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Help with kinetic energy/potential energy ratio

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the ratio of kinetic energy to potential energy, of a satellite orbiting a mass M around a radius R.

    2. Relevant equations

    ke = 1/2 (mv^2)/r, pe = -Gmm/r^2 (im not really sure of these, i dont have my book with me)

    3. The attempt at a solution

    this was a problem on my test, and i dont know if i got it right, so i would like to know what the answer is.

    i had no idea how to do this so i just did

    2ke = pe

    ke/pe = 1/2.

    what is the answer? i know i got it wrong hehe

    any input is appreciated! thanks!
     
  2. jcsd
  3. Apr 20, 2010 #2

    kuruman

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    Start from
    KE = (1/2)mv2
    PE = -GMm/r

    Write Newton's 2nd Law for a circular orbit and solve for the speed v.
    Replace v in the KE equation.
    Take the ratio and see what you get.
     
  4. Apr 20, 2010 #3
    all i want to know is if i have it right. were not gonna have a final exam so i wont see this stuff ever again.

    anyways ill try it cuz im bored...

    (1/2)mv2/-GMm/r

    thats its i have no idea what else
     
  5. Apr 20, 2010 #4

    kuruman

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    You don't have it right.
     
  6. Apr 20, 2010 #5
    haha i knew it

    so how do u do this problem? all i can get is to that first part i just did
     
  7. Apr 20, 2010 #6

    kuruman

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    In my posting #2 I gave you three steps. You skipped to the third without doing the first two. OK, once more

    Step 1
    Write Newton's 2nd Law for a circular orbit and solve for the speed v.

    Finish this and I will help you with the rest.
     
  8. Apr 20, 2010 #7
    i still dont know how to do it
     
  9. Apr 20, 2010 #8
    can someone help me?
     
  10. Apr 20, 2010 #9
    Maybe you should check your sign???
     
  11. Apr 20, 2010 #10
    im so lost, wat sign?

    im up to this part

    (1/2)mv^2/GMm/r
     
  12. Apr 20, 2010 #11

    kuruman

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    Newton's 2nd Law says F = ma. For a satellite of mass m orbiting the Earth in a circular orbit, what is F and what is a?
     
  13. Apr 20, 2010 #12
    F = m.v^2 / r ??
     
  14. Apr 20, 2010 #13

    kuruman

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    Correct for the ma part. Now what is the force F for the gravitational attraction?
     
  15. Apr 20, 2010 #14
    mg ?
     
  16. Apr 20, 2010 #15

    kuruman

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    No. It is the Law of Universal gravitational force that has upper case G not lower case g in it.
     
  17. Apr 20, 2010 #16
    F = GMm/r^2 ?
     
  18. Apr 20, 2010 #17

    kuruman

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    That's the one. Now let's go to step 2. You have F (posting # 16) and you have ma (posting #12). Set the two equal and solve for the quantity mv2.
     
  19. Apr 20, 2010 #18
    ok so

    F = ma
    GMm/r^2 = m . v^2 /r

    GMm/r = mv^2

    is that it? what next?
     
  20. Apr 20, 2010 #19

    kuruman

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    Step 3. You know that KE = (1/2)mv2. What is KE in terms of GMm/r?
     
  21. Apr 20, 2010 #20
    i dont understand lol

    is it GMm/2r?
     
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