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Acceleration due to gravity

  1. Nov 7, 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
    As energy is also conserved. mv(for perigee)/2-GMm/7.53*106=mv(for apogee)/2-GMm/6.81*106
    But how can I find v(for apogee)? Thus v(for perigee)?
    I can't find the MC answer which is 7840m/s


    Similar thread:
    https://www.physicsforums.com/threads/orbit-around-the-earth.867945/
    http://www.enotes.com/homework-help/sputnik-was-launched-into-orbit-around-earth-1957-566404

    Thanks
     
  2. jcsd
  3. Nov 7, 2016 #2

    gneill

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    There's another conservation law that will help here.
     
  4. Nov 7, 2016 #3
    Angular momentum? L=rp=rmv
    But still i need to find v for apogee... ehich i dont know how to get the answer avoiding this unknown variable...
     
  5. Nov 7, 2016 #4

    gneill

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    Yes, angular momentum is conserved over the whole orbit. What does that tell you about the angular momentum at both apogee and perigee? Can you express that as an equation?
     
  6. Nov 7, 2016 #5

    Janus

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    Can you determine the period of the orbit from what you are given? And if so, what does that give you?
     
  7. Nov 7, 2016 #6
    (a=apogee,p=perigee,r=radius,v=velosity)
    ra x va = rp x vp
    Right? Without vp, how to find va?
     
  8. Nov 7, 2016 #7
    T=2(pi)/(angular velosity) = 2(pi)r/(average v)?
     
  9. Nov 7, 2016 #8

    gneill

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    This gives you one relationship between va and vp. Energy conservation gives you another. Two equations, two unknowns.
     
  10. Nov 8, 2016 #9
    Do you agree with this?
    What is r and what is v?
    Thanks
     
  11. Nov 8, 2016 #10

    gneill

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    If the orbit were circular then I'd agree. But the velocity and radius both vary over time for an elliptical orbit.

    Use the conservation laws: energy and angular momentum.
     
  12. Nov 8, 2016 #11

    Janus

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    Not quite what I was looking for. There is a fairly direct way to get the answer to the problem, but it assumes some knowledge of the nature of orbits. I was trying to gauge just how much you knew about orbits in order to know in what direction to lead you in.
     
  13. Nov 9, 2016 #12
    Angular momentum: Rmva=rmvp
    Energy: 0.5ka^2 ; 0.5mv^2; -gMm/r => how should i arrange them?
    Thanks

    I have only little time left before my rxam is coming... hope you can tell me the sooner the better for me to try and understand them... thank you very much.
     
  14. Nov 9, 2016 #13

    gneill

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    Okay. Note that the m's can cancel.
    Apply energy conservation for the two orbit locations.
     
  15. Nov 9, 2016 #14
    I know i should use energy conservation.... but how to arrange them? If 0.5k(ap)^2+0.5m(vp)^2+gMm/r=0.5K(aa)^2+0.5m(va)^2+GMm/r
    How come there are so many variable?! Am i having something wrong?
    So how can i arrange them???
    Thanks...

    (Only little time left from test...)
     
  16. Nov 9, 2016 #15

    gneill

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    I don't know what your "k" terms are meant to represent. What type of energy do they correspond to? What is k?

    You need to get your signs right for the gravitational potential energy. Gravitational potential energy in this context takes its reference point at infinite distance, so its sign is always negative.

    You have two given locations, apogee and perigee, for which you have separate orbit radii. Be sure to identify them in your equation with separate variable names.
     
  17. Nov 9, 2016 #16
    K is spring constant, as such the planet is oscillating like a spring
    It is correspond to elastic potential energy
     
  18. Nov 9, 2016 #17

    gneill

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    There's no spring here. The "spring" action of the orbit is accounted for by the trading of gravitational potential energy and kinetic energy.
     
  19. Nov 9, 2016 #18

    haruspex

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    To the extent that the satellite oscillates in its orbit, it is a result of the gravitational attraction term you already have in the equation, not from some mysterious piece of elastic tethering it to Baikonur.
    Besides, gravitation follows an inverse square law of attraction, while elastic follows a linear law.
     
  20. Nov 9, 2016 #19
    Thank you very much.
    I can find the answer now.
     
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