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


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).

  • $\begingroup$ Where is the term undefined? $tan^{-1}(0)=0$ and $tan^{-1}(\pm\infty)=\pm \frac{\pi}{2}$, and it's monotonic in between $\endgroup$ Commented Aug 17, 2012 at 0:37
  • $\begingroup$ are we allowed to treat the term within the $tan^{-1}$ as $∞$ when $s=R$? if we are, i can carry on.. $\endgroup$ Commented Aug 17, 2012 at 0:43
  • 1
    $\begingroup$ Think of it as $\frac{\lim}{s->R}tan^{-1}\sqrt{\frac{s}{R-s}}$, which has the value $\frac{\pi}{2}$ $\endgroup$ Commented Aug 17, 2012 at 0:44
  • $\begingroup$ 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. $\endgroup$ Commented Aug 17, 2012 at 11:15
  • $\begingroup$ Related: physics.stackexchange.com/q/19388/2451 $\endgroup$
    – Qmechanic
    Commented Aug 17, 2012 at 19:53

4 Answers 4


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Johannes
    Commented Oct 30, 2014 at 17:00
  • $\begingroup$ 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). $\endgroup$ Commented Oct 30, 2014 at 18:08
  • $\begingroup$ 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! $\endgroup$
    – Simon S
    Commented Oct 30, 2014 at 18:19
  • $\begingroup$ @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. $\endgroup$ Commented Nov 4, 2014 at 0:30
  • $\begingroup$ Why is $ m = m_1 + m_2 $ ? $\endgroup$
    – Jon
    Commented Oct 16, 2021 at 17:13

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


I'd just like to cover solving this using the energy conservation method. It's funny how using newton's equations will result in an diffeq $y'' = \frac1y$ which is practically unsolvable, but anyways...

If we assume Earth is fixed (if it is not fixed, add in the Earth's kinetic energy as a term on the left), we have by conservation of energy $$-\frac{GM_Em}{x(t)} + \frac12 mv(t)^2 = -\frac{GM_Em}{R_i}$$ Here, $R_i \ge x(t) \ge R_E$ ($R_E$ is the radius of the Earth) and $v(t)$ are functions of time and represent the distance the object currently is at from the centre of the Earth and its velocity, respectively. $M_E$ is the mass of the Earth, and $m$ is the irrelevant mass of the object.

Solving for velocity gives $$ v = \frac{\text d x}{\text dt}= -\sqrt{\frac{2GM_E}{x}-\frac{2GM_E}{R_i}}$$ We have taken the negative square root, because in our case, $x(t)$ is decreasing from $R_i$ to $R_E$ and hence $v(t)$ should be negative (if we ignore this, our time is negative lol)

We can invert the differential expression and solve for the d$t$ differential (which is arguably very dubious in mathematical nature but hey I'll just hand wave and say something something nonstandard analysis something👁️), which yields

\begin{align*} \text dt &= \frac{\text dx}{-\sqrt{\frac{2GM_E}{x}-\frac{2GM_E}{R_i}}}\\ T = \int_{t_0}^{t_f}\text dt &= \int_{R_i}^{R_E}\frac{\text dx}{-\sqrt{\frac{2GM_E}{x}-\frac{2GM_E}{R_i}}} \end{align*}

The integral can be solved in closed form, as follows \begin{align*} T &= \frac1{\sqrt{2GM_E}}\int_{R_E}^{R_i}\frac{\text dx}{\sqrt{\frac{1}{x}-\frac{1}{R_i}}}\\ &= \left.\frac1{\sqrt{2GM_E}} \cdot \left(-R_i^{\frac32}\arctan\left(\sqrt{\frac{R_i}x - 1}\right) - \sqrt{R_ix(R_i-x)}\right)\right|^{R_i}_{R_E}\\ &= {\color{blue}{\frac{R_i^{\frac32}\arctan\left(\sqrt{\frac{R_i}{R_E} - 1}\right) + \sqrt{R_iR_E(R_i-R_E)}}{\sqrt{2GM_E}}}}\tag{1} \end{align*}

If we take the limit as $R_E$ goes to $0$ (as in approximating Earth as a point particle), we can see that this simplifies into \begin{align*} T &= \lim_{R_E \to 0^+} \frac{R_i^{\frac32} {\color{red}{\arctan\left(\sqrt{\frac{R_i}{R_E} - 1}\right)}} + {\color{green}{\sqrt{R_iR_E(R_i-R_E)}}}}{\sqrt{2GM_E}} \\ &= \frac{R_i^{\frac32} \cdot {\color{red}{\frac\pi2}} + {\color{green}{0}}}{\sqrt{2GM_E}}\\ &= {\color{blue}{ \frac{\pi R_i^{\frac32}}{2\sqrt{2GM_E}} }} \tag{2} \end{align*} which is consistent with Simon S's result.

To address Sameh Sayed's concerns, using $(2)$ will of course give a bad answer since it approximates Earth as a point particle and ignores the surface. We have to use $(1)$ instead. If we take the time it takes for an object dropped from $100$ meters as $4.52$ seconds from a kinematics approximation, we can see that the numerator is $1.276\times 10^8 \text m^{\frac32}$ from here, and the whole thing equals $4.519$ seconds from here, which is basically the same as the kinematics approximation.


I think that the equation $T=\frac{\pi}{2} * \sqrt(\frac{R^3}{2Gm})$ is incorrect equation.

If we tested the equation with a height of $h=100m$ away from the surface of earth. It could be good to use the kinematics for this height since it's not a very large one:


$T=\frac{\pi}{2} * \sqrt(\frac{R^3}{2Gm})=\frac{\pi}{2}\times\sqrt(\frac{(6371000+100)^3}{2\times6.67\times10^{-11}\times5.972\times10^{24}})\approx569.75s$ which is so far from the real answer.

Also i think that the real equation is the following: $T=\sqrt(\frac{2hr(r+h)}{Gm})$ which is equal to $4.5154$ which is so close to the kinematics.


The problem of $T=\sqrt(\frac{2h}{g})$ is $g$ and the trick here is to find the average acceleration due to gravity ($g_{avg}$) in the whole trip.

The whole trip is from the height from $h$ to $0$. We will find $g_{avg}$ by the following equation:


The integral $\int^0_h\frac{GM}{(r+h)^2}dh=GM\int^0_h\frac{1}{(r+h)^2}dh=GM\times\int^{r}_{h+r}\frac{1}{(u)^2}du=\frac{GMh}{r(r+h)}$

Then $g_{avg}=\frac{\frac{GMh}{r(r+h)}}{h}=\frac{GM}{r(r+h)}$

Now, we can replace $g$ with $g_avg$ then:

$T=\sqrt(\frac{2h}{g_{avg}})=\sqrt(\frac{2hr(r+h)}{GM})$ where $r$ is the radius of the planet (from the center of the mass by assuming that it's perfectly distributed) and $M$ is the mass of the planet and $h$ is the distance from the surface of the planet and the body (height).


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