A planet has twice the radius but the mean density is `1/4` th as compared to earth. What is the ratio of escape velocity from earth to that from the planet

Options

A. `3:1`
B. `1:2`
C. `1:1`
D. `2:1`

Answer

C

Explanation

Escape velocity from earth 's surface is
`v_(es)=sqrt(2GM_(e ))/(R_(e ))=sqrt(2g.4/3piR_(e)^(3)d_(e ))/(R_(e ))`
or `v_(es) prop sqrt(d_(e))xxR_(e)`
similary for a planet
`v_(es) prop sqrt(d_(p))xxR_(p)`
so `(v_(es))/(v_(es)=(d_(E ))/(d_(p))^(1//2)xx(R_(e))/(R_(p))`
given `d_(p)=1/4d_(e ),R_(p)=2R_(e )`
`(v_(es))/(v_(es))=(d_(e))/(d_(e)//4)xx(R_(e))/(2R(e))`
`=4^(1//2)xx1/2=2xx1/2=1=11:1`