Four particles each of mass M move along a circle of radius R under the action of their mutula gravitational attraction the speed of each paritcles 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

`F=F_(1)=F_(2)=(GM^(2))/(2R^(2))` `F_(3)=(GM^(2))/(4R^(2))` `therefore` `F_("Total")` towards center
`= sqrt(2) F+F_(3)` `= sqrt(2)(GM^(2))/(2R^(2))+(GM^(2))/(4R^(2))` `=(GM^(2))/(4_(R^(2)))[2 sqrt(2)+1]` `F_(T)=F_(CP)`
`(GM^(2))/(4_(R^(2)))[2 sqrt(2)+1]=(M_(V^(2)))/(R)` `therefore V= sqrt((GM)/(4 _(R))[2 sqrt(2)+1])`