Graviational acceleration on the surface of plane fo `(sqrt6)/(11)g.` where g is the gracitational 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 on the surface of the earht is taken to be `11 kms^(-1)` the escape speed on teh surface of the planet in `kms^(-1)` will be

Answer

`(3)`

Explanation

`g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) piGR rho`
`:. R = (3g)/(4pi G rho)` i.e. `R prop (g)/(rho)`
Hence, `(R_(p))/(R_(e)) = (g_(p))/(g_(e)) xx (rho_(e))/(rho_(p)) = (sqrt(6))/(11) xx (3)/(2)` ..(i)
Escape velocity, `upsilon = sqrt(2gR)`
`:. (upsilon_(p))/(upsilon_(e)) = sqrt((g_(p))/(g_(e)) xx (R_(p))/(R_(e))) = sqrt((sqrt(6))/(11) xx ((sqrt(6))/(11) xx (3)/(2)))`
or `upsilon_(p) = sqrt(6 xx (3//2))/(11) upsilon_(e) = (3)/(11) xx 11 = 3 km//s`