# Circular orbits: 3 measurements of the period

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Problem 10.12.

We can generalize an earlier problem by working out the period of a circular orbit as measured by three different observers. The first observer is attached to the orbiting object which circles the mass at a radius ${r}$. The second observer is at rest at radius ${r}$, while the third observer is at infinity.

From the formula for angular momentum as measured by the orbiting object:

$\displaystyle l^{2}=\frac{r^{2}GM}{r-3GM} \ \ \ \ \ (1)$

we can calculate the angular speed from ${l=r^{2}\omega}$, so

 $\displaystyle \omega$ $\displaystyle =$ $\displaystyle \frac{l}{r^{2}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r}\sqrt{\frac{GM}{r-3GM}} \ \ \ \ \ (3)$

The period is then

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{2\pi}{\omega}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\pi r\sqrt{\frac{r-3GM}{GM}} \ \ \ \ \ (5)$

Defining the velocity as ${2\pi r/T}$, we get

$\displaystyle v=\sqrt{\frac{GM}{r-3GM}} \ \ \ \ \ (6)$

The period and velocity as measured at infinity are found from the angular speed at infinity ${\Omega}$

 $\displaystyle \Omega$ $\displaystyle =$ $\displaystyle \frac{\sqrt{GM}}{r^{3/2}}\ \ \ \ \ (7)$ $\displaystyle T_{\infty}$ $\displaystyle =$ $\displaystyle \frac{2\pi}{\Omega}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\pi r\sqrt{\frac{r}{GM}}\ \ \ \ \ (9)$ $\displaystyle v_{\infty}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{GM}{r}} \ \ \ \ \ (10)$

Finally, the period ${T_{0}}$ as measured by the observer at rest at ${r}$ is found from the relation between ${\tau}$ and ${t}$, as before.

$\displaystyle \Delta\tau=\sqrt{1-\frac{2GM}{r}}\Delta t \ \ \ \ \ (11)$

In this case, ${\Delta t=T_{\infty}}$ and ${\Delta\tau=T_{0}}$ so

 $\displaystyle T_{0}$ $\displaystyle =$ $\displaystyle 2\pi r\sqrt{1-\frac{2GM}{r}}\sqrt{\frac{r}{GM}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\pi r\sqrt{\frac{r-2GM}{GM}}\ \ \ \ \ (13)$ $\displaystyle v_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{GM}{r-2GM}} \ \ \ \ \ (14)$

Thus ${T and ${v>v_{0}>v_{\infty}}$. The condition ${T could be explained by time dilation, since to the observer at rest, the orbiting clock would run slow, so less time would appear to elapse on it than on the stationary clock. The condition ${T_{0} is a consequence of the difference between the proper time for an observer at rest at a finite distance from the mass and the time at infinity, which is given by the time coordinate ${t}$.

Since the square roots must all be real, we must have ${r\ge3GM}$. This ensures that ${v_{\infty}, but for ${3GM, ${v>1}$. I’m not entirely sure what the resolution of this apparent paradox is, but in the frame of the orbiting object, its own velocity is, of course, zero, so at best the expression for ${v}$ above must be considered an artificial velocity which cannot be measured in any physical sense. If the speed of the object is measured by an external observer (such as the stationary observers at ${r}$ and infinity) the value always seems to be less than 1.

## 4 thoughts on “Circular orbits: 3 measurements of the period”

1. Chris Kranenberg

The condition that the proper velocity can be greater than 1 when 4GM<r<3GM can be approached using Moore's Eq. 10.32 and substituting into the effective energy equation. Then letting r = 3.5GM gives an effective energy greater than 1 indicating an unbound orbit. Therefore, this condition does not produce circular orbits.

1. Ben

I need to revisit this problem, but here is where I left off.

Ought not there to be a length contraction effect on the circumference of the orbit as observed by the spaceship?
Another point of view:
At the instant when the spaceship passes the stationary observer (who is at the same “r”), the two observers ought to agree about their relative velocity. And that velocity (they actually observe) won’t exceed c. (At this instant, I would think it should be possible to do something like a Lorentz transformation between the two frames.)

2. oliver

I think the reason for v>1 for 3GM<r<4GM is because v is defined as a tangential velocity v = r * Omega, where r is the Schwarzshild radius, which "does not actually specify the distance from the origin" (Th.Moore p.107).