Question

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

• A. V=
• B. V =
• C. V = $\frac { 1 } { 2 } \sqrt { \frac { \mathrm { Gm } } { \mathrm { R } } }$
• D. V = $\sqrt { \frac { 4 \mathrm { Gm } } { \mathrm { R } } }$

Answer 1

Answer: (C)

V = $\frac { 1 } { 2 } \sqrt { \frac { \mathrm { Gm } } { \mathrm { R } } }$

Answer 2

the gravitational force provide the necessary centripetal force for circular motion. so

f_g = f_e Ž $\frac { \mathrm { Gm } ^ { 2 } } { ( 2 \mathrm { R } ) ^ { 2 } }$ = $\frac { 1 } { 2 } \sqrt { \left( \frac { \mathrm { Gm } } { \mathrm { R } } \right) }$