Science and Astronomy Questions
Posted: 14 Sep 2020 19:09
Wait a minute... isn't the Schwarzschild radius the point were the escape velocity equals the speed of light? How could that be equal to 1g?
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JackDole wrote:Source of the post How do I calculate the gravitation at the Schwarzschild radius of a black hole?
JackDole wrote:Source of the post For example, there was the theory that the whole universe is a black hole!
JackDole wrote:Source of the post But what do you mean by rs. The upper r in the square root expression?
JackDole wrote:Source of the post Is that the distance from the center of the black hole in which I want to calculate the g-forces?
JackDole wrote:Source of the post Watsisname, okay, I think I got it. (Possibly!)
For a black hole of 1.5 trillion solar masses and a distance of 0.49 parsecs, I have a gravitation of 0.87118334.
Can you possibly check that?
JackDole wrote:Source of the post Then it seems right. Right on the horizon, about 0.14 parsecs, I have 10.67590883 g. Much the same as in your illustration.
gBH(r;M) = (G*M*mSol)/r^2*1/(sqrt(1-(G*M*mSol/(c^2*r))))
gBH(3.08567802e16*0.14360755;1.5e12) = 14.34163019
gBH(r;M) = (G*M*mSol)/r^2*1/(sqrt(1-(2*G*M*mSol/(c^2*r))))
G = 6.67408*10**(-11)
c = 299792458
MSol = 1.9891*10**30
pc = 3.086*10**16
M = 1.5*10**(12)*MSol
rs = 2*G*M/c**2
r = 0.49*pc
g_Newt = G*M/r**2
g_GR = G*M/r**2/np.sqrt(1-rs/r)
print('For a 1.5e12 solar mass =', MSol, 'kg black hole,')
print('horizon radius = ', rs, 'm =', rs/pc, 'pc')
print('At r =', r, 'meters =', r/pc, 'pc =', r/rs, 'horizon radii,')
print('the locally measured gravitational acceleration is:')
print('(Newtonian calculation) ', g_Newt, 'm/s^2 = ', g_Newt/9.81, 'g')
print('(General relativistic) ', g_GR, 'm/s^2 = ', g_GR/9.81, 'g')
g_Dole = (G*M)/r**2*1/(np.sqrt(1-(2*G*M/(c**2*r))))
print('Jacks calculation:', g_Dole, 'm/s^2 =', g_Dole/9.81, 'g')
For a 1.5e12 solar mass = 1.9891000000000002e+30 kg black hole,
horizon radius = 4431266548024232.0 m = 0.1435925647447904 pc
At r = 1.51214e+16 meters = 0.49 pc = 3.4124329548043497 horizon radii,
the locally measured gravitational acceleration is:
(Newtonian calculation) 0.8708739121207744 m/s^2 = 0.08877409909488015 g
(General relativistic) 1.03576139161 m/s^2 = 0.105582200979 g
Jacks calculation: 1.03576139161 m/s^2 = 0.105582200979 g
Watsisname wrote:JackDole wrote:Source of the post How do I calculate the gravitation at the Schwarzschild radius of a black hole?
The Newtonian formula for the acceleration due to gravity (by which we will mean the instantaneous acceleration measured for an object dropped by someone who is at rest at that location) fails very badly near black holes. It predicts a finite acceleration at the horizon, when in fact it must be infinite, since the only velocity an object can have at the horizon according to local observers is exactly c, and no observer can hover in place at or below the horizon, no matter how powerful their rockets.
The correct formula, which works just as well for small masses as for black holes, is:
where rs is the Schwarzschild radius, 2GM/c2. This formula closely matches the Newtonian formula for weak gravitational fields, like planets and stars, but then diverges dramatically near compact objects like neutron stars and black holes, and is infinite at the event horizon.
That the Newtonian expression for the gravitational acceleration at the horizon is finite and inversely proportional to the mass, while wrong, does at least carry some physical intuition: it is consistent with how the curvature and hence tidal forces at the horizon are weaker for a larger black hole.JackDole wrote:Source of the post For example, there was the theory that the whole universe is a black hole!
This is a popular idea, based largely on how if you calculate the Schwarzschild radius of a black hole with the mass of the observable universe, it is close to the radius of the observable universe. But the conclusion that the universe is literally a black hole is easily seen to be incorrect, since the universe's has the wrong spacetime metric (the uniform and expanding FLRW metric, as opposed to a black hole's Schwarzschild metric.) Basically, the universe is not a black hole because the matter and energy are distributed uniformly on large scales, whereas a black hole requires it to be concentrated somewhere.
And to really kill the idea, it turns out the apparent coincidence of the size of the observable universe and its Schwarzschild radius arises for a simple reason that has nothing to do with black holes. If the universe's expansion rate is constant over time from the Big Bang to present, then the radius of the observable universe will be equal to the Schwarzchild radius of the enclosed mass! I'll show the proof in a moment in the cosmology thread. The expansion history of our universe is not actually constant (it was faster at first, then slowed down, and now is speeding up again), but its present size coincidentally works out to be almost the same as if it had expanded at the present rate since the Big Bang.