How strong can a gravitational potential be?

[repoman]

This is not a very well defined question, more of a spitballing...

Some one at work asked this question, and neither of us have a great deal of technical knowledge.

So, I was thinking that the question itself is ill-defined. But let me put some bounds on it.

First though, I think that having the gravitational analogue to voltage is another very useful quantity to have. It is basically this for a round object in the newtonian sense, right? G(pot)=-GM/r^2...
G(pot) = Gravitational potential measured as acceleration.
M = mass of object, r = distance from center or the surface.
m is not used here.

For something like neutron stars,

The gravity at its surface is more than 300 billion times stronger than that on Earth

from http://www.dailygalaxy.com/my_weblog/2013/04/bizzare-binary-neutron-stars-with-gravity-300-billion-times-earth-confirm-theory-of-relativity.html

which works out to roughly -3*10^12 m/s^2, given that g=9.8 m/s^2 on earth.

Am I wrong that the gravity on the surface of a neutron star is even more than at any event horizon of a black hole of larger mass?
For smaller black holes the gravitational potential may (would?) be higher at the event horizon? Of course how can sub-stellar black holes be made, what could supply such a force?

This is not even addressing how high the G(pot) can be inside of the event horizon. I have no idea on any of that. How high can they get according to some models?

 

[skepticalbip]

I think your number for the gravitational potential at the surface of a neutron star is a bit off. The escape velocity from the surface of a neutron star is 100,000km/s to 150,000km/s (depending on the mass of the star) while the escape velocity from the event horizon of a black hole is 300,000km/s (two to three times that of a neutron star).

I didn't understand the difference you describe between a low mass and large mass black hole. The escape velocity from the event horizon of either is c. However, I suppose the gravitational gradient approaching a low mass black hole would be greater than the gradient approaching a high mass black hole.

 

[barbos]

This is not a very well defined question, more of a spitballing...

Some one at work asked this question, and neither of us have a great deal of technical knowledge.

So, I was thinking that the question itself is ill-defined. But let me put some bounds on it.

Ill-defined indeed.

First though, I think that having the gravitational analogue to voltage is another very useful quantity to have. It is basically this for a round object in the newtonian sense, right? G(pot)=-GM/r^2...
G(pot) = Gravitational potential measured as acceleration.
M = mass of object, r = distance from center or the surface.
m is not used here.

Gravitational potential is G(pot)=-GM/r.
GM/r^2 is gravitational field.

 

[repoman]

I think this may be a units issue as far as making a misunderstanding:

on the surface of a very dense neutron star the g-force (units is m/s^2 or acceleration) is apparently ~300 billion times stronger than on earth.

But if you could magically make a "surface" at the event horizon of a black hole to stand on, what would the g-force be?

Also, I guess that escape velocity is a good metric for surface gravity. But it is different from the g-force measured in acceleration for objects that are not black holes.

 

[skepticalbip]

The mass range for neutron stars is about 1.5 to 3 solar masses. More mass will collapse to a black hole. So the gravitational potential at the surface of a 3 solar mass neutron star will be (pot)=G * 3 solarmasses / radius of the neutron star. A black hole of 3 solar masses would have the radius of the event horizon about 25%-30% less (I think - just pulled from my arse) than the radius of a neutron star so its gravitational potential at the event horizon would be greater than the surface of a neutron star.

If you want numbers then you can look up the mass of a 3 solar mass neutron star (i think about 6*10^30 Kg) and its radius (I think about 11 Km) and plug them in.

 

[Bomb#20]

Also, I guess that escape velocity is a good metric for surface gravity.

It isn't really. Both escape velocity and surface gravity go down with radius and go up with mass, but with totally different power relationships. Earth has 10% higher surface gravity than Uranus, but Uranus has double the escape velocity.

 

[skepticalbip]

Also, I guess that escape velocity is a good metric for surface gravity.

It isn't really. Both escape velocity and surface gravity go down with radius and go up with mass, but with totally different power relationships. Earth has 10% higher surface gravity than Uranus, but Uranus has double the escape velocity.

It is a fair metric when comparing bodies with similar densities such as our system's rocky planets or comparing our system's gas giants or comparing neutron stars with black holes (assuming a density of a black hole is its mass divided by the volume of its event horizon) but then describing a black hole's density seems to be a bit problematic otherwise.

 

[Underseer]

potential energy = mgh

Obviously, that's assuming that you and whatever you're measuring is close to the surface of the Earth, but that is more than good for almost anything you're likely to deal with.

 

[bilby]

As gravitational potential is directly proportional to acceleration for a given mass, I think the OP question can be re-framed as 'Is there a maximum possible acceleration', in which case, the answer provided at https://physics.stackexchange.com/questions/3334/is-there-a-maximum-possible-acceleration seems right; It depends whether you assume a minimum length, and if so, what that minimum length is:

The spit horizon in a Rindler wedge occurs at a distance d = c²/g for the acceleration g. In spatial coordinates this particle horizon occurs at the distance d behind the accelerated frame. Clearly if d = 0 the acceleration is infinite, or better put indefinite or divergent. However, we can think of this as approximating the near horizon frame of an accelerated observer above a black hole. The closest one can get without hitting the horizon is within a Planck unit of length. So the acceleration required for d = ℓ_p = √(Gℏ/c²) is g = c²/ℓ_p which gives g = 5.6×10⁵³cm/s². That is absolutely enormous. The general rule is that Unruh radiation has about 1K for each 10²¹cm/s² of acceleration. So this accelerated frame would detect an Unruh radiation at ∼ 10³¹K. This is about an order of magnitude larger than the Hagedorn temperature. We should then use the string length instead of the Planck length √(4πα′) and the maximum acceleration will correspond to the Hagedorn temperature.