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Gravity fun

  1. Apr 25, 2008 #1
    [SOLVED] gravity fun

    1. The problem statement, all variables and given/known data
    1. Calculate the speed of objects on the earth's surface caused by the earth's own rotation around its own axis.

    2. Find the mass of the earth using its time period of revolution of the moon around the earth.

    3. Find the mass of the sun using the time period of the earth around the sun.

    4. What is the radius for a geo-synchronous satellite?

    2. Relevant equations
    The time period of revolution of the moon around the earth is 28 days?

    3. The attempt at a solution
    1. v= 2piR/T = 2pi (6.4E6) / (24hours*60min*60s) = 4.65E2

    2. I know how to solve this using F = GmM/(R^2) = mg but I don't know how to use the given 28 days. The answer to this problem should be around 6E24 kg

    3. The answer is around 2E31 kg

    4.

    Thanks in advance!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Apr 25, 2008
  2. jcsd
  3. Apr 25, 2008 #2

    G01

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    HINT: What type of motion is the moon in?
     
  4. Apr 25, 2008 #3
    oooh circular motion? I tried using V = 2piR/T, but I can't get to the right answer. What value would I use for R?
     
  5. Apr 25, 2008 #4
    Anyone? ):
     
  6. Apr 25, 2008 #5

    Dick

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    Maybe R should be the radius of the moon's orbit? You are working on 2), right?
     
  7. Apr 25, 2008 #6

    alphysicist

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    Hi smashingtime,

    This should be F = GmM/(R^2) = ma. (g is the gravitational acceleration at the earth's surface.) Once you have a in your equation, since it is circular motion, you have a formula for the acceleration in terms of speeds and orbital radius.

    You will still need to use your relation V = 2piR/T to get the period into the equation, and I think you will need either the orbital radius of the moon or the orbital speed of the moon.
     
  8. Apr 25, 2008 #7
    The radius of the moon's orbit is 3.84 * 10^8 m. (I just googled it haha).
    I plugged that in to v = 2piR/ T to find v, then used v^2 = Gm/R to find the mass, but my answers still don't match up :/
     
  9. Apr 25, 2008 #8

    Dick

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    What did you get for a number on v and m? That should have worked.
     
    Last edited: Apr 25, 2008
  10. Apr 25, 2008 #9
    v = 2piR/T
    = 2pi(3.84E8)/(28*60^60)
    =2.41 E 4

    v^2 = 5.82E8 = Gm/R
    5.82E8 = (6.67E-11)(m)/(3.84E8)
    m = 3.35 E27 kg
    What the answer is supposed to be: m = 5.98 E24 kg
     
  11. Apr 25, 2008 #10

    Dick

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    Well, there you go. 28*60*60sec is 28 hours. 28 days is 28*24*60*60sec. Oh, try to put units on everything, ok?
     
  12. Apr 25, 2008 #11
    ahaha *smacks forehead*
    thanks so much!
     
  13. Apr 25, 2008 #12
    I just worked out (3.) using the same method.
    I found the distance between the earth and the sun to be 1.5E11m.

    v=2piR/T
    = 2pi (1.5 E 11) / (365*24*60*60)
    = 2.99 E 4
    V^2 = 8.95 E 8 = Gm/R
    8.95 E 8 = (6.67E-11)(m) / (1.5E11)
    m = 2 E 30 kg
    Established answer: m = 2 E 31 kg
     
  14. Apr 26, 2008 #13
    oh nvm.
    yay, the established answer IS 2E30 kg.
     
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