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How fast must a satellite leave Earth's surface

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data

    How fast must a satellite leave Earth's surface to reach an orbit with an altitude of 895 km?

    2. Relevant equations

    v = √GM/r

    3. The attempt at a solution

    G = 6.67 x 10^-11
    M = 5.98 x 10^24
    r = (6.38 x 10^6) + (8.95 x 10^5) = 7.275 x 10^6

    v = √(6.67 x 10^-11)(5.98 x 10^24)/(7.275 x 10^6)
    v = 7404.5 m/s

    But this is the speed the satellite needs to stay in orbit. Is this the same speed that it needs to leave the Earth's surface? Do I need to use v = √2GM/r instead to find the velocity it needs to escape the gravitational field at Earth's surface and go into orbit?
     
  2. jcsd
  3. Jul 27, 2008 #2

    HallsofIvy

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    Science Advisor

    To the velocity required to orbit at that height, you need to add the velocity necessary so that the kinetic energy can be 'traded off' for the change in potential energy to get to that height.
     
  4. Jul 27, 2008 #3
    Would that be

    Ek = ΔEg
    1/2mv^2 = -GMm/r2 - (-GMm/r1)

    solve for v, then add it to 7404.5 m/s?
     
  5. Jul 27, 2008 #4
    Ohhh wait, conservation of energy applies, right?

    Ek1 + Eg1 = Ek2 + Eg2

    just plug v2 = 7404.5 m/s in
     
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