# Potential energy question

1. Jun 26, 2004

### KnowledgeIsPower

While i don't do physics i do study mathematics, and am very interested in the field of mechanics/pure/physics.
I already know that potential energy is defined by the formulae 'mgh' (mass x gravity x vertical distance).
I also know that gravity marginally varies at different points along the earth.
Something i was pondering was, if you moved a particle up vertically in an area of low gravity, onto a smooth plane, then moved it a relatively long distance to somewhere where gravity was much higher, then let it fall. Wouldn't you get more energy out than you originally put in?
My thought here is that mh(low gravity) when raising requires less energy than the mh(high gravity) will give out when it falls.
What's the problem in this train of thought?

2. Jun 26, 2004

### Kurdt

Staff Emeritus
you forget the energy that you must put into the particle to move it from one area to another. The gravity does not vary significantly, but on a smooth surface moving from one area to another the ball would slow slightly, assuming the surface were at constant height or deviate slightly from its original height to a lower height of equipotential.

3. Jun 28, 2004

### suffian

The purpose of a potential energy function is that its difference from one point in space to another is equal in magnitude to the work done on that body (by the forces that the function is meant to represent) as it travels along its path.

The gravitational potential energy of a body can be represented by Mgh if we assume that the weight of a body is the same everywhere is space. For typical physical problems (dropping a rock), this works out pretty well. For problems with rockets launching into space, this assumption doesn't hold up because the rocket's weight will decrease as it rises. Thus, the trouble with your scenario is simply that that potential function no longer applies.

4. Jun 29, 2004

### Gza

mgh is just an approximation to the potential energy of the (earth, object) system at close distance to the surface of the earth and the object.

5. Jun 29, 2004

### kuenmao

The gravitional accleration(g) depends on the strength of the gravitional field. The gravitational field depends on two things: the distance from the object that created the field, and the mass of that object. So, by saying "changing the 'g'", you would either mean
1. Changing the distance.
2. Changing the mass.

In case 1, changing the distance is like picking something up from the ground. Which means that the "smooth plane" you said is actually inclined.
In case 2, changing the mass....well, mass doesn't suddenly change, so you must be adding mass to the object which created the field. That already gives out energy, since you have to consider the gravitational field of the mass you are adding. After considering all this, you'll realize that you can't get around the law of conservation of energy.
Oh, and do remember that gravitational field lines radiates from one point, and there is no "smooth plane" of equivalent g unless you are reffering to a circle.

The main problem in your train of thought is that there is no simple way to change the "g" of a place without putting in energy.

6. Jul 10, 2004

### omin

"if you moved a particle up vertically in an area of low gravity, onto a smooth plane, then moved it a relatively long distance to somewhere where gravity was much higher, then let it fall."

To move the partical along the smooth plane, a force must be exerted on the partical, which will result in the partical moving at a contstant velocity along the smooth plane. When the partical reachs the point on the plane where the gravity changes, the slightest change is sufficient, the partical's velocity will change according to the equal and opposite change in gravity. If gravity is weaker at this instataneous position on the plane, the partical ascends to a higher plane. If gravity is stronger, the partical descends to a lower plane. The change in position of the partical relative to earth compesates for any percieved increase or decrease in instanteoous state of potential energy.

7. Jul 10, 2004

### Kurdt

Staff Emeritus
hence the lines of equipotential as already stated

8. Jul 11, 2004

### KnowledgeIsPower

Thanks for the explanations ^^.

9. Jul 12, 2004

### maverick280857

Yeah thats one of the common examples which shows how inaccurate elementary general physics is. When you say V = mgh, you approximate the gravitational potential energy expression, which actually is (mathematically)

$$\int_{\infty}^{r} Fdx = \int_{\infty}^{r} \frac{mgR_{e}^{2}}{r^{2}} = -\frac{mgR_{e}^{2}}{r} \qquad (r > R_{e})$$

(the work done to bring a mass m in the gravitational field of the earth, from a point located at an infinite distance to a point distant r from the center of the earth).

So if you were to use the concept of gravitational potential energy in gravitation/celestial mechanics, you would use this "accurate" expression and not mgh. The interesting thing is that mgh is used so commonly pre-gravitation that when we read this new relationship and about "gravitational potential", we begin to wonder what it is (at first sight).