- #1
travwg33
- 21
- 0
I am fairly new to astronomy, having just finished a course this past semester in high school, and I have a question about orbital velocity.
Now, I know that the event horizon is the boundary at which the escape velocity is equal to the speed of light, but is there any significance to the boundary at which the orbital velocity is equal to the speed of light?
I quickly messed around with the math to see where such a boundary would lie and determined this:
Vorb = √((Gm)/r)
substitute Vorb for C
C = √((Gm)/r)
r = (Gm)/(C2)
This is obviously very similar to determining the Schwarzschild radius and this boudary would actually be inside the event horizon, light orbiting the singularity, but never moving out past the radius once in orbit.
I looked online to see if such a boundary was important or existed and I think I found something similar the photon sphere, but the equation is different and this sphere is outside the event horizon.
Does the boundary where r = (Gm)/(C2) mean anything?
Now, I know that the event horizon is the boundary at which the escape velocity is equal to the speed of light, but is there any significance to the boundary at which the orbital velocity is equal to the speed of light?
I quickly messed around with the math to see where such a boundary would lie and determined this:
Vorb = √((Gm)/r)
substitute Vorb for C
C = √((Gm)/r)
r = (Gm)/(C2)
This is obviously very similar to determining the Schwarzschild radius and this boudary would actually be inside the event horizon, light orbiting the singularity, but never moving out past the radius once in orbit.
I looked online to see if such a boundary was important or existed and I think I found something similar the photon sphere, but the equation is different and this sphere is outside the event horizon.
Does the boundary where r = (Gm)/(C2) mean anything?