2 pt. masses @rest in space. 10m apart. Gravity is only acting force. Each weighs m=1kg. when do they meet

This one is tough. Under Newtonian gravity, the force of attraction between 2 masses m1 and m2 is given as F = m1*m2*G / r^2 where r is the distance between them. So for 2 equal point masses of m, the force of attraction would be:

F = 2*m*G / r^2

This force would act on each mass and drive them together. The tough part is that the force is time-dependent through the quantity r.

If we just look at one of the masses, we can write:

F = 2*m*G / r^2 = m*x''

where x'' is the acceleration of one of the masses. The quantity r can be rewritten as (R - 2*x) where R is the initial distance between the masses and x is the position of one of the masses. Therefore we have:

2*m*G / (R - 2*x)^2 = m*x''

x'' = 2*G / (R - 2*x)^2

So now, we have a differential equation for the position of one of the masses, x(t), as it relates to the acceleration x''(t), with the initial conditions x'(t) = 0 (starting from rest) and x(0) = 0. So we would essentially have to solve this system for x(t) and then solve for the time it takes for x to reach half the original distance between the masses (5 meters in this case).

I am not experienced with ODEs, but this is how you would solve it. Hope that helps some.

PS: Do not listen to Mike's answer. It is absolutely wrong. This cannot be solved like a kinematics problem, because the acceleration is NOT constant. Please refer to my explanation above for how to get started correctly.

** I just did the integration and solved for and explicit form of x(t). Too many steps to show you here, but if you integrate the equation above twice and apply the 2 initial conditions, you will get:
x(t) = R/2 - (1/2)*(48*G*t^2 + R^4)^(1/4)

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So if we are looking for the time the masses meet, we would put in 5 meters for x, 10 meters for R, and the gravitational constant 6.67300*10^(-11) m3 kg-1 s-2 for G), and solve for t. If you do this, you will get