# Trying to understand potential energy at infinity

1. Apr 25, 2010

### diagopod

A difficulty I'm having is that PE, at least in the context of gravity, is said to go up with altitude, or distance from the gravitational source (the analogy of pushing an object upward against the force of gravity and thus adding PE to it). Yet PE drops to zero at infinity, which would seem to be the maximum possible altitude, going against the notion that PE increases with altitude.

I think the gist of my misunderstanding is that U = mgh is just an approximation, only valid when g is constant, so that the true equation is U = G M1M2 / r.

Still, even if U = mgh is an approximation, how can U by any equation actually increase with R if the sum of all those changes (U = G M1M2 / r) is a steady decrease to zero at infinity.

Thanks for any guidance.

2. Apr 25, 2010

### Staff: Mentor

Under the convention in which U = 0 at r = infinity, U is negative. As r increases, U increases towards zero.

3. Apr 25, 2010

### diagopod

Thanks, makes sense after all.

4. Apr 25, 2010

### rcgldr

As implied in the previous post, the correct equation is:

U = -G m1 m2 / r

So U is a negative value (or 0 at ∞).

5. Apr 26, 2010

### diagopod

Thanks. The negative sign is puzzling though. Negative mass is considered exotic, right? But given e=mc2, negative energy is a measure of negative mass, or in all likelihood that kind of angle is invalid wrt potential energy. Either way, negative energy is an accepted concept? Or is this just a convention, and it's "really" positive energy?

6. Apr 26, 2010

### sophiecentaur

It's nothing like as esoteric as that. It's just that you define potential as work done bringing the mass in from infinity. As you get energy OUT of that process (for an attractive field), the sign of the work is negative.

7. Apr 26, 2010

### diagopod

8. Apr 26, 2010

### Staff: Mentor

Even with the "other" definition of gravitational potential energy, U = mgh, you can easily have negative potential energy. U = 0 at ground level, right? So what's U at the bottom of a mine shaft, or in your basement? (assuming your house has a basement, that is. :uhh:)

9. Apr 27, 2010

### diagopod

True, all depends on the reference. Of course, I'd like to think that PE is something real and that it actually has an actual positive value at each point even if we choose other units than the actual value for convenience. Then it would be analogous to there being an absolute zero regardless of whether we might choose 0 to be something more convenient in Celsius or something. I suppose in that case the "absolute zero" of PE would be -- for Earth's gravity -- the center of the earth, and the maximum PE would be infinity? Or no?

10. Apr 27, 2010

### curiousphoton

And U = - infinity at r = 0 ?

That kind of stuff always throws me off. Since U and r can never = 0, U and r can never equal -infinity or infinity, respectively. So since nothing can be at r = infinity, U can never equal 0, thus all physical problems / equations should technically have a U variable?

Probably an annoying response, sorry. I just find that interesting.

11. Apr 27, 2010

### Born2bwire

Not really. Potential by itself is rarely ever physically significant (I do not know of a case where it is actually). The physical manifestation is the change in potential. That is, the movement of an object from point A to point B through a potential is manifested as work that is equal to the change in the potental. Likewise, the spatial dependence of the potential is indicative of the force experienced by bodies within the potential (F = -\grad U). All of this is independent of the actual value of the potential but only dependent upon the relative changes in the potential. We can define our zero point anywhere we like and not change the physics though it certainly can obfuscate the mathematics.

When we say r = \infty, we are implicitly implying the limit of r as it approaches infinity. Still, like I stated above, we only really care about the changes in the potential. So having a non-zero potential is immaterial to our problem.

12. Apr 28, 2010

### rcgldr

Only if you could get gravity from a point source (an object with mass but not size). Once inside a uniform sphere of radius R and mass m, the formula for potential changes to

V = - G m (3 R2 - r2) / (2 R3), r ≤ R

I'm not sure how the radius of a sphere inside another sphere would affect the potential energy formula.

Last edited: Apr 28, 2010
13. Apr 28, 2010

### diagopod

thanks for your help on this, appreciate it

14. Apr 28, 2010

### rcgldr

Voltage is a potential, and commonly used in physics, but as mentioned, it's rarely used by itself.

15. Apr 28, 2010

### Born2bwire

Yes but we still work with voltage differences. Again we can offset the voltages in a problem by a constant offset and not change the physics. I do not meant that potential does not arise often in physics but that I cannot think of a physical manifestation that uses potential itself. Rather, we use the difference or change in potential to give rise to physical phenomenon.