Gravitational acceleration on the surface of a planet is `(sqrt(6))/(11)g` where g is the gravitational acceleration on the
surface of the earth. The average mass density of the planet is `(2)/(3)` times that of the earth. If the escape speed of
the surface of the earth is taken to be 11 `"kms"^(-1)` the escape speed on the surface of the planet in `"kms"^(-1)` will be

Answer

2

Explanation

`(g_(p))/(g_(E))=(G(4/3 pi R^(3)" "_(E)rho_(E))//R_(E)^(2))/(G(4/3 pi R^(3)" "_(p)rho_(p))//R_(p)^(2))=(R_(E))/(R_(p))xx (rho_(E))/(rho_(p)),` `(R_(E))/(R_(p))=(g_(p))/(g_(E)) xx (rho_(p))/(rho_(E))=(sqrt(6))/(11) xx (2)/(3)`
The ratio of escape velocities is given by `(v_(p))/(v_(E))=sqrt((2g_(p)R_(p))/(2g_(E)R_(E)))=sqrt((g_(p)/(g_(E)))(R_(p)/(R_(E))))`
= `[(sqrt(6))/(11) xx (2)/(3) xx sqrt(6)/(11)]^(1//2)=(2)/(11)` `therefore v_(p)=(2)/(11) xx 11 =2 " km/s"`