Two particles of the same mass $m$ are moving in circular orbits because of force, given by $F (r) = \frac{- 16}{r} - r^{3}$ . The first particle is at a distance $r = 1$ , and the second, at $r = 4$ . The best estimate for the ratio of kinetic energies of the first and the second particle is closest to

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$3 \times 10^{- 3}$

$6 \times 10^{2}$

$6 \times 10^{- 2}$

$10^{- 1}$

answer is A.

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Detailed Solution

Given force provides the centripetal force on the particle moving in circular orbits, $\frac{m v^{2}}{r} = \frac{16}{r} + r^{3}$

Kinetic energy, $K = \frac{1}{2} m v^{2} = \frac{1}{2} [16 + r^{4}]$

$\frac{K_{1}}{K_{2}} = \frac{\frac{16 + 1}{2}}{\frac{16 + 256}{2}} = \frac{17}{272} = 0.0625 or \frac{K_{1}}{K_{2}} ≃ 6 \times 10^{- 2}$

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