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vf321
Group Title
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?
 one year ago
 one year ago
vf321 Group Title
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?
 one year ago
 one year ago

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vf321 Group TitleBest ResponseYou've already chosen the best response.0
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": \[m_1\frac{d^2x_1}{dt^2}=G\frac{m_1 m_2 (x_2x_1)}{x_2x_1^3}=\pm G\frac{m_1m_2}{(x_2x_1)^2}\]So we get a system of DEs: \[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{(x_2x_1)^2}\]\[m_2\frac{d^2x_1}{dt^2}=\mp G\frac{m_1m_2}{(x_2x_1)^2}\]Still unsolvable to my knowledge...
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
EDIT: last equation should be\[m_2\frac{d^2x_2}{dt^2}=\mp G\frac{m_1m_2}{(x_2x_1)^2}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
consider one of the mass to be at rest and another is accelerating.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
You can't do that though! m_2 is accelerating too!
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
well if you want a dynamic system then ... it's going to be a system of second order DE.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
not sure ... first order system is fairly complicated.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
plus ... this is non linear :(( looks like there must be some other approach.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
Yeah but its not a question from a textbook... I just made this up. There might not be an approach :(.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let me try to ask mathematica. do you want general solution or ... do you have some boundary condition?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
didn't get any solution from mathematica.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
I think I have something that might help simplify a bit (N's 3rd law): \[m_2\frac{d^2x_2}{dt^2}=m_1\frac{d^2x_1}{dt^2}\]\[\int_0^tm_2\frac{d^2x_2}{dt^2}dt=\int_0^tm_1\frac{d^2x_1}{dt^2}dt\]\[m_2(\frac{dx_2}{dt}v_{i2})=m_1(\frac{dx_1}{dt}v_{i1})\]\[v_{i1}, v_{i2}=0 m/s\]\[\int_0^tm_2\frac{dx_2}{dt}dt=\int_0^tm_1\frac{dx_1}{dt}dt\]\[m_2(x_2(t)x_{i2})=m_1(x_1(t)x_{i1})\]\[x_2(t)=\frac{m_1(x_1(t)x_{i1})}{m_2}+x_{i2}\]That seems right. If it is, then we don't have a system of DE's anymore...
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
I think we get: \[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{((m_1x_1/m_2)x_1+c_1)^2}\]\[c_1=\frac{m_1x_{i1}}{m_2}+x_{i2}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let dx/dt = v ... dv/dt = dv/dx *dx/dt = v dv/dx not sure if you integrate it wrt dt ...
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
is \[\int v(t)dt=x(t)\]?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
Cause that's all i did.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this doesn't solve x(t) at all ... we need x as explicit function of time. sorry ... i think integration method would work.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Twobody_problem i remember doing this kinda (central force) problem using reduced mass :((
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
something like dw:1345580217154:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
still this system is interesting ... right now i'm working on first order systems. Maybe i'll start PDE right after it. If i encounter second order system ... i'll try to update.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i think using symmetricity we can simplify this problem for two identical bodies. dw:1345580536599:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
for non identical bodies , the acceleration produced is equal to their ratio of masses dw:1345580770508:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1345580931804:dw choosing initial condition, and choosing a suitable coordinate system, i think .... this is pretty easy to solve x(t) as explicit function of time.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
I'm not sure for general case ... with non rest IC.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
most probably ... it would be wise to .... page no 306 http://books.google.com.np/books?id=P1kCtNrpJsC&pg=PA305&lpg=PA305&dq=equation+of+orbit+classical+mechanics&source=bl&ots=ghSw0cFxyW&sig=IxK3JsJoSC79tCaTqeL9wPIAxhA&sa=X&ei=w_AzUKTUL43QrQfNsYHoCA&redir_esc=y#v=onepage&q=equation%20of%20orbit%20classical%20mechanics&f=false
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
Okay thx i looked at the wiki link. I'm on to something on my own and I'll see if that works.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
OK no one else is helping at physics so I'll repost the problem in the math section...
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
If any1's interested: http://math.stackexchange.com/questions/185593/solvingthedeforatwobodysystem
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
try asking this instead. How to solve for general solution of system of this differential equation? dw:1345676009886:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
more over this is just 1D case ... for general 2d case the system of DE could me much worse.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
Yeah I know it's 1D I was trying just to tackle this, let alone the 2D.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i guess CM frame of reference is a better choice to solve problem. later we can translate origin to any other choice after we find x and y. Try to simulate this graphically.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
since we know, the distance between the objects from CM with be at constant proportion of their masses.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
It's okay, I'll just do it out on my own using the CM. Don't worry about it.
 one year ago
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