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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: |
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Answer» `(1)/(2) sqrt((GM)/(R )(1+2sqrt(2)))`  `=(GM^(2))/((2R)^(2))+(GM^(2))/((R sqrt(2))^(2))cos45^(@)+(GM^(2))/((R sqrt(2))^(2))cos45^(@)` `=(GM^(2))/(R^(2))[(1)/(4 )+(1)/(sqrt(2))]` This force will be equal to centripetal force so `(Mu^(2))/(R )=(GM^(2))/(R^(2))[(1+2sqrt(2))/(4)]` `u=sqrt((GM)/(4R)[1+2sqrt(2)])=(1)/(2)sqrt((GM)/(R )[2sqrt(2)+1])` So correct choice is (a). |
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