An experiment is performed in deep space with two uniform spheres, one with mass

Question

An experiment is performed in deep space with two uniform spheres, one with mass 21.0 kg and the other with mass 108.0 kg . They have equal radii, r = 0.25 m . The spheres are released from rest with their centers a distance 36.0 mapart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres.

a. When their centers are a distance 29.0 m apart, find the speed of the 21.0 kg sphere.

b. Find the speed of the sphere with mass 108.0 kg .

c.

Find the magnitude of the relative velocity with which one sphere is approaching to the other.

d.

How far from the initial position of the center of the 21.0 kg sphere do the surfaces of the two spheres collide?

Explanation / Answer

a) conservation of energy would be a good start.
The force on each is the same at any point in time. F = ma, so the acceleration of each sphere is in inverse proportion to the mass. What's the total energy change in the system?
U = -GMm/r
Ui = -G*21*108/36 = -4.2 x 10^-9 J

Uf = -G*21*108/29 = -5.22 x 10^-9 J
dU = Uf - Ui = (-5.22 - (-4.2)) x 10^-9J
dU = -1.02 x 10^-9 J

The spheres have lost GPE, so they must have gained KE, and again in inverse proportion to their masses.
KE(21) = (108/129)KE(tot)
KE(108) = (21/129)KE(tot)

KE(21) = 8.5 x10^-10 J
KE(108) = 1.66 x10^-10J

a) speed you can just plug into the KE equation.
KE = 0.5 mv^2
v^2 = 2 KE/m
v = sqrt(2KE/m)
v = sqrt(2*8.5x10^-10/21) = 9*10^-6 m/s

b) v = sqrt(2KE/m)
v = sqrt(2*1.66x10^-10/108) = 1.75*10^-6 m/s

c) the relative velocity = 9*10^-6 + 1.75*10^-6 = 10.75*10^-6 m/s

d), by the distance formula (x = (1/2) a t^2), we see the distance is proportional to the acceleration. Since each sphere has a constant acceleration ratio, they have a constant total distance traveled ratio. Therefore of the total distance traveled by the pair, the 21 kg mass will have done (108/129) of it, and the other will have done (21/129).

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