Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is:

Options

A. `sqrt((GM)/(R))`
B. `sqrt(2 sqrt(2) (GM)/(R))`
C. `sqrt((GM)/(R) (1 + 2 sqrt(2)))`
D. `(1)/(2) sqrt((GM)/(R) (1 + 2 sqrt(2)))`

Answer

D

Explanation

2F cos `45^(@) + F. = (Mv^(2))/(R) ` (From figure )
where F = `(GM^(2))/((sqrt(2) R)^(2)) and F. = (GM^(2))/(4 R^(2))`
`rArr (2 xx GM^(2))/(sqrt(2) (R sqrt(2) )^(2) ) + (GM^(2))/(4R^(2)) = (Mv^(2))/(R)`
`rArr (GM^(2))/(R) [ (1)/(4) + (1)/(sqrt(2)) ] = Mv^(2)`
`therefore v = sqrt((GM)/(R) ( (sqrt(2) + 4)/(4 sqrt(2)))) = (1)/(2) sqrt((GM)/(R) (1 + 2 sqrt(2)))`
` (##DSH_NTA_JEE_MN_PHY_C07_E01_013_S01.png" width="80%">