MCQ
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 $r_1 \cdot r_2$ and their speeds are $v_1 \cdot v_2$, respectively. Then,
  • $v_1^2 r_1^n=v_2^2 r_2^n$
  • B
    $v_1^2 r_1^{-n}=v_2^2 r_2^{-n}$
  • C
    $v_1^2 r_1=v_2^2 r_2$
  • D
    $v_1^2 r_1^{2-n}=v_2^2 r_2^{2-n}$

Answer

Correct option: A.
$v_1^2 r_1^n=v_2^2 r_2^n$
a
(a)

Given,

Potential, $V=K r^{-n}$

So, magnitude of field,

$E=-\frac{d V}{d r}=-\frac{d}{d r}\left(K r^{-n}\right)$

$\Rightarrow E=\frac{n K}{r^{n+1}}$

Hence, force, $F=m E=\frac{m n K}{r^{n+1}}$

As particles are rotating in circular path,

$F=\frac{m v^2}{r}=\frac{m n K}{r^{n+1}}$

$\Rightarrow v^2 r^n=n K=$ constant

So, $\quad v_1^2 r_1^n=v_2^2 r_2^n$

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