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## Homework Statement

Two masses, M and m, are initially (at time t=0) at rest with respect to each other and separated by a distance d. They are affected only by their mutual gravitational attraction. Find the amount of time elapsed beginning at the instant the separation is equal to d/2 and ending at the instant the separation is equal to d/3. (Ignore relativity.)

## Homework Equations

The Vis Viva equation is relevant as a starting point. Solve it for the speed in orbit, and, since the orbit is a plunge orbit, change v into the differential dr/dt, creating an ordinary non-linear differential equation. This ODE is to be solved to give time elapsed from t=0 as a function of r, such that 0 ≤ r ≤ d.

## The Attempt at a Solution

This was a homework problem that I was given as an undergraduate, along with those who were my classmates at the time. We solved it, but it was somewhat more challenging that we'd at first expected. The solution is

t(r) = √{d/[2G(M+m)]} {√(rd−r²) + d Arctan[√(d/r−1)]}

t(d/2) = (1/2 + π/4) √{d³/[2G(M+m)]}

t(d/3) = {√(2)/3 + Arctan[√(2)]} √{d³/[2G(M+m)]}

Δt = {√(2)/3 + Arctan[√(2)] − 1/2 − π/4} √{d³/[2G(M+m)]}

Δt ≈ 0.141322975518 √{d³/[2G(M+m)]}

From t(r), you may notice that when r«d then, effectively, r ≈ 0, and

t ≈ π √[d³/(8GM)]

This approximate solution can be used to estimate the amount of time needed for a cloud of interstellar gas to collapse into a proto-star, where d is the initial radius of the cloud and M is the cloud's total mass.

Check my answer,

t(r) = √{d/[2G(M+m)]} {√(rd−r²) + d Arctan[√(d/r−1)]}