pot. energy

Kamataat

If h, the height of a body from the ground, approaches infinity, then the body's potential energy approaches 0? I'm assuming a non-constant value for g.

E_p = mgh. In this equation I get 0 times infinity, which is mathematically indeterminate, but I'm guessing that the physical interpretation takes it to equal 0?

- Kamataat

Doc Al

That equation--[itex]E_p = mgh[/itex]--is only good for measuring gravitational PE near the surface of the earth. (As you note, g is not constant as the height of a body increases, but that equation ignores that fact.) Also, the implied reference point in that equation takes the surface of the earth as the point where PE = 0.

If you want to go to infinity, you'll have to integrate the work done against gravity which varies with distance. Also, it's more convenient to take infinity as the PE = 0 point, which makes the potential energy negative for sub-infinite distances. Just for the record, the gravitational PE for an object of mass m at a distance of r from center of the earth (where r > radius of the earth) is given by:
[tex]-\frac{G M_{(earth)} m}{r}[/tex]

Meir Achuz

mgh is only valid near the surface of the Earth, that is when
h<<R(earth). The formula for larger h is E_p=(mgR^2)/(R+h), which-->0 as h-->infty.

Kamataat

So if a question asks to find the speed which a body must achieve at ground level to escape the Earth's grav. field, then I can take E_p@infinty = 0? From that |E_k|=|E_p| to find the speed?

- Kamataat

Doc Al

Sounds good. (Mechanical energy is conserved. The total energy at r = infinity that an object must have to just barely escape is zero.)

Kamataat

Thank you!

- Kamataat

vaishakh

I don't think what you told is correct. This only means that the Kinetic energy of a body that jumps out of earth's gravitational field increases suddenly. This doesn't mean that 0 kinetic energy is enough for a body to be propelled outside the earth's gravitational field.

Doc Al

I don't understand what you're saying. What do you mean by an object "jumping" out of the earth's field?

If you just make it to r = infinity with zero KE you have effectively escaped the earth's gravity. That condition will give you the minimum speed at the earth's surface that will get you to infinity. (This is called the escape velocity.) If you want to end up with some non-zero speed at infinity, you'll need a greater initial speed. The basic idea of energy conservation still applies.

vaishakh

sorry-Doc Al
yes, much better interpretation is that escape velocity depends on gravitational force. as r tends to infinity gravitational force tends to zero. thus that's true.
we were confused since we were comparing it with the potential energy.

Tuneman

ya i must ask what r going to infinity has to do with escape velocity, when one can simply assume its when KE=PE

Doc Al

"Going to infinity" is what escaping the earth means. To just barely make it, you'd need zero speed at r = infinity. Since PE = 0 (by definition) at r = infinity, the total energy needed is zero. At the earth's surface, that translates into: KE + PE = 0. Thus KE = - PE. (PE is negative at the earth's surface.)