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Energy of planetary orbit

  1. Oct 29, 2015 #1
    1. The problem statement, all variables and given/known data
    A 500-kg satellite is in a circular orbit at an altitude of
    500 km above the Earth’s surface. Because of air friction,
    the satellite eventually falls to the Earth’s surface,
    where it hits the ground with a speed of 2.00 km/s. How
    much energy was transformed into internal energy by
    means of air friction?

    2. Relevant equations


    3. The attempt at a solution
    Total Energy of satellite = PE = KE
    = G(mass of satellite)(mass of Earth)/ - 2 (Earth's radius + orbit's height)
    = -1.45 x 10^10 J

    KE before crash = 0.5 (500)(2000)^2 = 1 x 10^9 J

    Energy loss due to friction

    - 1.45 x 10^10 - 1 x 10^9 = 1.55 x 10^9 J

    But the answer is 1.58 x 10^9 J, what is missing?

    Thanks in advance!
     
  2. jcsd
  3. Oct 29, 2015 #2

    gneill

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    Staff: Mentor

    Did you account for all the energy? What about the potential energy change from orbit to surface?
     
  4. Oct 30, 2015 #3
    Why do I need to include PE change? Shouldn't all the PE energy be converted to KE right before it crashes? So the PE change will be included into KE.
     
  5. Oct 30, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    PE will be changed to KE continuously as the object changes its radial position. The KE it gains as a result must figure into what remains at the end of its journey.

    Suppose we consider a slightly different situation. Suppose an object with the same mass as the satellite is momentarily stationary above the surface at the same height as the satellite orbit and falls straight down to the surface. Ignoring air friction, what will be its KE and velocity at impact? Where did that KE come from? It's initial orbital KE was zero...
     
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