To all amateur physicists - a simple physics problem

Let's see if anyone likes to solve simple physics problems.We all know (I hope) that the standard Newtonian formula for the gravitational attraction of two (point like) masses is:F = G M*m/r^2 where G is the Gravitational Constant - M is the mass ofone object, m is the mass of the other and r is their separation.So far so good.Now imagine two point masses as above, separated by an initial distance 2r (it helps the algebra to make it 2r) and that they are initially at rest.They are then 'let go' and allowed to move together by gravity alone. I could couch this in some fancy language about assuming the space is maximally symmetric blah blah blah - but the initial conditions are as stated above - it doesn't need GR or such.Now the question is:How long does it take them to collide?Yep - that simple a question. Seems easy doesn't it? After all it's so basic it must be in any basic mechanics text - except it isn't in those books - or at least any I remember. In fact it is curiously absent - I wonder why?It really isn't that difficult but it is subtle enough to make you think a tad.Hint:There are two ways of solving this I can think of. One uses a very fundamental theorem that dates back 300 years or so - the other is a clever way of recasting the problem that actually dates back 400 years or so.(I don't know really what forum this belongs in - I thought Big Bang and Cosmology might be best just because most of the physics types post there but use your judgement.)

I specifically made them point masses so you can neglect anything related to finite source size effects.Actually where they collide is trivial. It's at the barycentre. But the time to do so is not quite as trivial - hence the question.The reason I asked for the collision time and not some other time to fall a given distance is that it simplifies the result a lot. The integral is pretty nasty then but for the time to collide at the barycentre the integral is trivial.Hint #2:Think Kepler

You cannot integrate twice like that anyway. The collision does occur at the barycentre but that is obvious since the centre of mass of this closed system cannot move because there are no external forces acting.You were somewhat on the right track initially though. That is what I called the first method of solving this. The second method is the 'cooler' one so to speak.Hint#3the final answer has Pi in it.

Restating it as an orbital problem is the 2nd method I referred to.

Imagine the two bodies are in orbit around each other. From this you can set up the equations of motion.Then just imagine the bodies are stopped in their orbits suddenly. Then they will fall inwards a la the described problem.Think about it: The equations of motion are describing the orbits in polar coordinates. By stopping the bodies and letting them fall you are fixing the angular coordinate thus theta(dot) is set to zero.Thus you can use the Keplerian equation of motion to solve the given problem.I'll post the solution in a day or so if no one else does.

Sorry.

I will post the correct solution in by the weekend.I will tell you the answer though.t = (Pi * r^3/2)/(G(M+m))Remember their initial separation was 2r. This gets rid of the numerical constants in the final answer - except the Pi of course.This message has been edited by Eta_Carinae, 07-28-2004 02:27 PM

You're right - I left the square root of the denominator. I did realise this I was just sloppy when I posted yesterday.You have the correct answer.There is a quicker way of doing it actually which I'll post sometime.And no - it is not cheating to use a table. I do it all the time - I have forgotten most of them it's much easier to use tables.This message has been edited by Eta_Carinae, 07-29-2004 04:47 PMThis message has been edited by Eta_Carinae, 07-29-2004 04:48 PM