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Homework Help: Physics: Gravitational Force Circular Motion

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A.) What is the acceleration due to gravity at an altitude of 1500 km above the earth's surface. Neglect the effect of the earth's rotation. The earth's radius is 6378100 m and its mass is 5.98 * 10[tex]^{24}[/tex]kg.

    B.) The moon has a mass of 7.35 * 10[tex]^{22}[/tex] kg and is located 3.84 * 10[tex]^{8}[/tex] m from the Earth (center to center). At some point between the Earth and the Moon, the gravitational force due to the Earth on any object cancels the gravitational force due to the Moon. Where is that position, relative to the center of the Earth?

    C.) The asteroid belt consists of so many particles, we haven't counted them all yet. However, more than half of the mass of the belt is contained in the four largest asteroids. The largest - Ceres - takes 4.6 years on average to circle the Sun. What is the orbital radius of Ceres in meters.

    1 year = 365 days, 1 day=24 hours, 1 hour=3600 seconds

    2. Relevant equations

    G=6.67 * 10[tex]^{-11}[/tex]
    Sun's Mass = 1.99 * 10[tex]^{30}[/tex]

    3. The attempt at a solution

    A.) (GMm)/r[tex]^{2}[/tex]=(mv[tex]^{2}[/tex])/r
    the m's cancel out and you are left with a (which was equal to v[tex]^{2}[/tex]/r) = (Gm)/r[tex]^{2}[/tex]
    so a = [tex]\underline{(6.67*10[tex]^{-11}[/tex])(5.98 * 10[tex]^{24}[/tex] }[/tex] (1500+6378100)

    B.) I really don't understand what this question is asking. I think that it means that there is a place between the moon and the earth that an object has no gravity. Would that mean that
    (GM[tex]_{earth}[/tex]m)/r[tex]^{2}[/tex][tex]_{x}[/tex] = (GM[tex]_{moon}[/tex]m)/r[tex]^{2}[/tex][tex]_{x}[/tex] then everything just cancels out.

    C.) (GMm)/r[tex]^{2}[/tex]=(mv[tex]^{2}[/tex])/r the m's cancel out. Then you replace v^2 with [tex]\underline{4[tex]\pi^{2}[/tex]r}[/tex]/T[tex]^{2}[/tex]. And then you move the r to the other side. So

    r=[tex]\sqrt[3]{GMT^{2}}[/tex]/ [tex]\sqrt[3]{4\pi^{2}}[/tex]

    Thanks for any help.
  2. jcsd
  3. Sep 26, 2009 #2
    Sorry, for my attempt

    A.) a=(GM)/r^2

    C.) r=the cube root of [(GMT^2)/4pi^2)]
  4. Sep 26, 2009 #3
    for C.) v=(2 pi r)/T so v^2 = (4 pi^2 r^2)/T^2
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