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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» 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` |
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