# Does lifting an object higher increase its potential energy?

1. Aug 5, 2010

### fluffy

I read that lifting an object higher increases its potential energy. Is this true? I can't see how, considering:

Ep = mgh
F = mg
so Ep = Fh

F = (G x m1 x m2)/ d2

so

Ep = (G x m1 x m2 x d)/ d2

so

Ep = (G x m1 x m2)/ d

So the greater the distance between the object and the earth, the less potential energy it should have. What am I missing here?

2. Aug 5, 2010

### Staff: Mentor

Yes. The higher you lift the object, the greater the gravitational potential energy of the system.
That's OK near the earth's surface, where the weight equals mg.

Since the force depends on the distance and is thus not constant, you cannot simply multiply force by distance. Instead you must integrate:

$$E = \int \frac{Gm_1m_2}{r^2} dr = - \frac{Gm_1m_2}{r}$$

That minus sign makes all the difference!

3. Aug 5, 2010

### fluffy

Thank you very much for your response. Except now I'm confused for a new reason.
So if a one kilo object is one meter away from the earth then it will have

-(6.67300 × 10-11 × 5.9742 × 1024 × 1)/(12) = -3.98658366 × 1014 potential energy? So it will have a huge amount of negative potential energy?

4. Aug 5, 2010

### Staff: Mentor

Two problems:
(1) The 'r' in that formula stands for the distance to the center of the earth, not the height above the surface.
(2) That second formula is useful when you are dealing with large distances where the gravitational force changes. The "0 point" of PE is at infinity. All other distances the PE is negative. (In the mgh formula, you are setting the zero of PE at some arbitrary height = 0. mgh is the PE compared to that point.)

Note: The absolute value of potential energy has no meaning. All that matters is the change in potential energy between locations.

5. Aug 6, 2010

### fluffy

Thank you very much, Doc.

6. Aug 6, 2010

### brainstorm

This is a strange thing to say, imo. If an object loses a certain amount of potential energy by moving from point A to point B, doesn't the remainder of potential energy imply a finite reserve of kinetic energy? Did you mean to say that potential energy is always relative to a chosen end-point of ultimate (or maybe relative) rest?

7. Aug 6, 2010

### Staff: Mentor

Only relative to some other point, where again it's the change in PE between points that matters, not the absolute value.

For example: A 1kg block of wood sits on a table. It's raised one meter. The change in PE is mgΔh ≈ 9.8 J. If I choose to measure PE using a reference point 100 m below the table (or anywhere else) it would not make a difference.

You're always measuring it with respect to some position, not necessarily anything ultimate about it. When you use the second formula there is a "natural" place to set PE = 0, namely where the distance goes to infinity. But again, nothing magic about setting PE = 0 there; it's still arbitrary, albeit very useful. When doing the integration to derive the potential energy, we just chose to set the integration constant to zero.

8. Aug 6, 2010

### brainstorm

So if you're dealing with weights driving a pendulum on a clock, you could measure PE from the lowest hanging-point of the weight and say that it expends half its PE to make it halfway to that point OR you could call the halfway point the zero-point and say it expends all its PE going from the highest point to the halfway point? Likewise, if you define the clock as hanging twice as far from the ground as the lowest point of its weights, you could say that weights expend half their PE relative to the ground to get to the lowest hanging point of the chain?

Couldn't you also say that there is a maximum amount of PE for any object in a gravitational field that would be exhausted if that object managed to reach the absolute center point of gravitation? (e.g. the center of the Earth) If something manages to reach the center of Earth, it does not have any PE in an absolute sense, does it? I suppose you could consider it as having the potential to eventually fall into the sun, the center of the galaxy, or just to Venus, etc. Well, actually could you really say that anything at the very bottom of a gravity well has any PE since it will be trapped in that gravity well as long as it exists without being imparted with new PE, correct?

9. Aug 6, 2010

### Staff: Mentor

All that matters is the change in the pendulum's PE as it moves from position to position. You could use a point a mile underground as your reference if you wanted to. Nothing special about the 'absolute' value of PE.

Again, PE in the 'absolute sense' doesn't have a well-defined meaning. The PE of something that made it to the center of the earth depends on what you are comparing it to. Only changes in PE make any physical difference.

10. Aug 6, 2010

### brainstorm

I still don't get it. What is the point of calculating change in PE if the remaining PE amount has no significance? If you say something lost half its PE, then you're also saying it kept half its PE, right? And if you say something still has half its PE, then you are saying that it still has the potential to express KE, right? So how can you claim something is out of PE if it still has the potential to exercise KE? Could you measure PE relative to a point beyond the center of the planet? E.g. if an object is 10,000km from the center, could you calculate PE to 20,000km depth?

11. Aug 18, 2010

### DLuckyE

I think the keyword is potential here, the amount of total PE can change depending on changes in the system. Don't know to much about it but it seems if you would want to know the real value of PE you'd have to take the whole universe into consideration?

12. Aug 18, 2010

### Staff: Mentor

Again, the 'remaining PE' only has significance when compared to something else. Only changes in PE are significant.
Half its PE compared to what?
Again, that's only a problem if you think that the 'absolute' value of PE has some significance. I'm sure you realize that it's often the case that the PE of an object is negative?
Now you're just being silly.

13. Sep 13, 2010

### Syed Arsalan

U r missing the negative sign. when u lift an object up, the work done is negative and hence u have to put a negative sign. hence greater the denominator( distance), smaller will be the over all negative value and our value will increase by increasing the distance from the centre of earth.

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