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Kepler's Laws and a satellite

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    a satellite of mass 200 kg is launched from a site on earth's equator into an orbit 200 km above the surface of earth. a) assuming a circular orbit , what is the orbital period of this satellite? b)what is the satellite's speed in it's orbit? c) what is the minimum energy necessary to place the satellite in orbit , assuming no air friction?

    2. Relevant equations
    PE= -G*Me*m/r

    how does
    -γMm/(R+r) become GmR²/(R+r)

    3. The attempt at a solution
    I did both parts a and b just 100% stuck on part c..no idea at all what to do :(
    ...This makes me frustrated...really!!!
    Last edited: Apr 21, 2009
  2. jcsd
  3. Apr 21, 2009 #2


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    Consider the difference in total energy of the satellite from before launch and then in orbit.
  4. Apr 21, 2009 #3
    that doesnt help me much i knew that i need to use the conservation of energy...i just dont know how to apply it here................
  5. Apr 21, 2009 #4
    would PEi= -GMem/ri???
    like i asked before how does -γMm/(R+r)= -gmR²/(R+r)
  6. Apr 21, 2009 #5


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    OK. So what is the conservation of energy?

    ΔPE + ΔKE = Energy supplied isn't it?
  7. Apr 21, 2009 #6
    yes but would this be it .5*Ms*vi^2-GMeMs/Re=.5*Ms*vf^2-GMeMs/Rt

    like what would vi be and vf be? does anything get canceled out? im not sure
  8. Apr 21, 2009 #7
    woud vi be the speed the earth is rotating and vf be my answer from part b?
  9. Apr 21, 2009 #8
    ARGGGGGGGG it still doesnt work what about -γMm/(R+r)= -gmR²/(R+r) how is this possible?!?!? idk im so confused. ive been working on this single part for over 3 hours now and still i have no answer..... i will never understand this :grumpy::mad:
  10. Apr 21, 2009 #9


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    Since you don't know where it would be launched from and ... I think you can assume that the rotational speed of the Earth is sufficiently slower than orbital speed, your initial speed would be zero.

    As to finding the Additional energy it's really then

    E = 1/2*m*Vf2 + G*Me*Ms( 1/6380 - 1/6580)

    (Pay careful attention to units. I've used km above.)
  11. Apr 21, 2009 #10
    Thank you!!!!!! I was messing up the units XD
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