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Kepler's Law and Non-terrestrial Orbits (Not Earth)

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A satellite orbits a planet at a distance of 6.80 multiplied by 10^8 m. Assume that this distance is between the centers of the planet and the satellite and that the mass of the planet is 3.08 multiplied by 10^24 kg. Find the period for the moon's motion around the earth. Express the answers in earth days

    2. Relevant equations
    Kepler's 3rd Law
    Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

    3. The attempt at a solution

    I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

    7779136.049 = Period.

    I'm incredibly sure this is a correct value, I've done multiple calculations and always got within 0.1% of this value

    I saw this question in the archive, but I don't believe anyone realized the answer was in Earth days or they just did not show their work (which is what I was hoping to see).

    Please help.

    Found my problem, needed to divide by days represented in seconds (Period / (60 * 60 * 24));

    Answer was 90 days. (decimal dropped out of laziness, forgive me significant digits.)
     
    Last edited: Feb 9, 2009
  2. jcsd
  3. Feb 9, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    HOW did you get it "in days of this planet" when nothing is said about the planet's rotation?
     
  4. Feb 9, 2009 #3

    D H

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    How did you go from Kepler's third law to 7779136.049? What are the units for this number?

    In other words, show your work. It's a bit difficult to find where your mistake is if you don't show us how you arrived at your result.
     
  5. Jan 26, 2012 #4
    How did you get the radius of planet X if the question on,y mentions the satellite's orbital distance?
     
  6. Jan 27, 2012 #5

    gneill

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    Staff: Mentor

    That's not the planet's radius, it's the satellite's orbital radius.
     
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