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Gravitational potential energy, orbital speed, binding energy.

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data
    a satellite having a mass of 1800 kg orbits the earth at a distance of 6.3 x 10^5 m above the surface find the gravitational potential energy of the satellite while in orbit, the orbital speed and the binding satellite.

    3. The attempt at a solution

    gravitational potential energy
    Eg = -GMm/Ro
    Eg = -(6.67x10^-11 Nm^2/kg^2)(5.98x10^24kg)(1800kg)/(6.37x10^6m)+(6.3x10^5m)
    Eg = -1.03 x 10^11 J

    the orbital speed

    Ek = 1/2(GMm/Ro)
    Ek = 0.5 x -1.03 x 10^11 J
    Ek = 5.128277145 x 10^10 J

    Ek = 1/2mv^2
    V = 7548.57 m/s

    binding satellite
    Eb = 1/2(GMm/Ro)
    Eb = 0.5 x -1.03 x 10^11 J
    Eb = 5.128277145 x 10^10 J
     
  2. jcsd
  3. Nov 25, 2012 #2

    haruspex

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    Whilst that is true, I would prefer to use the equivalence of the gravitational force and the centripetal force required to maintain the orbit. That seems to me to be a more fundamental principle.
    I'm not sure about that. Does binding energy take into account the KE? Maybe it does.
     
  4. Nov 25, 2012 #3
    You mean like this

    m(v^2/Ro) = GMm/Ro^2 ????

    Well binding energy is the energy required for the orbiting satellite to escape. So the total energy should be zero....

    Well yes Ek is included...

    Eg + Ek = Et1
    - GMm/Ro + 1/2GMm/Ro = - 1/2 GMm/Ro

    Et1 + Eb = Et2
    - 1/2 GMm/Ro + Eb = 0 J
    Eb = 1/2 GMm/Ro
     
  5. Nov 25, 2012 #4

    haruspex

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    Yes.
    OK.
     
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