The law of universal gravity

Newton built on his earlier insight that you have to associate a force with acceleration and not with velocity. If you follow a planet as it moves around, thinking it is force that causes velocity, you won’t get any defini-tive answer as to what is behind the force. On the other hand, if you calculate the acceleration of the planet, you will find that, at every instant, it points toward the sun. (This is obvious for the circular orbit.) If all bodies are accelerating toward the sun, it’s fairly clear the reason for the accelera-tion is the sun. You then postulate a force that’s exerted by the sun on the planet that bends it into a circle. Your job is to find that force.

The Kepler Problem 105

What’s the nature of the force? Again, it was Newton who figured out that the force that bends the planets around the sun is the same as the force that bends the moon around the earth or makes the apple fall to the earth. Now, the fact that the moon, orbiting at a radius Rmat a speed vm, is accelerating toward the earth at a rate

am=v_m² Rm

(7.4)

is something you have already learned in this course. As for the apple, it also is accelerating toward the earth at a rate

aa= g  9.8 ms⁻², (7.5)

independent of its mass ma.

Let us guess the formula for the force on the apple, looking at just the magnitude, the direction being obviously toward the center of the earth.

Near the earth, we all know, the acceleration of every body is the same.

Therefore from a=_m^F, we deduce that the force of gravity on a falling body is itself proportional to the mass of the body, so that the m may cancel out.

So we can write for the apple

Fa= maf (Me, Ra) (7.6)

where the unknown function f depends on Me, the mass of the earth, and its radius Re, which is also the distance Rabetween the apple and the center of the earth.

What else do we need? The third law says that if there’s a force on the apple exerted by the earth, the apple must exert an equal and oppo-site force on the earth. Consider the earth and the apple—but imagine the apple getting bigger and bigger. The formula is not going to change. Wait until the apple is huge compared to the earth. Then you will have to agree that the earth is falling toward the apple rather than the other way around.

Under this exchanged role, the force on the earth must then be propor-tional to Me. Thus, for any two bodies of mass m and M, we obtain an expression for the force compatible with Newton’s third law,

F= (Mm)f (R), (7.7)

106 The Kepler Problem

where R is the distance between their centers. We don’t know the distance dependence; we don’t know the function f (R).

To find it, let us compare the acceleration of the apple and the moon due to the pull of the earth:

aa=Mema million miles, that is not bad; the correct answer is 238, 000 miles. If you make an estimate that’s off by a factor of 4 in astrophysics, it’s fine, but if you say Rm= 1000 miles, we should have a very long talk. Anyway, let us round it off and say Rm= 240, 000 miles.

Next, what’s the radius of the earth? You have some idea, right?

How far is California? Three thousand miles. And how many hours is the time difference? Three hours. So that is one hour per thousand miles.

If you go all the way around the earth and come back, the accumulated time difference has to be 24 hours. That means the earth has a circum-ference of roughly 24, 000 miles. Dividing by 2π  6, we get Re 4000 miles.

I know that we should work with meters and kilometers but, like the rest of you, once I get on the freeway I’m watching how many miles per hour I’m driving, not how many meters per second. Nonetheless, we Americans do use a lot of British units. If you go buy insulation at Home Depot, it’s rated in BTUs per slug per poundal, right? Some-times you wonder why we fought the War of Independence if we’re still using those units. Anyway, my brain is split. When I do physics, I use the metric system. When I shop at Home Depot, I use the Home Depot units.

Back to finding the acceleration of the moon, am=_Rm^v2m. We’ll assume every orbit is a circle; that assumption turns out to be not so bad, even for planets. We already have Rm, which also gives us the length of the orbit

The Kepler Problem 107 2πRm, which it completes in roughly T= 28 days, yielding a velocity vm=

2πRm

T and an acceleration am=(2πRm/T)²

Rm

. (7.11)

If you plug in the numbers, you find aa

Combining this with Eqn. 7.7, we find the great law of universal gravity:

F= GMm

R² (7.14)

G= 6.67 · 10⁻¹¹Nm²kg⁻², (7.15)

where G is the universal gravitational constant that balances the units in Eqn. 7.14 and ensures that the numerical value we obtain at the surface of the earth reproduces g= 9.8ms⁻².

In this argument, we are assuming that the distance between the apple and the earth is Re, the radius of the earth. Why not use the height of the tree from which the apple fell? Because Newton’s formula is actually written down for two point-like objects, with an unambiguous distance between them. The correct way to handle the earth is to divide it into many small pieces and find the force on the apple due to each piece and add, or rather, integrate, over their contributions. The result will be that the earth acts as if all its mass were concentrated at its center. Newton knew this to be true, but he could not prove it to his satisfaction for many years, which is why he delayed publication. Even today it is a hard problem in integration.

Here is another similar result. Suppose you are inside a hollow spher-ical shell of some mass M. What force will you feel? It is clear that if you are at the center, you will feel no force because for every piece of matter

108 The Kepler Problem

in the shell pulling you one way, there is an identical one pulling you the opposite way. What is not obvious, but true, is that the gravitational force will be zero inside the entire shell. Of course, outside the shell the force will be that of a point mass M sitting at the center. In summary, for any spher-ically symmetric distribution of mass, the force felt by a body at radius r is due to all the mass inside a sphere of radius r, acting as a point mass at the center, while the mass outside contributes nothing.

Equation 7.14 is rightly called the law of universal gravitation. It was a tremendous leap of faith to believe that the laws that are operative near the earth also apply to the moon and beyond. This was the year 1687; peo-ple believed in witchcraft and harbored all kinds of superstitions. They were not thinking in modern scientific terms. They had a lot of illusions about what the heavens were made of. To believe they’re made of the same stuff, and controlled by the same laws, was far from obvious in those days.

Newton’s leap of faith has proven extraordinarily prescient: not only the law of gravitation but all the laws of physics that we deduce near the earth seem to work over the entire universe, not just now, but even in the distant past and, we hope, in the future. Indeed, given the long times light takes to get to us from far away galaxies and quasars, much of what we see in the heavens today happened a long time back, and yet we analyze them using the recently discovered laws. We have sampled a very tiny part of the universe, over a tiny period of time, but we apply the laws we deduce here and now to the far reaches of the universe and all the way back in time to the big bang. We confidently predict the future fate of the universe. It’s a great break for us that the laws we find seem to be universal and eternal. It need not have been so.