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Attraction of two spheres in deep space

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid led sphere of radius 10m has a mass of about 57 million kg. If two of these spheres are floating in deep space with their centers 20m apart, the gravitational attraction between them is only 540N. How large would this gravitational force be if the distance between the centers was tripled?


    2. Relevant equations
    F=ma
    W=mg
    Gm1m2 / d^2



    3. The attempt at a solution

    This is starting to piss me off.

    (G x 57,000,000^2) / 40^2 ≠ 540. It's not even close!
     
  2. jcsd
  3. Sep 21, 2012 #2

    HallsofIvy

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    Gravitational force is, as your formula shows, inversely proportional to the square of the distance. If the distance between centers is tripled, the force is multiplied by 1/9.

    (where did you get the "40" in the denominator? The given distance between centers is 20 m, not 40. And what did you use for "G".)
     
  4. Sep 21, 2012 #3

    Doc Al

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    Staff: Mentor

    Use ratios. Compare F1 when the distance = d1 to F2 when the distance = d2 = 3d1.
     
  5. Sep 21, 2012 #4
    The distance between the spheres was given as 20 but their radius was given as 10m, which makes the distance between the centers a total of 40m. It's still not even close.

    Did they pull these numbers out of their a-holes or something!?
     
  6. Sep 21, 2012 #5

    Doc Al

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    You are told that their centers are 20m apart.
     
  7. Sep 21, 2012 #6
    Woops, you're right. That makes the number even less plausible... Could you calculate it and see if you get 540N?
    I must be doing something wrong I get a HUGE number.
     
  8. Sep 21, 2012 #7

    Doc Al

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    Use ratios. If the distance doubles, for example, does the force get bigger or smaller? By what factor?

    I will check their numbers out to see if that quoted force makes sense, but you do not need to do that to answer the question.
    Edit: Their numbers work out just fine. For the given masses and distance, the force is about 540 N, just like they say.
     
  9. Sep 21, 2012 #8
    I did use ratios and got 60N. What threw me off was their numbers. They make absolutely no sense.Why would they confuse students by just arbitrarily making up numbers when there's a very specific and easy way to get real numbers!?
     
  10. Sep 21, 2012 #9

    Doc Al

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    Perfectly correct.
    There's nothing wrong with their numbers. You must have made an error somewhere.
     
  11. Sep 21, 2012 #10
    Could you show me how you calculated it, please?

    I'm having the same problem with a bunch of questions.
     
  12. Sep 21, 2012 #11

    Doc Al

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    You have the equation, just plug in the numbers. What did you use for G?
     
  13. Sep 21, 2012 #12
    G= 6.67 x 10^-11

    So what I did was (G(57,000,000)^2) / 20^2
     
  14. Sep 21, 2012 #13

    Doc Al

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    Good.
    Try it again. You can show your steps if you still can't get it to work.
     
  15. Sep 21, 2012 #14
    Actually this time it worked.... I'm apparently not calculator savvy.

    Thank you very much! Greatly appreciate the help!
     
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