The Problem
For a small mass a distance $R_i$ away from the center of the Earth, how long would it take for the object to fall to the surface of the Earth, assuming that the only force acting upon the object is the Earth's gravitational force?
Relevant Information
The following discussion seems to have solved exactly the same problem: http://www.physicsforums.com/showthread.php?t=555644
However, upon working out the mathematics, I'm not exactly sure how to evaluate the constant of integration.
A Partial Solution
$$ F=\frac {-GMm}{s^2} $$
$$ a=\frac {-GM}{s^2} $$
$$ \frac {dv}{dt} = \frac {-GM}{s^2} $$
Multiplying by $v$ and then integrating by $dt$ on both sides, we have
$$\frac {1}{2} v^2=\frac {GM}{s} +c_1$$
where $c_1$ is a constant of integration. Substituting initial conditions of $v=0, s=R$, we have
$$\frac {1}{2} v^2=GM(\frac {1}{s}-\frac{1}{R})$$
At this point of time, when I use Wolfram Alpha, I get
$$c_2+\sqrt{\frac{2}{R}}t=\frac{\sqrt{s}(s-R)+R\sqrt{R-s}\times{\tan^{-1}(\sqrt{\frac{s}{R-s}})}}{\sqrt{GM(R-s)}}$$
where $c_2$ is a constant of integration. Substituting initial conditions of s=R, t=0, we find that the term $$\tan^{-1}(\sqrt{\frac{s}{R-s}})$$ is undefined. At this point, I'm stuck. Any ideas on where I've made the mistake here?
(For those interested, this question was inspired by the Greek myth which states that a bronze hammer dropped from heaven would take 9 days to hit the Earth and would reach on the tenth).
Best Answer
For the record, here's a worked solution:
If $r$ is the distance between the two point masses $m_1$ and $m_2$--which start at rest--then as both accelerate towards each other, where
$$ \frac{d^2r}{dt^2} = -\frac{Gm}{r^2} \ \ \ , \ \ \ \ \ m = m_1 + m_2$$
The first step in solving this equation is the least obvious: Multiply both sides by $\displaystyle \frac{dr}{dt}$ and integrate from time $0 \rightarrow t$. Writing $v = dr/dt$, the limits in velocity are $0 \rightarrow v = v(t)$, we have on the left hand side
$\displaystyle \int_0^t \frac{d^2r}{dt^2} \frac{dr}{dt} dt \ = \ \int_0^t \frac{dv}{dt} . v \ dt \ = \ \int_0^t v . \frac{dv}{dt}\ dt \ = \ \int_0^v v \ dv \ = \ \frac{1}{2}v^2 \ = \ \frac{1}{2} \left( \frac{dr}{dt} \right)^2$.
On the right hand side, writing $R$ for the starting distance, we integrate $R \rightarrow r = r(t)$,
$\displaystyle \int_0^t - \frac{Gm}{r^2} \frac{dr}{dt} dt \ = \ -Gm \int_R^r \frac{dr}{r^2} \ = \ Gm \left( \frac{1}{r} - \frac{1}{R} \right)$.
Putting these two expressions together and choosing the negative square root as $v = dr/dt$ is negative, growing in magnitude, we have
$$ \frac{dr}{dt} \ = \ - \sqrt{2Gm} \sqrt{ \frac{1}{r} - \frac{1}{R} }.$$
We are now in familiar territory as this equation is separable. Integrate once more, with limits $t = 0 \rightarrow T$ and $r = R \rightarrow 0$, and we arrive at the collision time of
$$ T = \frac{\pi}{2} \sqrt{\frac{R^3}{2Gm}}.$$
The last step uses the integral
$$ \int \frac{\sqrt{r}}{\sqrt{R-r}} dr \ = \ -\sqrt{r}\sqrt{R-r} + R \arcsin \sqrt{\frac{r}{R}} \ \ \ \ (\ +C\ ) \ .$$
This integral is sometimes calculated and written with $\arctan$, but the form here with $\arcsin$ is slightly easier for our purposes as it avoids dealing with the limit discussed above.
When I arrived at the expression for collision time $T$, I was suspicious. I wrote a simple numerical simulation, and yep--it holds up.
I can't say I have ever seen this formula for collision time anywhere. If I ever have the privilege of teaching ODEs again, this would make a great problem.