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Planets & Period

  1. Oct 27, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Question: A Gak it a type of alien that lives on a planet in another galaxy. One day a Gak decides to find out a little more about his planet. He drops a ball (it starts at rest) with a mass of 6.18 kg and notes that it takes 0.928 s to fall a distance of 8.37 m.
    The Gak’s planet orbits its sun in a roughly circular orbit. The average distance to the Gak’s sun is 6.85 × 10^8 km. The Gak measures the force of attraction between his planet and his Sun and finds that this is 2.89 × 10^20 N.

    How long is a year on the Gak’s planet?


    My Attempt:

    So I started with the formula T^2 = 4(pi^2)(R^3)/GM

    R = 6.85*10^8 (from avg distance)

    Since on a previous question I found the mass of Gak's planet to be M = 2.12*10^24 (and got it right). Subbing it in & all relevant variables;

    So; T^2 = 4(pi^2)(6.85*10^11)^3/(6.67*10^-11)(2.12*10^24)

    T = 2.9956*10^11
    Then multiply this by 1/(60*60*24*365) for Earth Years, I get;

    T = 9499 yrs

    And this is apparently wrong.....


    Can someone tell me what I did wrong? I'm starting to suspect that R didn't include the radius of the planets and I might need to add it in... I've attached the quiz for reference.
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2016 #2

    PeroK

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    Are you sure you are using all the right variables in the formula for Kepler's law?
     
  4. Oct 27, 2016 #3
    I believe so from my understanding.
     
  5. Oct 27, 2016 #4

    PeroK

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    Why did you use ##M = ## mass of the planet?
     
  6. Oct 27, 2016 #5
    Isn't the "M" variable the mass of the planet?
     
  7. Oct 27, 2016 #6

    PeroK

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    In Kepler's 3rd law, no. A planet's orbit is independent of its mass.
     
  8. Oct 27, 2016 #7
    Oh it's the mass of the sun...
     
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