The ratio of escape velocity at earth `(v_(e))` to the escape velocity at a planet `(v_(y))` whose radius and density are twice

Options

A. `1:2 sqrt(2)`
B. `1:4`
C. `1: sqrt(2)`
D. `1:2`

Answer

A

Explanation

Since, the escape velocity of the earth can be given as
`v_(e)=sqrt(2gR)=Rsqrt((8)/(3)piGrho)" "[rho="density of the earth"]`
`rArrv_(e)=Rsqrt((8)/(3)piG rho)`….(i)
As it is given that the radius and mean density of planet are twice as that of the earth. So, escape velocity at planet will be
`v_(p)=2Rsqrt((8)/(3)pi G2 rho)`....(ii)
Divide, Eq. (i) by Eq.(ii), we get
`(v_(e))/(v_(p))=(Rsqrt((8)/(3)piG rho))/(2Rsqrt((8)/(3)pi Grho))rArr (v_(e))/(v_(p))=(1)/(2 sqrt(2))`.