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Derive Potential Energy from Newtons graviational law

  1. Feb 21, 2009 #1
    3. Beginning from Newton’s rule for the gravitational force = -GMm/r^2
    derive an expression for the potential energy of a particle of mass m above the surface of the earth, and show that it can be written as:

    U(r) = -g0 RE/r

    where r is the distance from the centre of the earth, RE is the radius of the earth, and g0 is the value of the
    gravitational field at the surface of the earth.

    To cut to the chase I am at this point.

    U(r) - U(RE) = -GMm(1/r - 1/RE) + C

    and rewrote it as U(r) - U(RE) = -GMm(RE-R/REr) + C

    But I can't seem to figure out how to get it in the form g0 RE/r.
     
  2. jcsd
  3. Feb 21, 2009 #2

    Doc Al

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    (a) You should be integrating from ∞ to r, not RE to r.
    (b) Since it's a definite integral, there's no integration constant.

    Once you fix these, then you can try to get the answer in terms of g0. (Start by figuring out what g0 equals!)
     
  4. Feb 21, 2009 #3
    Ok, just need to clarify something. Why is it from infinity to r?

    Also, the U(infinity) term goes to zero, since U(infinity) = -GMm/infinity as if this was the expression on its own.
     
  5. Feb 21, 2009 #4

    Doc Al

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    By convention, the potential is taken to be zero at infinity.

    That's true.
     
  6. Feb 21, 2009 #5
    Ok, but we care about the potential above the surface. Then if we subtract the potential energy at the surface from the potential energy at distance r from the centre of the earth, don't we get the expression I original had?
     
  7. Feb 21, 2009 #6

    gabbagabbahey

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    You seem to be misinterpreting "potential energy of a particle of mass m above the surface of the earth" as the difference in potential energy between the point r and the surface.

    I'm pretty sure that the question wants you to find the potential (at a point above the surface) relative to some reference point and then show that if the reference point is at infinity, you get the desired expression.

    The part about the point you are finding the potential at being above the surface is important because the force is only GmM/r^2 above the surface.
     
  8. Feb 22, 2009 #7
    Ah ok, I thought the reference point we were supposed to use was at the centre of the earth.

    So we should have that U(r) = -GMm/r, but I'm still clueless about getting it into that expression U(r) = g0 RE/r.
     
  9. Feb 22, 2009 #8

    gabbagabbahey

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    Well, if U(r) = -GMm/r , what is U(RE)? ....what does that make g0? (Remember, g0 is by definition the value of the gravitational potential at the surface)
     
  10. Feb 22, 2009 #9
    I thought U(RE) should be U(infinity), since were doing from infinity to r, which is zero.
     
  11. Feb 22, 2009 #10

    gabbagabbahey

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    That makes no sense. U(infinity)=0 since it is the reference point. U(RE) is what you get by plugging r=RE into your equation; U(r) = -GMm/r. Certainly, you don't get U(RE)=0 right?
     
  12. Feb 22, 2009 #11
    Lol, no I don't get that.

    In an earlier post Doc Al said we integrating from infinity to r, which I understand now. So I thought that if we were going from infinity to r for the right side, we should be doing the same for the left side a get:

    U(r) - U(infinity) = -GMm/r + GMm/infinity

    since infinity is the reference point, and then both these terms would go to zero, and we'd be left with:

    U(r) = -GMm/r

    Obviously its not since your telling me its wrong. Could you explain this for me?
     
  13. Feb 23, 2009 #12

    Doc Al

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    That's not wrong. Just be careful how you express things. When you wrote:
    That sure sounds like you are saying U(RE) = U(infinity) = 0, which is not what you meant at all. :wink:
     
  14. Feb 23, 2009 #13
    So are we going from infinity to r for the left side or from RE to r?
     
  15. Feb 23, 2009 #14

    Doc Al

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    From ∞ to r.
     
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