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Graviational acceleration on the surface...

Graviational acceleration on the surface of plane fo `(sqrt6)/(11)g.` where g is the gracitational 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 earht is taken to be `11 kms^(-1)` the escape speed on teh surface of the planet in `kms^(-1)` will be

Text Solution

Verified by Experts

The correct Answer is:
`(3)`
`g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) piGR rho`
`:. R = (3g)/(4pi G rho)` i.e. `R prop (g)/(rho)`
Hence, `(R_(p))/(R_(e)) = (g_(p))/(g_(e)) xx (rho_(e))/(rho_(p)) = (sqrt(6))/(11) xx (3)/(2)` ..(i)
Escape velocity, `upsilon = sqrt(2gR)`
`:. (upsilon_(p))/(upsilon_(e)) = sqrt((g_(p))/(g_(e)) xx (R_(p))/(R_(e))) = sqrt((sqrt(6))/(11) xx ((sqrt(6))/(11) xx (3)/(2)))`
or `upsilon_(p) = sqrt(6 xx (3//2))/(11) upsilon_(e) = (3)/(11) xx 11 = 3 km//s`
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Knowledge Check

  • The moon has a mass of (1)/(81) that of the earth and radius of (1)/(4) that of the earth. The escape speed from the surface of the earth is 11.2 km // s. The escape speed from surface of the moon is-

    A
    1.25`km//s`
    B
    `2.49km//s`
    C
    `3.7km//s`
    D
    `5.6km//s`

  • If the dentist of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :

    A
    0.2 g
    B
    0.4 g
    C
    2g
    D
    4 g

  • If the dentist of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :

    A
    0.2 g
    B
    0.4 g
    C
    2g
    D
    4 g