12 ORBITAL MOTION
12.6 Planetary orbits
Worked example 12.2: Acceleration of a rocket Question: A rocket is located a distance 3.5 times the radius of the Earth above the Earth’s surface. What is the rocket’s free-fall acceleration? Answer: Let R⊕ be the Earth’s radius. The distance of the rocket from the centre of the Earth is r1 = (3.5 + 1) R⊕ = 4.5 R⊕ . We know that the free-fall acceleration of the rocket when its distance from the Earth’s centre is r0 = R⊕ (i.e., when it is at the Earth’s surface) is g0 = 9.81 m/s2 . Moreover, we know that gravity is an inverse-square law (i.e., g ∝ 1/r2 ). Hence, the rocket’s acceleration is g1 = g 0
r0 r1
!2
=
9.81 × 1 = 0.484 m/s2 . 2 (4.5)
Worked example 12.3: Circular Earth orbit Question: A satellite moves in a circular orbit around the Earth with speed v = 6000 m/s. Determine the satellite’s altitude above the Earth’s surface. Determine the period of the satellite’s orbit. The Earth’s mass and radius are M ⊕ = 5.97 × 1024 kg and R⊕ = 6.378 × 106 m, respectively. Answer: The acceleration of the satellite towards the centre of the Earth is v 2 /r, where r is its orbital radius. This acceleration must be provided by the acceleration G M⊕ /r2 due to the Earth’s gravitational attraction. Hence, v2 G M⊕ = . r r2 The above expression can be rearranged to give (6.673 × 10−11 ) × (5.97 × 1024 ) G M⊕ = = 1.107 × 107 m. r= 2 2 v (6000) Thus, the satellite’s altitude above the Earth’s surface is h = r − R⊕ = 1.107 × 107 − 6.378 × 106 = 4.69 × 106 m. 276