# Circular orbit around a supermassive black hole

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Problem 10.9.

Suppose we are unfortunate enough to be orbiting a supermassive black hole with a mass of ${10^{6}}$ solar masses. For the sun, ${GM=1.477\mbox{ km}}$, so for the black hole, ${GM=1.477\times10^{6}\mbox{ km}}$. The radius of the orbit is ${r=10GM=1.477\times10^{7}\mbox{ km}}$. For comparison, the average earth-sun distance is ${1.5\times10^{8}\mbox{ km}}$, so this orbit is about 10 times closer to the black hole than Earth is to the sun. A rather frightening prospect.

From its radius, we can work out the energy and angular momentum per unit mass. From the formula for angular momentum as measured by the orbiting object:

 $\displaystyle l^{2}$ $\displaystyle =$ $\displaystyle \frac{r^{2}GM}{r-3GM}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{100\left(GM\right)^{2}}{7}\ \ \ \ \ (2)$ $\displaystyle l$ $\displaystyle =$ $\displaystyle \frac{10}{\sqrt{7}}GM\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.58\times10^{6}\mbox{ km} \ \ \ \ \ (4)$

The energy is

 $\displaystyle \tilde{E}$ $\displaystyle =$ $\displaystyle -\frac{GM}{2r}\left(1-\frac{3GM}{r}\right)^{-1}\left(1-\frac{4GM}{r}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{0.6}{20\times0.7}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -0.043 \ \ \ \ \ (7)$

The period of the orbit as measured by the spacecraft can be found from

 $\displaystyle l$ $\displaystyle =$ $\displaystyle r^{2}\omega\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi r^{2}}{T}\ \ \ \ \ (9)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{2\pi r^{2}}{l}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{200\sqrt{7}\pi}{10}GM\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 166GM\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.455\times10^{8}\mbox{ km}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 821\mbox{ s}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 13.7\mbox{ minutes} \ \ \ \ \ (15)$