# 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
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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 $$c=\lim_{s\rightarrow R^-}\arctan\left(\sqrt{\frac{R}{R-s}}\right)=\frac{\pi}{2}.$$