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Space shuttle energy problem

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    The space shuttle is in a 250 km high circular orbit. It needs to reach a 610 km high circular orbit to catch the Hubble telescope for repairs. The shuttle's mass is 75000 kg. How much energy is required to boost it to the new orbit?


    2. Relevant equations



    3. The attempt at a solution

    I keep coming up with answer that is twice the answer in the back of my book

    Uf= E + Ui
    (-6.67E-11)(5.98E24kg)(75000kg)/(610000m+6.37E6) = E + (-6.67E-11)(5.98E24)(75000kg)/(250000m+6.37E6)

    E= 2.33E11 J


    book gives 1.17E11
     
  2. jcsd
  3. Apr 18, 2008 #2

    D H

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    You are ignoring kinetic energy.
     
  4. Apr 19, 2008 #3
    In Uf= E + Ui,
    Ui should be the sum of kinetic energy mv^2/2 and potential energy -GmM/r
    using GmM/r^2 = mv^2/r, we can get mv^2/2 = GmM/(2r)
    so U = -GmM/(2r), not -GmM/r as you used.
    then we can get the half of your answer.
     
  5. Jul 26, 2009 #4
    I know for a fact that this solution is correct:

    We are looking for the change in mechanical energy.

    Emech = (1/2)Ug
    Therefore, ∆Emech = (1/2)∆Ug

    G = 6.67 X 10^-11
    Me = 5.98 X 10^24 kg
    Ms = 7.5 X 10^4 kg
    Re = 6.37 X 10^6 m
    r1 = Re + 2.5 X 10^5 m
    r2 = Re + 6.1 X 10^5 m

    Solve for Ug at r1. (Call this Ug1)
    Ug1 = (-G(Me)(Ms))/(r1) = -4.519 X 10^12 J

    Solve for Ug at r2. (Call this Ug2)
    Ug2 = (-G(Me)(Ms))/(r2) = -4.286 X 10^12 J

    ∆Ug = Ug2 - Ug1 = 2.33 X 10^11 J

    ∆Emech = (1/2)∆Ug = 1.17 X 10^11 J
     
    Last edited: Jul 26, 2009
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