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Work done to get to orbitGPE

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Materials to construct an 12000 kg lunar observatory are boosted from the earth's surface to an orbit of 4100 km. In orbit, the observatory is assembled and launched to the moon, 384000 km from Earth.
    Compare the work that must be done against earth's gravity on the two legs of the trip. (Hint: work done against gravity is change in potential energy)

    Work to get to orbit: ________ joules
    Work to get from orbit to moon: ______________ joules

    2. Relevant equations

    possibly U=-GMm/r or U=-GMm/Re + mgh?

    Me=5.97*10^24 kg
    Re=6.37*10^6 m

    3. The attempt at a solution

    for a)
    U=-G(5.97*10^24 kg)(12000 kg)/(6.37*10^6 m+4100*10^3 m)
    =-4.54*10^11 joules

    (both neg and positive of this answer did not work)

    mgh gives you almost the same thing, so i don't think that's the answer either (but feel free to prove me wrong). i've talked with my friend, and she tried -GMm/(Re+alt)-(-GMm/(Re)) and it also didn't work.

    didn't try b) because my answers for a were wrong...help?

  2. jcsd
  3. Jan 19, 2010 #2


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    Welcome to PF!

    Hi sour_lemon_1k ! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    erm :redface: … you've launched it from infinity. :rolleyes:

    (and it's never "+ mgh" … that only works when h << r :wink:)
  4. Jan 19, 2010 #3
    hey thanks for replying!! yes it's launched from infinity...but what next!! ?__?

    also when you say that only works when h<<r, doesn't that apply for all cases? because r = radius of planet + altitude (aka h) so r always > h...unless im misunderstanding what << means lol.

    thanks for helping again!
  5. Jan 20, 2010 #4


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    No, it's launched from the Earth's surface …

    you have to subtract that energy, don't you?​

    (my r was the Earth's radius, and my h was the height above it)
  6. Jan 20, 2010 #5
    wait...I GOT IT! thank you! :biggrin:
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