Question

With the initial conditions for a two mass system m1 and m2, where neither mass is significantly larger than the other, the resulting DE is unsolvable, right? Assume that they both start at rest:
(see DE below)
Do things get simpler if I give a starting velocity to both masses? What if I look at position relative to the com of the system?

Answer 1

The DE for the two masses in 2D space in general:\[m_1\frac{d^2\vec r_1}{dt^2}=G\frac{m_1 m_2 (\vec r_2-\vec r_1)}{|\vec r_2-\vec r_1|^3}\](switch necessary subscripts for m2)

Making the assumption that they're not moving at first, we know movement will be linear. Make the linear axis "x":
\[m_1\frac{d^2x_1}{dt^2}=G\frac{m_1 m_2 (x_2-x_1)}{|x_2-x_1|^3}=\pm G\frac{m_1m_2}{(x_2-x_1)^2}\]So we get a system of DEs:
\[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{(x_2-x_1)^2}\]\[m_2\frac{d^2x_1}{dt^2}=\mp G\frac{m_1m_2}{(x_2-x_1)^2}\]Still unsolvable to my knowledge...

Answer 2

EDIT: last equation should be\[m_2\frac{d^2x_2}{dt^2}=\mp G\frac{m_1m_2}{(x_2-x_1)^2}\]

Answer 3

consider one of the mass to be at rest and another is accelerating.

Answer 4

You can't do that though! m_2 is accelerating too!

Answer 5

well if you want a dynamic system then ... it's going to be a system of second order DE.

Answer 6

Yeah... But that's the problem. The DE seems to be unsolvable. My question is asking if there's some clever symmetry (like the com or something) that we can use to get around the DE.

Answer 7

not sure ... first order system is fairly complicated.

Answer 8

plus ... this is non linear :(( looks like there must be some other approach.

Answer 9

Yeah but its not a question from a textbook... I just made this up. There might not be an approach :(.

Answer 10

let me try to ask mathematica. do you want general solution or ... do you have some boundary condition?

Answer 11

didn't get any solution from mathematica.

Answer 12

I think I have something that might help simplify a bit (N's 3rd law):
\[m_2\frac{d^2x_2}{dt^2}=-m_1\frac{d^2x_1}{dt^2}\]\[\int_0^tm_2\frac{d^2x_2}{dt^2}dt=\int_0^t-m_1\frac{d^2x_1}{dt^2}dt\]\[m_2(\frac{dx_2}{dt}-v_{i2})=-m_1(\frac{dx_1}{dt}-v_{i1})\]\[v_{i1}, v_{i2}=0 m/s\]\[\int_0^tm_2\frac{dx_2}{dt}dt=\int_0^t-m_1\frac{dx_1}{dt}dt\]\[m_2(x_2(t)-x_{i2})=-m_1(x_1(t)-x_{i1})\]\[x_2(t)=\frac{-m_1(x_1(t)-x_{i1})}{m_2}+x_{i2}\]That seems right. If it is, then we don't have a system of DE's anymore...

Answer 13

I think we get:
\[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{(-(m_1x_1/m_2)-x_1+c_1)^2}\]\[c_1=\frac{m_1x_{i1}}{m_2}+x_{i2}\]

Answer 14

let dx/dt = v ... dv/dt = dv/dx *dx/dt = v dv/dx
not sure if you integrate it wrt dt ...

Answer 15

is \[\int v(t)dt=x(t)\]?

Answer 16

Cause that's all i did.

Answer 17

this doesn't solve x(t) at all ... we need x as explicit function of time.

sorry ... i think integration method would work.

Answer 18

It doesn't solve for x(t) as a function of time, but it makes x_2 represented as x_1. This way we have one DE to solve instead of a system, right?

Answer 19

http://en.wikipedia.org/wiki/Two-body_problem
i remember doing this kinda (central force) problem using reduced mass :((

Answer 20

something like