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