 𝑥̈2 cos Δ𝑥̈ 2 4 1, = 1.6816 . 29.3916 — 15.8551 cos Δ𝑥̈ 4 21

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

We then compute the minimum and maximum 𝑥̈𝑥̈𝑥̈𝑥̈ values, which occur for Δ𝑥̈= 0 and Δ𝑥̈ = 𝑥̈, respectively (one can more rigorously prove this by taking a derivative of 𝑥̈𝑥̈𝑥̈𝑥̈ with respect to Δ𝑥̈) # 𝑥̈𝑥̈𝑥̈𝑥̈ =

2.6014 LU 3.3633 LU

Δ𝑥̈ = 0 Δ𝑥̈ = 𝑥̈

Using the vis-viva equation we compute the energy and velocity associated with each Δ𝑥̈ value # 𝑥̈ —7.5878 LU2/TU2 Δ𝑥̈ = 0 — @ 𝑥̈𝑥̈𝑥̈ = 𝐸 = 2𝑎 𝑥̈𝑥̈𝑥̈ —5.8690 LU2/TU2 Δ𝑥̈ = 𝑥̈

d ˆ

˙

#

𝑥̈@ 6.053452 LU/TU Δ𝑥̈ = 0 𝑥̈1 = 2 𝑥̈𝑚𝑖𝑛 . = 𝑟1 6.331019 LU/TU Δ𝜃 = 𝜋 Thus, 𝑥̈1 < [6.053452, 6.331019] LU/TU. Notice that the minimum energy ellipse has a size (semimajor axis) based on the value of transfer angle Δ𝑥̈. So one should be mindful of the fact that the minimum energy ellipse for a transfer is not unique if only time of flight and the magnitudes of the initial and final positions are given.

The most notable minimum energy ellipse is the Hohmann transfer, which has transfer angle of Δ𝑥̈ = 180˝ along with the assumptions that the starting and final orbits are coplanar and circular.

22

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 12 A spacecraft is transferring from Earth to Mars. Assume a change in true anomaly of 120˝ (true anomaly from Earth departure to Mars arrival). Find the heliocentric transfer orbit characteristics (𝑥̈ and 𝑥̈) assuming a 300-, 200-, 100-, 50-, and 25-day transfer. Be sure to state all of your assumptions.

Make sure to compute the minimum transfer time you can have and still have an elliptical transfer to en-sure that you are solving Lambert’s equation using the correct conic type.

Solution In this problem, we use the physical constants in canonical units, as follows 𝑥̈@ = 4𝑥̈ LU3/TU2 𝑥̈1 = 𝑥̈C{@ = 1.00000 LU 𝑥̈2 = 𝑥̈D{@ = 1.523679 LU where here 1 LU is equivalent to 1 AU and 1 TU is equivalent to 1 year. We are also given the transfer angle Δ𝑥̈ = 120˝ = 2.09439 rad We start by computing the chord and semiperimeter

a

𝑥̈ = 𝑥̈21 . 𝑥̈ 22 — 2𝑥̈1𝑥̈2 cos Δ𝑥̈ = 2.20149 LU 1 𝑥̈ = (𝑥̈1 . 𝑥̈2 . 𝑥̈) = 2.36200 LU 2 Then, since Δ𝑥̈ c 180˝ we compute the parabolic time of flight as follows

,

‰ 2 “ 3{2 3{2 𝑥̈ 𝑥̈ — (𝑥̈ — 𝑥̈) = 97.695 days 𝑝 = , 3 𝜇@ which means that transfer times below 97.695 days will result in hyperbolic orbits. We solve Lambert’s problem following the solution method given in Section 5.3, making use of the elliptical or hyperbolic case, as appropriate. This gives the results summarized in the following table.

TOF (days)

SMA (AU)

Eccentricity

Conic Type

300

1.22027

0.4768

Elliptical

200

1.20035

0.2720

Elliptical

100

83.1556

0.9452

Elliptical

50

–0.224615

4.1741

Hyperbolic

25

–0.0418759

16.1385

Hyperbolic

Notice how the 100-day transfer has an extremely large semimajor axis and an eccentricity close to 1 since this transfer time is only about 3 days more than that for a parabolic transfer orbit.

23

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 13 Consider Lambert’s problem. Determine the expression for the parabolic transfer time 𝑥̈𝑥̈, i.e. the transfer time between 𝑥̈1 and 𝑥̈2 assuming that the orbit is a parabola, or

,

𝑥̈𝑥̈ =

2 “ ‰ 𝑥̈3{2 — sgn(sin Δ𝑥̈)(𝑥̈ — 𝑥̈)3{2

,

3 𝑥̈ Start with Lambert’s equation for an elliptical orbit,

,

𝑥̈(𝑥̈2 — 𝑥̈1) = 𝑥̈3{2 [𝑥̈ — 𝑥̈ — (sin 𝑥̈ — sin 𝑥̈)]

and proceed to the limit as 𝑥̈ → ∞. Be sure to account for the two cases, Δ𝑥̈ “ 𝑥̈ and Δ𝑥̈ > 𝑥̈.

Solution We start with Lambert’s equation

,

𝑥̈(𝑥̈2 — 𝑥̈1) = 𝑥̈3{2 [𝑥̈ — 𝑥̈ — (sin 𝑥̈ — sin 𝑥̈)]

As 𝑥̈ → ∞, then ( 𝑥̈— 2 𝑥̈1) → 𝑥̈𝑥̈. Let’s start the proof by considering the following term of Lambert’s equation 𝑥̈3{2(𝑥̈ — sin 𝑥̈) Using Equations (5.59) and (5.61) with the fact that the orbit is parabolic, we can rewrite 𝑥̈ as c 𝑥̈ 𝑥̈ = 2 arcsin 2𝑥̈ so we clearly see that as 𝑥̈ → ∞, then 𝑥̈ → 0. Furthermore, recall the Mclaurin series expansion for the sine and inverse sine functions for a generic function of x 8 ÿ (—1)𝑥̈ 2𝑥̈`1 𝑥̈3 𝑥̈5 sin = 𝑥̈ = 𝑥̈ ....

— . — 3! 5! ( 2𝑥̈ . 𝑥̈“0 3 5 1)!

8 (2𝑥̈)! 𝑥̈ ÿ 3𝑥̈ arcsin 𝑥̈ = 𝑥̈2𝑥̈`1 = 𝑥̈ . . ... 𝑥̈ 2 .

4 (𝑥̈!) (2𝑥̈ . 1) 6 40 𝑥̈“0

Using these expansions, we can rewrite 𝑥̈ — sin 𝑥̈ as , ˆ 22˙ 𝑥̈ —sin 𝑥̈ = . HOTs 3!𝑥̈3{2 where the first order terms in 𝑥̈ sum to zero, so we keep the next term in the series, while HOTs represent Higher Order Terms of the series expansion which approach zero as 𝑥̈ → ∞. Combining this term with the 𝑥̈3{2 term from Lambert’s equation and taking the limit as 𝑥̈ → ∞, we get , ˆ 22 ˙ 𝑥̈3{2 𝑥̈ sin 𝑥̈ 𝑥̈3{2 𝑥̈3{2 HOTs ( — )= .

3!𝑥̈3{2 24

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

𝑥̈3{2(𝑥̈ — sin 𝑥̈) →

,

2 3{2 𝑥̈ 3

25

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

as 𝑥̈ → ∞. Let’s now look at the term involving 𝑥̈ 𝑥̈3{2(𝑥̈ — sin 𝑥̈) Similarly to before, we can rewrite 𝑥̈ as 𝑥̈

=

2 arcsin

c 𝑥̈ — 𝑥̈ 2𝑥̈

which approaches zero as 𝑥̈ → ∞. Expanding and combining the 𝑥̈ and sin 𝑥̈ terms, along with using the McLaurin series expansions stated above, gives

„

3{2

𝑥̈ (𝑥̈ — sin 𝑥̈) = 𝑥̈

3{2

,

𝑥̈3{2(𝑥̈ — sin 𝑥̈) →

(𝑥̈ — 𝑥̈)3{2 8 , 3{2 . HOTs

ȷ

2 2𝑥̈

2 (𝑥̈ — 𝑥̈3{2 ) 3

as 𝑥̈ → ∞. Notice, however, that the sign of 𝑥̈ is determined by the value ofΔ𝑥̈ such that # 𝑥̈ = Combining all terms gives

.𝑥̈ if Δ𝑥̈ < [0, 𝑥̈) —𝑥̈ if Δ𝑥̈ < [𝑥̈, 2𝑥̈)

,

2 “ ‰ 𝑥̈ = , 𝑥̈3{2 — sgn(sin Δ𝑥̈)(𝑥̈ — 𝑥̈)3{2 3 𝑥̈ 𝑥̈

which is the equation for 𝑥̈𝑥̈ that we needed to prove.

26

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 14 Show that 𝑥̈𝑥̈ =

2(𝑥̈ — 𝑥̈1)(𝑥̈ — 𝑥̈2) 𝑥̈1𝑥̈2(1 — cos Δ𝑥̈) = 𝑥̈ 𝑥̈

Solution We can simplify the denominator in the problem statement so that 2(𝑥̈ — 𝑥̈1)(𝑥̈ — 𝑥̈2) = 𝑥̈1𝑥̈2(1 — cos Δ𝑥̈) Defining the left-hand side of this equation ‘LHS’ and the right-hand side as ‘RHS’, we thus need to prove that LHS = RHS. Seeing that the RHS has a cos Δ𝑥̈ term, we start with the definition of the chord using the law of cosines and simplify until we obtain the RHS of the above equation 𝑥̈2 = 𝑥̈12 . 𝑥̈ 22 — 2𝑥̈1𝑥̈2 cos Δ𝑥̈ 2𝑥̈1𝑥̈2 cos Δ𝑥̈ = 𝑥̈ 2 . 𝑥̈ 2 — 𝑥̈2 1

2

2

2

—𝑥̈1 — 𝑥̈2 .

2

𝑥̈1𝑥̈2 — . 𝑥̈1𝑥̈2 cos Δ𝑥̈ = 𝑥̈ 𝑥̈1𝑥̈2 2 . 2𝑥̈1𝑥̈2 . 𝑥̈ 2 —𝑥̈21 —2 𝑥̈ 2 𝑥̈ 𝑥̈ (1 — cos Δ𝑥̈) = 1 2 2 𝑥̈2 — (𝑥̈2 — 𝑥̈1)2 𝑥̈1𝑥̈2(1 — cos Δ𝑥̈) = 2 2 𝑥̈ — (𝑥̈2 — 𝑥̈1)2 RHS = 2 Thus, it is now sufficient to show that LHS is equal to 𝑥̈ ´p𝑥̈22´𝑥̈1q =𝑥̈ 𝑥̈. Since LHS has terms involving the semiperimeter 𝑥̈, we rewrite those terms using Equation (5.50) 2

2

𝑥̈1 . 𝑥̈2 . 𝑥̈ 𝑥̈ . (𝑥̈2 — 𝑥̈1) — 𝑥̈1 = 2 2 𝑥̈1 . 𝑥̈2 . 𝑥̈ 𝑥̈ — (𝑥̈2 — 𝑥̈1) (𝑥̈ — 𝑥̈ 2) = — 𝑥̈2 = 2 2

(𝑥̈ — 𝑥̈ 1) =

so that LHS can be rewritten as 1 LHS = 2(𝑥̈ — 𝑥̈1)(𝑥̈ — 2) = [ 𝑥̈ .

2 𝑥̈ (𝑥̈2 =

— 𝑥̈1)][𝑥̈ — (𝑥̈2

— 𝑥̈1)]

𝑥̈2 — (𝑥̈2 — 𝑥̈1)2 2

which is equal to 𝑥̈, thus concluding the proof – we used the difference of squares formula (𝑥̈ — 𝑥̈)(𝑥̈ . 𝑥̈) = 𝑥̈2 — 𝑥̈2 in the last step.

27

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 15 At time 𝑥̈0=0, the International Space Station (ISS) is ejecting a CubeSat through its CubeSat deployer mechanism. Assume the ISS is in a geocentric circular orbit with period of 92 minutes.

With respect to the Cartesian Hill-Clohessy-Wiltshire (HCW) frame of reference, the CubeSat is ejected with a relative velocity of 𝑥̈v(𝑥̈𝑥̈) = 5i . 10j . 0k m/s (a) Using the HCW relative motion equations, compute how far away (in km) the satellite is from the ISS at t = 46 minutes. (b) What happens to the relative distance as time increases? Does this make physical sense?

Briefly explain why or why not.

Solution (a) The CubeSat is ejected from the ISS at 𝑥̈ = 𝑥̈0 = 0 so 𝑥̈r0 = 0𝑥̈ . 0𝑥̈ . 0𝑥̈ km 𝑥̈v0 = 5 × 10´3𝑥̈ . 10 × 10´3𝑥̈ . 0𝑥̈ km/s Noting that the final time 𝑥̈ = 46 min = 𝑥̈2 2𝑥̈ = 𝑥̈ 𝑥̈𝑥̈ = 𝑥̈ 2 𝑥̈ 2𝑥̈ = 1.13826 × 10´3 rad/s 𝑥̈ = 𝑥̈ Using Equation (5.125), which is the state transition matrix definition of the HWC equations, we can write

"

*

„

ȷ

"

𝑥̈r(𝑥̈) Φ𝑥̈𝑥̈ (𝑥̈) Φ𝑥̈𝑥̈(𝑥̈) 𝑥̈r0 = = 𝑥̈v(𝑥̈) Φ𝑥̈𝑥̈ (𝑥̈) Φ𝑥̈𝑥̈(𝑥̈) 𝑥̈v0

*

so that c s0 𝑥̈r(𝑥̈) =c𝑟𝑟 Φ c(c 𝑥̈ )𝑥̈cr 0 .Φ 𝑟𝑣 (𝑥̈)𝑥̈v0

= Φ𝑥̈𝑥̈(𝑥̈)𝑥̈v0 In Cartesian coordinates

»

2 (1 — cos 𝑥̈𝑥̈) 𝑛

1 sin 𝑥̈𝑥̈ 𝑛

0

fi

Φ𝑥̈𝑥̈(𝑥̈) = –—𝑛2 (1 — cos 𝑥̈𝑥̈) 𝑛1 (4 sin 𝑥̈𝑥̈ — 3𝑥̈𝑥̈) 0 fl 1 sin 𝑛𝑡 0 0 𝑛 fi 4 » 0 0 𝑛 3𝑥̈ = –— 4𝑛 — 𝑥̈ 0fl 0 0 0 so that

»

04

𝑥̈r(46 min) = –— 𝑛 0

4 𝑛 3𝑥̈ — 𝑥̈

0

, fi $ , 0 &5 × 1 0 ´ 3 . $ & 35.1414 . = —100.371 km 0fl 10´4 % - % 0 0 0 28

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

(b) As 𝑥̈ → ∞, then |𝑥̈r | → ∞, which does not make physical sense since the spacecraft is not on an escape trajectory from Earth. However, this happens because the HWC equations are only valid for “small" |𝑥̈r|.

29

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 16 During the Apollo program during the lunar orbit rendezvous, the lunar module launched from the Moon and would rendezvous with the command module. Had the lunar module been able to get into orbit, but was not able to get to the command module, the command module could have maneuvered to rendezvous with the lunar module. (a) Assume that the command module is in a 100 km altitude circular orbit about the Moon.

The lunar module is a circular orbit 100 km behind and 10 km below the command module.

What is the Δ𝑥̈ necessary for the lunar module to rendezvous with the command in 1 hour? (b) If the command module had to go after the lunar module in its orbit, what Δ𝑥̈ would be needed to rendezvous in 1 hour (assuming the same relative positions in part (a)?

Solution From Appendix A, we can find the gravitational parameter and physical radius of the Moon 𝑥̈K = 4, 903 km3/s2 𝑥̈K = 1, 732 km (a) For part (a) we assume that the target is the Command Module (CM), which orbits the Moon in a circular orbit with radius 𝑥̈0 𝑥̈ =K 100 . km 1, 838 = km, and the chaser is the Lunar Module (LM) which has relative position and velocity given as follows 𝑥̈r0 = —10𝑥̈𝑥̈ . —100𝑥̈𝑥̈ . 0𝑥̈𝑥̈ km 𝑥̈v0 = 0𝑥̈𝑥̈ . 0𝑥̈𝑥̈ . 0𝑥̈𝑥̈ km/s Using the HCW definitions in cylindrical coordinates as given by Equations (5.131) - (5.136) along with the provided time of flight 𝑥̈=1 hour, we can compute the necessary Δ𝑥̈ maneuvers to initiate and complete the rendezvous between the LM and CM Δ𝑥̈1 = 19.7023 m/s Δ𝑥̈2 = 16.3806 m/s Δ𝑥̈𝑥̈𝑥̈𝑥̈ = 36.0899 m/s This rendezvous maneuver is shown in Figure 4. (b) For part (b) we switch the target and chaser roles, so that the target is now the LM and the chaser is the CM. This means that the relative motion of the (new) chaser with respect to the target is 𝑥̈r0 = 10𝑥̈𝑥̈ . 100𝑥̈𝑥̈ . 0𝑥̈𝑥̈ km 𝑥̈v0 = 0𝑥̈𝑥̈ . 0𝑥̈𝑥̈ . 0𝑥̈𝑥̈ km/s We again use the HCW definitions in cylindrical coordinates as given by Equations (5.131) - (5.136) along with the provided time of flight 𝑥̈ = 1 hour, to compute the necessary Δ𝑥̈ maneuvers to 30

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

initiate and complete the rendezvous between the LM and CM Δ𝑥̈1 = 19.6377 m/s Δ𝑥̈2 = 16.1805 m/s Δ𝑥̈𝑥̈𝑥̈𝑥̈ = 35.8182 m/s This rendezvous maneuver is shown in Figure 5.

The table below summarizes the results from parts (a) and (b).

Figure 4: Lunar Rendezvous between the CM (target) and LM (chaser)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 5: Lunar Rendezvous between the LM (target) and CM (chaser) Part Parameter Value (m/s) (a) (b) (a) (b) (a) (b) (a) (b) (a) (b)

|𝑥̈v`| 0

|𝑥̈v` | 𝑥̈

Δ𝑥̈1 Δ𝑥̈2 Δ𝑥̈𝑥̈𝑥̈𝑥̈

22.4770 22.4189 16.3806 16.1805 19.7023 19.6377 16.3806 16.1805 36.0899 35.8182

32

Reference Equation (5.133) Equation (5.135) Equation (5.134) Equation (5.136) Δ𝑥̈1 . Δ𝑥̈2

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 17 Compute the Δ𝑥̈1, Δ𝑥̈2, and Δ𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ (in km/s) and TOF (in years) required to perform an interplanetary Hohmann transfer from Earth to Jupiter. Assume that you start from a circular LEO having altitude of 300 km and you are targeting a circular orbit around Jupiter having altitude of 4,300 km. How could one lower the Δ𝑥̈ requirements for a mission targeting Jupiter? Explain, referring to maneuvers discussed in this chapter.

Solution Using Appendix A, we can find that the gravitational parameters and physical constants for this problem 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈C = 398, 600 km3/s2 𝑥̈E = 1.268 × 108 km3/s2 𝑥̈C{@ = 149, 597, 800 km 𝑥̈E{@ = 778, 327, 433.8 km 𝑥̈C = 6, 378.140 km 𝑥̈E = 71, 490 km We assume that the orbits of Earth and Jupiter are circular and coplanar. We start our analysis by computing the orbital speeds of the planets with respect to the Sun 𝑥̈ c @ = 29.7831 km/s = 𝑥̈ C{@ c 𝑥̈@ 𝑥̈E{@ = 𝑥̈ = 13.0563 km/s

𝑥̈C{@

E{@

We then compute the semimajor axis and energy of the Hohmann transfer ellipse between Earth and Jupiter 1 𝑥̈𝑥̈ = (𝑥̈C{@ . 𝑥̈E{@) = 464, 059, 200 km 2 — 𝑥̈𝑥̈ = —142.965 km 2s 2 𝑥̈@ = 2𝑥̈ 𝑥̈

We can therefore compute the required spacecraft orbital speeds at perhelion (Earth departure) and aphelion (Jupiter arrival) as follows

d ˆ 𝑥̈𝑥̈{@ = 2 𝑥̈𝑥̈ .

d ˆ

𝑥̈@ 𝑥̈ C{@

𝑥̈𝑥̈{@ = 2 𝑥̈𝑥̈ .

𝑥̈@ 𝑥̈ 33

˙ = 38.5763 km/s ˙ = 7.41205 km/s

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

E{@

34

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Then, we compute the speeds that the spacecraft will have at the boundary of sphere of influence with respect to Earth at departure and Jupiter at arrival 𝑥̈8{C = 𝑥̈𝑥̈{@ — 𝑥̈C{@ = 8.79325 km/s 𝑥̈8{E = 𝑥̈E{@ — 𝑥̈𝑥̈{@ = 5.64299 km/s Now we have enough information to compute the Δ𝑥̈ maneuvers required to leave Earth and be captured as Jupiter. We start by computing the radii and velocities of the starting and arrival orbits at Earth and Jupiter, respectively 𝑥̈1 = 𝑥̈C . 300 km = 6, 678.140 km c 𝑥̈ 𝑥̈1 = = 7.72584 km/s C 𝑥̈ 1

𝑥̈2 = 𝑥̈E . 4, 300 km = 75, 490 km c 𝑥̈ 𝑥̈2 = = 40.8867 km/s E 𝑥̈ 2

In order to compute Δ𝑥̈1, we perform an energy balance using the vis-viva equation, equating the energy immediately after Δ𝑥̈1 is performed and at the boundary of Earth’s sphere of influence

(𝑥̈1 . Δ𝑣1)2 2

𝑥̈C — 𝑥̈1

=

2 𝑣8{ C

2

which gives Δ𝑥̈1 = 6.29908 km/s (considering only the positive root of the quadratic equation).

Here, 𝑥̈3 = 8{C 𝑥̈2 = 77.3227 km2/s2. Although we do not use 𝑥̈3 in these calculations, 𝑥̈3 is generally used when determining launch vehicles for interplanetary missions – in fact, launch vehicle performance is often cited using 𝑥̈3. Furthermore, a few simple algebraic manipulations can be used to show that Δ𝑥̈1 can be computed as b 2 Δ𝑣1 = 𝑣 𝑥̈𝑥̈𝑥̈ { . 𝑥̈3 — 𝑥̈1 C

,

where 𝑥̈𝑥̈𝑥̈𝑥̈{C= 2𝑥̈1 is Earth’s escape velocity at 𝑥̈1. At arrival at Jupiter, we can perform a similar energy balance to that used for Earth departure. Alternatively, we can compute the velocity that the spacecraft has at perijove

c

𝑥̈𝑥̈{E = 𝑥̈2

.

8{E

2𝑥̈E 𝑥̈2

= 58.0973 km/s

b 2𝑥̈ , E = 2𝑥̈2, where the second term under the square root can be recognized to be 𝑥̈𝑥̈𝑥̈𝑥̈{E = 𝑥̈2 i.e. Jupiter’s escape velocity at 𝑥̈2. The required Δ𝑥̈2 to be captured into a Jovian circular orbit at altitude 4,300 km can be computed as Δ𝑥̈2 = 𝑥̈𝑥̈{E — 𝑥̈2 = 17.2106 km/s which, when summed to Δ𝑥̈1 gives a total Δ𝑥̈ of 35

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Δ𝑥̈𝑥̈𝑥̈𝑥̈ = Δ𝑥̈1 . Δ𝑥̈2 = 23.50968 km/s

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Time of flight can be computed as d 𝑥̈ 𝑥̈𝑥̈ = 𝑥̈

3

𝑎𝑥̈ = 998.0 days or 2.734 years 𝑥̈ @

In order to reduce the mission Δ𝑥̈, one could opt to enter an elliptical orbit at Jupiter instead of a circular orbit. Assuming a fixed perijove at an altitude of 4,300 km, the capture speed can be computed as a function of final orbital eccentricity d 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ =

𝑥̈E(1 . 𝑥̈) 𝑥̈ 2

so the minimum capture speed the spacecraft must achieve can be computed by setting 𝑥̈= 1 (parabolic orbit), which gives 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ = 57.8226 km/s. This means that the minimum Δ𝑥̈2 to be captured must be Δ𝑥̈2 > 𝑥̈𝑥̈{E — 𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈, 𝑥̈“1 = 0.274723 km/s in order to be captured into an elliptical orbit. Another option would be to perform planetary flybys of other planets, increasing TOF but reducing total Δ𝑥̈.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 18 Compute the Δ𝑥̈1, Δ𝑥̈2, and Δ𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈ (in km/s) and TOF (in years) required to perform an interplanetary Hohmann transfer from Earth to Venus. Assume that you start from a circular LEO having altitude of 300 km and you are targeting a circular orbit around Venus having altitude of 500 km.

Solution Using Appendix A, we can find that the gravitational parameters and physical constants for this problem 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈C = 398, 600 km3/s2 𝑥̈B = 325, 700 km3/s2 𝑥̈C{@ = 149, 597, 800 km 𝑥̈B{@ = 108, 204, 088.7 km 𝑥̈C = 6, 378.140 km 𝑥̈B = 6, 051.9 km We assume that the orbits of Earth and Venus are circular and coplanar. We start our analysis by computing the orbital speeds of the planets with respect to the Sun 𝑥̈ c @ = 29.7831 km/s = 𝑥̈ C{@ c 𝑥̈@ 𝑥̈B{@ = 𝑥̈ = 35.0214 km/s

𝑥̈C{@

B{@

We then compute the semimajor axis and energy of the Hohmann transfer ellipse between Earth and Venus 1 𝑥̈𝑥̈ = (𝑥̈C{@ . 𝑥̈B{@) = 128, 900, 944 km or 0.861650 AU 2 — 𝑥̈𝑥̈ = —514.7829 km 2s 2 𝑥̈@ = 2𝑥̈ 𝑥̈

We can therefore compute the required spacecraft orbital speeds at aphelion (Earth departure) and perihelion (Venus arrival) as follows

d ˆ 𝑥̈𝑥̈{@ = 2 𝑥̈𝑥̈ .

d ˆ 𝑥̈𝑥̈{@ = 2 𝑥̈𝑥̈ .

𝑥̈@ 𝑥̈ C{@ 𝑥̈@

˙ = 27.28891 km/s

˙

𝑟B{@ 38

= 37.72835 km/s

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Then, we compute the speeds that the spacecraft will have at the boundary of sphere of influence with respect to Earth at departure and Venus at arrival 𝑥̈8{C = 𝑥̈C{@ — 𝑥̈𝑥̈{@ = 2.495734 km/s 𝑥̈8{B = 𝑥̈𝑥̈{@ — 𝑥̈B{@ = 2.706970 km/s noting that since we are analyzing a transfer from an outer planet to an inner planet, we would need to have the spacecraft depart Earth in the opposite direction of Earth’s motion with respect to the Sun. Now we have enough information to compute the Δ𝑥̈ maneuvers required to leave Earth and be captured as Jupiter. We start by computing the radii and velocities of the starting and arrival orbits at Earth and Venus, respectively 𝑥̈1 = 𝑥̈C . 300 km = 6, 678.140 km c 𝑥̈ 𝑥̈1 = = 7.72584 km/s C 𝑥̈ 1

𝑥̈2 = 𝑥̈B . 500 km = 6, 551.9 km c 𝑥̈ 𝑥̈2 = = 7.050587 km/s B 𝑥̈ 2

In order to compute Δ𝑥̈1, we perform an energy balance using the vis-viva equation, equating the energy immediately after Δ𝑥̈1 is performed and at the boundary of Earth’s sphere of influence

(𝑥̈1 . Δ𝑣1)2 2

𝑥̈C — 𝑥̈1

=

2 𝑣8{ C

2

which gives Δ𝑥̈1 = 3.481531 km/s (considering only the positive root of the quadratic equation).

Here, 𝑥̈3 = 8{C 𝑥̈2 = 6.228687 km2/s2. Although we do not use 𝑥̈3 in these calculations, 𝑥̈3 is generally used when determining launch vehicles for interplanetary missions – in fact, launch vehicle performance is often cited using 𝑥̈3. Furthermore, a few simple algebraic manipulations can be used to show that Δ𝑥̈1 can be computed as b 2 Δ𝑣1 = 𝑣 𝑥̈𝑥̈𝑥̈ { . 𝑥̈3 — 𝑥̈1 C

,

where 𝑥̈𝑥̈𝑥̈𝑥̈{C= 2𝑥̈1 is Earth’s escape velocity at 𝑥̈1. At arrival at Venus, we can perform a similar energy balance to that used for Earth departure. Alternatively, we can compute the velocity that the spacecraft has at pericytherion

c

𝑥̈𝑥̈{B = 𝑥̈28{B .

2𝑥̈B 𝑥̈2

= 10.331952 km/s b

,

B where the second term under the square root can be recognized to be 𝑥̈𝑥̈𝑥̈𝑥̈{B = 2𝑥̈ = 2𝑥̈2, i.e. 𝑟2 Venus’s escape velocity at 𝑥̈2. The required Δ𝑥̈2 to be captured into a Venusian circular orbit at altitude 500 km can be computed as

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Δ𝑥̈2 = 𝑥̈𝑥̈{B — 𝑥̈2 = 3.281365 km/s

40

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

which, when summed to Δ𝑥̈1 gives a total Δ𝑥̈ of Δ𝑥̈𝑥̈𝑥̈𝑥̈ = Δ𝑥̈1 . Δ𝑥̈2 = 6.762896 km/s Time of flight can be computed as

d 𝑇 𝑂𝐹 = 𝜋

𝑥̈𝐻3 = 146.07 days or 0.399922 years 𝜇@

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 19 Create a porkchop plot for Earth-Venus transfers for the following sets of dates • Departure dates: 01 Sep 2052 through 01 Mar 2054 • Arrival dates: 01 Sep 2053 through 01 Jul 2054 using 1-day time steps. Use planetary ephemerides provided either by JPL Horizons or MATLAB’s planetaryEphemeris function. Assuming departure and arrival planetary altitudes of 400 km and 500 km, respectively, compute and plot a porkchop plot for total mission Δ𝑥̈. What is the minimum total Δ𝑥̈? For which departure and arrival dates does it occur? Is the transfer type I or type II? Solution Figure 6 shows the porkchop plot for Earth-Venus transfers during the desired time frame.

Figure 6: Earth-Venus Hohmann Transfer (outbound) The following table summarizes the minimum total Δ𝑥̈ case for the given time frame. A type I transfer is the optimal transfer for this time frame.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Value Δ𝑣𝑡𝑜𝑡 Δ𝑣1 Δ𝑣2 𝐶3 at launch 𝑣8 at arrival Departure date

30 Oct 2053

Arrival date TOF

days

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 20 Recreate the porkchop plot shown in Figure 5.22. Use similar contour levels for C3 at launch, arrival 𝑥̈8, and TOF. Solution In order to recreate the porkchop plot in Figure 5.22, the classical or unversal variable solutions for Lambert’s problem can be used. Refer to Section 5.3 and 5.8 for additional details.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 21 Assume that a spacecraft performs an Earth-Venus interplanetary Hohmann transfer – assume circular coplanar orbits for simplicity. However, at arrival, instead of performing an orbit insertion, the spacecraft does a leading edge flyby of Venus. Compute and plot the turn angle, hyperbolic flyby eccentricity, impact parameter, and post-flyby aphelion as a function of flyby altitude, from a minimum altitude of 200 km to a maximum altitude of 10,000 km.

Solution Using Appendix A, we can find that the gravitational parameters and physical constants for this problem 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈C = 398, 600 km3/s2 𝑥̈B = 325, 700 km3/s2 𝑥̈C{@ = 149, 597, 800 km 𝑥̈B{@ = 108, 204, 088.7 km 𝑥̈C = 6, 378.140 km 𝑥̈B = 6, 051.9 km Refer to the solution of Problem 18 to obtain the required 𝑥̈8{B = 2.706970 km/s.

For each flyby trajectory, we have a pericytherion ranging from 𝑥̈𝑥̈= 𝑥̈B .200 km 6, =251.9 km to 𝑥̈𝑥̈ 𝑥̈= B 10,.000 km 16,=051.9 km, which we use in combination with 𝑥̈8{B to compute flyby eccentricity, turn angle, and impact parameter as follows (for simplicity, we dropped the subscript /B) 𝑥̈𝐹𝐵 = 1 .

𝑥̈𝑥̈𝑥̈2

8

𝑥̈

ˆ 1˙ 𝑥̈ 2 arcsin = 𝑥̈𝑥̈𝑥̈

d

Δ = 𝑥̈𝑝 1 .

2𝑥̈ 𝑟𝑝 𝑣28

To compute aphelion, we follow the steps given in Section 5.12.1.1 to compute the post flyby heliocentric trajectory. Then, we compute aphelion as follows (all the following orbital parameters

45

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

are with respect to the post-flyby heliocentric trajectory) 𝑥̈@ 𝑥̈ = ` 2 2𝑥̈@ ( 𝑥̈ ) — 𝑥̈B{@ 𝑥̈ = 𝑥̈B{@ 𝑥̈𝜃 ` 𝑥̈ =

𝑥̈2 𝑥̈◉

c

𝑥̈ 𝑥̈ = 1 . 𝑎

𝑥̈𝑥̈{@ = 𝑥̈(1 . 𝑥̈) Figure 7 shows the turn angle, eccentricity, impact parameter, and aphelion as a function of flyby altitude at Venus.

46

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 7: Turn angle, Eccentricity, Impact Parameter, and Aphelion as a Function of Flyby Altitude

47

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 22 For a hyperbolic gravity assist (GA) of a target planet “ ⊙ 2𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈b and physical radius 𝑥̈b, the periapsis of the hyperbolic GA trajectory has eccentricity 𝑥̈, periapse radius 𝑥̈𝑥̈, and the inbound and outbound velocities at the sphere of influence of the planet are v´ and v`, 8 8

respectively. Determine the values of 𝑥̈𝑥̈ and magnitude of 8 v´ that yield the maximum possible magnitude of the gravity assist Δ𝑥̈𝑥̈𝑥̈. Express your answers for 𝑥̈𝑥̈ and 8 𝑥̈´ in terms of 𝑥̈b and 𝑥̈b (noting that 𝑥̈𝑥̈ě 𝑥̈b). Then, determine the maximum value of Δ𝑥̈𝑥̈𝑥̈ in terms of 𝑥̈b and 𝑥̈b. Lastly, determine the numerical values for the corresponding turn angle 𝑥̈ and eccentricity 𝑥̈.

Solution We use the law of cosines to compute Δ𝑥̈𝑥̈𝑥̈, using the fact that |v´| = |v`| = 𝑥̈8 8

a

8

Δ𝑥̈𝑥̈𝑥̈ = 𝑥̈28. 𝑥̈2 8— 2𝑥̈8𝑥̈8 cos 𝑥̈ a = 𝑥̈8 2(1 — cos 𝑥̈) 𝑥̈ = 2𝑥̈8 sin 2 where we used a trig identity in the last step. Turn angle 𝑥̈ and eccentricity can be written as

ˆ ˙ 1 𝑒

𝑥̈ = 2 arcsin 𝑥̈ = 1 .

𝑥̈𝑥̈ 𝑥̈2

8

𝑥̈b

Combining the above equations with the expression for Δ𝑥̈𝑥̈𝑥̈ we get Δ𝑥̈𝑥̈𝑥̈ = 2𝑥̈8𝑥̈b 𝑥̈ . 𝑥̈2 𝑥̈

8 𝑥̈

b

from which we can see that Δ𝑥̈𝑥̈𝑥̈ is maximized when 𝑥̈𝑥̈ is minimized, i.e. when 𝑥̈𝑥̈= 𝑥̈b. To find the value of 𝑥̈8 that maximizes Δ𝑥̈𝑥̈𝑥̈, we take a partial derivative of Δ𝑥̈𝑥̈𝑥̈ with respect to 𝑥̈8

BΔ𝑥̈𝑥̈𝑥̈ B𝑥̈8

=

2𝑥̈ 2 — 2𝑥̈b𝑥̈𝑥̈𝑥̈2 b

8

(𝑥̈b . 𝑥̈82 𝑥̈𝑥̈)2

Setting the above derivate equal to zero and solving for 𝑥̈8 gives c 𝑥̈ b 𝑥̈8 = 𝑥̈ 𝑥̈

We can easily show that the second derivative is negative at this value of 𝑥̈8, thus confirming that this value is a maximum. Since we know that 𝑥̈𝑥̈ = 𝑥̈b maximizes Δ𝑥̈𝑥̈𝑥̈, we can rewrite 𝑥̈8 as 𝑥̈8 =

c

48

𝑥̈b 𝑥̈b

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Using the first equation we derived for Δ𝑥̈𝑥̈𝑥̈ and plugging in this value results in the maximum value of Δ𝑥̈𝑥̈𝑥̈ Δ𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈ =

c

𝑥̈b 𝑥̈ b

Interestingly, this value, which is the same as that computed for 𝑥̈8, is equivalent to the circular orbital speed a spacecraft would need to have to orbit the targeted planet at a theoretical altitude of zero. Using these computed values, we can find the turn angle and eccentricity that maximize Δ𝑥̈𝑥̈𝑥̈ ´b ¯2 𝑥̈b 𝑥̈b 𝑥̈b =2 = . 𝑥̈𝑥̈ 𝑥̈8 = 1 . 𝑥̈b 𝑥̈

1

2

𝑥̈ b

ˆ ˙

ˆ ˙

1 1 𝑥̈ = 2 arcsin = 2 arcsin = 60˝ 𝑒 2

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Problem 23 A spacecraft is in orbit around Mars and is beginning to aerobrake. Initially, it starts out in an orbit with periapse and apoapse altitudes of 140 and 1000 km, respectively. Use the following atmospheric and spacecraft physical characteristics Density Model

Values

Reference Density

30 kg/km3

Reference Altitude

140 km

Scale Height

10 km

Drag Coefficient

2.2

Projected Area

23 m2

Mass

1100 kg

a Propagate the orbit and estimate how long it takes to reduce the apoapse to 140 km altitude. b If a propulsive maneuver of 1 m/s is added at apoapse on the first pass, how long does it take to reduce the apoapse to 140 km? c If a propulsive maneuver of 10 m/s is added at apoapse on the first pass, how long does it take to reduce the apoapse to 140 km?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Mars are 𝑥̈D = 43, 050 km3/s2 𝑥̈D = 3, 389.5 km In order to determine when the apoapse altitude reaches 140 km, we numerically integrate Equations (5.200) using the spacecraft parameters provided and the atmospheric model given by Equation (5.213). (a) When no propulsive maneuver is implemented, the time it takes to lower apoareion altitude to 140 km is 50.3512 hours. Figures 8, 9, and 10 show the aerobraking orbits along with the change of various orbital parameters over time. Note that while apoapse altitude decreases significantly, periapse altitude also decreases, although not as rapidly. (b) When a 1 m/s propulsive maneuver is implemented, the time it takes to lower apoareion altitude to 140 km is 32.7142 hours. Figures 11, 12, and 13 show the aerobraking orbits along with the change of various orbital parameters over time. Maneuvers at apoapsis lower periapsis, thus increasing the effects of drag and speeding up the aerobraking maneuvers. However, this is expected to increase the flux and temperature experienced by the spacecraft. (c) When a 10 m/s propulsive maneuver is implemented, the time it takes to lower apoareion show the aerobraking orbits along altitude to 140 km is only 1.05319 hours. Figures 14, 15, and with the change of various orbital parameters over time.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 8: Aerobraking Orbits (no Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 9: Altitude, Velocity, Energy, and Dynamic Pressure as functions of time (no Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 10: Semimajor Axis, Eccentricity, Apoapse Altitude, and Periapse Altitude as functions of time (no Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 11: Aerobraking Orbits (1 m/s Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 12: Altitude, Velocity, Energy, and Dynamic Pressure as functions of time (1 m/s Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 13: Semimajor Axis, Eccentricity, Apoapse Altitude, and Periapse Altitude as functions of time (1 m/s Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 14: Aerobraking Orbits (10 m/s Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 15: Altitude, Velocity, Energy, and Dynamic Pressure as functions of time (10 m/s Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 16: Semimajor Axis, Eccentricity, Apoapse Altitude, and Periapse Altitude as functions of time (10 m/s Propulsive Maneuver)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 24 A spacecraft is entering the Martian atmosphere, on its way to land. At 112,000 m altitude, it is traveling at 5588 m/s. a Using the atmospheric drag parameters in the table below, integrate the equations of motion and estimate the velocity 7 minutes later. Be sure to state your assumptions.

Density Model

Range

Values

Reference Density

0–25 km

0.0525 kg/m3

25–125 km

0.0159 kg/m3

0–25 km

11.049 km

25–125 km

7.295 km

N/A

65 kg/m2

Scale Height Ballistic Coefficient

b Provide three plots: altitude vs. time, heating vs. time, and dynamic pressure vs. time.

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Mars are 𝑥̈D = 43, 050 km3/s2 𝑥̈D = 3, 389.5 km We numerically integrate Equations (5.200) using the spacecraft parameters provided and the atmospheric model given by Equation (5.213), noting that here we have a piece-wise distribution for the Martian atmospheric density (which is what causes discontinuities in the final results). For this problem, we simulated the entry, descent, and landing (EDL) for entry flight path angles (EFPA) ranging from —10˝ to —5˝. Figure 17 shows the results of altitude, heat flux, and dynamic ˝ the spacecraft does not fly through pressure as functions of time. It can be seen that if EFPA >7— the atmosphere at a low enough altitude to land in one pass.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 17: Altitude, Heat Flux, and Dynamic Pressure as Functions of Time for Various EFPAs

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 25 Design a roundtrip trajectory for an Earth-Mars crewed mission during the 2050s, such that a crew is given a stay time between 50 and 100 days. What types of outbound and inbound transfers would you choose? Explain why, considering the time the crew spends in space and ensuring that your total mission Δ𝑥̈ is within a reasonable value. Use JPL’s planetary ephemeris data to obtain the position and velocities of Earth and Mars. (Just like real mission design, there is no unique answer for this problem) Solution Design problem.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 26 Compute the semimajor axis, eccentricity, radii of perihelion and aphelion for the Aldrin cycler discussed in Section 5.15.3. For this problem, assume that the orbits of Earth and Mars are coplanar and circular. Furthermore, assume that Earth and Mars orbit the Sun in exactly 1 year and 1.875 years, respectively. This means that the cycler would encounter Earth on successive passes that are 1/7 of an orbit (or 51.4˝) apart. Lastly, assume that Mars flybys do not provide any changes in the heliocentric orbit of the cycler, so that only Earth flybys can provide the necessary orbital changes to maintain the cycler’s orbit periodicity.

Hint: you will need to solve a multi-rev Lambert’s problem to find the semimajor axis of the cycler orbit.

Solution The problem statement tells us that 𝑥̈C = 1 year 𝑥̈D = 1.875 years So we can compute the synodic period of Earth and Mars as 𝑥̈C𝑥̈D 𝑥̈𝑥̈𝑥̈𝑥̈ = = 2.1429 years |𝑥̈C — 𝑥̈D | which is equivalent to 2 and 1/7 years. Since the cycler needs to rotate its heliocentric orbit by 51.4˝ at every Earth flyby, it means that the heliocentric pre-flyby and post-flyby true anomalies at every Earth flyby are 25.7 — ˝ and 25.7˝, i.e. half of 51.4˝ each. Thus, in order to compute the semimajor axis of the cycler orbit, we need implement multi-rev Lambert’s problem, or Equations (5.96) and (5.97)

,

𝑥̈(𝑥̈2 — 𝑥̈1) = 𝑥̈3{2 [2𝑥̈ 𝑥̈ . 𝑥̈ — 𝑥̈ — (sin 𝑥̈ — sin 𝑥̈)] where 𝑥̈ =1 since the spacecraft re-encounters Earth after one full revolution around the Sun. Therefore, the boundary conditions are 𝑥̈@ = 4𝑥̈2 km3/s2 𝑥̈1 = 𝑥̈C{@ = 1 AU 𝑥̈2 = 𝑥̈C{@ = 1 AU Δ𝑥̈ = 51.4˝ 𝑥̈2 — 𝑥̈1 = 𝑥̈𝑥̈𝑥̈𝑥̈ = 2.1429 years where we used canonical units for simplicity. Solving Lambert’s equation using one of the methods described in Sections 5.3.1 or 5.3.2 gives 𝑥̈ = 1.59959 AU, which corresponds to an orbital period of 2.024506 years. We use the orbit equation to solve for eccentricity 𝑥̈ 1 𝑥̈2 𝑥̈ = ( — ) 1 . 𝑥̈ cos 𝑥̈ 𝑥̈ . 𝑥̈𝑥̈ cos 𝑥̈ = 𝑥̈ — 𝑥̈𝑥̈2 𝑥̈ cos 𝑥̈ 𝑥̈ — 𝑥̈ 𝑥̈2 . 𝑥̈ . =0 𝑎 𝑎 63

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

which gives a quadratic equation for 𝑥̈. We use = 𝑥̈ 1.59959 AU, 𝑥̈ = 1 AU, and either 𝑥̈= —25.7˝ ˝ or 𝑥̈ = 25.7 to get 𝑥̈ 0.392283 by solving the quadratic equation and consider only the positive = root of the equation. We can now compute perihelion and aphelion of the cycler orbit as follows 𝑥̈𝑥̈{@ = 𝑥̈(1 — 𝑥̈) = 0.972097 AU 𝑥̈𝑥̈{@ = 𝑥̈(1 . 𝑥̈) = 2.227083 AU

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 27 Compute the minimum radial-only acceleration needed to transfer a spacecraft from a circular LEO with 400 km altitude to GEO. Then, using the acceleration value you computed, find the time at which the spacecraft would reach GEO. Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Earth are 𝑥̈C = 398, 600 km3/s2 𝑥̈C = 6, 378.140 km Therefore, the spacecraft starts in a LEO having radius and velocity 𝑥̈0 = 𝑥̈C . 400 km = 6, 778.140 km c 𝑥̈ 𝑥̈0 = = 7.66855 km/s C 𝑥̈ 1

and the minimum radial-only acceleration required to achieve GEO (or to escape) must be at least 𝑥̈ 𝑥̈𝑥̈𝑥̈ ,𝑥̈, 𝑥̈𝑥̈𝑥̈ = C = 1.0844917 × 10 ´3 km/s2 8𝑥̈ 0

Recall from Problem 1 that GEO has a radius of 𝑥̈𝑥̈𝑥̈𝑥̈=42, 164.15 km. In order to find the time the spacecraft takes to reach GEO, we need to numerically integrate the equations of motion, Equation (5.247), using the radial-only acceleration that we computed. The spacecraft takes approximately 1.332057 days to reach GEO with the above constant acceleration. Figure 18 shows a plot of the spacecraft’s low-thrust orbit along with its starting orbit in LEO and the final targeted orbit, GEO.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 18: LEO, GEO, and the Radial-only Low-thrust Orbit Connecting Them

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 28 Assume that a spacecraft starts from a circular heliocentric orbit equal to that of the Earth (𝑥̈0 =1 AU). The vehicle turns on its thrusters, producing a constant tangential acceleration of 1 /32 𝑥̈ /𝑥̈0.

How long will the spacecraft take to escape the Solar System? Compute this value by integrating the equations of motion and also by using the approximations given in Section 5.16.3.2. How do the values compare to each other?

Solution Using Appendix A, we can find the gravitational parameter of the Sun and the Earth’s distance from the Sun 𝑥̈@ = 1.327 × 1011 km3/s2 𝑥̈C{@ = 149, 597, 800 km or 1 AU The tangential-only acceleration the spacecraft receives is 𝑥̈◉ = 1.85315 × 10 ´7 km/s2 𝑥̈𝑥̈𝑥̈ ,𝑥̈ = 32𝑥̈0 With the analytical approximation discussed in Section 5.16.3.2 we get the following escape radius and escape time (we arbitrarily set 𝑥̈0 = 0 here) 𝑥̈0𝑥̈0 𝑥̈𝑥̈𝑥̈𝑥̈= = 2.674961 AU 2 (20𝑥̈𝑥̈𝑥̈ ,𝑥̈ 𝑥̈02 )1{4 « ˆ ff 20𝑎 2𝑥̈𝑥̈ 𝑥̈02 ˙1{8 𝑥̈ 1— 𝑥̈𝑥̈𝑥̈𝑥̈ = = 1.979047 years ,𝑥̈ 0 4 𝑥̈ 𝑥̈ 𝑥̈𝑥̈ ,𝑥̈

0

If we numerically integrate the equations of motion, Equation (5.229), using the tangential-only acceleration computed above, we get that the escape radius and escape time are 4.762696 AU and 3.455814 years, respectively. As mentioned in Section 5.16.3.2, the analytic approximation is valid 𝜇 only when 𝑥̈𝑥̈𝑥̈ ,𝑥̈ cc 0𝑥̈2 , which explains the discrepancy in the results. A plot of the numerically integrated trajectory can be seen in Figure 19.

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Figure 19: Tangential-only Low-thrust Trajectory from Earth’s Orbit to Escape

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Interplanetary Astrodynamics Chapter 6 Problem Solutions

Problem 1 Consider a Mars-orbiting spacecraft in a circular orbit at an altitude of 500 km. The spacecraft initiates a maneuver to land on Mars, targeting a periapsis equal to the radius of Mars at 180˝ from the location of the maneuver. However, the propulsion system malfunctions, causing the spacecraft to deliver a higher retrograde Δ𝑥̈ than expected. It is believed that the extra Δ𝑥̈ that the propulsion system delivered is somewhere between 1% and 5% of the nominal Δ𝑥̈. Based on this, what is the range of new landing location angles? What is the new range of times of flight for the spacecraft to hit the Martian surface? For this problem, ignore the atmospheric effects of Mars.

Solution As provided in Appendix A, the symbol for Mars is D, and Mars’s gravitational parameter and physical radius are 𝑥̈D = 43, 050 km3/s2 𝑥̈D = 3, 397 km Thus, the starting circular orbit has radius 𝑥̈1 = 𝑥̈D . 500km = 3,897 km, and an orbital speed of 𝑥̈1 =

b

𝑥̈D = 3.32369 km/s. Figure 1 𝑟1

shows a diagram of the starting orbit (green), the nominal

deorbit trajectory (blue), and an example of a deorbit trajectory whose Δ𝑥̈ was larger than the nominal Δ𝑥̈ (purple).

The nominal Δ𝑥̈ to deorbit the spacecraft is equivalent to the Δ𝑥̈1 of a Hohmann transfer whose apoapsis is 𝑥̈1 and peripasis is 𝑥̈D

ˆc Δ𝑥̈ = 1𝑥̈

˙

2𝑥̈D — 1 = —0.115941 km/s 𝑟1 . 𝑅D

where the negative indicates that the Δ𝑥̈ is performed in the retrograde direction, as expected.

The deorbit trajectory has the following semimajor axis and TOF 1

𝑥̈ = (𝑥̈1 . 𝑥̈D) = 3, 647 km 2

d

𝑇 𝑂𝐹 = 𝜋

𝑥̈3 = 55.58 min 𝜇D

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 1: Diagram of the Deorbit Trajectory (not to scale) We can therefore compute the 1% and 5% extra Δ𝑥̈ values to obtain a range of minimum and maximum apoapse velocities, 𝑥̈𝑥̈ that the spacecraft would have as a result of the additional velocity change Δ𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈ = 1.01Δ𝑥̈ = 0.117100 km/s Δ𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈ = 1.05Δ𝑥̈ = 0.121738 km/s so that 𝑥̈𝑥̈1 = 𝑥̈1 — Δ𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈ = 3.20659 km/s 𝑥̈𝑥̈2 = 𝑥̈1 — Δ𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈ = 3.20195 km/s where we used the subscript ‘a’ to denote apoapsis, ’1’ for the maximum, and ’2’ for the minimum values.

We can now compute various orbital parameters of the deorbit trajectories that received more Δ𝑥̈ than expected. We only need to compute these for the minimum and maximum Δ𝑥̈ to obtain a range of values. The following table summarizes these computed parameters.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Parameter

1

2

Reference

𝑥̈

3,644.7 km

3,635.5 km Equation (2.61)

𝑥̈𝑥̈

3,392.4 km

3,374.1 km Equation (2.33)

𝑥̈

0.069224

0.071915

𝑥̈

3,627.2 km

3,616.7 km Equation (2.30)

𝑥̈ (deg)

348.218˝

334.069˝

Equation (2.40)

𝑥̈ (rad)

6.07756

5.83060

Equation (2.40)

Equation (2.39)

Thus, the new landing location angles (true anomaly, 𝑥̈, as shown in Figure 1) are given by the range 𝑥̈ = (5.83060, 6.07756). Since the nominal landing location is at periapsis (𝑥̈ = 2𝑥̈), then the difference in landing location angles is Δ𝑥̈ = (0.205626, 0.452581). Converting these values into actual distances, 𝑥̈, on the surface can be done using the fact that 𝑥̈ = Δ𝑥̈𝑥̈D which gives values between 698.5 km and 1,537 km. This means that the spacecraft is predicted to be off target between 698.5 km and 1,537 km.

To compute TOF, we need to use Kepler’s time equation. However, we must first convert true anomaly into eccentric anomaly using Equation (2.91) keeping in mind that all calculations must be done using radians, not degrees,

„c

ˆ ˙ȷ

1 — 𝑥̈ 𝑥̈ 𝑥̈ = 2 arctan tan 2 1. 𝑒

where no quadrant check is needed since 𝑥̈ and 𝑥̈ always reside in the same quadrant. This gives 2

2

𝑥̈1 = —0.191939 rad or 6.09125 rad, and 𝑥̈2 = —0.422087 rad or 5.86110 rad. Then, from Kepler’s Time Equation, or Equations (2.96) and (2.97), we compute TOF

d 𝑇 𝑂𝐹 =

𝑥̈3 𝜇D (𝐸 — 𝜋 — 𝑒 sin 𝐸)

where the —𝑥̈ comes from the fact that we are computing time from apoapsis to the landing location, not from periapsis. This gives 𝑥̈ 𝑥̈𝑥̈1 = 52.37 min and 𝑥̈ 𝑥̈𝑥̈2 = 48.40 min, which correspond to a range of difference in TOF of Δ𝑥̈ 𝑥̈𝑥̈ = (3.21, 7.18) min. While the difference in TOF may not be significant, we earlier computed that even these relatively small time differences can lead to the spacecraft being off target by hundreds of km, or more.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 2 Consider Problem 1 from Chapter 5. Assume that the upper stage has a velocity deviation which can be modeled as a normal distribution with zero mean and 0.01% of the spacecraft speed in LEO in all x-, y-, and z-directions. Create a Monte Carlo simulation with these parameters, and propagate the orbit until apoapsis. How does the orbit differ from the nominal GTO? Create scatter plots similar to Figure 6.15 showing the variation of inclination, semimajor axis, and eccentricity.

Solution For reference, Problem 1 in Chapter 5 states: A spacecraft is launched into a circular orbit around the Earth at an altitude of 450 km. The upper stage of the launch vehicle initiates a Hohmann transfer to GEO through a GTO. How much Δ𝑥̈ is the launch vehicle expected to perform (Δ𝑥̈1) and how much Δ𝑥̈ is the spacecraft’s propulsion system expected to deliver in order to perform the adequate orbit insertion at GEO (Δ𝑥̈2)? How long does the transfer take?

Using the given parameters and the Earth’s radius and gravitational parameter from Appendix A, we get that the departing LEO has radius and speed 𝑥̈1 = 𝑥̈‘ . 450km = 6821km c = 𝑥̈1 𝑥̈ = 7.64442km/s ‘ 𝑥̈ 1

The spacecraft needs to transfer to GEO, i.e. at a radius of 𝑥̈𝑥̈𝑥̈𝑥̈ = 42, 164 km (see Chapter 5 for details). In order to do so, the nominal Δ𝑥̈ required is computed using Equation (5.12)

ˆc Δ𝑥̈1 = 𝑥̈1

˙

2𝑥̈𝑥̈𝑥̈𝑥̈ — 1 = 2.38553km/s 𝑟1 . 𝑟𝐺𝐸𝑂

resulting in a post-Δ𝑥̈ speed of 𝑥̈2 = 𝑥̈1 . Δ𝑥̈1 = 10.02995 km/s. The semimajor axis and TOF for this transfer are 1 𝑥̈ = (𝑥̈1 . 𝑥̈𝑥̈𝑥̈𝑥̈) = 24, 493km 2

d

𝑇 𝑂𝐹 = 𝜋

𝑥̈3 = 5.29820hrs 𝜇‘

Using these nominal parameters as initial conditions, we then perturb 𝑥̈2 in all Cartesian direction as given by the problem statement. Propagating 100 such perturbed orbits gives the variations of eccentricity, inclination, semimajor axis, and apoapsis that are shown in Figure 2. While these variations may not seem particularly significant, they would lead to the need of trajectory correction and stationkeeping maneuvers to maintain the spacecraft within its predefined allocated GEO location.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 2: Variation of Orbital Parameters from Monte Carlo Simulation of GTO Trajectories

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Problem 3 A spacecraft is arriving in the sphere of influence of Jupiter, where it is expected to perform a gravity assist of the planet at an altitude of 1 million km. The spacecraft has a nominal hyperbolic eccentricity of 2.5. Create a Monte Carlo simulation assuming that the spacecraft position and velocity at the sphere of influence can be modeled as provided in Equation (6.72) and that the nominal 𝑥̈8 has a deviation that can be represented as a normal distribution with zero mean and 1% of the nominal 𝑥̈8 (in all directions). Propagate 1,000 orbits, plot them and create a B-plane analysis similar to Figure 6.14, showing the 3𝑥̈ uncertainty ellipse resulting from your simulation.

How likely is it that the trajectory would impact Jupiter?

Solution As provided in Appendix A, the symbol for Jupiter is E, and Jupiter’s gravitational parameter and physical radius are 𝑥̈E = 1.268 × 108 km3/s2 𝑥̈E = 71, 492 km The following table summarizes the computed parameters for this problem, including the equations used for each computation.

Parameter

Value

Reference

𝑥̈

2.5

Given

𝑥̈𝑥̈{E

1,071,492 km

𝑥̈E + 1,000,000 km

𝑥̈E

1.268×108 km3/s2

Appendix A

𝑥̈

-714,328 km

Equation (2.37)

𝑥̈8{E

13.3233 km/s

Equation (2.50)

Δ

1,636,725 km

Equation (4.15)

𝑥̈𝑥̈𝑥̈𝑥̈

48,223,800 km

Equation (5.148)

𝑥̈

47.1564˝

Equations (5.176) and (5.177)

If 𝑥̈8{E is not computed using 𝑥̈𝑥̈𝑥̈𝑥̈ → ∞, but rather using the value indicated in the table above, then 𝑥̈8{E would be slightly larger, i.e. 13.5192 km/s. Similarly, Δ would instead be approximately 1,589,600 km.

Thus, the initial conditions of the spacecraft at the boundary of the sphere of influence of Jupiter are

a

r8{E = — 𝑥̈2𝑆𝑂𝐼 — Δ2𝑥̈ . Δ𝑥̈ . 0𝑥̈ km v8{E = 𝑥̈8{E𝑥̈ . 0𝑥̈ . 0𝑥̈ km/s Applying the deviations on the above v8{E with zero mean and 1% standard deviation means 6

Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

perturbing v8{E in all directions by 𝑥̈v8{E = 𝑥̈𝑥̈𝑥̈𝑥̈ . 𝑥̈𝑥̈𝑥̈𝑥̈ . 𝑥̈𝑥̈𝑥̈𝑥̈ Running a Monte Carlo simulation with 1,000 orbits gives the orbits shown in Figure 3, where the black orbit is the nominal trajectory while the blue orbits are the perturbed trajectories. The 3𝑥̈ ellipse shown in this figure has semimajor axis of approximately 709,200 km and semiminor axis of approximately 517,200 km. Numerically, we get the results shown in the following table.

Parameter Minimum Value

Maximum Value

𝑥̈

1.86363

3.10975

𝑥̈

-729,261 km

-699,915

𝑥̈

37.5157˝

64.9032˝

𝑥̈𝑥̈{E

617,339 km

1,504,493

Due to the inherent randomness of the 𝑥̈𝑥̈𝑥̈, 𝑥̈𝑥̈𝑥̈, and 𝑥̈𝑥̈𝑥̈ deviations, one would unlikely get these exact numerical results, though they should be similar in value. Using these results, we can create the plot requested for the B-plane, as shown in Figure 4. In Figure 4, the black dot represents the nominal trajectory, the blue crosses are the Monte Carlo simulation trajectories and the blue dot is their average, and the blue ellipse represents the 3𝑥̈ ellipse based on this particular simulation.

Although it is nearly impossible for any of these trajectories to impact Jupiter, some get very close to the planet, varying greatly from the nominal trajectory.

Although not requested by the problem statement, we are also reporting the variation on various orbital elements, as shown in Figure 5.

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 3: Monte Carlo Simulation of Jupiter Flyby Trajectories (SB plane)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 4: Monte Carlo Simulation of Jupiter Flyby Trajectories (B-plane)

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Solution Manual for Interplanetary Astrodynamics 1st Edition by David B. Spencer, Davide Conte.

Figure 5: Variation of Orbital Parameters from Monte Carlo Simulation of Jupiter Flyby Trajectories (Bplane)

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