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Gravitational acceleration on the surfac...

Gravitational acceleration on the surface of a planet is `(sqrt(6)//11)` g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is `2/3` times that of the earth. If the escape speed on the surface of the earth is taken to be 11 `km*s^(-1)` , the escape speed on the surface of the planet in `km*s^(-1)` will be

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Knowledge Check

  • The acceleration due to gravity on the surface of the earth is 9.8 m*s^(-2) . The size of a planet is the same as that of the earth, but its density is twice the density of the earth. The value of the acceleration due to gravity on that planet is

    A
    `19.6 m*s^(-2)`
    B
    `9.8 m*s^(-2)`
    C
    `4.9 m*s^(-2)`
    D
    `2.45 m*s^(-2)`

  • Escape speed on the surface of a planet varies with the mass m of a body as

    A
    `m^0`
    B
    m
    C
    `m^(-1)`
    D
    `m^2`

  • The mass of a planet is 4 times the mass of the earth and its radius is 2 times the radius of the earth. The acceleration due to gravity on that planet is

    A
    `9.8 m*s^(-2)`
    B
    `19.6 m*s^(-2)`
    C
    `4.9 m*s^(-2)`
    D
    `39.2 m*s^(-2)`

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    At what height above the surface of the earth is the gravitational potential energy of a body equal to that on the surface of the moon ? Assume that the mass of the earth is 80 times that of the moon and the radius of the earth is 4 times that of the moon. Given ,Radius of the earth =6400 km.

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