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For particles of equal masses M that mov...

For particles of equal masses M that move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Text Solution

Verified by Experts

The correct Answer is:
`sqrt((GM)/(R )((2sqrt(2)+1)/(4)))`
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Knowledge Check

  • Four particles, each of mass M 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(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)`
    B
    `sqrt((4GM)/R)`
    C
    `1/(2R) sqrt(1/(Gm))`
    D
    `sqrt((Gm)/(2R))`

  • 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

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

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