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Gravitational Force-Need Help Fast

  1. Nov 7, 2007 #1
    Gravitational Force---Need Help Fast!!!

    1. The problem statement, all variables and given/known data

    You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8*10^7 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 12.2 m/s^2.

    What is the mass of the planet? in kg

    What is the mass of the star? in kg

    Equations:
    Mass(of planet)= g*R^2/G

    3. The attempt at a solution

    Ok, so I calculated the mass of the planet to be 1.48*10^25 kg using the above equation. M=[(12.2m/s^2)*(9*10^6m)^2]/6.67*10^-11 Nm^2/kg^2.

    Now I am stuck with how to find the mass of the star and which equation to use to get it..

    Thank you
     
  2. jcsd
  3. Nov 7, 2007 #2
    Anybody???
     
  4. Nov 7, 2007 #3

    Astronuc

    User Avatar

    Staff: Mentor

  5. Nov 7, 2007 #4
    It states that the planet orbits 2.2*10^11 m from its star. This would them be the distance, correct??
     
  6. Nov 7, 2007 #5
    So, then I use T^2=4pi^2/GM somehow??? I guess I'm not exactly sure how...
     
  7. Nov 7, 2007 #6
    oops, I forget r^3 in the equation..
    so:

    T^2=(4pi^2/GM)r^3
     
  8. Nov 7, 2007 #7
    T should = 34732800 sec
    r should be 2.2*10^11 m

    correct??
     
  9. Nov 7, 2007 #8
    Using that I calculated the mass of the star to be 1.39 * 10^64...
    but that is wrong...
    so where did I go wrong??
     
  10. Nov 7, 2007 #9
    I calculated it again and this time I got 1.44*10^37, but that is also wrong...

    What am I doing wrong?
     
  11. Nov 7, 2007 #10
    If I rearrange the equation to find the mass of the star this is what I get:

    M=[(2pir^3/2)^2]/GT

    Is this right??? It is what I have been using, but I can't come up with the correct answer.
     
  12. Nov 8, 2007 #11

    Astronuc

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    Staff: Mentor

    Yes, that is a or r in Kepler's forumula relating period with distance for one mass orbiting another mass, e.g. moon around a planet or planet about a star.

    In the simplest form, one may assume that the mass of the star greatly exceeds the mass of the planet.

    So [tex]T^2\,=\,\frac{4{\pi^2}{a^3}}{GM}[/tex], which can be rearranged to get

    [tex]M\,=\,\frac{4{\pi^2}{a^3}}{GT^2}[/tex]

    So substitute in the appropriate numbers

    G = 6.67 x 10-11 N m2/kg2, the universal gravitational constant,

    a = 2.2 x 1011 m

    T = 402 d * 24 h/d * 3600 s/h = 3.47328 x 107 s

    Now compare the mass of the star with the mass of the planet. The mass calculated for the star might need adjusting for the mass of the planet since M is the sum of the masses, but if M >> m(planet), the M is approximately the mass of the star.

    This is not correct. Be careful about moving terms and exponents. T should be squared and if one brings r^3 inside the parentheses and squares those terms, then one must use r^(3/2) within the parentheses.
     
    Last edited: Nov 8, 2007
  13. Nov 8, 2007 #12
    Ok... I see.

    Thank you very much!!
     
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