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Calculating minimum energy to remove satellite from orbit

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A satellite of mass 4500 kg orbits the Earth in a circular orbit of radius of 7.6 x 10^6 m (this is above the Earth's atmosphere).The mass of the Earth is 6.0 x 10^24 kg.
    What is the speed of the satellite?

    What is the minimum amount of energy required to move the satellite from this orbit to a location very far away from the Earth?

    2. Relevant equations
    F = mSatellite*a = G*mEarth*mSatellite/r²
    a = G*mEarth/r² = v²/r -> v=SQRT(G*mEarth/r)
    r = 7.6 x 10^6 km
    mEarth = 6 × 10^24 kg

    v = 7272.88 m/s

    minimum amount of energy = increase of potential energy = 0 - (-GMm/r) = GMm/r = 6.67*10^(-11)*6.0*10^24*4500/7.6*10^6 = 2.38e11 J

    3. The attempt at a solution
    I know that I obtained the correct velocity, but when I try to solve for the energy required I get the wrong answer. Is the above equation for calculating the minimum amount of energy incorrect? What should I be doing?
    Thanks in advance!
  2. jcsd
  3. Feb 22, 2009 #2


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    Homework Helper

    Escaping the Earth (to very far away) means having kinetic energy equal to GMm/r.
    You have some KE already, so you need GMM/r - the present KE.
  4. Feb 22, 2009 #3
    Ahhhh I see, so I would need to do something such as

    2.38e11 J - (4500 kg)(7272.88 m/s)^2 =-209,935 J

    But I cannot have a negative number for this, as energy must be put in to remove the satellite from the Earths orbit.
    If I am correct, what I did above was total energy - energy of satellite... Shouldn't this be correct, since the energy needed to remove the satellite plus the energy of the satellite must equal total energy.
    I feel that I am making an elementary mistake, so could someone please help clarify this for me?
  5. Feb 22, 2009 #4
    You forgot to divide (4500 kg)(7272.88 m/s)^2 by 2!

    It looks like we're working on the same homework questions:)
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