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Space Shuttle- Gravitation

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A space shuttle of 68000 kg is placed in orbit at an altitude of 1000 km from the surface of the Earth. What energy must we provide?

    3. The attempt at a solution

    I will need to provide energy in order to decrease the potential energy and an additional energy for the kinetic energy of the rotation.

    [tex]\Delta E_p=-G\frac{Mm}{R+h} + G\frac{Mm}{R}[/tex]

    [tex]E_c=\frac{1}{2}mv^2=\frac{1}{2}mGM/(R+h)[/tex]

    So,
    [tex]E=-G\frac{Mm}{R+h}+ G\frac{Mm}{R}+\frac{1}{2}mGM/(R+h)[/tex]

    Am I right?
     
    Last edited by a moderator: Jan 15, 2007
  2. jcsd
  3. Jan 15, 2007 #2

    Hootenanny

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    You are almost correct, but you have made one 'small' mistake in the first line which has carried through.
     
  4. Jan 15, 2007 #3
    Thank you for answering, Hootenanny!:rolleyes:

    I have just corrected my mistake. My doubt was whether to consider or not the kinetic energy of rotation. Because when the space shuttle reaches an height of 1000 km relative to the surface of the Earth it will need an extra impulse from the fuel engines.
     
  5. Jan 15, 2007 #4

    Kurdt

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    The potential energy equation looks fine except you do not square the distance and you have the correct overall idea. The kinetic energy term looks fine as you have used circular motion to derive the velocity.

    EDIT: Beaten by Hoot and the coffee machine again!
     
  6. Jan 15, 2007 #5
    Kurdt, thank you anyway for your kind help!
     
  7. Jan 15, 2007 #6

    D H

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    Gyroscope:

    The Shuttle has kinetic energy prior to launch. You have solved for the final kinetic energy, not the change in kinetic energy. The distinction is important: It is why the Shuttle is launched to the East.
     
  8. Jan 15, 2007 #7
    D_H thanks for your reply!

    Could you be more specific, please? The kinetic energy prior to launch of the shuttle is the energy associated with the rotation motion of the Earth?
     
  9. Jan 15, 2007 #8

    BobG

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    Yes, it is.

    Although I'm wondering if you're supposed to worry about that for this problem. Your initial velocity will depend on your latitude. At the equator, the launch site is moving 465 m/sec. Your speed anywhere else is equal to the cosine of the latitude times 465. Even at the equator, the rotation of the Earth will contribute less than 2% of the shuttle's energy.
     
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