Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each partial with respect to their centre of mass is :

Question

Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each partial with respect to their centre of mass is : (A) √(Gm/4R) (B) √(Gm/3R) (C) √(Gm/2R) (D) √(Gm/R)

Tardigrade Admin · Accepted Answer

(Gm2/(2R)2) = mω2R (Gm2/4R3) = ω2 ω = √(Gm/4R3) v = ω R v = √(Gm/4R3)× R = √(Gm/4R)

Q. Two particles of equal mass $^{'} m^{'}$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each partial with respect to their centre of mass is :

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

$\frac{G m}{3 R}$

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

$\frac{G m}{R}$

$\frac{G m ^{2}}{( 2 R ) ^{2}} = m ω^{2} R$
$\frac{G m ^{2}}{4 R ^{3}} = ω^{2}$
$ω = \frac{G m}{4 R ^{3}}$
$v = ω R$
$v = \frac{G m}{4 R ^{3}} \times R = \frac{G m}{4 R}$

Q. Two particles of equal mass ′m′ go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each partial with respect to their centre of mass is :

4RGm​​

3RGm​​

2RGm​​

RGm​​