How long does it take for two masses to collide with each other in space, if all other masses were deleted?

Question

F=gmm(r^-2)
F=ma
-substitution-
a=Gm(r^-2)
The problem is as the to objects get closer, this equation says there acceleration increase, thus making the time quicker then assumed constant acceleration. I have worked on the problem for weeks and come up dry with summantions after summantions. Need some serious help.

I recently talked with my physics teacher and calculus teacher and we all agree with Dr. Spock's approach, but one issue we keep discussing is the equation of time in relation of orbit involves the imput of the masses. We calculated some pratical calculutions of real models and kept inferring and observing that the masses do not effect the time it takes for the mass to meet. If the masses are revelent then we will continue this rout, but if they are not then this equation is invalid for this problem.

Answer 1

I presume that the objects start from rest. Otherwise, in general you're left with an integral that doesn't really have a particularly clean answer.

There are two ways to do this problem:

1. The FIRST way is to do it in the usual, grunge way that PHYSICISTS with absolutely no insight into the problem would do it. That can take quite some time!

That's probably not only the way that you're expected to do it, but also basically the way that you've been taught, perhaps. It's easiest if either: (i) one body is so massive it can be treated as FIXED, or (ii) both masses are equal. (You didn't clarify this issue.)

In any case, whether it's (i), (ii), or the remaining case (iii), that of general, unequal masses, here is a quick run-down.

(a) At first sight it looks like you have to integrate the equation(s) of motion in order to obtain v(r) as a fuction of r, where r is the separation and v(r) is the relative rate at which the distance between the two bodies is changing (it's NEGATIVE, of course). However, THIS step can be shortened since the Conservation of Energy applies, so the sum of Kinetic Energy/Energies and the gravitational Potential Energy is constant (thank goodness)! (I hope you know the form for the Potential Energy.) Using that, you can AVOID this/these first integration(s).

(b) Then, reminding yourself that v(r) is the NEGATIVE root of [v(r)]^2, you formally extract that root (or those separate roots for the two different masses - ugh!) from what you already have, and recognizing that v(r) = dr/dt, you invert that result to obtain ' t ' as an integral of [1/v(r)] dr between the initial separation and r = 0 (with appropriate sign). The integral is solvable with an appropriate substitution involving a sine function, if memory serves me correctly.

This will probably involve you in hours of work unless you're absolutely on top of your physics and calculus. You'll regret not taking a class in astronomy.

2. The SECOND, much better way : You may be glad to know that the whole thing can be solved in a few lines and a few minutes, when done the way it should be done, as an ASTRONOMER would do it:

All particles in ANY attractive inverse-square force law (gravity, electricity for oppositely charged particles, etc. ) follow paths obeying "Kepler's Laws" or their equivalent. (In this case it IS Kepler's Laws, as explained by Newton.) One of those is that the Period of a complete orbit, P, satisfies:

P^2 = 4 pi^2 D^3 / [G (m1 + m2)],

where ' D ' is the semi-major axis of the relative orbit (their separation, if the orbit is circular), and m1, m2 are the two masses.

THIS WORKS no matter WHAT the shape of the (necessarily elliptical orbit) is, even if it becomes skinnier and skinnier, provided all the while that it has the same ' D ' ! ALL elliptical orbits with the SAME sum of masses and the SAME semi-major axis have the SAME Period.

So: how can you view an object falling in from rest (or two objects falling in from rest) along a straight line as NEVERTHELESS pursuing (part of) a true "Keplerian orbit" of some kind?

THAT is the key to doing this problem VERY quickly.

[ HINTS : (1.) Think "Degenerate ellipse !" and (2.) Ask yourself "In that case, what's the semi-major axis for THAT ?! " ]

Good luck with this second approach.

Live long and prosper.

P.S. To help you check that you understand this method, it would take the Earth a fraction 1/2^(5/2) of a year, or just over 64 1/2 days to fall into the Sun, if it started at rest from its present distance. See if you can see why. (This entire solution takes two lines, looked at the right way!)