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The ratio of escape velocity at earth (v...

The ratio of escape velocity at earth `(v_e)`to the escape velocity at a planet `(v_p)` whose radius and mean density are twice as that of earth is

A

`1:2 sqrt(2)`

B

`1:4`

C

`1:sqrt(2)`

D

`1:2`

Text Solution

Verified by Experts

The correct Answer is:
A
Escape velocity =`sqrt((2GM)/R)=sqrt(8/3piGR^2rho)`
`therefore (v_e)/(v_p)sqrt((R_2/R_p)^2(rho_e/rho_p))=sqrt((R_e/(2R_e))^2(rho_e/(2rho_e)))=sqrt(1/4*1/2)=1/(2sqrt(2))`
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Knowledge Check

  • The escape velocity from the earth is v_e . If both the mass and the radius of a planet are twice that of the earth, then the escape velocity from that planet will be

    A
    `v_e`
    B
    `2v_e`
    C
    `4v_e`
    D
    `16 v_e`

  • The value of the escape velocity from the earth is v_e . If the radius of a planet is 4 times that of the earth and its density is 9 times the density of the earth, then the value of the escape velocity from that planet will be

    A
    `6 v_e`
    B
    `12 v_e`
    C
    `20 v_e`
    D
    `36 v_e`

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