Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :

Question

Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is : (A) √(G/2 R3) (B) (1/2) √(G/R3) (C) (1/2 R) √(1/G) (D) √(2 G/R3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2424d6b5a68760c4f2b37411b6c2212e-.png" /> F=(G m2/(2 R)2)=m R ω2 ω=(1/2) √(G/R3)

Q. Two identical particles of mass $1 k g$ each go round a circle of radius $R$ , under the action of their mutual gravitational attraction. The angular speed of each particle is :

$\frac{G}{2 R ^{3}}$

$\frac{1}{2} \frac{G}{R ^{3}}$

$\frac{1}{2 R} \frac{1}{G}$

$\frac{2 G}{R ^{3}}$

$F = \frac{G m ^{2}}{( 2 R ) ^{2}} = m R ω^{2}$
$ω = \frac{1}{2} \frac{G}{R ^{3}}$

Q. Two identical particles of mass 1kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :

2R3G​​

21​R3G​​

2R1​G1​​

R32G​​