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Question

(A) Given the potential $V(r) = K r^{-n}$.The force $F$ acting on the particle is given by the negative gradient of the potential:$F = -\frac{dV}{dr}

Vedclass · Accepted Answer

$v_1^2 r_1^n = v_2^2 r_2^n$

Vedclass · Answer

(A) Given the potential $V(r) = K r^{-n}$.The force $F$ acting on the particle is given by the negative gradient of the potential:$F = -\frac{dV}{dr} = -\frac{d}{dr}(K r^{-n}) = -K(-n)r^{-n-1} = \frac{nK}{r^{n+1}}$.For a particle of mass $m$ moving in a circular orbit of radius $r$ with speed $v$,the centripetal force is provided by this central force:$\frac{mv^2}{r} = F = \frac{nK}{r^{n+1}}$.Rearranging the terms to isolate the variables:$v^2 = \frac{nK}{m} \cdot \frac{r}{r^{n+1}} = \frac{nK}{m} \cdot r^{-n}$.Thus,$v^2 r^n = \frac{nK}{m}$.Since $n$,$K$,and $m$ are constants for both particles,the product $v^2 r^n$ must be constant for any orbit under this potential.Therefore,$v_1^2 r_1^n = v_2^2 r_2^n$.

Two particles of identical mass are moving in circular orbits under a potential given by $V(r) = K r^{-n}$,where $K$ is a constant. If the radii of their orbits are $r_1$ and $r_2$ and their speeds are $v_1$ and $v_2$,respectively,then:

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