It'll probably make this easier if we consider the scenario where a satellite above the surface area of the earth at distance (X), with the earth having a radius of (R). Earth having a mass (M) and the satellite having a mass of (m).
I was looking at another thread a while back where someone mentioned when calculating the value of the GPE is the change in GP x m. Calculating the change in ending up with finding both values of GP, so when calculating the GP for the satellite at height X in comparison to the GP at height R:
G(M)/(R+X) < G(M)/(R)
BUT, if we times out G(M)/(R+X) with the mass of the satellite:
(Back to the thread I was reading) someone also mentioned that this was the work done to move mass (m) to this point from infinity? And likewise with:
..for when the satellite was on earth. BUT the gravitational field should have the opposite effect surely? Because no work is being done against the Grav field if we're not working against it?
So the main question is:
Is Vxm (not change in, just a value of V at height (R+X) the GPE required to move mass m from infinity to that point, or is it something else? If not, what is it? I'm totally confused.
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- Thread Starter
Last edited by TSB01; 25-01-2010 at 23:32.
- 25-01-2010 23:24
- 26-01-2010 10:50
GPE at X is by definition the work done moving the mass from infinity to that point. However, because gravity is an attractive force, you don't need to do work, instead, work is done by the field. To put it another way, you actually have to do work to move the mass from X to infinity. This is expressed by saying that the GPE at the point X is actually negative. = - GMm/x
So to put some imaginary numbers in it just for illustration.
The satellite in its orbit at x has, say, -100J of GPE. On the surface of the Earth, radius r, it would have, say -150J. [-GMm/r compared with GMm/x when r is less than x)
So to move the satellite from the surface of the Earth up into its orbit, you have to go from -150J of GPE to -100J of GPE. You have to give it GPE.