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Homework Help: Gravity weight orbit

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A satellite mass 481 is in circular orbit of 9.29x10^6 m above the surface of the Earth
    the weight of a 65.1 kg astronaut inside the satellite
    What work was done (ignoring friction) to put the satellite into this orbit. Assume it started at rest on the surface of the Earth


    Also, A satellite of mass 730 kg is in circular orbit around the Earth, 3.12x10^6 m above the surface of the Earth. Find the minimum work needed to move the satellite outward and put it in circular orbit at a new height of 6.05x10^6 m above the surface of the Earth

    2. Relevant equations
    U = -GMm/r
    fg = GMem/r^2


    3. The attempt at a solution
    For the first problem, I tried fg = GMem/r^2, (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6)^2 and got 105.69 N

    For the second, U = -GMm/r; W = Ui-Uf
    - (6.67x10^-11) x (5.97x10^24) x 481 / (6.371 x 10^6) - (- (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6) = -1.78E10 Should the work be negative? Is that the work done by gravity, so the work done to move the satellite is the same magnitude, opposite sign?

    I tried a similar strategy for the third problem, and it did not work there either.

    Hopefully you can help me understand what I am doing wrong. Thanks in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 16, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi cobrasny! :smile:

    Hint: for a circular orbit, the work done is the change in U plus the change in KE. :wink:
     
  4. Nov 16, 2008 #3
    So..
    - (6.67x10^-11) x (5.97x10^24) x 481 / (9.29 x 10^6 + 6.371 x 10^6) - (- (6.67x10^-11) x (5.97x10^24) x 481 / (6.371 x 10^6) = 1.78E10
    is just the change in potential energy.

    The change in kinetic energy, 1/2mv^2;
    v^2 = GMe/r
    Re = 6.371x10^6
    v=2pi r/T
    Vsurface = 2pi 6.371x10^6/(24hrx60minx60sec)

    1/2m (vf^2-vi^2)...
    .5 (481) [((6.67x10^-11) x (5.97x10^24) / (9.29 x 10^6 + 6.371 x 10^6)) - (2pi 6.371x10^6/ (24x3600))^2] = 6.063E9

    Do I then add the change in Ek and change in Eu? To get 1.78E10 + 6.063E9 = 2.386E10 Joules of work?
     
  5. Nov 16, 2008 #4
    All right, that worked for the second problem.

    Two to go:
    A satellite of mass 730 kg is in circular orbit around the Earth, 3.12x10^6 m above the surface of the Earth. Find the minimum work needed to move the satellite outward and put it in circular orbit at a new height of 6.05x10^6 m above the surface of the Earth

    Delta U = - (6.67x10^-11) x (5.97x10^24) x 730 / (6.05 x 10^6 + 6.371 x 10^6) - (- (6.67x10^-11) x (5.97x10^24) x 730 / (3.12 x 10^6 + 6.371 x 10^6) = 7.2247E9

    delta k = .5 (730) [((6.67x10^-11) x (5.97x10^24) / (6.05 x 10^6 + 6.371 x 10^6)) - ((6.67x10^-11) x (5.97x10^24) / (3.12 x 10^6 + 6.371 x 10^6))] = -3.6124E9

    So work done = 7.2247E9 + -3.6124E9 = 3.6123E9 Joules (if you could check this I would appreciate it)

    A satellite mass 481 is in circular orbit of 9.29x10^6 m above the surface of the Earth
    the weight of a 65.1 kg astronaut inside the satellite
    I tried fg = GMem/r^2, (6.67x10^-11) x (5.97x10^24) x 65.1 / (9.29 x 10^6 + 6.371 x 10^6)^2 and got 105.69 N
    I still don't see what I did wrong here.
     
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