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Radius of Orbits

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A GPS satellite is put in a high circular orbit around the earth. The period of revolution is 8 hours. Calculate the radius r of the orbit, compared to the radius R of the earth.

    2. Relevant equations

    Rearth= 6.4 x 10^6 m
    mearth= 6 x 10^24 kg

    3. The attempt at a solution

    I have no idea how to attempt this problem. Can someone please help?
     
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 24, 2007 #2

    Dick

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    Write down the acceleration of an object moving in circular orbit of radius r with period T. Equate that to the acceleration caused by gravity at radius r.
     
  4. Oct 24, 2007 #3
    It is not clear to me what you mean. Don't you mean velocity? How does the mass of the earth come into play? So, step-by-step, what do I do? Whats first? Please Help.
     
  5. Oct 24, 2007 #4

    Dick

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    No, I mean acceleration. An object moving in a circular path experiences a centripetal acceleration. There's a formula for it in terms of v and r. Now equate that to the gravitational acceleration. There's a formula for that as well. Find those formulas and set them equal.
     
  6. Oct 24, 2007 #5
    Ohhh... I see now that it is ac. mv^2/r right? I'll try to work it out and respond if i get stuck, thanks.
     
  7. Oct 24, 2007 #6

    nrqed

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    Right (m is the mass of the satellite). Now set this equal to the force of gravity of the Earth on the satellite (using the universal law of gravitation). You will also need to use that for circular motion, [tex] v = \frac{2 \pi r }{ T} [/tex]
     
  8. Oct 24, 2007 #7
    Do I find the velocity for earth's orbit AND the satellite with this equation? Or just the satellite? Isn't the mass of the satellite negligible, so set it to 1?
     
  9. Oct 24, 2007 #8

    Dick

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    I think finding the radius of the satellite is the question. You can set the mass of the satellite equal to one if you wish, it will cancel from both sides of the equation you have yet to write down. And sure, you'll want to assume it's negligible compared with the earth so you can assume the earth is stationary while the satellite orbits. WRITE DOWN THE EQUATION, ok?
     
  10. Oct 24, 2007 #9
    mv^2/r = Gm1m2/R^2

    Is this the equation? And all values are known except r, correct? And you find the v from [tex] v = \frac{2 \pi r }{ T} [/tex] correct?

    So this would mean that the centripetal force from the satellite equals the gravitational force between the earth and the satellite??
     
    Last edited: Oct 24, 2007
  11. Oct 24, 2007 #10

    Dick

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    Finally. Yes. One of the m1 and m2 on the right is equal to m on the left. Cancel it. The remaining m is mass of the earth, right?
     
  12. Oct 25, 2007 #11
    How are you supposed to find the v of the satellite using [tex] v = \frac{2 \pi r }{ T} [/tex] if you don't know what r is???
     
  13. Oct 25, 2007 #12

    D H

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    From post #9, you have [itex]v^2/r=GM_e/r^2[/itex]. You don't know [itex]v[/itex] but you do have an expression for it: [itex]v=2\pi r/T[/itex]. So substitute this expression in the equation from post #9 and solve for [itex]r[/itex].
     
  14. Oct 25, 2007 #13
    I keep getting 8.08 x 10^14 m for r, but this is incorrect. Is my math just wrong? You set [2(pi)r/(T)]^2 = GME/RE^2 right???
     
  15. Oct 25, 2007 #14

    Dick

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    No. i) the quantity on the left should be v^2/r. I think you are missing the r. And the R_E on the right should be r of the satellite, right?
     
  16. Oct 25, 2007 #15
    Sorry. When I did the calculation I had the r on the left side and the right side, but I still keep getting the answer incorrect.

    The equation is [itex]v^2/r=GM_e/r^2[/itex] with [tex] v = \frac{2 \pi r }{ T} [/tex]

    I know this and keep getting the wrong answer of 8.08 x 10^14. What are you getting??
     
  17. Oct 25, 2007 #16

    Dick

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    I get something round about 10^7m. Can you post your final equation for r and maybe some intermediate results?
     
  18. Oct 25, 2007 #17

    Dick

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    You should be getting that r^3 is proportional to T^2. That's one of Kepler's laws.
     
  19. Oct 25, 2007 #18
    I am getting [2(pi)r]^2 x [1/T]^2 = GME/r^2

    From here I can't simplify and I am not getting 10^7. I'm getting around 10^14 instead.

    When I tried to simplify, I got [2(pi)r]/(8.29x10^8) = [4.002x10^36]/(r^2). Then I crossed multiplied and got the wrong answer.

    I am not simplifying correctly, can someone help me with this, I feel dumb.
     
    Last edited: Oct 25, 2007
  20. Oct 25, 2007 #19

    Dick

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    The 1/r in v^2/r has disappeared again. You should be able to simplify it to r^3=(something)*T^2. What's the (something)?
     
  21. Oct 25, 2007 #20
    Is it GME/ 2(pi) ???
     
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