Timeline for What is the escape velocity of a Black Hole?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 20, 2022 at 14:52 | comment | added | ProfRob | Is this only true for radial motion? | |
| Sep 14, 2019 at 9:59 | comment | added | jinawee | After coming back to this question, your answer is valid for a Schwarzschild BH when the escaping particle travels radially and you measuring the speed from the starting point. The formula I wrote assumes you measure the distance from a point at infinity, which is not the same since speed is coordinate dependent. | |
| Aug 15, 2017 at 21:29 | comment | added | jinawee | I don't buy your derivation. The energy of a particle is not given by the second formula you've written since it neglects the gravitational field, you are considering a free particle. The correct energy is $E=\gamma mc^2 \sqrt{1-2GM/R}$, which leads to $v_e=\sqrt{2GM/R - (2GM/R)^2}$. | |
| Aug 11, 2012 at 20:24 | comment | added | voix | @AlanSE - Yes, for a point-like object. | |
| Aug 11, 2012 at 12:36 | comment | added | Alan Rominger | So the escape velocity at the event horizon would go to the speed of light as the ratio in the radical goes to 1? Am I reading that right? That would still quite simply imply infinite energy, so it seems consistent. | |
| Aug 10, 2012 at 22:21 | history | edited | voix | CC BY-SA 3.0 |
deleted 2 characters in body
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| Aug 10, 2012 at 22:08 | history | answered | voix | CC BY-SA 3.0 |