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A dumbbell has a mass m on either end of a rod of length 2a. The center of the dumbbell is a distance r from the center of the Earth, and the dumbbell is aligned radially. If r≫a, the difference in the gravitational force exerted on the two masses by the Earth is approximately 4GmMEa/r3. (Note: The difference in force causes a tension in the rod connecting the masses. We refer to this as a tidal force.)
Suppose the rod connecting the two masses m is removed. In this case, the only force between the two masses is their mutual gravitational attraction. In addition, suppose the masses are spheres of radius a and mass m=43πa3ρ that touch each other. (The Greek letter ρ stands for the density of the masses.)

Write an expression for the gravitational force between the masses

Find the distance from the center of the Earth, r, for which the gravitational force found in part A is equal to the tidal force (4GmMEa/r3). This distance is known as the Roche limit.

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  1. Re is distance from earth center

    first checking your paragraph
    force on each = F = G m ME /Re^2
    dF/dr = -G m ME 2 Re / Re^4
    = - 2 G m ME/Re^3

    so if dr = 2a then
    dF = -4 G m ME a/Re^3 agree

    Hey, surely you mean
    mass = ρ(4/3) pi a^3 !!!!!!!!!

    Mutual Force
    = G m^2 /(2a)^2 = G m^2/(4 a^2)

    so
    G m^2/(4 a^2) = 4 G m ME a/Re^3
    m = 16 Me a^3/RE^3 well the units check :)
    ρ(4/3) pi a^3 = 16 Me a^3/Re^3
    ρ(4/3) pi = 16 Me/Re^3
    so
    ρ(4/3) pi Re^3 = 16 Me
    in other words the mass of a sphere of radius Re (the Roche limit) would be 16 times that of earth
    16^(1/3) = 2.5
    so your Re = r = 2.5 the radius of earth :)

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