Mass points in equilibrium
In 3D Euclidian space, find a collection of mass points lying on the line [0,1]
such that:
1) There is more than 1 point.
2) Each point has non-zero mass.
3) The total mass is 1.
4) Each point attracts each other point with inverse square gravitational
force.
5) The entire system is in static equilibrium - that is, the total force on
each point exists and is zero.
Mass points in equilibriuim
topsquark and Captain Black
I didn't say the number of points was finite - in fact, your two responses
showed that it couldn't be. For an infinite number of points, of course,
there doesn't have to be a left-most point.
I'm not trying to trick anybody - this is a serious question.
I don't know how to place the points, but there is a way to do so and this fact was stated by Ulam in one of his books. I think it was "A Collection of
Mathematical Problems". I can't find my copy right now, or I would give you
the page number.
I posted this to see if anybody knows how to do it.
Thanks for your responses.
Bob
Mass points in equilibriuim
It's not an infinite line, it is a line of length 1.
And don't you really think that the sum of an iinfinite set of non-zero masses
can add up to 1?
Bob
Infinite number of mass points
But what if the masses are NOT all the same - like 1/2+1/4+1/8+ ... =1?
I'm sure Ulam's solution didn't have all the masses equal!
Bob
Mass points in equilibriuim
Quote:
Originally Posted by
CaptainBlack
Take three masses two of them equal, with the two equal masses either
side of the other on a straigt line segment at a distance $\displaystyle R$ from the
central mass. Then if the central mass is $\displaystyle m_1$ and the other two are
both of mass $\displaystyle m_2$ and (if I have done my sums correctly) the line segment
rotates about the central mass at an angular velocity of:
$\displaystyle
\dot{\theta}=\sqrt{\frac{G}{R^3}\left(m_1+\frac{m_ 2}{4}\right)}
$
we will have a stable configuration.
(there should be a simpler configuration with just two masses but I
can't be bothered to look for it just now - I take that back, just let
$\displaystyle m_1=0$. This is the simplest version of the rotating string of pearls
like configuration)
RonL
Is this supposed to be a solution to the problem I posed? If so, congratulations! You did it!
Mass points in equilibriuim
But it is! Why don't you try to find the solution instead of saying it's
impossible?
Bob