1. Jun 11, 2004

### dcl

Heya's
Im a bit confused reguarding gravitational potential energy.
I've seen 2 different calculations

U(h) = mgh
where h is the height above the ground.

and
U = -GMm/r (off the top of my head, havnt got my notes here with me right now)

Could anyone help clear this up for me? Why does one expression have a negative potential?

2. Jun 11, 2004

### pmb_phy

The former, U(h), applies to a uniform gravitational field where the zero of potential is at h = 0. The later, U = -GMm/r, applies to the gravitational field of a spherical body where the zero potential is at r = infinity. Plot each on a graph and you'll notice that each is a decreasing function of r, only the shape of the curve is different.

Keep in mind that only changes in potential are physically meaningful. You can always add a constant without changing the physics. The constant is usually chosen to make the math as simple as possible.

Pete

3. Jun 11, 2004

### dcl

I see, makes sense. Thanks for clearing that up.

4. Jun 11, 2004

### baffledMatt

Just to add to what pmb_phy said, we use mgh much of the time in calculations because it is a good approximation to the change in the gravitational potential (given by U = -GMm/r) for small distances. i.e. when h << r.

Matt

5. Jun 11, 2004

### Gokul43201

Staff Emeritus
$$r=R+h$$

So,

$$U =\frac {-GMm} {r} = \frac {-GMm} {R+h} = \frac {-GMm} {R} (1 + h/R)^{-1}$$

For h<<R, you can expand the last term binomially, and neglect terms of second order and up. So,

$$U = \frac {-GMm} {R} (1 - h/R) =\frac {-GMm} {R} + \frac {GMmh} {R^2}$$
$$= Constant + (m)*(\frac {GM} {R^2})*h = Constant + mgh$$

Since we are interested only in changes in potential, we can throw away the constant term.