Question

Two particles of equal mass $m$ go around in a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is $v$. Find value of $v$.

• A. $v=\frac{1}{2R}\sqrt{\left(\frac{1}{Gm}\right)}$
• B. $v=\sqrt{\left(\frac{Gm}{2R}\right)}$
• C. $v=\frac{1}{2}\sqrt{\left(\frac{Gm}{R}\right)}$
• D. $v=\sqrt{\left(\frac{4Gm}{R}\right)}$

Answer 1

The correct option is C $v=\frac{1}{2}\sqrt{\left(\frac{Gm}{R}\right)}$

$\text{As,}\frac{Gmm}{{4R}^2}=\frac{m{v}^2}{R}$
$v^2=\frac{Gm}{4R}$
$v=\frac{1}{2}\sqrt{\frac{Gm}{R}}$