A ring of radius R = 4 m made of a highly dense material has a mass $m_{1} = 5 .4 \times 10^{9} kg$ distributed uniformly over its circumference. A highly dense particle of mass $m_{2} = 6 \times 10^{8} kg$ is placed on the axis of the ring at a distance 3 m from the center of the ring. Find the speed of the particle (in cm/s), when the particle is at the center of the ring. Except mutual gravitational interaction of the two, neglect all other forces.

Moderate

Solution

When the particle is at the center, let their respective speeds
are v₁ and v₂.
From conservation of linear momentum,
$\begin{array}{l} 0 = 5 .4 \times 10^{9} v_{1} - 6 \times 10^{8} v_{2} \\ ∴ v_{2} = 9 v_{1} \end{array}$
From conservation of energy, $(U + K)_{i} = (U + K)_{f}$
$\begin{array}{l} - \frac{{Gm}_{1} m_{2}}{\sqrt{4^{2} + 3^{2}}} + 0 = - \frac{{Gm}_{1} m_{2}}{4} + \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} \\ {Gm}_{1} m_{2} (\frac{1}{4} - \frac{1}{5}) = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} (81 v_{1}^{2}) \\ G \frac{(9 m_{2}^{2})}{20} = \frac{1}{2} (90 m_{2}) v_{1}^{2} \end{array}$ $∵ m_{1} = 9 m_{2}$
$Now m_{1} = 9 m_{2} (given)$
$\begin{array}{l} v_{1} = \sqrt{\frac{{Gm}_{2}}{100}} = \sqrt{\frac{6 .67 \times 10^{- 11} \times 6 \times 10^{8}}{100}} = 0 .02 {ms}^{- 1} \\ ∴ v_{2} = 9 v_{1} = 0 .18 {ms}^{- 1} = 18 {cms}^{- 1} \end{array}$

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