To all amateur physicists - a simple physics problem

Since Eta said he'd come with a solution to this problem I thought I'd bump it.In doing so I'll also propose a new attempt at a solution.
I'll start off from the realisation that the problem could be reduced to the form:
a = G*M/r^2.This can be rewritten:r^2*dv = G*M*dtWhich can be rewritten as:(A) dt = (1/GM)*r^2*dvNow we know (of course) that:dr/dt = vWhich can be rewritten as:(B) v*dt = dr.Now inserting (A) into (B) we gain:(1/(G*M))*r^2*v*dv = drWhich can be rewritten as:(1/(G*M))*v*dv = dr/r^2Integrating from v = 0 to v = u on the left hand side and from r = A to r = R (R < A) on the right hand side we gain:(1/(2*G*M))*u^2 = 1/R - 1/AWhich gives us "u":u = sqrt(2*G*M)*sqrt(1/R - 1/A) = dR/dtWhich yields:dt = dR/(sqrt(2*G*M)*sqrt(1/R - 1/A) = K*dR*R/sqrt(1 - R/A){Where K = 1/sqrt(2*G*M)}What follows is a rather tedious calculation where I find the antiderivative of the function on the right hand side.
In this I perform a substitution of variables such that R = p^2 (don't ask why ):I gain:- sqrt(1 - p^2/A)*(A*p^2 + A^2*p + A^3)Which is the primitive function expressed in p.Integrating (C) from t = 0 to t= t and from p = P to p = 0 we now get:t = 2*K*[- sqrt( 1 - P^2/A)*(A*P^2 + A^2*P + A^3) + A^3]Where P^2 = A, so t = 2*K*A^3t = 2*A^3/sqrt(2*G*M);I think this is right... but then again I thought my other solution was correct as well, so I remain skeptic.

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

I re-checked my calculations and realised my final integration was in error. I think you missed a square root in your answer though.
I'll take my solution from the top.Solution:If we are allowed to assume that the collision will occur at the centre of gravity for the two point masses, then we can define two vectors, r1 and r2, such that m1*r1 + m2*r2 = 0.The vectors are therefore the vectors leading from respective point to the centre of gravity.Also defining r = r1 - r2 we can write the equation of motion for one of the particles as:m1*r1'' = Gm1m2/r^2 (where ' denotes the derivative with respect to time).Using that r2 = -(m1/m2)r1, and that [b]r = r1 - r2 we can eliminate r1 from the equation of motion.We get:r'' = G(m1 + m2)/r^2, where I'll let M = m1 + m2 from now on.This can be rewritten as:(1) dv = dt*G*M/r^2But we also know that:(2) dt = dr/v (since v = dr/dt)We can eliminate dt by insering (2) into (1).
We get:dv = (G*M/v)*dr/r^2Which can be rewritten:dv*v = (G*M)*dr/r^2We intregrate once from v = 0 to v = u on the left hand side and from r = 2a to r = R on the right hand side.(1/2)*u^2 = (G*M)*(1/R - 1/2a)Which gives us:u = dR/dt = sqrt((2*G*M)*(1/R - 1/2a))This gives usdt = dR/sqrt((2*G*M)*(1/R - 1/2a)).Integrating this from t=0 to t=t' on the left hand side and from R = 2a to R = 0 on the right hand side we get:sqrt(2*G*M)*t(R) = (2a)^(3/2)(ArcSin(1) - ArcSin(-1))Which gets us:t(R) = pi*a^(3/2)/sqrt(G*M)There... That has to be right!
Oh, I used a table of indefinate integrals to solve the final integral. I hope that's not considered cheating.

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