# Gravitational field idea

1. Oct 29, 2005

### daniel_i_l

Lets imagine that there is a gravitational field and you are in a position that if you fall a meter your g increases by 2m/s/s (I'm pretty that thats possible). In this case, the more you fall the more P energy you get and you get more K energy. In order for this not to contradict the conservation of energy law, I thought that the extra energy had to do with the gravitational energy, but that also increases. Were does the extra energy come from?

2. Oct 29, 2005

### Staff: Mentor

You have potential energy backwards. Potential energy increases as you go up.

3. Oct 29, 2005

### daniel_i_l

I know that normally potential energy decreases as you go down because
Ep = mgh and h gets smaller. But in this case h gets smaller as you fall
(Hf = Hi-1), but the g gets bigger (gf = gi + 2) so the total Ep get bigger?
This can happen with a very dense mass were a small change in the H can make an even bigger change on the g (R^2).

4. Oct 29, 2005

### rbj

maybe somewhere close to a neutron star or black hole. that's gotta be a pretty scary graviational field i'm in if it increases by 20% in the space of a meter!

5. Oct 29, 2005

### daniel_i_l

Thank for the reply, well it doesn't have to be by that much, and even if it was, were did all that energy come from?

6. Oct 29, 2005

### Q_Goest

There's no "extra" energy. The equation E=mgh assumes g=constant. For the case where the factor g changes with h, you will need to integrate. As you drop, h gets smaller and g gets bigger, so the amount of potential energy per unit h increases as you look at a point closer to the 'black hole'.

7. Oct 30, 2005

### daniel_i_l

Thanks, could you tell me what the integral is (I'm just curious)

8. Oct 30, 2005

### Q_Goest

I don't have it handy. Maybe someone else does or would like to figure it out, it can't be that tough.

9. Oct 30, 2005

### Staff: Mentor

The gravitational PE of an object of mass m at a distance R from the center of a spherically symmetric mass M (with radius < R) is:
$$- \frac{G M m}{R}$$

This results from integrating $$\frac{G M m}{r^2} \ dr$$. (Note that the PE is taken to be zero when m is infinitely far from M.)

10. Oct 30, 2005

### daniel_i_l

Thanks Doc Al, I was happy to see that resualt cause after Q's reply I worked it out and got to that answer!
But doesn't that just complicates things more!? According to that equation, $$E_{p}$$ is inversly related to R, so the closer you get the more Ep you get, not less, that in addition to getting more $$E_{k}$$:surprised ?

11. Oct 30, 2005

### Staff: Mentor

Don't neglect the minus sign! Gravitational PE gets smaller (more negative) as R decreases.

12. Oct 30, 2005

### daniel_i_l

How can there be negative energy? Dosen't the (-) just show that we are looking at the energy relative to infinity, what if I'm looking for the potential energy relative to the ground?
How can something be "closer" or "further" from infinity?
Sorry for my misunderstanding Doc Al.

13. Oct 30, 2005

### Staff: Mentor

Potential energy is measured relative to an arbitrary zero point. (The most convient zero point is at infinity, when considering large distances where the gravitational field strength varies. Near to the earth, one simply uses PE = mgy, choosing any level as zero.) Of course the PE can be negative. If you measure the PE with respect to ground level, what's the PE of a object sitting in a hole?

To find the difference in PE between any two points, plug the distances into the formula. (Compare the PE for R = radius of earth to the PE for R = radius of earth + h.) You'll find, as expected, that the PE increases as the object is raised above the ground.