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Call them $m_1,m_2$. They are compressed to their center of masses, if you wish. If the initial distance at $t=0$ is $d$, is there a formula or an efficient way to calculate the distance between them at time $t$, provided, of course, that they haven't collided?

The problem is the circularity of the speed. It's not mere acceleration that pushes the masses together. If the masses approach each other with acceleration, the force will increase faster than it does, and then the masses are pushed into each other with more than just acceleration. This also means that it is not mere change in acceleration, because you could argue similarly. It must be something more "exponential" not polynomial.

But my knowledge of calculus doesn't help me here.

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    $\begingroup$ "the masses are pushed into each other with more than just acceleration" There is an issue with terminology here; accelerations don't "push", only forces do. Acceleration is the result of a push. The force causes the acceleration. So, your starting point should simply be to figure out the force function, from which you can easily (with Newton's 2nd law) extract the acceleration function. And with an acceleration function, you just need some integration to reach the distance function. $\endgroup$
    – Steeven
    Commented Nov 7, 2018 at 9:49
  • $\begingroup$ At $t=0$ are the bodies at rest relative to one another? $\endgroup$
    – PM 2Ring
    Commented Nov 7, 2018 at 13:22

1 Answer 1

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The force of attraction between $m_1$ and $m_2$ is given by

$$F = G\frac{m_1m_2}{r^2},$$ and so the acceleration of $m_1$ with respect to $m_2$ is given by

$$a = a_1 + a_2 = G\frac{m_1 + m_2}{r^2}.$$

Thus our problem becomes to solve the initial value problem

$$\frac{d^2r}{dt^2} = G\frac{m_1 + m_2}{r^2}, \frac{dr}{dt}(t = 0) = v_1(t = 0) + v_2(t = 0), r(t = 0) = d.$$

So if you can solve this IVP, you're done.

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