This is in-fact a physics question but seems as how I only want to know how to treat it mathematically I thought I would ask it here(Note: this is not a homework question, I'm just studying for an exam that's in a few days):

I want to show that the free fall time ($t_{ff}$) for a body starting at rest from a radius R ,to land on a body of mass M is $t_{ff}^2=\tfrac{\pi^2}{8}(\tfrac{R^3}{GM})$.

Here's what I've done so far :

Using conservation of energy $E=\tfrac{1}{2}mv^2-\tfrac{GMm}{r}$ we know that this is conserved , so the change in kinetic enrgy is equal to the difference in potential energy $\Rightarrow \tfrac{1}{2}mv^2=\tfrac{1}{2}m\tfrac{dr}{dt}=\tfrac{Gmm}{r}-\tfrac{GMm}{R}\Rightarrow \tfrac{dr}{dt}=-\sqrt{2GM(\tfrac{1}{r}-\tfrac{1}{R})}$

So now we have a first order seperable ode

$\Rightarrow -\int^R_0(2GM(\tfrac{1}{r}-\tfrac{1}{R}))^{\tfrac{-1}{2}}dr=\int dt$

Here is my problem:

I can't figure out how to solve the integral on the right I tried substitution but it didn't work, any suggestions anyone ?

  • $\begingroup$ Is your body with mass $M$ a black hole? In that case, you need to account for relativistic effects. Otherwise, what is the radius? $\endgroup$
    – Andrei
    Jan 7, 2019 at 3:14

1 Answer 1


$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$The desired answer clearly is meant for a practice in classical mechanics without relativistic effects.

Presumably the OP wants to see his/her method comes to fruition so this answer continues the approach in the question post. There exists analysis that some might deem more physical and nice (see this answer or that).

First of all, your original expression is off by a negative sign. It is true that $r$ decreases along time $t = 0$ to $t = t_{ff}$ so the derivative is negative (taking the minus square root). Accordingly, the integration limit should start at $r = R$ at $t = 0$ and ends at $r = 0$ at $t = t_{ff}$. Hence the negative sign cancels and we have an integration of positive square root from $r = 0$ to $R$.

$$\int \dd{t} = \frac1{ \sqrt{2GM} }\int^R_0 \left( \frac1r - \frac1R \right)^{\frac{-1}2} \dd{r} = \sqrt{\frac{R}{2GM} }\int^R_0 \left( \frac{R}r - 1 \right)^{\frac{-1}2} \dd{r} $$ Upon seeing "something minus one" in the square root, one should think of $\sec^2\theta$ for the very useful $\sec^2\theta - 1 = \tan^2\theta$.

To have $\frac{R}r = \sec^2\theta$ one considers the change of variable $\theta \equiv \arccos \bigl(\sqrt{ \frac{r}R } \bigr)$, a well-defined one-to-one mapping where the lower integration limit $r = 0$ becomes $\theta = \frac{\pi}2$ and at upper $r = R$ it is $\theta = 0$. \begin{align} \frac{R}r = \sec^2\theta & \implies \frac{-R}{r^2} \dd{r} = 2 \sec\theta \cdot \sec\theta\tan\theta \dd{\theta} \\ &\implies \frac{-1}R \bigl( \frac{R}r \bigr) ^2 \dd{r} = 2 \sec^2\theta\tan\theta \dd{\theta} \\ &\implies \frac{-1}R \sec^4\theta \dd{r} = 2 \sec^2\theta\tan\theta \dd{\theta} \\ &\implies \dd{r} = -2R \cos^2\theta\tan\theta \dd{\theta} \end{align} Meanwhile, the integrand $\sqrt{ \frac{R}r - 1} = \tan\theta$ is the positive root by definition. Putting things together: \begin{align} t_{ff} = \int_{t = 0}^{t_{ff} } \dd{t} &= \sqrt{\frac{R}{2GM} }\int^R_0 \frac1{ \sqrt{ \frac{R}r - 1} } \dd{r} \\ &= \sqrt{\frac{R}{2GM} }\int_{\theta = \frac{\pi}2 }^0 \frac1{\tan\theta} (-2R)\cos^2 \theta \tan\theta \dd{\theta} \\ &= R \sqrt{\frac{R}{2GM} }\int_{\theta = 0}^{ \frac{\pi}2 } 2\cos^2\theta \dd{\theta} \\ &= R \sqrt{\frac{R}{2GM} }\int_{\theta = 0}^{ \frac{\pi}2 } \bigl( \cos(2\theta) + 1 \bigr) \dd{\theta} \\ &= R \sqrt{\frac{R}{2GM} } \frac{ \sin(2\theta) + 2\theta }2\Bigg|_{\theta = 0}^{ \frac{\pi}2 } \\ & = R \sqrt{\frac{R}{2GM} } \frac{ \pi }2 \end{align} Squaring this elapsed-time $t_{ff}$ yields the desired expression of $t_{ff}^2$ seen in the question post.$\endgroup$

  • $\begingroup$ wow ! thank you for your great answer :) $\endgroup$
    – excalibirr
    Jan 8, 2019 at 16:19

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