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Force of Gravity Question

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A person weighing 800N on Earth travels to another planter with twice the mass and twice the radius of the Earth. The person's weight on this other planet is most nearly...

    earth radius is 6,380,000m
    earth mass is 5,970,000,000,000,000,000,000,000kg

    2. Relevant equations
    What is the person's weight on the other planet?


    3. The attempt at a solution

    I thought it was still 800N because everything was doubled so it would proportionally be the same. It turns out the answer is 400N and I was hoping someone could explain why that is.

    thanks to anyone who tries.
     
  2. jcsd
  3. Oct 29, 2007 #2

    learningphysics

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    suppose m1 is the mass of the person. me is the mass of the earth... re is the radius of the earth

    so Fe = Gm1me/re^2, where Fe is the gravitational force on the earth.

    What is the gravitational force on the person when he's on the other planet?

    Fp = Gm1mp/rp^2

    we know that mp = 2me. rp = 2re

    Fp = Gm1(2me)/(2re)^2

    take all the constants out to the left... so that this has the form:

    Fp = k*Gm1me/re^2
    Fp = k*Fe

    what is k?
     
  4. Oct 29, 2007 #3
    thanks, so you're supposed to answer the question without any numbers first? i understand what you wrote up until the k part. is k 2?
     
  5. Oct 29, 2007 #4

    learningphysics

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    No, k isn't 2. Try to manipulate:

    Fp = Gm1(2me)/(2re)^2

    simplify this... We don't want those 2's like that... we just want a numerical constant out to the left...
     
  6. Oct 29, 2007 #5
    is k the person's mass? Because that is supposed to be constant.
     
  7. Oct 29, 2007 #6

    learningphysics

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    Another way to approach this is: take the ratio of Fp/Fe.

    Fp/Fe = [Gm1(2me)/(2re)^2]/[Gm1me/re^2]

    simplify the right side... try to cancel everything that you can... what is Fp/Fe come out to?
     
  8. Oct 29, 2007 #7

    learningphysics

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    no k isn't the person's mass... m1 is the person's mass...

    simplify:

    Fp = Gm1(2me)/(2re)^2
     
  9. Oct 29, 2007 #8
    can't you cancel everything out except the 2s, then that's just 2/2?
     
  10. Oct 29, 2007 #9

    learningphysics

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    don't forget about the squared part... (2re)^2 etc...
     
  11. Oct 29, 2007 #10

    learningphysics

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    don't take any shortcuts... work through it...
     
  12. Oct 29, 2007 #11
    So I could get rid of the 2s and have that:

    Fp = Gm1(me)/(re)^2

    but then what else could i simplify?
     
  13. Oct 29, 2007 #12

    learningphysics

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    Fp = Gm1(2me)/(2re)^2

    Fp = Gm1(2me)/(4re^2)

    Fp = Gm1me/(2re^2)

    Fp = (1/2)[Gm1me/r^2]

    Fp = (1/2)*Fe
     
  14. Oct 29, 2007 #13
    oh my god! that's so good. thank you so much!
     
  15. Oct 29, 2007 #14

    learningphysics

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    no prob. careful of those squares. :wink:
     
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