Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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...
EDIT: last equation should be\[m_2\frac{d^2x_2}{dt^2}=\mp G\frac{m_1m_2}{(x_2-x_1)^2}\]
consider one of the mass to be at rest and another is accelerating.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

You can't do that though! m_2 is accelerating too!
well if you want a dynamic system then ... it's going to be a system of second order DE.
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.
not sure ... first order system is fairly complicated.
plus ... this is non linear :(( looks like there must be some other approach.
Yeah but its not a question from a textbook... I just made this up. There might not be an approach :(.
let me try to ask mathematica. do you want general solution or ... do you have some boundary condition?
didn't get any solution from mathematica.
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...
I think we get: \[m_1\frac{d^2x_1}{dt^2}=\pm G\frac{m_1m_2}{(-(m_1x_1/m_2)-x_1+c_1)^2}\]\[c_1=\frac{m_1x_{i1}}{m_2}+x_{i2}\]
let dx/dt = v ... dv/dt = dv/dx *dx/dt = v dv/dx not sure if you integrate it wrt dt ...
is \[\int v(t)dt=x(t)\]?
Cause that's all i did.
this doesn't solve x(t) at all ... we need x as explicit function of time. sorry ... i think integration method would work.
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? i remember doing this kinda (central force) problem using reduced mass :((
something like |dw:1345580217154:dw|
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.
i think using symmetricity we can simplify this problem for two identical bodies. |dw:1345580536599:dw|
for non identical bodies , the acceleration produced is equal to their ratio of masses |dw:1345580770508:dw|
|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.
I'm not sure for general case ... with non rest IC.
most probably ... it would be wise to .... page no 306
Okay thx i looked at the wiki link. I'm on to something on my own and I'll see if that works.
OK no one else is helping at physics so I'll repost the problem in the math section...
If any1's interested:
try asking this instead. How to solve for general solution of system of this differential equation? |dw:1345676009886:dw|
more over this is just 1D case ... for general 2d case the system of DE could me much worse.
Yeah I know it's 1D I was trying just to tackle this, let alone the 2D.
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.
since we know, the distance between the objects from CM with be at constant proportion of their masses.
It's okay, I'll just do it out on my own using the CM. Don't worry about it.

Not the answer you are looking for?

Search for more explanations.

Ask your own question