MCQ
Three particles of equal mass $m$ are situated at the vertices of an equilateral triangle of side $l$. They are moving in a circle in influence of their mutual gravitational interaction. Then their time period of revolution is directly proportional to
  • A
    $l^{1/2}$
  • B
    $l^{-1/2}$
  • $l^{3/2}$
  • D
    $l^{-3/2}$

Answer

Correct option: C.
$l^{3/2}$
c
$\frac{m V^{2}}{\frac{\ell}{\sqrt{3}}}=\frac{\sqrt{3} G m^{2}}{\ell^{2}}$

$V=\sqrt{\frac{G m}{\ell}}$

$\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}=\frac{2 \pi \times \frac{\ell}{\sqrt{3}}}{\sqrt{\frac{\mathrm{Gm}}{\ell}}}$

T $\propto \ell^{3 / 2}$

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