How much does time slow down with gravitational time dilation, on a black hole with these graph trajectories.?

If you watch a journey into a black hole on Youtube before reading this question, it only takes four minutes to watch the whole video, you will understand better what I mean by graph trajectory zones. Orbiting a black hole is dangerous. There are four graph trajectory zones when orbiting a black hole. The green zone is a safe zone, the yellow zone is a risky zone, the orange zone is a danger zone where there are no orbits, neither stable nor unstable. Then there is the red zone, the Event Horizon. So if you had a spacecraft and wanted to slow yourself down with gravity from the black hole, so that you could visit Earth in the future in your lifetime. I heard from this show you can orbit a black hole for a year and four years will have passed on Earth from your perspective. How much would time slow down if you orbited in the clour coded regions?

So Question 1: If you orbited in the green zone for a year, how much time would pass on Earth?

Question 2: If you orbited in the yellow risky zone for a year, how much time would pass on the Earth?

Question 3: If you orbited in the orange zone, where there are no orbits stable, or unstable, this area is called the photon shere where light can just stay in orbit. So the spacecraft would be orbiting in the orange zone, as close to the Event Horizen as possible. Is this orange zone the best place to be in terms of slowing down a space shuttle in time, to kind of travel in time relative to outside observers watching? Also will the spacecraft get sphagettified in this orange region?

  • 1
    $\begingroup$ Please take the time to read help center to learn how this site works. Please also rewrite the question to make it readable. $\endgroup$ – Deer Hunter Jun 28 '14 at 5:01
  • $\begingroup$ You need to add a link to the video. $\endgroup$ – Hobbes Jun 28 '14 at 9:25

I don't get your colors, but the time dilation (how much more slowly your time passes relative to the outside observer, so greater than one) is $1/\sqrt{1-{r_S\over r}}$, where $r$ is your distance from the center of the mass, and $r_S$ is the Schwarzschild radius for that mass.

| improve this answer | |

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