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Find the Mass of Mars

  1. Sep 15, 2007 #1
    The Martian satellite Phobos travels in an approx circular orbit with r=9.4*10^6 meters and period 7h 39min. Find the mass of mars.

    I am supposed to use the concept of gravitational F=centripital force[tex]=m\frac{v^2}{r}[/tex] and the fact that [tex]v=\frac{2\pi r}{T}[/tex]

    so this is my attempt:

    [tex]F_g=\frac{GMm}{r^2}=m(\frac{v^2}{r})[/tex]

    implies [tex]\frac{GMm}{r^2}=\frac{m4\pi^2r}{T^2}[/tex]

    implies [tex]M=\frac{4\pi^2r}{GT^2}[/tex]

    Should this not work? or am I just putting these numbers into my calculator wrong. I have T in seconds=275405

    Text says it is 6.5*10^23 kg

    Casey

    ps I am more concerned with learning the method here.
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2

    Kurdt

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    You reasoning is fine just two small mistakes. Look at the radius in your final equation again and ask if it should be to that power, and calculate your time in seconds again.
     
  4. Sep 15, 2007 #3

    cristo

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    You should have a factor r^3 on the RHS instead of r

     
  5. Sep 15, 2007 #4
    should it be r^3 ....I am messing up my algebra here....wow...
    T=7h 39 min= (7*3600)+(39*60)=27540...dont know where that last 5 came from! Thanks

    Casey
     
    Last edited: Sep 15, 2007
  6. Sep 15, 2007 #5

    Kurdt

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    Yes. Go to the previous step and ask yourself how you move the r2 term from the left to the right.
     
  7. Sep 15, 2007 #6

    cristo

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    Sorry for butting in-- I was rather late! Glad you've got it though, Casey :smile:
     
  8. Sep 15, 2007 #7
    Yeah, I'm a stooge. I treated the equals sign in this[tex]\frac{GMm}{r^2}=\frac{m4\pi^2r}{T^2}[/tex]
    like it was a multiplication sign!

    I love making up my own rules!

    Casey
     
  9. Sep 15, 2007 #8
    Feel free to butt in anytime:wink:
     
  10. Sep 15, 2007 #9

    Kurdt

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    No problem. We've all made that mistake and we all will again unfortunately. :smile:
     
  11. Sep 16, 2007 #10

    dynamicsolo

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    If you solve your expression for T^2, you'll see that you've derived Kepler's Third Law for Mars, so you have that a check on your algebra. (I'm assuming you've seen Kepler's Law of Celestial Motion in your course by this point.)
     
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