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Orbit around the Earth

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Sputnik I was launched into orbit around Earth in 1957. It had a perigee (the closest approach to
    Earth, measured from Earth's center) of 6.81 × 106 m and an apogee (the furthest point from
    Earth's center) of 7.53 × 106 m. What was its speed when it was at its perigee? The mass of Earth
    is 5.97 × 1024 kg and G = 6.67 × 10- 11 N ∙ m2/kg2



    2. Relevant equations
    F=GMm/r2 F=mv2/r

    3. The attempt at a solution
    I tried to use the formula above which is GM/r=v2, however I substituted both 6.81x106 ,7.53x106 and took the average of two of them , the answer still did not match with the MC answer which is 7840m/s
    I am curious about what should be the radius, and actually the orbit is not a perfect circle with a perigee and an apogee , is it acceptable to use the equation in circular motion . Thank you.
     
  2. jcsd
  3. Apr 20, 2016 #2

    PeroK

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    Note that the centripetal acceleration is related to the curvature, so you can't use the equation for a circular orbit.

    What do you know about angular momentum?
     
  4. Apr 20, 2016 #3
    The angular momentum is conserved as the gravitational pull by the Earth is always perpendicular to the motion, no matter where is the position of the Sputnik I . And L=mvr
     
  5. Apr 20, 2016 #4

    PeroK

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    Yes. And total energy (kinetic & potential) is conserved also. Can you do something with that?

    To be precise, the angular momentum is ##L = mv_ar## where ##v_a## is the angular component of velocity.

    And, ##v_a = v## only at the turning points, where the radial component of the velocity is 0.
     
    Last edited: Apr 20, 2016
  6. Apr 20, 2016 #5
    OK Thanks a lot.
    As angular momentum is conserved. mv(for apogee)*7.53*106=mv(for perigee)*6.81*106
    Then v(for apogee)=0.90428v(for perigee)
    M is equal to the mass of earth
    As energy is also conserved. mv(for perigee)/2-GMm/7.53*106=mv(for apogee)/2-GMm/6.81*106
     
  7. Apr 20, 2016 #6

    PeroK

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    Now you've reached this level of physics, you ought to consider working more algebraically. I would say:

    ##mv_1r_1 = mv_2r_2## where ##r_1## is the perigee etc.

    Numbers like ##6.81*10^6## are just going to clutter up your equations and hide the physics.

    Solve for ##v_1## in terms of ##r_1, r_2## then plug the numbers in at the end.
     
  8. Apr 20, 2016 #7

    ehild

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    The v-s have to be squared in the kinetic energies.
     
  9. Apr 21, 2016 #8
    Thank you very much. Get the answer finally.
    May I ask how to quote the formula for the angular momentum in a more systematic way like you?
     
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