The escape Velocity from the earth is `11.2 Km//s`. The escape Velocity from a planet having twice the radius and the same mean density as the earth, is :

Options

A. `11.2 Km//s`
B. `22.4 Km//s`
C. `15.00 Km//s`
D. `5.8 Km//s`

Answer

B

Explanation

At a certain Velocity of projection, the body will go out of the gravitational field to the earth and will never returns,this initial Velocity is called escape velocity
`v_(e)=sqrt((2GM_(e))/(R_(e)))`
Where `G` is gravitational constant, `M_(e)` is mass of earth and `R_(e)` is radius
For planet `R_(p)= 2R_(e),d_(p)=d_(e)`
Also since earth is assumed spherical in shape its mass is given by
`M=4/3piR^(3)xxd`
`:. v_(e)=sqrt((2G)/R_(e)xx4/3piR^(3)d)`....(i)
`v_(p)=sqrt((2G)/(2R_(e))xx4/3pi(2R_(e))^(3))d` ....(2)
Dividing Eq.(i) by Eq.(2),we get
`(v_(e))/(v_(p))=1/2`
`implies v_(p)= 2v_(e)= 2xx11.2=22.4 km//s`