# How to find the free fall time of a particle starting at rest at a distance R from a mass M.

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 ?

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

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