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dcl
Jun11-04, 06:40 AM
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?

pmb_phy
Jun11-04, 07:19 AM
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?

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

dcl
Jun11-04, 07:56 AM
I see, makes sense. Thanks for clearing that up.

baffledMatt
Jun11-04, 10:49 AM
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

Gokul43201
Jun11-04, 10:54 AM
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.