My friend and I have been wracking our heads with this one for the past 3 hours... We have 2 point masses, $m$ and $M$ in a perfect world separated by radius r. Starting from rest, they both begin to accelerate towards each other. So we have the gravitational force between them as:

$$F_g ~=~ G\frac{Mm}{r^2}$$

How do we find out at what time they will collide? What we're having trouble with is this function being a function of $r$, but I have suspected it as actually a function of $t$ due to the units of $G$ being $N(m/kg)^2$. I've tried taking a number of integrals, which haven't really yielded anything useful. Any guidance? No, this is not an actual homework problem, we're just 2 math/physics/computer people who are very bored at work :)

  • $\begingroup$ FYI, if everything is in SI units with magnitude 1, except G ofc, it would take 1 day 2 hours 42 minutes 47 seconds 198 milliseconds... approximately. $\endgroup$
    – sato
    Commented Jun 1, 2022 at 9:40

5 Answers 5


You should be able to use energy conservation to write down the velocities of the bodies as a function of time.

$$ \textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right) $$


$$ \frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right) $$

Momentum conservation ensures that the magnitude of the momenta of both masses is the same. Does this help?

Substituting into the second equation from the first you should be able to solve for: $$ \int_0^T dt = -\int_{r_0}^0 dr \sqrt{\frac{rr_0}{2G(M+m)(r_0-r)}} = \frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}} $$

  • $\begingroup$ What are $r$ and $r_0$ above ? The OP only mentioned an $r$. $\endgroup$
    – Kantura
    Commented Oct 19, 2019 at 14:33
  • $\begingroup$ And what about p? What is that? $\endgroup$ Commented Feb 26, 2020 at 2:31
  • $\begingroup$ $p$ is the combined momentum of the two bodies. The reduced mass of the system has been utilized. $r_0$ seems to be initial distance of the two point masses. $r$ seems to be a variable, that denotes the distance between the two point masses. $\endgroup$
    – McSuperbX1
    Commented Aug 15, 2020 at 18:07
  • $\begingroup$ Just a matter of signs. But, shouldn't be $ KE = -PE $? $\endgroup$
    – Jon
    Commented Oct 16, 2021 at 17:53

Since this old question got bumped I might as well add my own answer. In classical mechanics, two masses that interact gravitationally define a two-body problem, which follows Kepler's laws: they will orbit each other in an ellipse, and Kepler's Third Law states that $$ n^2a^3 =\mu, $$ where $a$ is the semi-major axis, $\mu=G(M+m)$ and $n=2\pi/T$, with $T$ the orbital period, i.e. the time taken to go from the apocentre (the point of greatest distance) to the pericentre (the point of smallest distance) and back.

Now, the case of two colliding masses is a special case: here, the ellipse becomes a line, so the semi-minor axis $b$ is zero, the apocentre is the initial distance $r_0$ and the pericentre is zero. Also, $r_0=2a$. So the masses only complete half an orbit (from apocentre to pericentre), which takes a time $$ t = T/2 = \frac{\pi a^{3/2}}{\sqrt{\mu}} = \pi\sqrt{\frac{r_0^3}{8G(M+m)}}. $$


For completeness, here's an another solution (though not so elegant as Shanth's solution).

The equations of motion (from the 2nd Newton's Law) of the two point masses $m_1$ and $m_2$, respectively, are:

$$G\frac{m_1 m_2}{(r_2-r_1)^2}=m_1 \ddot{r_1}\Leftrightarrow \ddot{r_1}=G\frac{m_2}{(r_2-r_1)^2}$$

$$-G\frac{m_1 m_2}{(r_2-r_1)^2}=m_2 \ddot{r_2}\Leftrightarrow \ddot{r_2}=-G\frac{m_1}{(r_2-r_1)^2}$$

Subtracting these two equations we get:


And setting $r=r_2-r_1$ we finally get the following second-order non-linear differential equation:


Here (link) you can find how to solve this ODE.

  • $\begingroup$ Yep, this is the solution I used in the other answer that I linked to from my comment on the question. $\endgroup$
    – David Z
    Commented Sep 16, 2011 at 8:40
  • 1
    $\begingroup$ Note that multiplying this by $\frac{\dot r}{r^{2}}$ and integrating with respect to time gives you the energy equation, so it's easy to show that these answers are equivalent. $\endgroup$ Commented Dec 18, 2013 at 18:41

Similar task: A body is thrown straight up with an escape velocity $v_e$ from a height $R$. How long it takes for the body to attain a height $h$?

As well known $\large v=\frac{dr}{dt}$, so $\large dt= \frac{1}{v}dr$

$\large{ v_e^2= \frac{2GM}{r}}$, so $\large \frac{1}{v_e} = \sqrt{\frac{r}{2GM}}$

$\large T=\int\limits_R^h \sqrt{\frac{r}{2GM}}dr=\frac{2}{3}(h^{3/2}-R^{3/2})\frac{1}{\sqrt{2GM}}=\large \frac{2}{3}\Big(\frac{h}{v_{eh}}-\frac{R}{v_{eR}}\Big)$

where $\large{ v_{eh}= \sqrt{\frac{2GM}{h}}}$ and $\large{ v_{eR}= \sqrt{\frac{2GM}{R}}}$ - escape velocities at $h$ and $R$

Escape velocity can be derived from the conservation of energy:

$\large -\frac{GMm}{r} + \frac{mv_e^2}{2}=0$

  • 1
    $\begingroup$ This is a neat idea, and is correct in the combined limits of very high separation and one body much more massive than the other. $\endgroup$ Commented Sep 15, 2011 at 23:43
  • $\begingroup$ @dmckee, it's a hint. The decision is the Shanth's answer. $\endgroup$
    – voix
    Commented Sep 16, 2011 at 0:16

I think the answer is simpler:

$$F = m.a $$ but $$ F= G. \frac{Mm}{r^2} $$ then: $$ ma =G. \frac{Mm}{r^2} $$ then:

$$ a = \frac{GM}{r^2} $$ on the other hand: $$V_f= V_i +at $$ but as the objects start from rest: $$ V_0 =0 => V_f = a.t $$ then: $$t = \frac{V_f}{a} $$ then, clearing the time we obtain: $$ t = \frac{V_f.r^2}{G.M} $$ Now, $$V_f^2 = 2.a.r $$ then: $$ V_f = \sqrt{2r} . \sqrt{\frac{GM}{r}} $$ then: $$ t = \frac{r^2}{GM} . \sqrt{2r}. \sqrt{\frac{GM}{r}}$$ ending: $$ t = \sqrt{\frac{2.r^3}{GM}}$$

  • 9
    $\begingroup$ This doesn't work because the acceleration is time dependent. The velocity formula invoked assumes constant acceleration. It would be true if the time averaged acceleration were used, but computing that is computing the answer. $\endgroup$ Commented Sep 15, 2011 at 22:48
  • 1
    $\begingroup$ Unfortunately, this won't work due to the acceleration increasing as the objects come closer. $\endgroup$
    – MGZero
    Commented Sep 15, 2011 at 23:38

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