As we said in class, the acceleration due to gravity, near the Earth’s surface, is

The gravitational potential energy

where we’ve written $r=R_E+y$ and $y$ is the height above the Earth’s surface. For $\mid y\mid \ll R_E$, we can approximate this by a power series in $y$.

If you know Taylor series, you can compute it directly. If not, consider writing

for some choice of constants, $a_1,a_2,a_3,\dots$. If we multiply by $(R_E+y)$, we should get $1$:

So, to make the equation true, we must have

So (2) can be approximated

The first term is an irrelevant constant. The second is the gravitational potential energy we wrote down before, and the third term is the first correction, due to the fact that the gravitational force, due to the Earth, falls off with distance (rather than being constant).

In class, we considered only the case where one of the masses is much greater than the other, $M\gg m$, and so is more-or-less stationary. What happens when that isn’t the case?

Let’s consider the extreme opposite situtation, namely two equal masses, $m$, in a circular orbit about the midpoint between them.

The distance between them is $r=2R$, so the force on the first one is

(and, similarly, for $F_2$). So, for a circular orbit, we have

or

To compute the period, $T=\frac{2\pi R}{v}$, we get something that looks just like Kepler’s 3^rd Law.

but the constant is 4 times as big as before.