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Homework Help: Calculate the work done to locate a 150kg satellite

  1. Nov 11, 2005 #1


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    calculate the work done to locate a 150kg satellite into a geostationary orbit.

    I'm guessing i need to use ep=- GMmG/r
    but it would be possitive right?

    the radius will be 35800 km? or 42,245 cause the 35800 is the altitude for the satellite?????
    m and M is mass of earth and the satellity
    so is this how to work out the right answer

    6.67*10^-11 (mass of eath)(150)
    35800000 or 42,245 000
    Is this the correct way??/
  2. jcsd
  3. Nov 11, 2005 #2


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    Homework Helper

    It depends on what the question means, exactly. To loft the satellite, you'd need to do work against gravity. This is what you're calculating. You'd need to use the height above the Earth's surface for the distance, since this is the distance through which you're lifting it.

    However, to maintain it in orbit, you'd also need to give it the proper speed, which implies you'd need to add kinetic energy. From the way the question is worded, I'm not sure whether you'd be expected to include that or not. It all turns on what is meant by "locate [it] into orbit". If you do need to include it, just calculate the necessary speed for the satellite and it becomes a plug and chug.
  4. Nov 11, 2005 #3


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    GMmG/r should only have one G, i.e. GMm/r.

    Calculate the change in potential energy from earth's surface to GEO, and as Diane indicated the change in rotational kinetic energy.
  5. Nov 13, 2005 #4


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    The equation that you mention is the potential energy, at a distance r from earth centre, relative to an infinite distance from earth - that is at infinity the potential energy is zero (which you can see from the equation). To put the satellite in a geostationary orbit you would need to change (raise) it's potential energy, [itex]\Delta U[/itex], and give it the appropiate orbital speed to stay in orbit at this distance. The change in potential energy is given by
    [tex]\Delta U = U_{final} \ - \ U_{initial}[/tex]
    where the initial potential energy is at a distance equal to the radius of the earth. This can be calculated by using your equation to be
    [tex]= GMm\left(\frac{1}{r_o} \ - \ \frac{1}{r_1}\right)[/tex]
    where [itex]r_o[/itex] is the radius of the earth and [itex]r_1[/itex] is the geocentric radius. You can see that this change in potential energy is positive as one would expect.
    Last edited: Nov 13, 2005
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