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Can two particles be in equilibrium unde...

Can two particles be in equilibrium under the action of their mutual gravitational force? Can three particle be? Can one of the three particles be?

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Two particles of equa mass (m) each move in a circle of radius ® under the action of their mutual gravitational attraction. Find the speed of each particle.

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Knowledge Check

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

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

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

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

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

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

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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. Calculate the speed of each particle

    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. Calculate the speed of each particle

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    Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles are their mutual gravitation find the initial acceleration of the two particles.

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