Time for two point masses to collide under gravity

1. Aug 23, 2011

KingNothing

As a thought experiment, I have attempted to calculate the time it takes two point masses to collide solely due to the force of gravity. The final answer I got was this:

$$t=\frac{R^\frac{5}{2}2\sqrt{2}}{5\sqrt{G}\sqrt{m_1+m_2}}$$

Where G is the gravitational constant, R is the distance (at the beginning) and the m's are the two masses. It is assumed they begin from rest. (relative to eachother).

Is this correct? If not, I can post my steps when I'm done with work. How sad is it that I've spent my lunch break doing this?

2. Aug 23, 2011

elfmotat

$$r''(t)r(t)^2 + GM = 0$$

I have no idea how I would go about solving that, so I'm assuming that's not what you did.

EDIT: Your equation doesn't seem to satisfy the diff eq I posted.

Last edited: Aug 23, 2011
3. Aug 23, 2011

bp_psy

The units seem off too ${\frac {{m}^{5/2}}{\sqrt {{\frac {{m}^{3}}{{\it kg}\,{s}^{2}}}}}}{ \frac {1}{\sqrt {{\it kg}}}}.$

4. Aug 23, 2011

K^2

Hint, Kepler's Law holds even for extremely elliptic orbits.

5. Aug 23, 2011

elfmotat

Yes, this simplifies to $ms$.

6. Aug 23, 2011

rcgldr

Last edited: Aug 23, 2011
7. Aug 23, 2011

elfmotat

8. Aug 23, 2011

K^2

Answer's right, but why bother with all that algebra?

$$T^2 = \frac{4\pi^2 a^3}{G(m_1+m_2)}$$

$$\tau = \frac{T}{2} = \frac{\pi a^{3/2}}{\sqrt{G(m_1+m_2)}} = \frac{\pi r^{3/2}}{2\sqrt{2G(m_1+m_2)}}$$

9. Aug 23, 2011

K^2

a is the semi-major axis, which is half of the longest separation between two bodies. Since in the question above they start at distance r away from each other, a=r/2.

10. Aug 23, 2011

KingNothing

Are you saying my original answer is correct, or are you referring to someone else here? Sorry :)

And if my answer is correct, I prefer it over anything with pi in it.

11. Aug 23, 2011

K^2

Your answer is completely wrong, including the units. I'm referring to result that rcgldr linked to. It gives correct derivation of the fall time via integration of the forces. I merely show a much simpler way to arrive at the result by taking Kepler's Laws for granted.

12. Aug 23, 2011

rcgldr

I think the point was to derive the formula. The OP could have asked to derive the formula for Kepler's law.

13. Aug 23, 2011

K^2

From practical perspective, you need to solve the problem by simplest means using the sort of things you should just know. I'd count Kepler's Laws as "should just know" material, meaning they can be used gratis for solving problems.

From perspective of complete knowledge, you are absolutely right, but that still leaves two options. Either do it the hard way, or derive Kepler's Laws first. I'd take the second route, since T²~a³ law can be derived from conservation of angular momentum and energy without having to deal with the differential equations.

And at very least, use it to verify your hard work. There is no need to run numeric integration to check your solution when there is such a simple analytic tool available. I'd also say, "Look what happened to OP," but he didn't even check the units, so it's a moot point.

14. Aug 23, 2011

KingNothing

Please, explain what happened to me. Sorry I got the units wrong - no need to get snappy about it.

15. Aug 23, 2011

K^2

I'm not being snappy. It's just a trivial check. See if units of your answer make sense. Yours are meters*seconds. Once you see that, there is no need to ask if the formula is right or wrong. It can't possibly be right if the answer isn't in seconds. The fact that all of the coefficients are also wrong is irrelevant at that point.

If you don't want to use Kepler's Law for computation, at least use it to check yourself. But like I said, if you don't even bother to check units, then that's a moot point.

16. Aug 24, 2011

KingNothing

It's not that I "don't even bother to check units", I just happened to not do so in this case. This wasn't an academic exercise where I'm obligated to be extremely thorough.

It was me playing around and brushing up on math during my lunch break over a chicken sandwich. Thank you for the physics/math help, but I would politely ask that you consider how your extra comments might be received. I don't want you to think (or worse, imply) that I'm somehow dumb or lazy for not figuring that out myself.

I'll check the math and see where I went wrong when I have time later this week.

Last edited: Aug 24, 2011
17. Aug 24, 2011

rcgldr

True, but I already have numeric integration methods setup in spreadsheets, so it wasn't difficult and doing this was more like a hobby (like riding a bicycle instead of a motorcycle), instead of a task I needed to accomplish where the time factor was important. Plus I thought Arildno's integration trick was a nice way to solve what seemed to be a difficult integral.

18. Aug 25, 2011

KingNothing

I want to do this with integration. Where did I go wrong in setting up my integral?

$$F=G\frac{m_1m_2}{r^2}=m_1a_1=m_2a_2$$
$$a_1=G\frac{m_2}{r^2}, a_2=G\frac{m_1}{r^2}$$
$$a_{total}=G\frac{m_1+m_2}{r^2}$$
$$d=\frac{1}{2}at^2, d=r, a=\frac{2r}{t^2}$$
$$\frac{2r}{t^2}=G\frac{m_1+m_2}{r^2}$$
$$t=\sqrt{\frac{2r^3}{G(m_1+m_2)}}$$
$$T=\int^R_0{\sqrt{\frac{2r^3}{G(m_1+m_2)}}}\,dr$$

I've always thought of integrals as summing up a bunch of (infinitely) small pieces. So in this case I am trying to sum up all the little amounts of time from r=0 to r=R. I know it's wrong, but where did I go wrong?

19. Aug 25, 2011

bp_psy

From what I know this formula works only for constant acceleration which is not the case in this problem. You should look at the derivations in the threads posted above they are very nice.

20. Aug 25, 2011

K^2

KingNothing, read some introductory text on differential equations. Or if possible, take a course.

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