Question

Four particles of equal mass M move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is –

• A. $\frac { \mathrm { GM } } { \mathrm { R } }$
• B.
• C. $\sqrt { \left[ \frac { \mathrm { GM } } { \mathrm { R } } ( 2 \sqrt { 2 } + 1 ) \right] }$
• D. $\sqrt { \left[ \frac { \mathrm { GM } } { \mathrm { R } } \left( \frac { 2 \sqrt { 2 } + 1 } { 4 } \right) \right] }$

Answer 1

Answer: (D)

$\sqrt { \left[ \frac { \mathrm { GM } } { \mathrm { R } } \left( \frac { 2 \sqrt { 2 } + 1 } { 4 } \right) \right] }$

Answer 2

Gravitational force on each due to other three particles

provides the necessary centripetal force.

\ $\sqrt { 2 }$ cos 45⁰ + =

Simplifying it, we get

v = $\sqrt { \frac { \mathrm { GM } } { \mathrm { R } } \left( \frac { 2 \sqrt { 2 } + 1 } { 4 } \right) }$