Circular orbits: relation between radius and angular momentum

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Problems 10.7-8.

For a circular orbit, the relation between the angular momentum and radius is

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

We can invert this to get ${l}$ in terms of ${r}$:

 $\displaystyle \mp\sqrt{1-12\left(GM/l\right)^{2}}$ $\displaystyle =$ $\displaystyle \frac{6GM}{r}-1\ \ \ \ \ (2)$ $\displaystyle -12\frac{G^{2}M^{2}}{l^{2}}$ $\displaystyle =$ $\displaystyle \left(\frac{6GM}{r}-1\right)^{2}-1\ \ \ \ \ (3)$ $\displaystyle l^{2}$ $\displaystyle =$ $\displaystyle \frac{12r^{2}\left(GM\right)^{2}}{r^{2}-\left(6GM-r\right)^{2}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{r^{2}GM}{r-3GM} \ \ \ \ \ (5)$

From this we can get an expression for ${\tilde{E}}$ for a circular orbit where ${dr/d\tau=0}$:

 $\displaystyle \tilde{E}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{l^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GM}{r}+\frac{r^{2}GM}{r-3GM}\left(\frac{1}{2r^{2}}-\frac{GM}{r^{3}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GM}{2r}\left[2-\frac{2r^{3}}{r-3GM}\left(\frac{1}{2r^{2}}-\frac{GM}{r^{3}}\right)\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GM}{2r}\left[\frac{r}{r-3GM}\left(\frac{2r-6GM}{r}-\left(1-\frac{2GM}{r}\right)\right)\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GM}{2r}\left(1-\frac{3GM}{r}\right)^{-1}\left(1-\frac{4GM}{r}\right) \ \ \ \ \ (10)$

The energy per unit mass ${e}$ is then, using ${\tilde{E}=\frac{1}{2}\left(e^{2}-1\right)}$:

$\displaystyle e=\left[1-\frac{GM}{r}\left(1-\frac{3GM}{r}\right)^{-1}\left(1-\frac{4GM}{r}\right)\right]^{1/2} \ \ \ \ \ (11)$

For ${r=6GM}$, this comes out to ${e=\sqrt{\frac{8}{9}}}$.

As an example of the use of these formulas, suppose we start an object at infinity with no radial velocity, but with an infinitesimal tangential velocity which gives it an angular momentum ${l}$. Since the object is at an infinite distance, the formula ${l=r^{2}\omega}$ means that in the limit as ${r\rightarrow\infty}$, ${\omega\rightarrow0}$ so that the product ${r^{2}\omega=l}$, thus the tangential motion really is infinitesimal.

As the object falls in towards the central mass, it spirals in, keeping ${l}$ constant. Since the object started off essentially at rest, ${e=1}$ so one solution is for the object to end up in a circular orbit with ${r=4GM}$. From the above formula, this corresponds to

 $\displaystyle l$ $\displaystyle =$ $\displaystyle 4GM\sqrt{\frac{GM}{4GM-3GM}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4GM \ \ \ \ \ (13)$

12 thoughts on “Circular orbits: relation between radius and angular momentum”

1. Chris Kranenberg

The third line of calculating E tilde, second right hand term should be a 2 instead of a 1 after factoring out -GM/(2r). The final displayed result is correct for E tilde .

2. Jean-Luc Piedanna

Hello
You could simplify (11) up to : e = (1 – 2GM/r) (1 – 3GM/r)^(-1/2)

proof :
say x = GM/r
e² = 2E + 1 = -x(1-4x)/(1-3x) + 1 = (-x + 4x² + 1 – 3x)/(1-3x) = (1-2x)²/(1-3x)