If Earth has mass nine times and radius twice that of the planet Mars, calculate the velocity required by a rocket to pull out of the gravitational force of Mars. Take escape speed on surface of Earth to be `11.2km//s`

Explanation

Here `M_(e)=9M_(m)` and `R_(e)=2R_(m)`
`v_(e)` (escape speed on surface of Earth) =11.2 km/s
Let `V_(m)` the speed reauired to pull out of the gravitational force of mars.
We know that
`v_(e)=sqrt((2GM_(e))/(R_(e)))` and `v_(m)=sqrt((2GM_(m))/(R_(m)))`
Dividing we get `(v_(m))/(v_(e))=sqrt((2GM_(m))/(R_(m))xx(R_(e))/(2GM_(e)))`
`=sqrt((M_(m))/(M_(e))xx(R_(e))/(R_(m)))=sqrt(1/9xx2)=(sqrt(2))/3`
`impliesv_(m)=(sqrt(2))/3(11.2km//s)=5.3km//s`