Question

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`

Answer 1

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)` be the speed required 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`
`rArr v_(m)=(sqrt(2))/3(11.2km//s)=5.3 km//s`