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

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  9. 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 .

    Reply
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  11. 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)

    Reply

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