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Four similar particles of mass m are orb...

Four similar particles of mass m are orbiting in a circle of radius r in the same direction and same speed because of their mutual gravitational attractive force as shown in the figure . Speed of a particle is given by

A

`[(Gm)/(r)((1+2sqrt2)/(4))]^(1/(2))`

B

`root(3)((GM)/(r))`

C

`sqrt((Gm)/(r)(1+2sqrt2))`

D

zero

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • Two particles of same mass m 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))`
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    `V =1/2sqrt((Gm)/R)`
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    `V = sqrt((4Gm)/R)`

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    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)))`

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

    A
    `1/2 sqrt((Gm)/R)`
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    `sqrt((4GM)/R)`
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