Two masses under mutual attraction

[shakgoku]

1.The question
Two point masses $$m_{1}$$ and $$m_{2}$$ are initially at rest. Distance between them is 'd'. How much time does it take for them to collide due to mutual gravitational attraction?2. Relavant Equations:
$$F = \frac{Gm_{1}m_{2}}{r^2} \ \mu = \frac{m_{1}m_{2}}{m_{1}+m_{2}}$$Attempt at a solution:
3. force is varying with time. My first guess is to try to solve the differential equation
$$\mu \frac{d^2x}{dt^2} = F(x)$$
and tried to solve it for $$x(t)$$
But , was not successful so far. There should be a better approach..

 

[tiny-tim]

hi shakgoku!

that should work …

show us how far you got ​

 

[shakgoku]

hi shakgoku!

that should work …

show us how far you got ​

Hello tiny tim! This is how I proceeded,

$$\mu \frac{d^2x}{dx^2} = - \frac{Gm_{1}m_{2}}{x^2} \\$$
.

.$$-\mu \frac{x^2}{Gm_{1}m_{2}} = \frac{d^2t}{dx^2} \\$$

by guess general solution will be of the form,

$$t = Cx^4 +C_{1}$$.

.$$\frac{d^2t}{dx^2} = 12Cx^2 = -\mu \frac{x^2}{Gm_{1}m_{2}}$$.

.$$C = \frac{-\mu}{12Gm_{1}m_{2}} \\$$

And at t = 0, x = d,

$$C_{1} = \frac{\mu d^4}{12Gm_{1}m_{2}}$$

..$$t = \frac{\mu (d^4-x^4)}{12Gm_{1}m_{2}}$$

When the masses collide, d = 0,
using it I got,
$$t = \frac{d^4}{12G(m_{1}+m_{2})}\\*$$

I doubt if my 2nd step is wrong.

 

[tiny-tim]

eugh! :yuck:

you've used d²t/dx² = 1/(d²x/dt²),

but that's not true …

d²t/dx² = d(dt/dx)/dx = 1/(dx/d(dt/dx)), and that's about as far as you can go

instead, use d²x/dt² = dv/dt = dv/dx dx/dt = v dv/dx ​

 

[shakgoku]

Thanks. I had doubts but
I just got carried away because its easy to solve after making that wrong step :D

I finally made the substitution you suggested and got an integral of form

$$A \int_{0}^{d} \sqrt{\frac{x}{d-x}} dx =t$$

And I solved it to get

$$t = \frac{\pi d}{2} \frac{1}{\sqrt{2G(m_{1}+m_{2})d}}$$

Which is the answer. I throughly liked it because, $$\pi$$ got involved even though point masses separated by a distance, Newton's gravitational law do not contain $$\pi$$
:)

 

[tiny-tim]

π gets everywhere!

 

[shakgoku]

I get the same answer by considering,

$$F = -kx$$

and using

$$k = \frac{dF}{dx}$$

$$==> k = \frac{2Gm_{1}m_{2}}{x^3}$$

and

$$T = {2 \pi} \sqrt { \frac{\mu}{k}}$$

At

$$\frac{T}{4}$$

Collison occurs, so $$t = \frac{T}{4} = {\frac{\pi d}{2}}{\frac{1}{\sqrt{2G(m_{1}+m_{2})d}}}$$ Which is same as the above answer. How does this approach work even though k is not a constant and this is not a simple harmonic motion?
Can I use this approach for any time varying force?

 

[tiny-tim]

i think it's just a coincidence