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

`(V_(C2))/(V_(erho)) = (rho_(2)R_(2))/(rho_(1)R_(1))`
`= sqrt((rho_(2)R_(2))/(rho_(1)R_(1)))`
`(g_(2))/(g_(1)) = ((M_(2))/(M_(1)))((R_(1))/(R_(2)))^(2)`
`(rho_(2))/(rho_(1))=((M_(2))/(M_(1)))((R_(1))/(R_(2)))^(3)`
`(g_(2)//g_(1))/(rho_(2)//rho_(1)) = ((R_(2))/(R_(1)))`
`(V_(C_(2)))/(V_(C_(1)))=sqrt(((g_(2))/(g_(1)))((g_(2))/(g_(1)))//(rho_(2))/(rho_(1)))`
`=(g_(2))/(g_(1)) sqrt((rho_(2))/(rho_(1))) = (sqrt(6))/(11)`
`sqrt((3)/(2))`