2

In the movie Interstellar, there is a planet called Miller that orbits a supermassive black hole (Gargantua) so close that each hour is 7 years on Earth.

Miller's planet orbits Gargantua at ISCO with a speed of 0.5$c$. Doing a naive calculation using the circumference of the orbit ($r$ in Boyer–Lindquist coordinates) yields an orbital period of 1.72 hours. These are the numbers I used:

$$a = 1.00 - 1.33\times10^{-14}$$

$$M = 10^8 M_\odot$$

$$r_{\rm ISCO} = 1.00003763264895$$

But that naive calculation doesn't take into account the Kerr metric, which describes rotating black holes and all the spacetime warping that it entails. Also, would 1.72 h be the orbital period measured by an stationary observer at infinity? If that's so, knowing that the time dilation on Miller's is 61000, the orbital period measured by people on Miller would be 0.1 seconds.

Is Miller's orbital period really 1.72 hours? Does it really rotate around Gargantua 10 times each second from its perspective? How can the orbital period of Miller be calculated using the Kerr metric?

Camillus
  • 719
Stellar_Enginner
  • 105

1 Answers1

5

The orbital period of an equatorial circular orbit around a Kerr black hole as measured by a distant observer is given by

$$ P = 2\pi\frac{G M}{c^3} (\tilde{r}^{3/2} + \tilde{a}),$$

where $\tilde{r}$ and $\tilde{a}$ are the radius of the orbit and spin of the black hole adimensionalized by the gravitational radius $\frac{G M}{c^2}$.

Putting in

\begin{align} \tilde{a} &= 1- 1.33\times 10^{-14}\\ M&=10^8 M_{\odot}\\ \tilde{r} &=r_{\rm ISCO} \end{align}

gives a period of 1.719 hours.

Note that a circular orbit at this radius with this spin accounts for the time dilation experienced on Miller, but not for the extreme tides.

TimRias
  • 13,497