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Gravitation Potential Energy Help

  1. Jan 30, 2005 #1
    Calculate the ratio of the energy that would be required to put a mass M into orbit near the earth's surface (if there were no friction) and the energy to put the same into distant space.

    I'm stuck. How am i suppose to calculate the ratio. I know the that the energy to put M into orbit near the earth's surface would be a small positive energy while putting it into distant space would be a large positive energy.
     
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  3. Jan 30, 2005 #2

    dextercioby

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    The signs are purely conventional.Actually for gravitational potential energy,it is MINUS INSTEAD OF PLUS...
    The gravity field of a spherically symmetric celestial body (i.e.Earth) is CONSERVATIVE,which means u can apply the law of conservation of total energy.

    Can u compute the energy required to put a mass into orbit near the surface of the Earth??

    Daniel.
     
  4. Jan 30, 2005 #3
    There are no values given to me so i dont know how to compute it
     
  5. Jan 30, 2005 #4
    I'm not sure how to find the answer, but I can tell you this: Binding energy is the additional energy you have to supply a satellite (or anything in orbit) to escape the Earth's gravitational field. This is the formula for it:
    [tex]E_{binding}=\frac{Gm_1m_2}{2r}[/tex]

    When a satellite is in orbit, it's total energy (total mechanical energy) is:
    [tex]E_{mechanical}=-\frac{Gm_1m_2}{2r}[/tex]

    So I guess the ratio would be 1:1?
     
  6. Jan 30, 2005 #5

    dextercioby

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    Not really.The second formula doesn't hold for bodies on the earth (which do not orbit,hence do not have KE),so i guess the OP is correct.The problem is missing some data...

    Daniel.
     
  7. Jan 30, 2005 #6
    Wouldn't the "r" be different values so I dont think they can be 1:1?
    I was thinking more like 1:1/r ratio would make sense.
     
  8. Jan 31, 2005 #7

    Andrew Mason

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    For near earth orbit:

    [tex]mv^2/R = F = GMm/R^2[/tex]

    Which is:
    (1)[tex]2KE = GMm/R[/tex]
    where KE is the kinetic energy of the orbiting body, which is just -1/2 * its gravitational potential at radius R.

    The condition for distant space (escape) is:

    [tex]PE \ge 0[/tex]

    You should see from (1) that [itex]2KE_{orbit} + PE_{orbit} = 0[/itex], which means that the kinetic energy required for escape is ______ the KE for orbit. I think that should help you answer the question.

    AM
     
  9. Jan 31, 2005 #8
    So ratio would be 1:2?
     
  10. Jan 31, 2005 #9

    Andrew Mason

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    If you mean: Kinetic energy of escape = 2 KE of orbit, then yes.

    AM
     
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