Newton reasoned that the easiest dependence to consider is one where the strength of the gravitational acceleration (ag) decreases with distance (r) raised to some power. In equation form, we use “n” to denote “some power,” where n stands for an integer that we still need to determine here. The symbol ∝ means “proportional to.”
\begin{equation} a_{g} \propto r^{n} \end{equation}
We expect that the value of n will be negative because only negative values will cause the gravitational acceleration to get smaller with r. By comparing the gravitational acceleration at Earth’s surface to the acceleration of the Moon toward Earth, it is possible to determine the value of n.
The r here is understood to be the distance between the centers of the objects involved. So, for example, if we are standing on the surface of Earth, our distance is taken from Earth’s center, not its surface. Similarly, the distance of the Moon from Earth is taken to be the distance from Earth’s center (not its surface) to the center of the Moon. This is not obvious, and it was one of Newton’s great insights into gravity; he was able to demonstrate its validity mathematically. In the case of the Moon–Earth distance, the distinction between the center and surface is not very important because their separation is much larger than their sizes. However, for we who stand upon Earth’s surface, the difference is vital: we are not zero meters from Earth, we are about 6400 km away from its center, at least as far as gravity is concerned.
You should have recognized in the previous interactive activity that the value of n turns out to be -2. However, Newton had to derive this, as we will now show.
We can set up a ratio between the acceleration at Earth’s surface and at the Moon’s orbit.
aE / aM = (rE / rM )n
Then we just have to plug the appropriate numbers into this expression.
( 9.8 m/s2 ) / ( 2.685 × 10-3 m/s2 ) = ( ( 6.368 × 106 m ) / ( 3.844 × 108 m ) )n
Note how units on both sides cancel, leaving no units in any resulting expression. We are left with the following…
( 3.650 ) = ( 1.657 × 10-2 )n
Now, to solve for n, it will help if we take the log of both sides. Then we have an expression we can easily solve algebraically. Here are the steps:
| log ( 3.650 ) | = log ( ( 1.657 × 10-2 )n ) |
| = ( n ) ( log ( 1.657 × 10-2 ) ) | |
| n | = log ( 3.650 ) / log ( 1.657 × 10-2 ) |
| = 3.562 / -1.781 | |
| = -2 |
This is essentially the reasoning that Isaac Newton used to deduce the 1/r2 dependence of gravity. Such a dependence had been suspected by others, but Newton was the first to be able to demonstrate it in this manner.