What is the time to fall in a plunge orbit?

Two bodies having a total mass M (i.e., M=M₁+M₂) are initially at rest, separated by a distance d, in vacuum, and isolated from all forces except their mutual gravitational attraction. Find the time elapsed from the initial time, t₀, to the instant, t₁, at which the separation is r₁, such that 0<r₁<d.

Update:

Extra credit.

With reference to the original question, how fast are the two bodies approaching each other when r=r₁ ?How much time elapses beginning with the instant that r=d/2 and ending with the instant that r=d/3 ?What fraction of the initial separation, d, is closed at the instant when half of the total time to fall has elapsed?

2 Answers

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  • 7 months ago
    Favourite answer

    The time to fall (t) from a height "d" to a height "r" is:

    p = r/d

    mu = G (M1 + M2)

    t = { arccos(sqrt(p)) + sqrt(p(1-p)) } d^(1.5) / sqrt (2 mu)

  • Clive
    Lv 7
    7 months ago

    Do your own homework.  You learn better that way.

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