# Circular orbit: Schwarzschild vs Newton

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Box 10.3; Problems 10.10, 10.11.

For a circular orbit, the Schwarzschild angular speed is obtained from the angular momentum:

 $\displaystyle \omega=\frac{d\phi}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{l}{r^{2}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\sqrt{\frac{r^{2}GM}{r-3GM}} \ \ \ \ \ (2)$

We can write this as

$\displaystyle \omega^{2}=\frac{GM}{r^{2}\left(r-3GM\right)} \ \ \ \ \ (3)$

Comparing this with the angular speed measured at infinity ${\Omega=\frac{d\phi}{dt}}$ we have

$\displaystyle \Omega^{2}=\frac{GM}{r^{3}} \ \ \ \ \ (4)$

Thus ${\omega>\Omega}$.

The relation between ${r}$ and ${l}$ for a circular orbit is

$\displaystyle r=\frac{6GM}{1\mp\sqrt{1-12\left(GM/l\right)^{2}}} \ \ \ \ \ (5)$

In the Newtonian case, we can use Kepler’s law at all distances, so we have

 $\displaystyle \Omega^{2}$ $\displaystyle =$ $\displaystyle \frac{l^{2}}{r^{4}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{3}}\ \ \ \ \ (7)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{l^{2}}{GM} \ \ \ \ \ (8)$

Note that we can get the same result by defining the Newtonian energy

 $\displaystyle E_{N}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\frac{dr}{dt}\right)^{2}+V_{N}\ \ \ \ \ (9)$ $\displaystyle V_{N}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{l^{2}}{r^{2}}-\frac{GM}{r} \ \ \ \ \ (10)$

and then solving

$\displaystyle \frac{dV_{N}}{dr}=-\frac{l^{2}}{r^{3}}+\frac{GM}{r^{2}}=0 \ \ \ \ \ (11)$

In the Newtonian case, the potential energy ${V_{N}}$ has only one minimum.

For large ${l}$, we can expand 5 to get (using the minus sign)

 $\displaystyle r$ $\displaystyle \rightarrow$ $\displaystyle \frac{6GM}{1-\left(1-\frac{1}{2}\frac{12\left(GM\right)^{2}}{l^{2}}\right)}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{l^{2}}{GM} \ \ \ \ \ (13)$

Thus the Schwarzschild case reduces to the Newtonian case for large ${l}$.