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Physics Gravitation Question?

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the change in gravitational potential energy of a 64.5-kg astronaut, lifted from
    Earth’s surface into a circular orbit of altitude 4.40 3 102 km?



    3. The attempt at a solution
    Ok, i know to find this i have to find the gravitational energy on earth using earths radius and find the gravitational energy in orbit using the earths radius + the altitude. Therefore, the equation at earth would be Eg = -GMm/re and the equation in orbit would be
    Eg = -0.5(GMm/re + altitude).

    However, in the solutions manual it uses the same equation Eg = -GMm/re when the object is in orbit.
    I just want to know if the book is wrong or if i am wrong. I thought that when objects are in orbit you have to play in the value of kinetic energy which creates the equation of Eg = -0.5(GMm/re + altitude). Please explain
     
  2. jcsd
  3. Apr 29, 2012 #2

    Andrew Mason

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    Is Eg the total energy or the potential energy?
    Why would kinetic energy affect potential energy? Potential energy is a function of r (distance from the centre of the earth) only. It looks like you may be confusing total energy with potential energy.

    AM
     
  4. Apr 29, 2012 #3
    Ok, so the formula of -0.5(GMm/re + altitude) is for the total energy?
     
  5. Apr 29, 2012 #4
    This is correct, for something in a stable orbit. You would lose the factor of 1/2 if you just wanted the contribution from gravitational potential energy.
     
  6. Apr 29, 2012 #5
    So when it asks for just the gravitational potential energy, i use Eg = -GMm/r ?
     
  7. Apr 29, 2012 #6
    and also, does binding energy = -0.5(Gmm/r) or 0.5(Gmm/r)?
     
  8. Apr 29, 2012 #7

    Andrew Mason

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    Correct. But I would suggest you use PE or U to distinguish it from total energy. This is the potential energy relative to r = ∞.

    Binding energy is the work that must be done on an object of mass m to get it from R = r to R = ∞. It is positive.

    AM
     
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