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Help with orbital period Problem please

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?


    2. Relevant equations
    T2=(4∏2r3/GME)


    3. The attempt at a solution
    I used Kepler's third law with r being the distance. I took (27.3*86400)2, which is the period squared and set it equal to the expression (4∏2r3/GME). I got 19404m, which is wrong. I think the problem is the fact that r is cubed and the period is squared, that screws me.
     
  2. jcsd
  3. Nov 20, 2011 #2
    r^3 = GMeT^2/4pi^2
     
  4. Nov 20, 2011 #3

    sophiecentaur

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    This satellite needs to orbit the Moon every 27.3 days as well as orbiting the Earth, does it not? So that it is always on this straight line.

    I think you need to include both gravitational forces and equate these to the required centripetal force to keep it in this orbit. A bit fiddly, but do-able, I think.
     
  5. Nov 20, 2011 #4

    gneill

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    I think you'll end up with a 4th order polynomial, so probably a candidate for numerical solution.
     
  6. Nov 21, 2011 #5

    Filip Larsen

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    You may want to read up on the three body problem and Lagrangian points.
     
  7. Nov 21, 2011 #6
    Sorry, but that didn't work.
     
  8. Nov 22, 2011 #7

    Filip Larsen

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    Have you tried to do as sophiecentaur suggested in post #3? If so, what result did you get? You should end up with something similar to (but not quite) what whoareyou blurted out and you may find it requires some kind of iteration or approximation to get a numerical answer.

    For some theoretical background of what is going on in this problem you may also, as I've noted earlier, want to read up on the math behind the three body problem and the concepts of Lagrangian points. If you find that this is too advanced for you then don't worry and just stick to the approach suggested by sophiecentaur.
     
  9. Nov 22, 2011 #8

    sophiecentaur

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    That's because you aren't involving theMoon's mass too. You can hardly ignore a massive body like that, in the vicinity.

    They don't just want a 28 day orbit around the Earth (that would be at the same radius as the Moon); they want it in a particular position with respect to the Moon.
     
  10. Nov 22, 2011 #9
    Well, I'm able to get this far, but with all the denominators, I'm not sure what to do now:

    1020/x = 3.982e14/x2 - 4.902e12/(1.466e17 - x2)
     
  11. Nov 22, 2011 #10

    Filip Larsen

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    This is not quite right. You may have the right idea, but something has gone wrong along the way, both with the numbers and with your x-values.

    If you haven't already, I suggest you make a small diagram of the x-axis where you plot the position of the Earth at zero and the Moon at R with a small dot (the satellite) in between and then for each acceleration you think will apply to the satellite in this rotating system draw a small arrow above the dot that points in the direction on the x-axis you believe this acceleration works (e.g. acceleration from the gravity of Earth point towards the Earth). Next, given that the satellite is at position x you may want to write up an equation for each acceleration in isolation as a function of x (e.g. acceleration on the satellite due gravity from earth is aE=GME/x2). Take special care to get the distance from the center of the Moon correct as a function of R and x. Finally you can write up your equation by adding all accelerations that point towards the Earth and equate that with the sum of all those that points towards the Moon.

    You should now end up with an equation in x that unfortunately is not so easy to solve directly. To solve it you may want to rely on a numerical solution provided by a graph calculator, a solver like matlab or WolframAlpha, or perhaps by a spreadsheet yourself. You should probably choose the method of numerical solution that you think will be accepted by your teacher.
     
  12. Nov 24, 2011 #11

    sophiecentaur

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    In order to get the signs right (to avoid a negative answer somewhere along the line), it may be better to put your dot on the Other Side of the Moon. That will be where the stable point is. A bit of, literal, arm waving should show you that this is true.
     
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