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Homework Help: Gravitational Potential Energy Problem

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    The magnitude of the attractive force of gravity between two massive bodies is F=GMm/r^2, where G is a constant, M and m are the masses, and r is the distance between the centers of the two bodies. The radius of the Earth is 6.38×10^6 m and its mass is 5.97×10^24 kg. A satellite of mass 1.13e+3 kg is propelled from the surface of the Earth to a height of 35,786 km above the surface of the Earth. What is its change in gravitational potential energy?

    2. Relevant equations


    3. The attempt at a solution

    I plugged all the numbers into the given equation to get F=1.6268x10^15
    I thought this represented mg in PE=mgh, but after multiplying it with h (and changing h to m rather than km), I didn't come up with the correct answer (which should be 5.96x10^10). Can you tell me where I went wrong?
  2. jcsd
  3. Feb 20, 2010 #2
    Well g is not constant over such large distances.
    Instead you can use

    delta(P.E)= - integral(F.dx) where F is the conservative force.(gravity here).
  4. Feb 20, 2010 #3
    the problem is that g in high altitudes is significantly different.
    Work is F*h when F is constant. But F varies with height, and in this case the work is the area below F in a graph where F is drawn against h. Formally:
    [tex]\int_{r1}^{r2} F dr = \int_{r1}^{r2} \frac{G m_{1} m_{2}}{r^{2}} dr[/tex]
    The area under [tex]\frac{1}{r^2}[/tex] from r=1 to r=r1 is [tex] 1 - \frac{1}{r1}[/tex]
    From now on you can compute it even if you don't know what an integral is.
  5. Feb 20, 2010 #4
    Okay, so my integral will be GMm x integral 1/rinitial - 1/rfinal

    which gives me GMm (1/6.38e6)-(1/4.138e7), so GMm (1.326e-7), but if I times this by the masses I get G(8.945e20)...but I'm stuck, and this doesn't seem anywhere near the correct answer :/ alas
  6. Feb 20, 2010 #5
    Hang on- that IS the correct answer.
  7. Feb 21, 2010 #6
    Unfortunately my book says the right answer is 5.96x10^10.

    How would I possibly get rid of the G in the answer I got (G(8.945e20)) if I don't know what it is? Could I set this equal to something appropriate and make it cancel out?
  8. Feb 21, 2010 #7
  9. Feb 21, 2010 #8
    BLAST! Textbook semantics have tricked me again! I assumed we weren't supposed to be able to know this since it's not mentioned in the book. However, it is obviously necessary. Thank you!
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