Four particles, each of mass M and equidistant "from each other, move along a circle of radius R under the action of the mutual gravitational attraction. The speed of each particle is ......... (G = universal gravitational constant)

Options

A. `sqrt((GM)/R (1+2sqrt2))`
B. `1/2sqrt((GM)/R (1+2sqrt2))`
C. `sqrt((GM)/R)`
D. `sqrt(2sqrt2 (GM)/R)`

Answer

B

Explanation

`implies` The force of attraction due to to particles B and D on particle A is
`F_1 = F_2 = (GM^2)/(2R^2)`

and force attraction exerted by particle at C on particle at A
`F_3 = (GM^2)/(4R^2)`
`:.` The total force on particle A
`F= F_1 + F_2 + F_3`
but resultant force of `F_1 +F_2`
`F.=sqrt(F_1^2 +F_2^2+2F_1^2 cos 90^@) `
`F. = sqrt(F_1^2 +F_1^2) = sqrt(2F_1^2)`
`:. F. = sqrt2F_1`
`:. F. = sqrt2 F_1 +F_3`
`=sqrt2 (GM^2)/(2R^2) + (GM^2)/(4R^2)`
`:. (Mv^2)/R = (GM^2)/(2R^2) [sqrt2 +1/2]`
`v^2 =(GM)/(2R) [(2sqrt2 +1)/2]`
`:. v = 1/2 sqrt((GM)/R (1+ 2sqrt2)`