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The angular velocity of rotation of a pl...

The angular velocity of rotation of a planet of mass M and radius R, at which the matter start to escape from its equator is

A

`sqrt((2GM)/(R^(3)))`

B

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

C

`sqrt((2G^(2)M)/(R ))`

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`V_(E )=sqrt((2GM)/(R ))`
`therefore omega=(V_(E ))/(R )=(1)/(R )sqrt((2GM)/(R ))=sqrt((2GM)/(R^(3)))`
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Knowledge Check

  • The angular velocity of rotation of star (of mass M and radius R ) at which the matter start to escape from its equator will be

    A
    `sqrt((2GM^(2))/R)`
    B
    `sqrt((2GM)/g)`
    C
    `sqrt((2GM)/R^(2))`
    D
    `sqrt((2GR)/M)`

  • If v_(e) is the escape velocity of a body from a planet of mass 'M' and radius 'R . Then the velocity of the satellite revolving at height 'h' from the surface of the planet will be

    A
    `v_(e)sqrt(R/(R+h))`
    B
    `v_(e)sqrt((2R)/(R+h))`
    C
    `v_(e)sqrt((R+h)/R)`
    D
    `v_(e)sqrt(R/(2(R+h)))`

  • The escape velocity form the centre of a unifrom ring of mass M and radius R is :

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

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