Two particles of the same mass are moving in circular orbits beca… - Vidyadip

MCQ

Two particles of the same mass 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

A
$10^{-1}$
✓
$6 \times {10^{-2}}$
C
$6 \times {10^2}$
D
$3 \times {10^{-3}}$

✓

Answer

Correct option: B.

$6 \times {10^{-2}}$

b
As the particles moving in circular orbits, So

$\frac{{m{v^2}}}{r} = \frac{{16}}{r} + {r^3}$

Kinetic energy, $K{E_0} = \frac{1}{2}m{v^2} = \frac{1}{2}\left[ {16 + {r^4}} \right]$

For fist particle, $r = 1,$ ${K_1} = \frac{1}{2}\left( {16 + 1} \right)$

Similarly, for second particle, $r = 4,$

${K_2} = \frac{1}{2}\left( {16 + 256} \right)$

$\therefore \,\frac{{{K_1}}}{{{K_2}}} = \frac{{\frac{{16 + 1}}{2}}}{{\frac{{16 + 256}}{2}}} = \frac{{17}}{{272}} \simeq 6 \times {10^{ - 2}}$

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