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Gravitational Force

  1. Aug 30, 2009 #1
    1. What is the gravitational force exerted by the Earth on the same 1.7 kg rock resting on the surface of the Moon?



    2. Relevant equations
    F= GMeM2/r^2
    G = 6.67*10^-11 Nm^2/kg^2
    Me = 5.98*10^24 kg
    re= 6400 km

    3. Attempt at a solution:
    6.67 *10^-11 Nm^2/kg^2 * 5.98 *10^24 kg * 1.7 kg/(6400 *10^3 m)^2 =1.655 * 10^27 m <---this is what I get and I know this number isn't right. Could someone please explain to me what I am doing wrong b/c I do not understand this problem
     
  2. jcsd
  3. Aug 30, 2009 #2

    ideasrule

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    The "r" in Newton's gravity formula is the distance from one mass to the other (in this case, from the Earth's center to the rock). It's not the radius of the Earth.
     
  4. Aug 30, 2009 #3

    Pengwuino

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    The radius of the Earth is 6400km. However, this is NOT the radius you are looking for. The radius, r, is the distance between the center of the Earth and the object in question. In this case, you're looking for the distance to the moon.
     
  5. Aug 30, 2009 #4
    Ok, If I'm looking for the distance to the moon. How would I go about solving for that?
     
  6. Aug 30, 2009 #5

    ideasrule

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    Do you have to solve for that? Are you sure it's not given?

    If not, you know that the Moon orbits once every 29.5 days. There's a formula for centripetal acceleration, and using that you can calculate the distance from the Moon to the Earth. You probably haven't learned this yet, though, so I think the Earth-Moon distance should be given somewhere in a table of constants at the front/back of your book.
     
  7. Aug 30, 2009 #6
    Ok in the back of my text book it gives the mean distance of the moon from the earth = 3.845 * 10^8 m and the mean radius of the moon is = 1.737 * 10^6 m
     
  8. Aug 30, 2009 #7

    Pengwuino

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    Use the mean distance to the moon. They didnt explicitly tell you to calculate it. I honestly wouldn't bother with including the radius of the moon since the problem wasn't specific as to where the object is on hte surface (ie, is it on the side of the moon pointing towards the Earth making it as close to the Earth as possible, or is it on the backside making it as far as possible to the Earth). Of course, the ratio of the radius of the moon to the distance to the moon is ~1/300 so it's not really significant.
     
  9. Aug 30, 2009 #8
    How would you set up the equation to solve for that?
     
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