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Find the satellite's altitude above the earth's surface

  1. Nov 28, 2006 #1
    When in orbit, a communication satellite attracts the earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

    Find the satellite's altitude above the earth's surface.

    We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

    We also know the gravitational force to be Fg = (GMeM)/r^2

    I am not sure, but can we set the two equations equal to each other and solve for r
  2. jcsd
  3. Nov 28, 2006 #2

    Forgive me I have just started here. You might be on the right track but (and my physics is rusty) r would likely include the earth's radius as r should be measured from the centre of gravity.

    Why not F=mg U=mgh U=Fh h=U/F

    Sorry what does this (below) mean?
    ""and the earth-satellite gravitational potential energy (relative to zero at infinite separation)""

    i suspect this needs to be understood again as to whether the radius of earth is included or not.
  4. Nov 28, 2006 #3
    Yes, I have tried F=mg U=mgh U=Fh h=U/F and substituting in 1.49×10^11 J/ 15.1 kN. But I've got the wrong answer.

    Can someone comment more on this

  5. Nov 28, 2006 #4
    I think we have to use the radius of the earth to calculate for the satellite's altitude

    So the above equation F = mgU = mgh U=Fh h= U/F is not correct
  6. Nov 29, 2006 #5


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    Staff Emeritus
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    No, you can't "set the two equations equal to each other"! They measure different things and are not equal.

    Have you copied the problem correctly? You have both gravitational force and potential energy equal to -1.49 x 10^11 (one in Newtons and the other in Joules). It is possible that they happen to be the same but rather peculiar!

    Assuming that is correct, you can use the fact that Fg= (GMeM)/r^2= 1.49 x 10^11 to determine that GMeM= 1.49 x 10^11 r^2. Now put that value into GMeM/r= 1.49 x 10^11 and solve for r. Hmm, you seem to have, then,
    1.49 x10^11r^2/r= 1.49 x 10^11 from which we determine that r= 1 m. That satellite is "orbiting" one meter more than 10^6 m deep in the earth! Check the problem again!
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