mgh at infinity.

PrincePhoenix

According to the eq. U=mgh the gravitational potential energy should keep increasing with height. But it actually reaches zero at infinity.
At what point does the g.p energy start decreasing when going away from earth?
Why does this eq. seem to contradict -GMm/r where g.p energy decreases with distance?
How does one solve mgh for infinity?

Thanks in advance.

DrewD

[itex]mgh[/itex] is the gravitational field if you have an infinite flat surface that is the source. When you are close to the earth's, it is a very good approximation, but [itex]-\frac{GMm}{r}[/itex] is always the correct formula. The issue is that when you are very close to earth, you can no longer approximate the earth as a perfect, homogeneous sphere. In reality you need to use Newton's gravity formula and sum over every particle, but that is impossible, so we approximate the shape of the earth. When very close, a flat surface is a better approximation.

To start with you last question one does not. There is no need to solve [itex]mgh[/itex] at infinity.

[itex]-\frac{GMm}{r}[/itex] does not decrease as [itex]r[/itex] gets large. It increases to zero.

never

nasu

It doesn't. The energy increases when r decrease. Don't forget the minus sign.
The value of zero at infinity is a maximum, when compared with the negative values at finite r.

PrincePhoenix

So mgh is fundamentally incorrect and it only works in a particular case? And the g.p. energy actually does keep increasing the farther the body goes?

russ_watters

Why would you think that would make mgh incorrect? The fact that gravitational force goes to zero at infinity does not contradict the increasing gravitational potential energy with altitude. You'd have to integrate the changing g, but ultimately what you'd discover is the total potential energy of the object. Set equal to kinetic energy and solve for V and you get escape velocity.

nasu

"mgh" is an approximation that works pretty well for the cases where this approximation applies. To say that is incorrect does not make sense.
It would be, maybe, "wrong" if the approximation and the (more) general formula will predict different things , which is not the case. Both show that the PE energy increases as you move further from the Earth, if you measure h from the surface.

The other formula is an approximation as well, good for point mass or spherical body. If you apply the -GMm/r to the actual planet (Earth) it is just an "approximation".

Unfortunately the introductory physics courses do not, usually, spend time to discuss the domain of validity for various equations and sometime (quite often) students get the wrong idea that they are "absolutely" true whereas most of them are just "approximations".

K^2

I think you are missing the point that potential energy is defined up to a constant (in Classical Mechanics.) U=mgh is just as valid as U=mgh+C for any arbitrary value of C. In definition where U goes to zero at infinity, at ground level the approximation you are really using is U=mgh - GMm/R, where M is mass of the Earth, R - it's radius, and G is gravitational constant. But that extra constant in the end doesn't change any physics, so you can just use U=mgh. Also, this is an approximation, and it's only valid for h<<R.

sophiecentaur

I can't write this out on my iPod but mgh is what you get when you work out the difference in potential for two distances r1 and r2, using the full definition of GPE for a point Earth mass M, which is -mGM/r. Take the difference between two nearly equal values of r ( where the difference is h, being small compared with the Earth radius). The algebra gives you (approx) mgh where g is a shorthand expression for the 'rest of' the answer. First year A level maths, I reckon.

DaleSpam

As other posters have mentioned, the mgh equation is an approximation which is valid for small distances. To see this, just do a Taylor series expansion of -GMm/r about r=r0.

[tex]-\frac{GMm}{r} = -\frac{GMm}{r_0} + \frac{GMm}{{r_0}^2}(r-r_0) + O(r-r_0)^2[/tex]
The first term is just a constant, so you can zero it out simply by arbitrarily setting our value at infinity to be something other than 0 (specifically GMm/r0). So then we can make the following substitutions:
[tex](r-r_0)=h[/tex]
[tex]\frac{GM}{{r_0}^2}=g[/tex]
which gives
[tex]\frac{GMm}{{r_0}^2}(r-r_0)=mgh[/tex]

sophiecentaur

Has anyone actually pointed out that gpe is always negative. The nearer Earth you get, the more negative. Going 'up' makes your gpe less negative - adding potential energy by approximately mgh. By the time you get to infinity it has reached zero.

nasu

True, but It is "always" when you use some specific reference point, like (the usual) zero at infinity. I don't think that the "negativity" is a general property of gpe. You can add some constant to make it always positive or even sometimes positive, sometimes negative . This would be equivalent to changing the reference point.
What should not depend on the reference point is the fact that it increases (or decreases) when you move between two specific points.

sophiecentaur

If you can think of a better reference point then write a text book and deal with the comments.

nasu

Oh, a variation of the the typical argument: "You should write a new text book on ..."?

Any reference point is as "good" as any other. There is no such think as "better reference point". Some may be more convenient that others.
Fortunately I don't need to write a new text book. Many existing ones will mention that the reference point is arbitrary, as is the absolute value of GPE (sign included).
But I suppose you are just teasing. And I agree that sometimes may be better to keep it simple and consistent with introductory text books, to avoid (extra) confusion.

sophiecentaur

Yes I am, a bit - but if you were on Mars and I were on Earth or Alpha Centauri and we wanted to come to terms about what reference to use for our basic GPE reference, I think that the only reference we would be likely to agree on would in fact have to be ∞.

rcgldr

GPE is always negative for GPE = K - G M m / r, where K is defined to be zero so that GPE is zero when r is ∞. If GPE is defined as m g h, then it's positive when h is positive. In both cases, GPE increases (becomes less negative or becomes more positive) as r or h increases.

Defining GPE to be zero at ∞ uses the source of the gravitational field as the reference point for r, which is usually a single object. For a n-body situation, you'd need to choose some common reference point, perhaps the center of mass of the n-body system, as well as secondary reference points in order to map space in 3 dimensions.

sophiecentaur

Why? I don't see how a common point between you and me (on Alpha Centauri) would be any more use than using ∞. Even if we were on Earth and Moon, I couldn't see any occasion when the CM of the two would be a useful reference for 'local' calculations. And then, how would a three-sided discussion (Earth / Moon / α.C.) of GPE work?
We need to get this in context, I think. A reference at ∞ has just got to be the least arbitrary one and the most likely to be chosen by most Physicists in the Cosmos - that is after they have stopped using their own particular version of mgh. We're back to my 'text book' argument here.

sophiecentaur

Bearing in mind that the (constantly varying) equipotential surfaces would all be the same shape whatever reference you chose then I don't see how having one particular surface labelled as 'zero' would help a lot. (Or would it be a Lagrange Point?- I'm not sure)