4 ms·
I can't say for sure on this particular case, but the usual reasoning is this: Using Newtonian Mechanics (I'll skip the derivation.) you can calculate escape v
by Enginerrrd 5y ago
I can't say for sure on this particular case, but the usual reasoning is this:
Using Newtonian Mechanics (I'll skip the derivation.) you can calculate escape velocity as follows: v = sqrt(2GM / r)
r is the distance to the center of mass, G is the usual gravitational constant and M is the mass of the body.
So, now we can reason about an object whose escape velocity is c.
c^2 = 2GM / r and rearranging a bit:
r* = 2GM / c^2.
Thus, IF we packed a mass M into the radius r*, we would have a Newtonian blackhole.
It turns out, quite incredibly, that this r* is in perfect agreement with the swarzchild radius from a non-rotating blackhole calculated using general relativity. Exactly why these two are in such agreement has never been explained to me other than as a coincidence. But it's quite incredible since usually r^3 terms pop up prior to the event horizon (at the same radius) which make GR make different predictions about orbits. (Such as the precession of mercury)
- mr_toad 5y agoDoesn’t GR always agree with Newtonian gravity when objects aren’t accelerating?
- Enginerrrd 5y agoFamously, not quite. See https://en.wikipedia.org/wiki/Tests_of_general_relativity#Perihelion_precession_of_Mercury https://en.wikipedia.org/wiki/Tests_of_general_relativity#Pe...
- Beldin 5y agoWhat's "not accelerating" about going in an ellipse? Doesn't that imply constant (and varying) acceleration?