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Mass of Planet

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Io, a satellite of jupiter has an orbital period of 1.77 days and an orbital
    radius of 4.22 x 10^5 km. From this Data determine the mass of Jupiter


    2. Relevant equations
    4pie^2/GM
    Kepler's Third Law

    3. The attempt at a solution
    I keep getting turned around. I know the answer but I
    keep finding different ways to start
    I also used T^2 = Ka^3
    But that seems independ of Mass?
    Thanks,
    Kevin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2008 #2
    A generic equation for orbitting bodies that you may want to commit to memory or have handy for some quick calculations is:

    [tex]T=2\pi a\sqrt{a/GM}[/tex].
     
  4. Nov 9, 2008 #3
    That was nowhere to be found in my book. Thanks...
    Should I next find acceleration by taking 1.77 days
    and making it 152928s and then find the circumference
    to be 2.65 x 10^9 and then velocity is 17338 m/s?
    Is this the right direction to go.
    Kevin
     
  5. Nov 9, 2008 #4
    The a given in that equation is the radius. I remember it from long ago, but, if it's not found in your book, don't bother with it. We can derive it from scratch, can't we? ;)

    You want to find the mass of the planet, given by the equation [tex]\vec{F}=\vec{G}Mm/\vec{R}^2[/tex]. You know that this system follows uniform circular motion, too: [tex]\vec{a}=\vec{v}^2/\vec{r}[/tex]. And, that period is time, which can be related with displacement and velocity: [tex]\vec{x}/\vec{v}=T[/tex]. Do you agree?
     
  6. Nov 9, 2008 #5
    I agree
     
  7. Nov 9, 2008 #6
    Then you can solve for M. :) Let me know what you try.
     
  8. Nov 9, 2008 #7
    Except how can I use your first equation when I don't know the mass of Io?
     
  9. Nov 9, 2008 #8
    Because [tex]\vec{f}=m\vec{a}[/tex] so the small masses cancel. ;)
     
  10. Nov 9, 2008 #9
    So a = G(M/r^2)
     
  11. Nov 9, 2008 #10
    Yup, which is also equal to the quotient between the square of velocity and radius. Just relate the equations, you'll end up with M.
     
  12. Nov 9, 2008 #11
    So I get v^2 = G(M/r)
    I'm not sure how to get M =
     
  13. Nov 9, 2008 #12
    [tex]\vec{v}=2\pi\vec{r}/T[/tex]...
     
  14. Nov 9, 2008 #13
    I'm just not seeing it sorry
     
  15. Nov 9, 2008 #14
    [tex]\frac{4\pi^2\vec{r}^2}{T^2}=\frac{\vec{G}M}{\vec{r}}\rightarrow M=\frac{4\pi^2\vec{r}^3}{\vec{G}}[/tex].
     
  16. Nov 10, 2008 #15
    Thanks alot for the help but using that equation I don't get the right answer for some reason.
    Kevin
     
  17. Nov 10, 2008 #16
    You forgot the T^2 in the formula.
    And never mind the vectors. G is not a vector and r^3 is the magnitude cubed and not the vector cubed.

    M=4Pi^2 r^3/(G T^2)

    I've got about 1.9 x 10^27 and it's very close to the accepted mass of Jupiter.
    The period should be in seconds, right?
     
  18. Nov 10, 2008 #17
    asleight's posts are so full of errors, you're better off ignoring them.
     
  19. Nov 10, 2008 #18
    I figured that out. Thanks again. You were a great help.
    Talk to you soon.
    Kevin
     
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