Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :)

share|cite|improve this question

closed as off-topic by ACuriousMind, MAFIA36790, Gert, Asher, CuriousOne Jun 17 at 4:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, MAFIA36790, Gert, Asher, CuriousOne
If this question can be reworded to fit the rules in the help center, please edit the question.

This is partially addressed by another answer I posted, but now that I look at it there seem to be some errors in that answer. I don't have time to check it through now but I'll come back to this later. – David Z Sep 15 '11 at 18:33
Would it be sufficient if I said "express Newton's law as a differential equation in r and t, and solve it" or is the problem that you haven't seen the math yet? Second do-you-have-the-math question: are you familiar with the canonical transformation to the reduced mass form? – dmckee Sep 15 '11 at 18:41
@dmckee 1) No, I have not been able to figure the math out yet. 2) I'm not familiar with it (my advanced math only goes up to Calc 3). I'll look into it, though! – MGZero Sep 15 '11 at 23:39
@David, your old answer appears to be correct through "where I've chosen the negative square root." Wolfram Alpha gives me a different looking answer for the integral, though... – sevenofdiamonds Sep 16 '11 at 0:19
@sevenofdiamonds: thanks for the tip, I found the same thing working with Mathematica. I want to see if I can figure out why I originally wrote what I did before I edit it though. – David Z Sep 16 '11 at 0:40
up vote 17 down vote accepted

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: $$ \int_0^T dt = -\int_{r_0}^0 dr \sqrt{\frac{rr_0}{2G(M+m)(r_0-r)}} = \frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}} $$

share|cite|improve this answer

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.

share|cite|improve this answer
Yep, this is the solution I used in the other answer that I linked to from my comment on the question. – David Z Sep 16 '11 at 8:40
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. – Jerry Schirmer Dec 18 '13 at 18:41

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 $$ t = T/2 = \frac{\pi a^{3/2}}{\sqrt{\mu}} = \pi\sqrt{\frac{r_0^3}{8G(M+m)}}. $$

share|cite|improve this answer

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:

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

share|cite|improve this answer
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. – dmckee Sep 15 '11 at 23:43
@dmckee, it's a hint. The decision is the Shanth's answer. – voix Sep 16 '11 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: $$ t = \sqrt{\frac{2.r^3}{GM}}$$

share|cite|improve this answer
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. – sevenofdiamonds Sep 15 '11 at 22:48
Unfortunately, this won't work due to the acceleration increasing as the objects come closer. – MGZero Sep 15 '11 at 23:38

protected by Qmechanic Dec 19 '13 at 9:37

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.