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.
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 :)
To write an expression for the gravitational force between the masses, we can use Newton's law of universal gravitation. The formula for gravitational force between two masses is given by:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
In this case, both masses are spheres of radius a and mass m = (4/3)πa^3ρ. Since they touch each other, their centers are at a distance of 2a from each other.
Applying the formula, the gravitational force between the masses is:
F = G * [(4/3)πa^3ρ] * [(4/3)πa^3ρ] / (2a)^2
F = G * (16/9) * π^2 * (a^3 * ρ^2) / 4a^2
F = G * (4/9) * π^2 * a * ρ^2
Now, to find the distance from the center of the Earth, r, at which this gravitational force is equal to the tidal force (4GmMEa/r^3), we can set the two expressions equal to each other:
4GmMEa/r^3 = G * (4/9) * π^2 * a * ρ^2
Canceling out common terms and simplifying, we get:
mME/r^3 = (1/9) * π^2 * ρ^2
mME = (1/9) * π^2 * ρ^2 * r^3
Now we can solve for r by isolating it on one side:
r^3 = (9 * mME) / (π^2 * ρ^2)
Taking the cube root of both sides:
r = (9 * mME / (π^2 * ρ^2))^(1/3)
This expression gives the distance from the center of the Earth, r, for which the gravitational force is equal to the tidal force. This distance is known as the Roche limit.