# Gravitational effect of a black hole based on distance

1. Nov 26, 2015

### Gerinski

Hi,

I have often read that at large enough distances, the gravitational effect of a black hole is no different than the gravitational effect of a star or other body of the same mass.

But that at close distances the difference shows up, for example the notion of the photon sphere at 1.5 times the Schwartschild radius which does not appear for non-black-hole objects.

At which distance precisely do the gravitational effects of black holes become undistinguishable from the effects of a non-black-hole object of the same mass, and why?

TX !

2. Nov 26, 2015

### Buzz Bloom

Hi Gerinski:

I confess that I am not a GR expert, but I will share what I believe I understand about this question.

Massive bodies that are not black holes occupy a much larger volume that a black hole of the same mass. The smallest such bodies are neutron stars, although I understand that the smallest black holes previously detected are much more massive than a neutron star. In principle though it is possible for a black hole to have the mass as small as that of a neutron star. One possible interpretation of your question could be that for a mass equal to that of a neutron star, at distances greater than the radius of a neutron star the gravitational effects are indistinguishable from a black hole. For this interpretation, the reason why it is unreasonable to compare conditions at a smaller distance is that for the neutron star there is matter present which is not present near a black hole.

If you would post a specific different interpretation of your question, I will try to answer that interpretation also.

Regards,
Buzz

3. Nov 26, 2015

### Gerinski

Thanks Buzz B,

I clearly understand that at distances smaller than the physical boundary there are distinguishing effects. The Sun has a certain radius, and a black hole of 1 Sun mass has a much smaller (Schwardschild) radius, so in all the distance scales between the 1 solar mass black hole radius and the actual Sun's radius the gravitational effects differ.

So my question goes like, in this precise case for example, at which distances from the center of such a 1 solar mass black hole or the actual Sun, would the gravitational potential characteristics become indistinguishable from each other.

4. Nov 26, 2015

### Buzz Bloom

Hi Gerinski:

At distances greater than the sun's radius, say R, (ignoring the sun's rotation and oblateness which makes sun's shape not quite a sphere), there is no gravitational difference between the sun and a BH of the same mass. At a radius r < R the gravitational effects are completely different. A major reason why this is so is that the sun's mass in the spherical shell between r and R has no gravitational effect on a particle at radius R.

Regards,
Buzz

5. Nov 26, 2015

### Staff: Mentor

I would say that the distance where a black hole is noticeably different from a non black hole would be the Schwartschild radius itself! Otherwise, gravity is gravity. It works the same for everything.

6. Nov 26, 2015

### rootone

If you are external to the massive object, you experience it's gravity as a force pulling you towards where the center of it's mass is..
The closer you get to the center of the mass the stronger the gravity is.
Whether the massive object is a black hole or a star or something else makes no difference, the gravity field for a given mass will be the same.

Last edited: Nov 26, 2015
7. Nov 26, 2015

### Buzz Bloom

Hi russ:

Your statement is of course correct, but but I think my interpretation of Gerinski's original question is different than yours. I may be wrong, but I think Gerinski question was seeking a description of the nature of a some specific non-BH object with the same mass as a BH, and with respect to those two objects, a radius which could be clearly defined separating BH gravitational effects from non-BH gravitational effects.

Applying this idea to your statement, suppose we consider a BH of mass M, and Schwartschild radius r. Now consider a non-BH object of mass M and radius, say for example,
R = 1.1 × r.​
Could such an object as this non-BH object possibly be stable? If not, then I don't think it would be a good example to use to answer Gerinski's question. If so, then it would be very strange indeed - a non-BH with photons moving around it in circular orbits with a radius of 1.5 × r.

Regards,
Buzz

8. Nov 27, 2015

### stevebd1

The full equation for gravity for a non-(or very slow)rotating, non-charged object in GR is-

$$g=\frac{Gm}{r^2}\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}$$

where the first part of the equation $(Gm/r^2)$ is an approximation of the stress energy tensor and the second part represents the metric tensor.

Though technically it will always apply, the metric tensor part of the equation can be dismissed when approximating/calculating gravity at the surface of non-relativistic objects such as planets and non-exotic stars. The difference the second part makes becomes apparent when considering the surface of white dwarfs, neutron stars and black holes (note as r approaches 2M (the Schwarzschild radius or event horizon), gravity begins to increase exponentially, becoming infinite at 2M (M=Gm/c2)).

It's also worth noting that the metric tensor produces the time dilation equation, which in the case of a non-rotating, non-charged object is $d\tau=dt\sqrt(1-2Gm/(rc^2))$.

It's predicted that the smallest radius an object of mass can collapse to before the runaway effect turns it into a black hole is 9/5M which is 2.25M (or 1.125 the Schwarzschild radius) which means technically, some neutron stars can have a photon sphere at 3M.

9. Nov 27, 2015

### DaveC426913

Consider a hypothetical star and a hypothetical black hole.
The star's mass is 1 zillion kg* and its diameter is one million miles.
The BH's mass is also 1 zillion kg and its diameter is ten miles.
* It doesn't matter, as long as they're both the same

If you got in your spaceship and brought it to within 500,001 miles of the centre of the star (i.e. one mile above its surface), you would feel the same pull as if you brought it to 500,001 miles above the centre of the BH (i.e.499,996 miles above its Schwartschild radius). So, without windows, you could not tell what body you are hovering near.

Your distance to the surface of an object is irrelevant to its gravitational pull on you; the only relevant distance is to its centre.

So now you drop 2 miles closer to the BH, and you are now hovering 499,999 miles from its centre. Gravity increases slightly, as it should.

But if you drop 2 miles closer to the star your're in trouble, because now you're inside the star. Assuming your spaceship can survive, you will still experience a gravitational attraction to the star, but it will actually be less because the outer shell of the star is no longer a gravitational factor.

The point here is that a black hole's gravity is identical in every way to a star's (or any other body's) gravity. The only thing about a black hole is that you can get closer to its centre.

But be clear on this: gravity does not behave any differently around a black hole than anywhere else.

10. Nov 30, 2015

### Gerinski

Thanks a lot to all !