Four identical particles of equal masses 1kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

Options

A. `sqrt((G)/(2)(2sqrt(2)-1))`
B. `sqrt(G(1+2sqrt(2)))`
C. `sqrt((G)/(2)(1+2sqrt(2)))`
D. `sqrt((1+2sqrt(2))G)/(2)`

Answer

D

Explanation

implies By resolving force `F_(2)`, we get
`implies F_(1)+F_(2)cos45^(@)+F_(2)cos45^(@)`
`implies F_(1)+2F_(2)cos45^(@)=F_(c)`
`F_(c)` = centripetal force = `(Mv^(2))/(R)`
`implies (GM^(2))/((2R)^(2))+[(2GM^(2))/((sqrt(2)R)^(2))cos45^(@)]=(Mv^(2))/(R)`
`implies(GM^(2))/(4R^(2))+(2GM^(2))/(2sqrt(2)R^(2))=(Mv^(2))/(R)`
`implies (GM)/(4R)+(GM)/(sqrt(2).R)=v^(2)impliesv=sqrt((GM)/(4R)+(GM)/(sqrt(2).R))`
`implies v=sqrt((GM)/(R)[(1+2sqrt(2))/(4)])`
`implies v=(1)/(2)sqrt((GM)/(R)(1+2sqrt(2)))`
(Given : mass = 1kg, radius = 1m)
`implies v=(1)/(2)sqrt((GM)/(R)(1+2sqrt(2)))`