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Planetary Motion - Energy Totals and Binding Energy

  1. Jun 2, 2007 #1
    The following question deals with planetary motion: energy totals and binding energy. I have solved the question to the best of my ability, but I cannot get the right answer! I don't know what I am doing wrong, and would greatly appreciate it if someone could point out my mistakes.

    1. The problem statement, all variables and given/known data
    This is a two part question.

    1. a) What is the total amount of energy needed to place a 2000 kg satellite in circular Earth orbit, at an altitude of 500km?

    b) How much additional energy would have to be supplied to the satellite once it was in orbit, to allow it to escape from the Earth's gravitational field?

    2. Relevant equations

    a) Etot = 1/2 Eg
    Etot = 1/2 -(GMm)/d

    b) Etot > Eg

    3. The attempt at a solution

    a) m = 2000 kg
    M = mass of Earth = 5.98x10^24 kg
    d = 500 000 m + 6.38x10^6 (radius of earth)
    d = 6.88x10^6 m

    Etot = 1/2 Eg
    Etot = 1/2 -(GMm)/d
    Etot = 1/2 -(6.67x10^-11)(5.98x10^24)(2000)/6.88x10^6
    Etot = - 6.0x10^7 J

    However - the correct answer is 6.7x10^7. That seems like a big difference - does anyone know what I am doing wrong?

    b) I know I am being asked for the binding energy - the extra amount of energy required to break free of the gravitational field of the earth, also known as Etot. My total energy is - 6.0x10^7 J, so I assume I would need + 6.0x10^7 J to overcome it.
    However - the correct answer is 5.8-x10^10 J.

    Please point me in the right direction if you can.
     
    Last edited: Jun 2, 2007
  2. jcsd
  3. Jun 3, 2007 #2

    Andrew Mason

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    Check your math. I get 5.8x10^10 for the gravitational potential in orbit. You have found the gravitational potential energy of the satellite in orbit but you have not determined the amount of energy added to get it from the earth's surface to orbit.

    Your method is right for b).

    AM
     
    Last edited: Jun 3, 2007
  4. Jun 3, 2007 #3
    Thank-you for your reply!

    a) Oh, so is this question asking how much work is required to move the satellite from the surface of the earth to an orbit 500 000 m away? That makes sense.

    W = Egf - Egi
    W = mgh - mgh
    W = (2000)(9.8)(6.88x10^6) - (2000)(9.8)(6.38x10^6)
    W = -1.35x10^11 - (-1.25x10^11)
    W = -1.0x10^10??

    Sorry, I still do not understand what I am doing wrong - I would appreciate further help!

    I tried this question again using the quicker "Eg = mgh" method, since I have the constant ag (9.8 m/s'2), but I calculated the gravitational potential in orbit to be -1.35x10^11. Argh!
     
    Last edited: Jun 3, 2007
  5. Jun 3, 2007 #4

    Andrew Mason

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    (I may have confused you with my previous response. I should have said the orbital energy is -5.8x10^10J, not the gravitational potential in orbit. The gravitational potential is double that, or -11.6x10^10J.)

    Determine the gravitational potential energy of the satellite on the surface:

    Us = -GMm/R = -6.67x10^-11 x 5.98x10^24 x 2x10^3/6.38x10^6 = -1.25x10^11 J

    Determine the potential energy in orbit at a distance of earth's radius + 500 km:

    Uo = -GMm/(R+5e5) = -6.67x10^-11 x 5.98x10^24 x 2x10^3/6.88x10^6 = - 11.6x10^10 J

    Its kinetic energy is 0 while on the earth (ignoring motion due to rotation of the earth).

    Putting it in orbit of 500 km requires giving it a kinetic energy of GMm/2r (for orbit, F = GMm/r^2 = mv^2/r so GMm/2r = mv^2/2 = KE), so when in orbit its kinetic energy is -half of the potential energy: 5.8x10^10 J.

    The total energy, therefore in orbit is -5.8x10^10J.

    The total energy on the surface is -1.25x10^11 J, so the difference is: ____

    For b), the total energy when it just escapes is 0. What is the difference between that an the total energy in orbit?


    You cannot use mgh because g decreases as r increases. It would be close though.

    AM
     
    Last edited: Jun 3, 2007
  6. Jun 3, 2007 #5
    YES!! : ) It makes sense!!! First, thank-you for clarifying the orbital vs. gravitational potential energy; second, thank-you for guiding me through the calculations for part A. I was having a difficult time keeping my concepts and formulas straight while working with large numbers.

    Here is my final work for the question:

    a) I now know I am *not* being asked for work (the question would have needed to use the word "work" if this was the case). I simply need to find the *difference* in *total* energies.

    Total energy in orbit is equal to 1/2 Eg (which is the same as Kinetic energy, only the total energy is negative).

    Etot in orbit = 1/2 Eg
    Etot = 1/2 -(GMm)/d
    Etot = 1/2 -(6.67x10^-11)(5.98x10^24)(2000)/6.88x10^6
    Etot = -5.88x10^10 J

    (originally, I made a silly calculation error here and rounded up)

    Etot on earth = Eg (no kinetic energy from satellite yet)
    Etot = -(GMm)/d
    Etot = -(6.67x10^-11)(5.98x10^24)(2000)/6.38x10^6
    Etot = -1.25x10^11

    But I want to find the amount of energy required to put satellite into orbit, and I already have gravitational potential on earth, so I calculate the difference between the energy totals.

    Etot = Etot_final - Etot_initial
    Etot = -5.88x10^10 - (-1.25x10^11)
    Etot = 6.62x10^11

    If I had used -5.8x10^10, it would have worked out perfectly.

    ------

    I think what confused me the most was the concept of "work" vs. "amount of energy required to put object into orbit." The formula for work is Egf - Egi, but I was not asked for work, so I needed to deal with energy totals, *and* take the surface of the earth into account as there is energy to begin with. Hooray! It makes sense!

    Thank-you once again for your help! :smile:
     
    Last edited: Jun 3, 2007
  7. Jun 3, 2007 #6

    Andrew Mason

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    Check your arithmetic here. I get -5.80x10^10 J.
    Check your math again and use -5.80x10^10 J for the orbital energy. You are out by a factor of 10. It should be 6.7x10^10 J.

    Careful. The amount of work done on the satellite and the amount of energy added are identical.

    It is not sufficient to simply lift the satellite to 500 km above the earth. You also have to give it enough kinetic energy that it stays out there. Lifting it and giving it kinetic energy requires doing an amount of work that is equal to the change in total energy of the satellite (work = change in kinetic and potential energy: [itex]W = \Delta U + \Delta KE[/itex])

    AM
     
    Last edited: Jun 3, 2007
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