Two balls of mass m each are hung side by side by side by two long threads of equal length l. If the distance between upper ends is r, show that the distance r' between the centres of the ball is given by `g r'^(2)(r-r')=2l G m`

Explanation

The situation is shown in figure

Following force act on each ball
(i) Weight of the ball m g in downward direction
(ii) Tension in thread T along string
(iii) Force of gravitation attraction towards each other
`F = G(mm)/(r'^(2))`
Here for equilibrium of balls we have
`T sin theta = (Gm^(2))/(r'^(2))` ....(i)
`T cos theta = mg` .....(ii)
Dividing equation (i) and (ii), we get
or `tan theta = (Gm^(2))/(mgr'^(2))` .....(iii)
In `Delta ACP" tan " theta = (r-r')/(2l)` .....(iv)
From equation (iii) and (iv)
`(r-r')/(2l) = (Gm^(2))/(mgr'^(2))`
`g r'^(2)(r-r')=2l G m`
For partical consideration, it is viable to assume that the two masses. will be attracted by very small distance as gravitational force acting between then will be very less. Hence, the balls will be at almost the same horizontal level as before or `theta` will be very small.