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Kepler's Law

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    (a) Determine the forces that the moon and the sun exert on a mass, m, of water on Earth. Your answer will be in terms of m with units of N. (Use data from table 8-1 in your textbook for this question. The mass of the moon is 7.3 x 10 ^22 kg, and it can be assumed to be 3.9 x 10^5 km from the Earth's center.)

    ______ m N (force exerted by moon)

    ______ m N (force exerted by sun)

    2. Relevant equations

    Fgr = G M1m2 / r^2

    mass of earth = 5.98 x 10^24

    3. The attempt at a solution

    I thought that I had everything for the equation, but I still can't seem to get it. Check out my work and see if there's something I'm missing. I'm not sure on what to use for the radius.

    Fgr = (6.67e-11)(7.3 x 10^22)(5.98 x 10^24) / (3.9 x 10^5)^2
  2. jcsd
  3. Dec 16, 2009 #2
    Say the earth is not there but replaced by a bit of water of mass "m". Then?
  4. Dec 16, 2009 #3
    Thanks for the response Bright Wang.

    So I replaced my equation with m, and this is what I got

    Fgr = (6.67e-11)(7.3 x 10^22)(m) / (3.9 x 10^5)^2


    Fgr = (4.860e12)(m) / (3.9 x 10^5)^2


    (3.9 x 10^5)^2 = 4.869e12m

    does that look right?
  5. Dec 16, 2009 #4
    The given data for radius is in KM.
  6. Dec 16, 2009 #5
    Alright I multiplied the radius side by 1000 to make up for the KM units. But one more question. Is that number supposed to be divided by 2 since I want radius, and isn't that diameter? Anyways, i got .03 for an answer, does that seem too small?
  7. Dec 16, 2009 #6


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    Homework Helper

    Which number? There's no diameters in this question.
  8. Dec 16, 2009 #7
    i need r^2 for the Force of gravity equation. The statement said that the moon is 3.9 x 10^5 km from the earth. Is that not diameter? or is that the radius? Regardless, i still got .03 for the answer, which doesnt seem to make sense to me
  9. Dec 16, 2009 #8
    Actually, im not sure what i got was .03. But here's my equation anyway.

    (3.9 x 10^5)^2 = 1.521e11 x 1000 (for km) = 1.521e14

    (6.67e-11)(7.3 x 10 ^22)(m) = 4.8691e12(m)


    4.8691e12(m) / 1.521e14 = m????
  10. Dec 16, 2009 #9
    Distance is radius. It doesn't even matter, r represents the distance between the objects.

    4.8691e12(m) / 1.521e14 = m???? "

    What? Your just using the equation.

    You want to find Fg. Fg = GMm/r^2.... Why do you have mass on both sides? You trying to find force not acceleration.

    Its just plug and chug. I didn't get 0.3m N.
  11. Dec 16, 2009 #10
    I thought you said to use "m" for a variable ???

    "Say the earth is not there but replaced by a bit of water of mass "m". Then? "

    or do i just ignore that number
  12. Dec 16, 2009 #11
    m can be whatever.

    Fg = GMm/r^2 = G*(Mass of the moon)*m / distance{in meters}^2 = {some number}*m
  13. Dec 16, 2009 #12
    wasnt able to get it in time. Thanks for the help though. Hopefully my teacher will explain it better
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