Newton’s Law of Gravitation and Kepler’s Third Law We showed already this week that Newton was able to prove Kepler’s third Law, the Harmonic Law, from his own gravitation law. If the force between the Sun and a planet is given by F = G M m/r 2, where M is the mass of the Sun and m the mass of the planet, then, if the planet is moving in a circular motion (which is almost true), then F = m r  2 must also be true. Therefore we have m r  2 = G M m/r 2, which gives us r 3 = G M /  2. We know that  = 2  /T, so r 3 = (G M/4  2 ) T 2. So (G M/4  2 ) is the constant in Kepler’s third law. In SI or mks units, it is 3.36 x m 3 /s 2, but in units in which time is measured in years and distance is measured in AU (astronomical units, the distance from the Earth to the Sun), G M/4  2 = 1 AU 3 /year 2.

Suppose we discover a new planet and we observe its motion through the sky, until it becomes clear that it takes 8 years to make one orbit of the Sun. In other words, its year is equal to 8 Earth Years in length. In terms of the Astronomical Unit, the distance of the Earth from the Sun, how far away is this new planet from the Sun? (Clue: Such a “planet” probably exists, as it would be in the asteroid belt). a)1 A. U. b)2 A. U. c)4 A. U. d)8 A. U. e)16 A. U.

Correct Answer – C (4 A. U.) If we use units of Earth years and A. U., then R 3 = T 2 holds for any planet, where T is its year and R is its distance from the Sun. If this planet has a year of 8 of our years, then R 3 = 8 2 = 64. The cube root of 64 is 4, which means that R = 4 A. U.

Gravitational Potential Energy We have used the concept of gravitational potential energy before, but now we want to apply it to the Moon rather than to Newton’s apple, i.e. to objects in space rather than objects on Earth. Why would it not be a good idea to use our previous formula U = m g h? For instance, if we put the apple up in orbit, would the formula still give a correct result for its potential energy? a)No, because the mass of the apple would be different if it was further from the Earth. b)Yes, the formula should work fine for apples. It doesn’t work for the Moon because the formula breaks down for very large masses c)No, because g, the acceleration due to gravity, only applies to objects on the surface of the Earth. Thus the formula only works well when h is small

Correct Answer – C The mass of a body is a constant which, for most purposes, never changes (though as we will see next semester, Einstein was able to prove that the mass of a body can change in certain very extreme circumstances). But we recall that g is defined as the acceleration due to gravity on the surface of the Earth. While G, the gravitational constant, is the same everywhere in the Universe, g only applies here on Earth. So we need a new definition of the gravitational potential energy written in terms of G, rather than g.

The new definition of gravitational potential energy will be U = - G M m/r For the energy of a body of mass m a distance r from another body (say a star or planet) of mass M. In our old definition, zero potential energy was at the surface of the earth (if we took h=0 to be that point). If the body of mass M is the Earth, and m is the apple, where in the Universe will the apple be when U = 0? a)At the center of the Earth b)On the surface of the Earth c)About where the Moon is d)Infinitely far from the Earth

Correct Answer – D If U = - G M m/r then if U = 0 it follows that 0 = G M m/r which means that r = G M m/0 If you divide a number by zero the answer is infinity. So the zero point of our new definition of potential energy is an infinite distance away from the star or planet in question. As the apple approaches the Earth from infinity, what should happen to its potential energy? a)It should decrease, so it will become negative b)It should increase, so it will become positive c)It should stay the same, so it will be zero d)It should eventually become infinite

Correct Answer – A We recall from our previous experience with potential energy, that as a body falls it loses potential energy which is converted into kinetic energy. Even if it is falling towards the Earth from deep space the same rule should apply, so if we start with U = 0 very far from Earth, then U will have to be negative for all non-infinite distances. This is why the formula has a minus sign in it U = - G M m/r It is true that in principle when r = 0, U will be infinite (actually -  ). When will r=0? Only when the body is at the center of the Earth! Will its potential energy really be infinite there? That’s the topic of the next question.

If an apple falls towards the Earth from Space, how does its weight change? We know its weight is simply the gravitational attraction of the Earth, so W = F = G M m/r 2 As r gets smaller W will get larger at a rapid rate, because it goes as the inverse square power of r. Suppose the apple reaches the Earth’s surface and happens to fall down a very deep mine shaft, so it can continue to fall deep inside the Earth. Will its weight continue to increase the closer it gets to the center of the Earth? a)Sure, the inverse square law still applies! b)No, some of the Earth’s mass is now trying to pull it up, it will actually weigh less as it gets deeper. c)It should weigh the same, as being in the Earth counts as the same thing as being on the Earth.

Correct Answer – B When you are above the Earth, it appears as if the whole mass of the Earth is pulling you towards the center of the Earth. This can be proven mathematically. But suppose you were at the center of the Earth. In that case, if there was still a force on you, which way would it act? It wouldn’t know which way to pull, different parts of the Earth would pull you equally in all directions. The force would be zero. Suppose you were half-way to the center. In that case you would still feel a force pulling you down, but the part of the Earth above you would in fact be trying to slow your descent and even pull you back up! Therefore in real life there is no infinite acceleration or infinite potential energy at the center of the Earth. So our formula U = - G M m/r is only useful outside the Earth, in space.

Escape Velocity Now let’s look at a typical case where potential energy can help us. Suppose you wish to take off from Earth in a rocket ship and fly to a distant planet. How fast will you have to fly upwards to escape the Earth and get into space? This speed is called the escape velocity. How can we define the escape velocity at the surface of the Earth? a)It is the speed at which your weight becomes zero (no gravitational force) b)It is the speed at which your aerodynamic buoyancy equals your weight c)It is the speed at which your kinetic energy equals twice your potential energy d)It is the speed at which your kinetic energy plus your potential energy equals zero.

Correct Answer – D If you think back to earlier questions on potential energy, the height you can reach depends on how much energy you have. If you want to reach a height h, you have to start with a total energy equal to m g h. In this case we want to escape the Earth altogether, so we want to be able to reach infinity if we have to. But we just defined the potential energy at infinity, so the minimum initial energy we need to have to get there is zero.

Given that the total energy of a rocket ship taking off must be zero if it is to escape the Earth without further acceleration, which of the following is the correct equation for escape velocity? a)½ m v 2 – G M m/r = 0 b)½ m v 2 + m g h = 0 c)M g h + G M m/r = 0 d)½ m v 2 = 0

Correct Answer – A The escape velocity is defined as the velocity at which the kinetic energy of the escaping body (K = ½ m v 2 ) plus the potential energy (G = - G M m/r) is equal to zero. ½ m v 2 – G M m/r = 0 Which means that ½ v 2 = G M/r So that v 2 = 2 G M/r So the escape velocity for a body on the surface of a planet is v =  (2 G M/R) Where M is the mass of the planet and R is the radius of the planet. The escape velocity at the surface of the Earth is 11,200 m/s or 25,000 miles/hr.