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Four particles, each of mass M and equi...

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

A

`sqrt((GM)/R)`

B

`sqrt(2sqrt(2)(GM)/R)`

C

`sqrt((GM)/R(1+2sqrt(2)))`

D

`1/2sqrt((GM)/R(1+2sqrt(2)))`

Text Solution

Verified by Experts

The correct Answer is:
D
Centripetal force on the particle at A Fig.1.31, i.e., the force along AO due to the particle at B,C and D, is
`F=(GM*M)/((sqrt(2)R)^2) cos 45^@+(GM*M)/((2R)^2)+(GM*M)/((sqrt(2)R)^2) cos45^@`
`=(GM^2)/(4R^2)+2*(GM^2)/(2R^2)*1/sqrt(2)`
`=(GM^2)/(4R^2)(1+4/sqrt(2))=(GM^2)/(4R^2)(1+2sqrt(2))`
Again ,`F=(Mv^2)/R`
`therefore (Mv^2)/R=(GM^2)/(4R^2)(1+2sqrt(2))`
or, `v=1/2sqrt((GM)/R(1+2sqrt(2)))`
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Knowledge Check

  • Four particles of mass M move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle.

    A
    `sqrt((GM)/R)`
    B
    sqrt(2 sqrt 2 (GM)/R)`
    C
    `sqrt((GM)/R)((2sqrt2+1)/4)`
    D
    `sqrt((GM)/R)((2sqrt2+1))`

  • Two particles of equal mass go around a circle of radius r under the action of their mutual gravitational attraction. The speed of each particle is

    A
    `v = 1/(2R)sqrt((1/(GM))`
    B
    `v = sqrt(((GM)/(2R))`
    C
    `v = 1/2 sqrt(((GM)/(2R))`
    D
    `v = sqrt((4GM)/R)`

  • Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction the speed of each particle with respect to their centre of mass is

    A
    `sqrt((Gm)/R)`
    B
    `sqrt((Gm)/(4R))`
    C
    `sqrt((Gm)/(3R))`
    D
    `sqrt((Gm)/(2R))`

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