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)

Four particles, each of mass M and equidistant "from each other, move

1.	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)
Answer» `sqrt((GM)/R (1+2sqrt2))` `1/2sqrt((GM)/R (1+2sqrt2))` `sqrt((GM)/R)` `sqrt(2sqrt2 (GM)/R)` Solution :`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)`