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Attractive force between Venus and Sun

  1. Oct 29, 2012 #1
    I just need someone to double check my answer and confirm they get the same answer as I do. Thanks.
    Question: Find the attractive force between Vneus and the sun. The distance between their centers is 1.08x10^11.

    Equation: F = Gm1m2/d^2
    G= 6.67x10^-11 N.m^2/kg^2
    m1 (mass of sun)= (1.98x10^30kg)
    m2 (mass of venus) = (4.83x10^24kg)
    d= (1.08x10^11)

    = (6.67x10^-11 N.m^2/kg^2)(1.98x10^30kg)(4.83x10^24kg)/(1.08x10^11)^2
    =5.4687 x 10^22

    Please let me know if this is the correct answer asap. Thanks!
     
  2. jcsd
  3. Oct 29, 2012 #2

    Borek

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    Staff: Mentor

    Your answer has no units.
     
  4. Oct 29, 2012 #3

    PeterO

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    Homework Helper

    You have just omitted the next digits rather than round off to the 5 figures you have chosen to specify.

    besides, your answer should only be to 3 figures!!

    And as per the previous post - you need some units.
     
  5. Oct 29, 2012 #4
    Thanks folks, appreciate it!
     
  6. Oct 29, 2012 #5

    mfb

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    2016 Award

    Staff: Mentor

    WolframAlpha gets a slightly different result: 5.54*10^22 N

    It uses a mass of venus which is larger by 1% and a mass of sun which is larger by .5%.
     
  7. Oct 29, 2012 #6

    PeterO

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    So you should have been saying 5.47 (correctly rounded) not 5.46 (merely truncated)
     
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