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Escape Speed of a Satellite

  1. Mar 13, 2013 #1
    A satellite of mass 7500 kg orbits the Earth in a circular orbit of radius of 7.3(10^6) m (this is above the Earth's atmosphere). The mass of the Earth is 6.0(10^24) kg.

    What is the minimum amount of energy required to move the satellite from this orbit to a location very far away from the Earth?

    We are supposed to employ the Energy Principle to solve this problem, so we start with:

    K_i + U_i = K_f + U_f

    We know that K (at low speeds) = (1/2)*m*(v^2) and U = -GmM/r

    Using the Energy Principle, we know that

    K_f - K_i = U_i - U_f

    ΔK = -ΔU = -GmM[(1/r_f) - (1/r_i)]

    Since r_f is very large, ΔK = GmM(1/r_i)

    Using accepted and aforementioned values,
    ΔK = [6.7(10^-11) * 7500 * 6(10^24)]/[7.3(10^6)]

    This got me approximately 4.13(10^11)J, which is apparently incorrect. What am I doing wrong?
  2. jcsd
  3. Mar 13, 2013 #2

    Andrew Mason

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    Homework Helper

    You are forgetting that it already has kinetic energy. If it gained GmM/R in kinetic energy it would have more than enough energy to escape.

    In order to escape, it has to have just enough kinetic energy to get to an arbitrarily large distance from the earth ie. where it has 0 kinetic and 0 potential energy.

    So the condition for escape is: KE + U = 0

    Can you write the equation for KEr+ Ur (hint: if there are no external forces acting, does KE + U change?)

    Welcome to PF by the way!

  4. Mar 13, 2013 #3
    the binding energy of a satellite moving in an orbit is GMm/2r. so i think you ought to divide your answer by 2. [4.13*(10^11)] will be the answer if it is at rest on the earth's surface.
    PS: i am new here. could you please tell me how to start a new thread?
  5. Mar 13, 2013 #4
    Ah, okay, that makes sense now. Thanks for your help, and the warm welcome!

    To start a new thread I just went to the specific forum (Introductory Physics in this case) and clicked "New Topic" at the top right. It's in the same location as "New Reply" on this page.
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