# Radial fall in a Newtonian gravitational field [duplicate]

This question already has an answer here:

Suppose an object of mass $m$ starts at rest at a radial distance $r_0$ from a perfectly spherical mass $M$ (where $m << M$), $r_0 > R =$ radius of $M$.

Can we analytically determine when $m$ will hit the surface of the $M$?

In other words, can we analytically solve this initial value problem:

$$\frac{d^2r}{dt^2} ~=~ - \frac{GM}{r^2} ,$$ $$\dot{r}(0) ~=~ 0 ,$$ $$r(0) ~=~ r_0?$$

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See e.g. Wikipedia. –  Qmechanic Jan 11 '12 at 20:57
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## marked as duplicate by Qmechanic♦May 7 at 19:49

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## 1 Answer

I believe that is covered by this answer I posted some time ago to a related (but not quite the same) question. Adapting it to your notation,

$$t = \frac{1}{\sqrt{2G(m + M)}}\biggl(\sqrt{r_0 R(r_0 - R)} + r_0^{3/2}\cos^{-1}\sqrt{\frac{R}{r_0}}\biggr)$$

The same formula is given in the Wikipedia article Qmechanic mentioned in a comment.

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