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How to find the point between Mercury and the Sun so force = 0?

  1. Jun 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the point between Mercury and the sun at which an object can be placed so that the net gravitational force exerted by Mercury and the sun on this object is 0.


    2. Relevant equations
    Mass of sun = 1.991 x 10^30kg
    Mass of Mercury = 3.18 x 10^23kg
    distance between mercury and sun = 5.79 x 10^10m

    GMm/R^2

    3. The attempt at a solution
    I've tried setting the force between the object and the sun = force between the object and mercury but got stuck in figuring out what R is on either side.
     
  2. jcsd
  3. Jun 23, 2013 #2

    Borek

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    Pay attention to what R is. You have two unknowns - distance between object and Sun, and distance between object and Mercury. What is their sum?
     
  4. Jun 23, 2013 #3
    If the distance from mercury to the point of interest = x, what is the distance from the point of interest to the sun?
     
  5. Jun 23, 2013 #4
    Ms - x ?
     
  6. Jun 23, 2013 #5
    And you know the distance from the sun to mercury....some algebra involved in the force equation....tricky....can you do it?
     
  7. Jun 23, 2013 #6
    Here's what I have so far...

    Mm/x^2 = Ms/(d-x)^2
     
  8. Jun 23, 2013 #7

    Borek

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    What is x, what is d, what is Ms, what is Mm? You seem to be randomly juggling symbols - earlier you wrote Ms - x, which suggests you were trying to subtract some distance from the Sun mass. List all symbols you use, NAME them, and follow these definitions.
     
  9. Jun 23, 2013 #8
    Mass of Mercury/x^2 = Mass of Sun/(distance from Mercury and Sun - x)^2

    Where x is the distance I'm trying to find
     
  10. Jun 23, 2013 #9

    Borek

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    How many unknowns?
     
  11. Jun 23, 2013 #10
    Just x. I ended up getting 23225331.66m using that proportion
     
  12. Jun 23, 2013 #11
    agentlee: I think your equation in post 8 is correct. You must have made a math error.
    You are working with numbers that are vastly different so make sure you are using your calculator correctly.
     
  13. Jun 23, 2013 #12

    D H

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    Not a math error so much a physics error. agentlee's answer is far too precise.

    There is a math error somewhere, but it's small, about 0.4% relative error.

    agentlee, could you show how you got that value?
     
  14. Jun 23, 2013 #13

    PeterO

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    deleted
     
    Last edited: Jun 23, 2013
  15. Jun 23, 2013 #14

    D H

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    The problem isn't asking for center of mass of the Sun and Mercury, PeterO. It's asking for the point between the Sun and Mercury where the gravitational forces toward those two objects on some test object counterbalance one another. Qualitatively that point has to be much, much closer to Mercury than it is to the Sun. So yes, agentlee's answer does make sense from a qualitative perspective.
     
  16. Jun 23, 2013 #15
    I did the quadratic formula on post 8 and did not get what agen got.
    I hate to see this getting out of control Agents equation should be..
    (Ms-Mm)r^2 - 2*d*Ms*r + (Ms^2)*d = 0
    plug in the numbers and solve, yes?
     
  17. Jun 23, 2013 #16

    D H

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    Neither did I. You should however get an answer that is very, very close to what agentlee got.
     
  18. Jun 23, 2013 #17
    I didn't...Also, there are two answers
     
  19. Jun 23, 2013 #18

    D H

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    That's not right. It looks like you made some errors in clearing the denominators.
     
  20. Jun 23, 2013 #19

    D H

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    No, there's only one. There are two solutions to the quadratic but one of them is not a solution to the problem.
     
  21. Jun 23, 2013 #20
    OOPS, it should be (Ms-Mm)r^2 - 2*d*Ms*r + Ms*d^2 = 0
    When solved you will get a position between Mercury and the Sun and another on the other side of Mercury away from the Sun.
     
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