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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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