Question

Two identical hard spheres, each of mass m and radius r, are released from rest in otherwise empty space with their centers separated by the distance R. They are allowed to collide under the influence of their gravitational attraction. (a) Find the magnitude of the impulse received by each sphere before they make contact. (b) Find the magnitude of the impulse each receives during their contact if they collide elastically.

Answer 1

Answer:

a)   I = √ (2G m³ (1/2r³ - 1/R)), b) I = √ (8 G m³ (1/2r -1/R))

Explanation:

.a) The relation of the Impulse and the moment is

                    I = Δp = m $v_{f}$ - m v₀

We can use Newton's second law with force the force of universal attraction

                  F = ma

                  G m m / r² = m a

                  dv / dt = G m / r²

Suppose re the direction where the spheres move is x

                 dv/dx  dx/dt = G m / x²

                  dv/dx  v = G m / x²

                  v dv = G m dx / x²

We integrate

                   v² / 2 = Gm (-1 / x)

We evaluate this integra from the lower limit v = 0 for x = R to the upper limit, where the spheres v = v and x = 2r are touched

                  v² / 2-0  = G M (-1 / R + 1 / 2r)

                  v = √ [2Gm (1 /2r - 1/ R) ]

The impulse on the sphere is

                 I = m vf - m v₀

                 I = m vf - 0

                 I = m √ (2Gm (1 / 2r-1 / R)

                 I = √ (2G m³ (1/2r³ - 1/R))

b) during the crash each sphere arrives with a velocity v and leaves with a velocity –v, the same magnitude but opposite direction

                      I = m $v_{f}$- m v₀

                      I = m v - m (-v)

                      I = 2mv

                      I = 2m √ (2Gm (1 / 2r-1 / R)

                      I = √ (8 G m³ (1/2r -1/R))