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Mass of Jupiter using its moon Sinope

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Using a modified version of Kepler's third law and data about Sinope calculate the mass of Jupiter.

    Keplers Third Law: p^2=a^3
    Newton's Version: p^2=a^3 / M + M
    p^2 = a^3/M
    M=a^3/p^2
    Sinope's period of orbit = 2.075 years
    Average orbital distance of Sinope is 0.158 AU

    2. Relevant equations

    Keplers Third Law: p^2=a^3
    Newton's Version: p^2=a^3 / M + M
    p^2 = a^3/M
    M=a^3/p^2

    3. The attempt at a solution

    I need Jupiter in solar mass units and then in kg by multiplying it by the mass of the sun (2.00*10^30).

    Here is what I have so far but its not making sense.
    p=2.075
    p^2=4.305625
    a^3=4.305625
    a=1.62684209

    if m=a^3/p^2 and p^2=a^3 I am just going to come up with the same answer of 2.075 which doesn't seem to help me at all. I think the problem is worded bad and I am confused as to where the orbital distance of Sinope comes into play.

    Any tips on how to proceed would be appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 20, 2011 #2
    Let us use F = ma for the moon of Jupiter.

    What is F?
     
  4. Nov 20, 2011 #3
    Keplers Third Law: p^2=a^3

    This is only valid for orbits around the sun (or a star with the same mass),
    so you can't use this.

    You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)
     
  5. Nov 20, 2011 #4
    Can we use:

    F = ma ( for the moon of Jupiter)

    [itex]\frac{GMm}{R^{2}}[/itex] = mR[itex]\omega[/itex][itex]^{2}[/itex]

    Gm = [itex]\frac{R^{3}4\pi^{2}}{T^{2}}[/itex]

    T[itex]^{2}[/itex] = [itex]\frac{4\pi^{2}R^{3}}{GM}[/itex]
     
  6. Nov 20, 2011 #5
    The mass of the Juipter moon Sinope is 8*10^16 but I looked that up, didn't figure it from what was given.
     
  7. Nov 20, 2011 #6
    You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)

    so would it be...
    M= .158/2.075
    M= .076144578313253 AU
    .076144578313253(2*10^30) = 1.52289157 × 10^29kg

    That's off by over 37% so that doesn't seem right.
     
    Last edited: Nov 20, 2011
  8. Nov 20, 2011 #7
    It's about a hundred times too heavy. You used M = a/p, and not M = a^3/p^2
     
  9. Nov 20, 2011 #8

    sophiecentaur

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    Gold Member

    Why would you need to know the mass of Sinope?
    Which is the relevant mass?
     
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