# Time taken for object in space to fall to earth

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).

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Where is the term undefined? $tan^{-1}(0)=0$ and $tan^{-1}(\pm\infty)=\pm \frac{\pi}{2}$, and it's monotonic in between – Jerry Schirmer Aug 17 '12 at 0:37
are we allowed to treat the term within the $tan^{-1}$ as $∞$ when $s=R$? if we are, i can carry on.. – Vincent Tjeng Aug 17 '12 at 0:43
Think of it as $\frac{\lim}{s->R}tan^{-1}\sqrt{\frac{s}{R-s}}$, which has the value $\frac{\pi}{2}$ – Jerry Schirmer Aug 17 '12 at 0:44
thank you! i've done the computation, and if I haven't made an error, it seems that the distance from heaven to earth based on the Greek myth is 6.226*10^8 m, or 1.6 times of the mean moon-earth distance. – Vincent Tjeng Aug 17 '12 at 11:15
– Qmechanic Aug 17 '12 at 19:53

You're allowed to treat the argument of $\tan^{-1}$ as $\infty$ at the initial point, provided of course you use the appropriate limit $\tan^{-1}(\infty)=\frac{\pi}{2}$. More formally, change "evaluate the function at $s=R$ to find the constant" to "take the limit $s\rightarrow R$ to find the constant" (which you should do since the function is indeed formally undefined). Then the constant is $$c=\lim_{s\rightarrow R^-}\arctan\left(\sqrt{\frac{R}{R-s}}\right)=\frac{\pi}{2}.$$

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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.

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Nice (+1). So in summary: the free falling time is 1/8th of the orbiting time. Or: orbiting to an antipodal position takes twice as long as falling to the same position. – Johannes Oct 30 '14 at 17:00
Thanks for taking the time to write all that down. It's a great solution. I'd just like to point out that the question you're solving is slightly different from the way I've phrased it; you're working with two point masses, while my question asks for the behavior of a small (point) mass away from a large mass with non-zero radius. Nevertheless, your solution generalizes - the acceleration term will scale by a factor of 2, and the final integration will take different limits (from $R$ to $R_E$, where $R_E$ is the radius of the large object). – Vincent Tjeng Oct 30 '14 at 18:08
Yes, fair point. Using the more general form I've been calculating a few dystopian scenarios, e.g., if the moon suddenly stopped rotating about the earth, how long would it take for it to crash into the earth (70.9 terrifying hours). Longer than I would have guessed! – Simon S Oct 30 '14 at 18:19
@SimonS I bet that would make for a fun examination question. Once again, thank you for taking an interest in my question - and hope you had fun working with it. – Vincent Tjeng Nov 4 '14 at 0:30