Two particles of the same mass are moving in | vedclass

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Question

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} =

Vedclass · Accepted Answer

$6 	imes {10^{-2}}$

Vedclass · Answer

As the particles moving in circular orbits, So

$\frac{{m{v^2}}}{r} = \frac{{16}}{r} + {r^3}$

Kinetic energy, $K{E_0} = \frac{1}{2}m{v^2} = \frac{1}{2}\left[ {16 + {r^4}} ight]$

For fist particle, $r = 1,$ ${K_1} = \frac{1}{2}\left( {16 + 1} ight)$

Similarly, for second particle, $r = 4,$

${K_2} = \frac{1}{2}\left( {16 + 256} ight)$

$	herefore \,\frac{{{K_1}}}{{{K_2}}} = \frac{{\frac{{16 + 1}}{2}}}{{\frac{{16 + 256}}{2}}} = \frac{{17}}{{272}} \simeq 6 	imes {10^{ - 2}}$

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