Force of Gravity

[BobbyJoeCool]

Ok... let's say you have a mass of 1kg, contained in a single point. by the equation: [math]F=-G\frac{m_{1}m_{2}}{r}[/math]

 

the smaller the r, the larger the force...

 

In fact, [math]\lim_{r\to 0^{+}}{F=-G\frac{m_{1}m_{2}}{r}}=-\infty[/math]

 

So, that means that any mass, if you're close enough to it (as in, like a trillionth of a nanometer) the force will be so great that once again, light would not escape it and an event horizon exists that is astronomically small.

 

Is there something I'm missing here or is this true?

[mezarashi]

Well yes, technically there's nothing wrong with that. However, you have to keep in mind that first of all, objects like the Earth are not composed of a singularity. That equation more or less is assumed to be true if the force originates from a point in space or a distance from a perfect sphere. If you've done your calculus exercises you will see that it cancels out to where a point mass estimate will be okay.

 

Another factor to consider is that as r->0, there are 3 other forces (electromagnetic, weak, strong) that start to come into play which are MUCH MUCH MUCH stronger than gravity. Once you bring two atoms close to each other, their electron clouds (electromagnetic force which is about 10^39 times stronger than gravity) will interact repulsively.

[timo]

There´s several thing coming to my mind:

1) Classical mechanics is simply not valid at very small distances.

2) Since 1) is a rather useless answer for you I´ll continue a bit:

3) I wouldn´t know how to treat light within the scope of newtonian gravity. If you set the mass of light to zero, your limit also approaches zero and your problem vanishes.

4) The "tightness" of binding is usually described by the binding energy, not the binding force. In fact, the formula you gave is that for the binding energy (force would be 1/r²). There´s several resons for using energy instead of force but that´s not the topic here.

5) For all distances r>0, the binding energy will be finite. Therefore, there is always a kinetic energy which is sufficient for that the other particle (I don´t really want to call it light) can escape. This is in gross contrast to a black hole where there is a finite r>0 at which nothing can escape, regardless of it´s energy. That´s probably the main point you missed.

 

 

astronomically small

^^ Can´t help but liking that expression.

[gib65]

Essentially, the problem comes down to this: r represents the distance between the 2 objects from their centers of mass. Since all applicable objects are, in practice, made of matter, their volumes will always be greater than a singularity, and therefore they will "hit" each other before their centers of mass coincide, thereby putting a lower limit on how small r can become and thus how great the force of gravity can be.

 

I think this is what the other posters were saying, not in so many words.

[timo]

^^ It wasn´t what I was saying, at least. Might still be true but my main point was that in a black hole scenario you hit the singularity before you reach r=0.

[BobbyJoeCool]

and my point was that you are a singlularity, and the other object is light (which it has to be in order to determine an event horizon). and say that this light is occupying only a single point in space. if the light gets close enough to the singularity of mass 1kg, the force on the light created by gravity will make the escape velocity greater than that of the speed of light. Now, this never happens in real space because 1kg of mass has electromagnetic force much greater than 1kg of mass can generate.

 

So, in effect, you need the MASSIVE force of a black hole in order for any of it to work, at which point, the EH is so large that it's notciable with the naked eye...

[Ice Demon]

Scientifically that is true, but its just thoeretical. It is in some ways true, but depending on circumstances r could be less and still have more force. Its like reverse magmetism.

[timo]

and my point was that you are[/i'] a singlularity, and the other object is light (which it has to be in order to determine an event horizon). and say that this light is occupying only a single point in space. if the light gets close enough to the singularity of mass 1kg, the force on the light created by gravity will make the escape velocity greater than that of the speed of light.

How do you calculate the escape velocity? And what mass do you asuume for light? And why?

[BobbyJoeCool]

How do you calculate the escape velocity? And what mass do you asuume for light? And why?

 

[math]v_{e}=\sqrt{\frac{2Gm}{d}}[/math]

 

Source named Wikipedia (I love that site)

 

The event horizon exists at the point where the escape velocity is the speed of light.

 

[math]c=\sqrt{\frac{2Gm}{d}}[/math]

 

[math]c^2=\frac{2Gm}{d}[/math]

 

[math]dc^2=2Gm[/math]

 

[math]d=\frac{2Gm}{c^2}[/math]

 

as long as the object has mass=/=0, d=/=0, because G and c are constants...

 

You don't need a mass for light to calculate the EH location...

[timo]

My question actually was a bit of rhetorical because I was hoping that you´d actually do the calculation to see that your assumption of an escape velocity runs into problems. I didn´t really expect you to copy-paste a solution from wikipedia. Nevertheless, if you inspect the line above the one you quoted, you might see where the problems lie.

[BobbyJoeCool]

you mean that the mass of light cancils out with itself and that if light has no mass, you're in essence dividing both sides by 0, which is illegal?

 

but a photon has mass, it's just that it's mass is relative (because it's moving at the speed of light, it seems to have no mass to us because we're not moving at at speed of light.)

 

But if that's true, how can you determine where the EH is on a black hole?

[timo]

Well, let´s forget about the lightmass for a second. There is no way to treat light correctly in classical mechanics and I just realize that I run into problems imagining how a semi-correct treatment would look like.

A much more important point is this one: The event horizont is not characterized by the radius where the escape velocity would exceed the speed of light. It does, but this is only a result of a relativistic mechanism. The condition set by the event horizont is a much stricter one: Nothing can escape this radius.

This is in contrast to the "escape velocity" you used above since the latter gives the velocity to escape to infinity. So even if you limited yourself to velocities <c, it would be possible to escape your "event horizont". Simply kick your particle long enought till it´s close to d. Then, kick it so that it´s outside of d (d is the distance you calculated above, in case you didn´t notice it). Then, kick it to it´s escape velocity and say goodbye.

The "kick it till it´s outside of d" will not be possible in a black hole scenario.

 

But if that's true, how can you determine where the EH is on a black hole?

Huh, how much background of general relativity should I suppose here? Within spacetime, particles can only travel on a certain subset of all possible directions. These are the so-called light-cones. Gravitiy can bend these light-cones. At the event horizont, the light-cone is bent in such a way that is completely points inward. No such thing as limited directions of movement exist in classical mechanics which makes the analogy a little hard.

[5614]

Possibly the mass in that equation refers to the rest mass in which case for a photon this is 0.

 

Light doesn't easily fit into classical mechanics.

 

Classical mechanics fails at subatomic distances.

 

Gravity has never been observed on a subatomic scale, that's one of the problems with detecting the graviton.