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Can I assume this?

  1. Jul 16, 2008 #1
    If the question reads:
    Space explorers land on a planet that has the same mass as Earth, but they find they weigh twice as much as they would on Earth
    What is the radius of the planet?


    Do I assume the 'g" = 9.8(2) ?
    Since they weight twice as much...does that mean the gravity is doubled?

    Im going to use the eqn. F(ma)=GMm/r^2
     
  2. jcsd
  3. Jul 16, 2008 #2

    Janus

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    Yes, you can.
     
  4. Jul 16, 2008 #3
    Can anyone plz check the math?
    I get 4.50e6 which sounds right meters....
    But masteringphysics is telling me im wrong
    I did:

    (6.67e-11)(5.98e24)/19.62=r^2

    19.62 came from 9.8(2) since they wieght twice as much...
    Thankyou for your help.
     
  5. Jul 16, 2008 #4

    Kurdt

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    Seems ok to me.
     
  6. Jul 16, 2008 #5

    tiny-tim

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    Hi Lance! :smile:
    No … you needn't assume anything about g …

    this will work for any two planets of the same mass where the weight differs by a factor of 2.
    ok … but you'll have to write it twice, won't you, with an r1 and an r2? :smile:
     
  7. Jul 16, 2008 #6

    arildno

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    Lance:
    It just gets messy with writing digits!
    Instead, use better symbols like this:
    [tex]m_{e},r_{e},w_{e}, m[/tex]
    which means mass of earth, radius of Earth, weight on Earth and mass of explorer, respectively..
    These quantities are related by the following equation:
    [tex]w_{e}=\frac{Gm_{e}m}{r_{e}^{2}}(*)[/tex]
    On new planet "p", we also have the equation:
    [tex]w_{p}=\frac{Gm_{p}m}{r_{p}^{2}}(**)[/tex]
    You are given the following information:
    [tex]m_{p}=m_{e},w_{p}=2w_{e}[/tex]
    Inserting these into (**), we get:
    [tex]2w_{e}=\frac{Gm_{e}m}}{r_{p}^{2}}(***)[/tex]
    Now, perform the division (***)/(*), and we get:
    [tex]\frac{2w_{e}}{w_{e}}=\frac{\frac{Gm_{e}m}{r_{p}^{2}}}{\frac{Gm_{e}m}{r_{e}^{2}}}[/tex]
    which simplifies to:
    [tex]2=(\frac{r_{e}}{r_{p}})^{2}[/tex]
    Now, you can solve this equation for the planet radius in terms of the Earth radius, only THEN introduce digits!
     
  8. Jul 16, 2008 #7
    I get 4.51e6 which is still wrong.....I hate online homeowrk...:(
     
  9. Jul 17, 2008 #8

    alphysicist

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    I think that answer is correct, if the units are supposed to be meters. Are you supposed to use a different unit?
     
  10. Jul 17, 2008 #9
    the units they want just say R_p_=(answer) R_e_
     
  11. Jul 17, 2008 #10

    Kurdt

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    Ahh right that explains it. What do you multiply the Earth's radius by to get the planets radius?
     
  12. Jul 17, 2008 #11
    So my (answer) times 2?
     
  13. Jul 17, 2008 #12

    Kurdt

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    No, the radius of the planet is definitely not twice the Earth's radius. You can work it out since you have both quantities. Just rearrange the equation in post # 9 or follow arildno's post for hints.
     
  14. Jul 17, 2008 #13
    .707(R_e_) Got it.
    Thankyou!
     
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