APSI 2018 AP Physics 1: Chapter 3.3 by Edvantage Science

3.3 Newton’s Law of Universal Gravitation Warm Up The Moon orbits Earth about once every 28 days. Why doesn’t the Moon crash into Earth? ___________________________________________________________________________________________ ___________________________________________________________________________________________ ___________________________________________________________________________________________

From Kepler to Newton

Kepler’s three laws describe the orbits of the planets around the Sun. Newton used Kepler’s laws to derive a law describing the nature of the gravitational forces that cause the planets to move in these orbits. Newton concluded that the force that keeps planets in orbit is the same force that makes an apple fall to the ground. He stated that there is a gravitational force between any two bodies in the universe. Like all other forces, gravity is a mutual force. That is, the force with which Earth pulls on a falling apple is equal to the force with which the apple pulls on Earth, but in the opposite direction. Earth pulls on your body with a force of gravity that is commonly referred to as your “weight.” Simultaneously, your body exerts a force on planet Earth of the same magnitude but in the opposite direction. Relative to each other, Earth “weighs” the same as your body! Newton was able to use Kepler’s laws as a starting point for showing that the force of gravity between the Sun and the planets varied as the inverse of the square of the distance between the Sun and the planets. He was convinced that the inverse square relation would apply to everyday objects near the surface of Earth as well. He produced arguments suggesting that the force should depend on the product of the masses of the two bodies being attracted to one another. The mathematical details of how Newton arrived at his famous law of universal gravitation can be found in many references, but are too lengthy to reproduce here. Newton’s Law of Universal Gravitation Every body in the universe attracts every other body with a force that is (a) directly proportional to the product of the masses of the two bodies and (b) inversely proportional to the square of the distance between the centres of mass of the two bodies. The equation for Newton’s law of universal gravitation is: F = G

Mm R2

Chapter 3 Circular Motion and Gravitation 169

The constant of proportionality G is called the universal gravitation constant. Isaac Newton was unable to measure G, but it was measured later in experiments by Henry Cavendish (1731–1810). The modern value for G is 6.67 × 10–11 Nm2/kg2.

Cavendish’s Experiment to Measure G

You can imagine how difficult it is to measure the gravitational force between two ordinary objects. In 1797, Henry Cavendish performed a very sensitive experiment, which was the first “Earthbound” confirmation of the law of universal gravitation. Cavendish used two lead spheres mounted at the ends of a rod 2.0 m long. The rod was suspended horizontally from a wire that would twist an amount proportional to the gravitational force between the suspended masses and two larger fixed spherical masses placed near each of the suspended spheres (Figure 3.3.1).

m1 m2 m2 m1

Figure 3.3.1 Cavendish’s apparatus for confirming Newton’s law of

universal gravitation

The forces involved in this experiment were extremely small (in the order of 10–6 N), so great care had to be taken to eliminate errors from air currents and static electricity. Cavendish did manage to provide confirmation of the law of universal gravitation, and he arrived at the first measured value of G. To calculate the force of gravity on a mass m, you simply multiply the mass by the Gravitational Field gravitational field strength g (F = mg.) You could also use the law of universal gravitation: Strength of the Earth Mm F = G 2 , where M is the mass of Earth. R This means that mg = G

Mm and, therefore, R2 g = G

170 Chapter 3 Circular Motion and Gravitation

M R2

Thus, the gravitational field strength of Earth depends only on the mass of Earth and the distance, R, from the centre of Earth to the centre of mass of the object that has mass m.

Sample Problem 3.3.1 — Newton’s Law of Universal Gravitation What is the force of gravity on a 70.0 kg man standing on Earth’s surface, according to the law of universal gravitation? Check your answer using F = mg.

What to Think About

How to Do It

1. What data do you need?

G = 6.67 × 10–11 Nm2/kg2 g = 9.80 N/kg Earth’s mass = 5.98 × 1024 kg Earth’s radius = 6.38 × 106 m

2. Find the force of gravity using the law of universal gravitation.

Fg = G

m 1m 2 r2

( 70.0 kg)( 5.98 10 24 kg) F g = (6.67 10 11 Nm 2 kg2 )( (6.38 106 m)2 F g = 686 N 3. Calculate the force of gravity using F = mg.

F g = mg F g = ( 70. kg)( 9.80 m s 2 ) F g = 686 N

Practice Problems 3.3.1 — Newton’s Law of Universal Gravitation 1. What is the force of gravitational attraction between a 75 kg boy and a 60.0 kg girl (a) when they are 2.0 m apart?

(b) when they are only 1.0 m apart?

Practice Problems continued

Chapter 3 Circular Motion and Gravitation 171

Practice Problems 3.3.1 — Newton’s Law of Universal Gravitation (Continued) 2. What is the force of gravity exerted on you if your mass is 70.0 kg and you are standing on the Moon? The Moon’s mass is 7.34 × 1022 kg and its radius is 1.74 × 106 m.

3. What is the force of gravity exerted on you by Mars, if your mass is 70.0 kg and the mass of Mars is 6.37 × 1023 kg? The radius of Mars is 3.43 × 106 m, and you are standing on its surface, searching for Mars bars.

Weightlessness

According to Newton’s law of universal gravitation, the force of gravity between any two bodies varies inversely as the square of the distance between the centres of mass of the two bodies. Figure 3.3.2 is a graph showing how the force of gravity (commonly called “weight”) changes with distance measured from Earth’s centre. To be truly “weightless” (experience no force of gravity) an object would have to be an infinite distance from any other mass. According to the law of universal gravitation, as R → ∞, F → 0. F

Figure 3.3.2 The force of gravity decreases dramatically with

distance from Earth.

Obviously, true weightlessness is not likely to be achieved! When people talk about “weightlessness,” they usually are referring to apparent weightlessness. Apparent weightlessness is experienced when you feel zero force from supporting structures like

172 Chapter 3 Circular Motion and Gravitation

your seat, the floor, or Earth’s surface. This will happen when your supporting structure has the same acceleration as you have. Of course, if there is no supporting structure as when you jump off a cliff or a ladder or when you are in the middle of a jump you will also experience apparent weightlessness.

Examples of Apparent Weightlessness

1. A Falling Elevator Imagine a person standing on a scale in an elevator (Figure 3.3.3(a)). When the elevator is standing still, the scale will give the true weight of the person, which is the force of gravity exerted by Earth on the person (F = mg). In the illustration, this is shown as the pair of forces of the person’s weight (w = mg) and the normal force (N) from the scale exerting a force back on the person. With no acceleration, the scale reads the true weight: w = N = mg.

∑F = 0

∑ F = ma

w = N = mg true weight

mg – N = ma w = N = m(g – a)

w' = N = 0 a=g

“weightless”

less than true weight

not accelerating (a = 0)

mg N = 0

Descending with accelerating a < g

(a) (b)

Descending with a = g

(c)

Figure 3.3.3 A person in an elevator can experience apparent weightlessness.

In Figure 3.3.3(b) the elevator is accelerating down at a rate less than gravity. Now the person’s weight is greater than the normal force being exerted back, so the person appears to be lighter than their true weight based on the reading on the weigh scale. Imagine now that the cable breaks, as in Figure 3.3.3(c). The elevator will accelerate down at rate a = g. The person in the elevator will also fall, accelerating at rate a = g. The scale, placed in the elevator by the person, will read the apparent weight of the person, which is zero. In this situation, w = N = 0. At this moment, the person experiences true weightlessness. You get a similar feeling when you go over a large bump while driving in a car or take a quick drop on an amusement park ride. 2. Orbiting Astronauts Astronauts in an orbiting space vehicle feel weightless for the same reason as a person in a falling elevator. Both the astronauts and the vehicle they occupy are in free fall. The astronauts feel no resistance from any supporting structure, so they feel weightless. 3. Momentary Weightlessness You experience brief sensations of weightlessness during everyday activities. If you are running, you will experience apparent “weightlessness” during those intervals when both feet are off the ground, because you are in a momentary free fall situation. Jumping off a diving board or riding your bike swiftly over a bump, you will experience brief moments of apparent weightlessness. © Edvantage Interactive 2017

Chapter 3 Circular Motion and Gravitation 173

Quick Check 1. Why would a BMX rider feel momentarily “weightless” during a jump?

2. A 70.0 kg man is in an elevator that is accelerating downward at a rate of 1.0 m/s2. What is the man’s apparent weight?

3. How fast would Earth have to move at the equator before a person standing on the equator felt “weightless”? Earth’s radius is 6.37 × 106 m.

4. Earth is in free fall toward the Sun. Why do you not feel “weightless” when you are an occupant of this satellite of the Sun?

What orbital (tangential) speed must a space vehicle have to achieve a circular orbit at a

Satellites in Circular given altitude (Figure 3.3.4)? Since the centripetal force (mv 2/R) is provided by gravity, we Orbits — Orbital can write: Velocity 2 mv GMm = 2 R R

Therefore,

GM , and R GM v orbital = R v2 =

where M is the mass of Earth, R is the distance from Earth’s centre to the space vehicle, and G is the universal gravitation constant.

174 Chapter 3 Circular Motion and Gravitation

satellite

Earth

Figure 3.3.4 A space vehicle must achieve and

maintain a minimum speed to remain in orbit.

Quick Check 1. What orbital speed is required by an Earth satellite in orbit 1.0 × 103 km above Earth’s surface? Does it make a difference what the mass of the satellite is? Explain your answer. Earth’s mass = 5.98 × 1024 kg — Earth’s radius = 6.37 × 106 m

2. What is the speed of a satellite orbiting Earth at a distance of 9.0 × 105 m above Earth’s surface?

Gravitational Potential Energy

Newton’s law of universal gravitation tells us the force between a space vehicle and planet Earth at any distance R from the centre of mass of Earth: F = G

Mm R2

where M is the mass of Earth, m is the mass of the space vehicle, and R is the distance from Earth’s centre to the space vehicle. If a space vehicle is to travel to other parts of the solar system, it must first escape the grasp of Earth’s gravitational field. How much energy must be supplied to the vehicle to take it from Earth’s surface (where R = Re) to a distance where the gravitational force due to Earth is “zero” (R → ∞)? The force on the space vehicle is continually changing. Figure 3.3.5 shows how the force changes with distance. To get the vehicle out of Earth’s gravitational field, enough energy must be supplied to the space vehicle to equal the amount of work done against the force of gravity over a distance between Earth’s surface and infinity where the force approaches zero.

F The area under this graph between Re and inﬁnity is Mm ΔEp = G R e

Figure 3.3.5 A large force is needed at first to move the space vehicle out

of Earth’s gravitational field.

Chapter 3 Circular Motion and Gravitation 175

The amount of work that must be done to escape Earth’s gravitational field is equal to the area underneath the force vs. distance graph, between R = Re and R → ∞ (Figure 3.3.5). Using calculus, it can be shown that the area under the graph (shaded portion) is equal to GMm/Re. This area represents the change in gravitational potential energy of the vehicle as it moves from distance Re out to infinity. EP = G

Mm Re

In routine, Earthbound problems involving gravitational potential energy, we often arbitrarily take the “zero” level of potential energy to be at Earth’s surface. In situations involving space travel, it is conventional and convenient to the let the “zero” of Ep be at infinity. Mm Since the change in Ep between R = Re and R → ∞ is +G , the Ep at Earth’s Re surface is negative. EP ( at R = Re ) = – G

Escape Velocity

Mm Re

How fast must a spaceship travel to escape the gravitational bond of the planet Earth? In order to escape Earth’s gravitational pull, the space vehicle must start with enough kinetic energy to do the work needed to bring its potential energy to zero. Sitting on the launch pad, the total energy of the space vehicle is EP = – G

Mm Re

If the space vehicle is to escape Earth’s field, it must be given sufficient kinetic energy so that: Mm –G + Ek = 0 Re Mm 1 2 –G + mv = 0 Re 2 Mm 1 2 mv = G Re 2 M v 2 = 2G Re v=

This means that v escape =

2GM Re

2GM is the minimum speed the space vehicle must be Re

given in order to escape Earth’s gravitational pull. This minimum speed for escape is called the escape velocity for planet Earth. It depends only on Earth’s mass and Earth’s radius. Obviously different planets or moons would have different escape velocities, as questions in the rest of the chapter will show. 176 Chapter 3 Circular Motion and Gravitation

Quick Check 1. A 2.0 x 104 kg satellite orbits a planet in a circle of radius 2.2 x 106 m. Relative to zero at infinity, the gravitational potential energy of the satellite is –6.0 x 109 J. What is the mass of the planet?

2. Ganymede, one of Jupiter’s moons, is larger than the planet Mercury. Its mass is 1.54 x 1023 kg, and its radius is 2.64 x 106 m. (a) What is the gravitational field strength, g, on Ganymede?

(b) What is the escape velocity for a space vehicle trying to leave Ganymede?

Chapter 3 Circular Motion and Gravitation 177