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)

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