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Gravitation question

  1. Nov 23, 2004 #1
    A uniform solid sphere with radius R produces a gravitational acceleration a(g) on its surface. At what two distances from the center of the sphere is the gravitational acceleration a(g)/3?

    I know that gravitational acceleration = GM/r^2, and that on the surface of the sphere, a(g) = (4 pi G rho /3)R. Beyond that...I'm kinda stumped. (I managed to find an explanation of this somewhere, but it didn't really help. I get that you can replace (4 pi G rho/3) with a constant, but...)
  2. jcsd
  3. Nov 23, 2004 #2


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    You should realize or be able to figure out that

    [tex]g(r) = g_0 \left(\frac {R}{r} \right)^2[/tex]

    when r > R and

    [tex]g(r) = g_0 \frac {r}{R}[/tex]

    when r < R.
  4. Nov 23, 2004 #3


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    One of the position is gonna be outside the sphere and the other is gonna be inside.

    For outside: One of the distance is going to be outside the sphere because the gravitationnal force, and therefor the gravitationnal field (=F/m) decreases continuously with the distance. So there must be a distance [itex]r_1 >R[/itex] somewhere where the field is g/3.

    We know [itex]MG/R^2 = g[/itex], and we want to find [itex]r_1[/itex] such that [itex]MG/r_{1}^2 = g/3 = MG/3R^2[/itex]. Solve for [itex]r_1[/itex].

    For inside: You have to know that a shell of uniform matter density produces no net gravitationnal field inside of it. With that in mind, you can regard a point a distance [itex]r_2[/itex] inside a uniform sphere as being inside a shell of thickness [itex]R-r_2[/itex] and at the surface of a sphere of radius [itex]r_2[/itex]. Therefor, only the matter of the sphere exerts a net gravitationnal force at [itex]r_2[/itex]. You must also know that a sphere of uniform density produces the exact same gravitationnal field at every distance at its surface (and beyond) as a point particle located at its center would. Work out a formula for the mass of the sphere. How does it relate to M? Solved for [itex]r_2[/itex] just like for outside.
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