Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Question

Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is (A) v=(1/2 R) √(1/G m) (B) v=√(G m/2 R) (C) v=(1/2) √(G m/R) (D) v=√(4 G m/R)

Tardigrade Admin · Accepted Answer

Centripetal force provided by the gravitational force of attraction between two particles <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/84c55d931edf78345f8d99f0253c13e7-.png" /> i.e. (m v2/R)=(G m × m/(2 R)2) ⇒ v=(1/2) √(G m/R)

Q. Two particles of equal mass go round a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

$v = \frac{1}{2 R} \frac{1}{G m}$

$v = \frac{G m}{2 R}$

$v = \frac{1}{2} \frac{G m}{R}$

$v = \frac{4 G m}{R}$

Centripetal force provided by the gravitational force of attraction between two particles

i.e. $\frac{m v ^{2}}{R} = \frac{G m \times m}{( 2 R ) ^{2}}$
$\Rightarrow v = \frac{1}{2} \frac{G m}{R}$

Q. Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

v=2R1​Gm1​​

v=2RGm​​

v=21​RGm​​

v=R4Gm​​