Gravitational Potential Energy and work

[DocZaius]

Setting gravitational potential energy's zero at infinity distance, is it safe to say that the limit as r approaches zero of gravitational potential energy is negative infinity? Thus is it safe to say any object at r=0 is necessarily bound to the system, no matter what kinetic energy it has?

Also, I keep reading the zero is arbitrary (but most useful at infinity). Yet it is apparently impossible for me to create a mathematical function that describes gravitational potential energy with its zero at r=0, since it is obvious that as r approaches zero, the slope of the curve needs to approach infinity. The curve ln(1+x) seems to me to be the one whose shape would best approximate such a curve but it obviously does not accurately enough describe the behavior of -GMm/r because of the slope problem I stated above. Can someone explain why it is not possible to have a function of gravitational potential energy with its zero at r=0, when one would think the zero should be arbitrary (as stated by many sources!).

The reason I am interested in such a version of the function is because it seems to me to be best at describing the phenomenon of gravity. A mass has more and more potential energy the farther it is from an object (a positive quantity here would be quite adequate!) and has none when it is at no distance from the object (no work can be done at r=0!).

I have always looked at potential energy as a way of describing how much work would be done by the force in question based on the distance of the object acted upon. My version of gravitational potential energy would give the amount of potential work (positive quantities!) based on distance as well as its potential energy.

 

[CompuChip]

While you are editing, let me already reply to this part

Setting gravitational potential energy's zero at infinity distance, is it safe to say that the limit as r approaches zero of gravitational potential energy is negative infinity?

Yes, the expression for gravitational energy has a singularity for r approaching zero.
In practice, objects are always of finite size (so this limit cannot be reached) or they are so small that other forces will play a role and/or the "standard" (Newtonian) theory no longer holds anyway. In that case we enter the realm of "quantum gravity" (a theory we seem to have not quite found yet).

 

[CompuChip]

Also, I keep reading the zero is arbitrary (but most useful at infinity). Yet it is apparently impossible for me to set gravitational potential energy to zero at zero distance. You would need a mathematical function that is 0 at r= 0, and infinity at r->infinity. The curve ln(1+x) seems to me to be the one whose shape would best approximate such a curve but it obviously does not accurately enough describe the behavior of -Gmm/r.

Yes, by "it is arbitrary" they mean that you can choose the value at a single point (but then at all other points, it is fixed). After all, the value of the energy is not physical, only differences in the energy are. So if you replace E(r) = -Gmm/r by E'(r) = -Gmm/r + E₀ for some constant value E₀, then for any two values r and r', you will have that
E(r) - E(r') = E'(r) - E'(r')
(note that I am using the prime here to indicate a second value, not a derivative) - i.e. the E₀ drops out.

For example, suppose that R is the radius of the earth, and you want to set the gravitational energy to zero at the Earth's surface. With the usual choice of E(r) = - GMm/r, the value would be E(R) = -GMm/R (note that this is just a constant, for any given mass m). So if you take E₀ to be E(R), you can use
E(r) = - GMm/r + GMm/R
instead. Now for r = R, E(r) = 0: you have chosen it to be zero there. Of course, it is now no longer zero at infinity (it is GMm/R instead) - if you fix it at one point it will change at all other points. But that does not matter, if you want to know - for example - how much energy an object will gain by falling to the Earth's surface from infinity, you can subtract the two values of the energy (at r = infinity and at r = R) and will find the same value of GMm/R in both cases (try it!)

 

[DocZaius]

Sorry for the additional edits I made before I saw your replies. Thanks for your replies!

So looks like the singularity at r=0 makes it impossible to formulate a version of potential energy with any finite number at r=0. The reason is that the kinetic energy gained from going from an arbitrary r to r=0 must be infinity:

\Delta KE = WORK = -\Delta U = - \left(\left(\lim_{x\to0+}\frac{-GMm}{x}\right)-\frac{-GMm}{r}\right) = \infty

(despite its unphysicality), and my version would have to give the same answer. Any version with GPE=0 at r=0 cannot give that answer.

 

[rcgldr]

Setting gravitational potential energy's zero at infinity distance, is it safe to say that the limit as r approaches zero of gravitational potential energy is negative infinity?

Only if the source of the gravitational field is a point source. Normally the source of a gravitational field has some finite size. For a spherical object with radius R, once r < R, then the formula for gravitational potential energy changes to

- G M m (3 R² - r²) / (2 R³), r < R

and gravitational potential energy at r = 0 is a finite negative value.

 

[ZealScience]

Setting gravitational potential energy's zero at infinity distance, is it safe to say that the limit as r approaches zero of gravitational potential energy is negative infinity? Thus is it safe to say any object at r=0 is necessarily bound to the system, no matter what kinetic energy it has?

This sentence can only be applied to extreme cases like black holes or any other form of singularities. If you consider the case of the Earth as you are going down the Earth mass is all above you, when you reach the centre of the Earth where r=0, there is no mass at all! Only when you arrive at the black hole, the mass is concentrated at r=0. That's why even light cannot escape the balck hole.

I think this is an important idea that you need to interpret in order to understand the question. In addition, you need to know Gauss Law (or Gauss theorem, more directly) which states that the field is only caused by mass inside in a symmetrical system...

 

[DocZaius]

This sentence can only be applied to extreme cases like black holes or any other form of singularities. If you consider the case of the Earth as you are going down the Earth mass is all above you, when you reach the centre of the Earth where r=0, there is no mass at all! Only when you arrive at the black hole, the mass is concentrated at r=0. That's why even light cannot escape the balck hole.

I think this is an important idea that you need to interpret in order to understand the question. In addition, you need to know Gauss Law (or Gauss theorem, more directly) which states that the field is only caused by mass inside in a symmetrical system...

I was aware that everything I was saying was in regards to a point mass, sorry for not saying so explicitly.

 

[rcgldr]

A bit off topic, but since it was mentioned, the mass of a black hole doesn't have a radius of zero, but it is small enough that escape velocity at the "surface" of the mass is greater than the speed of light.

 

[ZealScience]

A bit off topic, but since it was mentioned, the mass of a black hole doesn't have a radius of zero, but it is small enough that escape velocity at the "surface" of the mass is greater than the speed of light.

Yes, there might be small black holes that have radius, probably like quark stars that we don't know. But what about super massive black holes (like ones that reside in the centre of the galaxis, where event horizon extend millions of miles arround not just on the surface)? There would be a point where gravitational force is much greater than degeneration pressure, I didn't analysed it quantitatively, but then what is maintaining the radius (Pauli's Exclusion also has limit to it)?

And to the thread starter's question , I think it's quite clear that even speed of light cannot escape, and escape velocity is independent of mass here (though this is classical mechanical way of saying this). Then how can anything with whatever the kinetic energy escape? If you are referring to exactly point mass then the force when r→0 the force is infinity thus infinitive amount of work.