 Mr. Abner
Pioneer Posts: 443
Joined: 08 Jun 2017
Location: Mississauga

### Science and Astronomy Questions

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? JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

To be honest, I don't know how that is possible either. But fact is, the bigger a black hole, the lower the gravity on the 'surface'.
For example, there was the theory that the whole universe is a black hole! https://en.wikipedia.org/wiki/Black_hole_cosmology
JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum) Watsisname
Science Officer Posts: 2012
Joined: 06 Sep 2016
Location: Bellingham, WA

### Science and Astronomy Questions

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. JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

I know that G is the gravitational constant and I assume that M is the mass of the black hole. But what do you mean by rs. The upper r in the square root expression? It's so small I can't read it. Then what would the other three r be?
JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum) Watsisname
Science Officer Posts: 2012
Joined: 06 Sep 2016
Location: Bellingham, WA

### Science and Astronomy Questions

JackDole wrote:
Source of the post But what do you mean by rs. The upper r in the square root expression?

Yes, sorry that was a bit small. rs is the Schwarzschild or event horizon radius, equal to 2GM/c2. The other r by itself is the radius of the observer who is at rest (which can only be possible outside the horizon). I'll make it more readable:   JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

Yes, but what is the other r, like the r2?

Is that the distance from the center of the black hole in which I want to calculate the g-forces?

And if so, which units do I use? Parsec, light years, kilometers?

(Sorry, but I'm not a mathematician!)
JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum) JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

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's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum) Watsisname
Science Officer Posts: 2012
Joined: 06 Sep 2016
Location: Bellingham, WA

### Science and Astronomy Questions

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?

Exactly. If this distance you want to calculate it at is on the event horizon, then no math is required. The answer is infinite, and no object can remain in place there.

Otherwise, for locations close to but outside the black hole, the g-forces will be finite, but stronger than you would expect from Newtonian physics. To calculate it, first find the size of the event horizon (2GM/c2), in whatever units you like. A nice shortcut is to multiply 2953 meters by the mass of the black hole in solar masses. Plug that result in for rs. Then, in the same units, plug in your desired radius for r.

I saw your post in the image dump thread so I understand the context now. Let's apply this to a 1.5 trillion solar mass black hole, such that the Newtonian formula predicts the gravitational pull you would feel at the horizon is close to 1 Earth gravity: Newton predicts the gravitational acceleration is a comfortable 1g at the horizon. The truth is that it is infinite, and it approaches infinity extremely quickly as you get very close to the horizon. For instance at 1.0001 radii it is a over 103g. At larger distances it is less extreme and gets closer to the Newtonian prediction. At 10 radii Newton predicts 0.01037g of gravity, while general relativity predicts 0.01093g. Watsisname
Science Officer Posts: 2012
Joined: 06 Sep 2016
Location: Bellingham, WA

### Science and Astronomy Questions

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?

Yes, that looks good!  JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

Then it seems right. Right on the horizon, about 0.14 parsecs, I have 10.67590883 g. Much the same as in your illustration.
JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum) Watsisname
Science Officer Posts: 2012
Joined: 06 Sep 2016
Location: Bellingham, WA

### Science and Astronomy Questions

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.

Mmm, that's not right.  For 1.5 trillion solar masses and exactly on the horizon, Newton predicts 10.17m/s2 or about 1.03g, while in reality and according to general relativity, it is infinite. Maybe you're not plugging in exactly the right distance for the horizon radius, since 10.6759g is what GR predicts at about 10.0092 horizon radii.

edit:  At 1.0092 radii, I mean. JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

SpaceEngine gives 0.14 parsecs for a black hole of 1.5 billion solar masses.
If I do the math myself, I get 0.14360755 parsecs.

I changed the function a little and I entered now this function in my calculator:
``gBH(r;M) = (G*M*mSol)/r^2*1/(sqrt(1-(G*M*mSol/(c^2*r))))``

(mSol = 1.9891e30 kilograms solar mass.)
(c = 299792458 in meters.)
(G = 6.67408e-11 m3/(kg-s2))

Now at 0.49 parsecs I get 0.94287451 g and at 0.14 parsecs I even get 15.28853412 g.

(With 0.14360755 parsecs I get 14.34163019 g.)
``gBH(3.08567802e16*0.14360755;1.5e12) = 14.34163019``

(Parsecs in meters.)
Am I doing anything wrong?
JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum) Watsisname
Science Officer Posts: 2012
Joined: 06 Sep 2016
Location: Bellingham, WA

### Science and Astronomy Questions

JackDole, okay, I define all my constants the same as yours and tested it out. Your formula looks right except there was a factor of 2 missing within the square root. The correction is:
``gBH(r;M) = (G*M*mSol)/r^2*1/(sqrt(1-(2*G*M*mSol/(c^2*r))))``

Using 1.5e12 solar masses for the black hole, I get a horizon radius of 0.14359 parsec. (Close enough to yours, I think our values of a parsec may be slightly different, but that doesn't matter.) At 0.49 parsec, or 3.41243 horizon radii, the gravitational acceleration should be about 1.03576 m/s2 or 0.1055822 g. (I define 1g as 9.81m/s2.)

If I plug in exactly the horizon radius for r, then the code spits out 'inf', or infinity, as it should.

For 0.14 parsecs, there should be no answer, since this is inside the horizon, where our notion of a local gravitational acceleration ceases to have meaning, because no observer can remain at rest there to make such a measurement. The gravitation is overwhelming and pulls all objects and allowed frames of reference inward.

Hope that helps!

Python code:
``G = 6.67408*10**(-11)c = 299792458MSol = 1.9891*10**30pc = 3.086*10**16M = 1.5*10**(12)*MSolrs = 2*G*M/c**2r = 0.49*pcg_Newt = G*M/r**2g_GR = G*M/r**2/np.sqrt(1-rs/r)print(' ')print('For a 1.5e12 solar mass =', MSol, 'kg black hole,')print('horizon radius = ', rs, 'm =', rs/pc, 'pc')print(' ')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')print(' ')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')``

Output:
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 A-L-E-X
Star Engineer Posts: 2305
Joined: 06 Mar 2017

### Science and Astronomy Questions

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.

But what about black hole cosmology, which states that the universe isn't a black hole but exists inside a black hole inside a larger universe?  Poplawski worked this out mathematically as a way to combine relativity and quantum mechanics and others like Lee Smolin have also put forth this conjecture. JackDole
Star Engineer Posts: 1871
Joined: 02 Nov 2016
Location: Terra

### Science and Astronomy Questions

With my calculator I have slightly different values:
1.03599996 m/s = 0.10560652 g
My calculator may be a little inaccurate.

I use Speedcrunch.
Attachments JackDole's Universe 0.990: http://forum.spaceengine.org/viewtopic.php?f=3&t=546
JackDole's Archive: http://forum.spaceengine.org/viewtopic.php?f=3&t=419
JackDole: Mega structures ... http://old.spaceengine.org/forum/17-3252-1 (Old forum)

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