How does the speed (and kinetic energy) necessary to orbit a black hole vary with the radius of orbit?

Question

I understand that closer orbits around Schwarzschild black holes require faster speeds, culminating in a required speed of $c$ at the photon sphere ($r = 3M$ in Schwarzschild coordinates). I'd love to be able to work out the orbital speed at the ISCO ($6M$) and marginally bound orbit ($4M$) and others. Is there a formula for this orbital speed as a function of radius? (I don't mind that these orbits are unstable.)

I gather that there are some challenges of interpretation here for subluminal velocities as to which reference frame the speed is being measured with respect to. I suppose I'm most interested in measurement with respect to a stationary observer at that same radius and to a stationary observer a distance infinity.

I'd also love to have a formula for the amount of kinetic energy required to be in that orbit in terms of the rest mass of the object. I gather that these are all going to be relativistic speeds and thus that this can be a substantial fraction of the rest mass -- approaching an infinite multiple as radius approaches $3M$.

Answer 1

There are a couple of expressions you could consider for circular orbits in the Schwarzschild metric (at $r \geq 3r_s/2$) - see here.

A "shell" observer, at the same radial coordinate as the orbiting body, but at a fixed $\theta$ and $\phi$, will observe its speed as it passes to be given by $$ \frac{v^2}{c^2} = \frac{r_s}{2r-2r_s}\ ,$$ where $r_s = 2GM/c^2$. Thus $$(\gamma -1)mc^2 =\left[ \left(\frac{2r-2r_s}{2r-3r_s}\right)^{1/2} -1 \right]mc^2\ . $$ This expression tends to infinity as $r \rightarrow 3r_s/2$. This would be the kinetic energy with which something would need to be launched tangentially from a stationary platform at that radius to be put into a circular orbit.

A distant observer would infer an orbital speed given by $$ \frac{v^2}{c^2} = \frac{r_s}{2r}$$ and thus $$(\gamma -1)mc^2 = \left[ \left(\frac{2r}{2r -r_s}\right)^{1/2} - 1\right]mc^2\ . $$

Note that these two expressions do not agree since kinetic energy is not the same in different frames of reference.

Answer 2

I believe for a circular orbit the orbital velocity is always going to be $$v=\sqrt{GM/r}\tag{1}$$ in "coordinate time", as per a distant observer at rest. This is the same as in Newtonian gravitation. At the photon sphere you will no longer be able to reach this speed because you would have to travel at the speed of light which in a graviational field in coordinate time, but not in proper time, is less than c.

Regarding the energy of the orbiting body if I remember things correctly you could, assuming $E=mc^2$ at rest at infinity, write:

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\right).\tag{2}$$

($\hat{r}=\bar{r}/r,\hat{v}=\bar{v}/v$)

Notice that this is basically both the kinetic and the potential energy baked into the same expression. For a circular orbit $\hat{r}\cdot\hat{v}=0$ and $|\hat{r}\times\hat{v}|=1$. and the expression becomes simpler. Notice that this is basically both the kinetic and the potential energy baked into the same expression. For a circular orbit we can insert (1) into (2) and get: $$E_{circularorbit}=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{3GM}{rc^2}}}\right).\tag{3}$$

From (2) and (3) we see that we need infinite energy to sustain a circular orbit at the photon radius. You can take the derivatve of (3) with respect to $r$ and find that the expression have a minimum at the "innermost stable circular orbit". At closer radial distances than that circular orbits closer to the central mass require more energy to uphold.

I like (2) because you can by "inspection" see that the energy for a "hoovering" object close to the Schwarzschild radius goes to zero but that even a miniscule motion will push the expression towards infinity, thus suggesting the known property of objects slowing down and "freezing" at the Schwarzshild radius, when using coordinate time.

Now in general relativity we have a phenomenon causing time, frequency and energy to be perceived different within a gravitational field. Energy and frequency will be perceived to be a factor: $$1\over{\sqrt{1-{2GM\over{rc^2}}}}\tag{4}$$ higher for a stationary observer within the gravitational field than at infinity. Time will slow down with the same factor. It so happens that the slowing down of time for a local observer, measuring "proper time", exactly balances the observed slowing down of light that a distant observer notices so the locally measured velocity of light will always be c.