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How much energy per unit mass is needed

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    How much energy per unit mass (E/m) must you give a rocket to put it into a geosynchronous orbit three earth radii above the surface of the Earth?

    Radius of earth : 6380 km
    GM/r = 6.25 x 10^7 m^2/ s^2

    2. Relevant equations

    U(r) = - GMm/r

    3. The attempt at a solution

    I am not really sure where to start with this one. I think it should be at equilibrium 3radii up.

    1/2mv^2 = Gm/r ??
     
  2. jcsd
  3. Apr 3, 2013 #2
    Start by writing the equation for conservation of energy. Say E were the energy given to it on the ground. What would be its energy in orbit as a function of the radius of the orbit (Firstly what would be the radius of the orbit)?
     
  4. Apr 3, 2013 #3

    haruspex

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    Since it's all in multiples of Earth's radius, you can simplify things by expressing things in terms of g, rather than G. Your equilibrium equation is wrong. What's the formula for centripetal acceleration?
     
  5. Apr 4, 2013 #4
    What is GM/r ? Is that the gravitational force of the entire earth? What's throwing me off is how to use this in E = K +U.

    1/2mv^2/r - GMm/r = E ? If that is correct would I then multiply by three?
     
    Last edited: Apr 4, 2013
  6. Apr 4, 2013 #5
    -GM/r is the gravitational potential due to the earth at a distance of r from its center. The gravitational potential energy possessed by a mass m at a distance r from the center (r > R) is -GMm/r. The gravitational force at this point is GM/R2 . Like haruspex told it is easier to write the potential energy in terms of g as GM/r = gR2/r. (this relation is be derived from newton's law of gravitation)

    What you need to do in the energy equation is to equate the initial and final mechanical energy right?
    The initial energy is the sum of initial kinetic energy and gravitational potential energy on the surface. The final energy is the kinetic energy due to orbital motion and the gravitational potential energy at the orbit.
     
  7. Apr 4, 2013 #6
    Yes, this is correct:

    1/2mv^2/r - GMm/r = E ? If that is correct would I then multiply by three?

    Except you should not divide K by r!
    This gives the total energy when it is in orbit - so no need to multiply by three. You also want E/m.
    So you need to determine v when it is in the specified orbit.

    U(r) = - GMm/r

    is the potential energy of an object, mass m, with respect to the earth, mass M, when it is at a distance r form the middle of the earth.
     
    Last edited: Apr 4, 2013
  8. Apr 4, 2013 #7

    gneill

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    The total specific mechanical energy (where here specific means "per unit mass") of a body in a gravitational field is given by:
    $$\xi = \frac{v^2}{2} - \frac{\mu}{r}$$
    where v is the speed of the object, and ##\mu## is the gravitational parameter of the central gravitating body, ##\mu = GM##. You should be able to spot the kinetic and gravitational potential energy terms in the expression.

    If you can determine ##\xi## for the two locations of interest then their difference will be the energy per unit mass that needs to be added or subtracted to move a unit mass from one state to the other.
     
  9. Apr 4, 2013 #8
    so I divide out m and get 1/2v^2 - GM/r = E/m

    so to replace v I would use mv^2/r = v = √r/m but I don't know what to do with that last m
     
  10. Apr 4, 2013 #9

    gneill

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    mv^2/r is a force, not a velocity. And your expression / manipulation doesn't make sense to me.

    There's another expression for the velocity of a body in circular orbit that depends only on the gravitational parameter and the radius of the orbit...
     
  11. Apr 4, 2013 #10
    you mean a = v^2 / r
     
  12. Apr 4, 2013 #11

    gneill

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    That's the centripetal acceleration. You can derive the required expression by equating it to the gravitational acceleration at the same radius -- centripetal and gravitational force are equal for a circular orbit (gravity provides the "string" that holds the body in circular motion).
     
  13. Apr 5, 2013 #12
    Yes you can get v from this relationship. Use Newton's Universal law of gravitation to replace the acceleration according to

    a = F/m

    where F is supplied by the gravitational attraction at that point in the orbit. This means your equation above says that the centripetal acceleration is then supplied by the gravitational attraction.
     
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