Ravi can throw a ball at a speed on the earth which can cross a river of width `10 m`. Ravi reaches on an imaginary planet whose mean density is twice that of the earth. Find out the maximum possible radius of the planet so that if Ravi throws the ball at the same speed it may escape from the planet. Given radius of the earth `= 6.4 xx 10^(6) m`.

Answer

`4km`

Explanation

Let the speed of the ball which cross `10 m` wide river be `V`.
`Then R=(V^(2)sin(2xx45^@))/g=10, v=sqrt(10g)`
Let the radius of plane be `R_(p)`, then mass of the planet
`M_(p)=4/3piR_(p)^(3)xx2sigma=4/3piR_(p)^(3)xx(2xxM_(e))/(4//3piR_(e)^(3))=(2M_(e)R_(p)^(3))/(R_(e)^(3))`
Escape velocity of the planet
`V_(e)=sqrt((2GM_(p)))/(R_(p))=sqrt((2Gxx2xxM_(e)R_(p)^(3))/(R_(p)xxR_(e)^(3)))`
`=sqrt((10GM_(e))/(R_(e)^(2))) [:.g=(GMe)/(R_(e)^(2))]`
`2R_(p)=sqrt(10R_(e))implies2R_(p)=sqrt(10xx6.4xx10^(6))`
`implies R_(p)=(8xx10^(3))/2=4xx10^(3)=4km`