0
$\begingroup$

If we observe from Earth a planet in very close orbit around a supermassive black hole (as close as possible to the black hole without the planet being swallowed up or destroyed by tidal forces), would we see the planet in question have a period of revolution around the black hole slower than it should because of the phenomenon of clocks slowing down indicated by general relativity?

If so, would it be possible to observe a planet in a distant circular orbit around a supermassive black hole having a shorter period of revolution than another planet in a circular orbit much closer to the same supermassive black hole (this which is the opposite of what we observe for example around the Earth or the Sun, the further away we are, the longer the orbital period)?

$\endgroup$
1
  • $\begingroup$ What you "see" is caused by relativistic Doppler effect and light deflection, not just by the Lorentz transformations and gravitational redshift. It's complicated and people have made actual videos of the observed effects. $\endgroup$ Nov 6, 2022 at 15:48

1 Answer 1

1
$\begingroup$

Sebastyen Laroche asked: "would it be possible to observe a planet in a distant circular orbit around a supermassive black hole having a shorter period of revolution than another planet in a circular orbit much closer to the same supermassive black hole?"

That would not be possible, since the required local orbital velocity around a Schwarzschild black hole is higher by the same amount the time slows down due to gravity, so that cancels out and the observed angular velocity is the same as with Newton:

ASTR498 wrote: "By a lovely coincidence, in Schwarzschild coordinates the angular velocity observed at infinity is exactly the same as it is in Newtonian physics."

So what remains is that the radius by which we multiply gets smaller and the angular velocity by which we divide larger, so the period gets smaller with smaller radius as it would under Newton.

If we plot the observed time period for one revolution (in units of GM/c³) by the radial coordinate (in GM/c²) we see that the period gets shorter with smaller radius, and is shortest at the photon sphere at r=3GM/c² where the required velocity is 1c and the proper time therefore goes to 0:

orbital period around a black hole

The red curve is the orbital period t in the frame of the coordinate bookeeper far away from the black hole, and the blue one in terms of the proper time τ of the orbiting clock.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.