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A planet in a distant solar systyem is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is `11kms^(-1)`, the escape velocity from the surface of the planet would beA. 110km/sB. 11km/sC. `1.1km//s`D. `0.11km//s` |
Answer» Correct Answer - A Let `V_(P)` and `V_(e)` be the escape velocities from the planet and the earth. Then `(V_(P))/(V_(e))=sqrt((GM_(P))/(R_(P))xx(R_(e))/(GM_(e)))=sqrt((M_(P))/(M_(e))xx(R_(e))/(R_(P)))` `therefore (V_(P))/(11)=sqrt(10xx10)` `therefore V_(P)=11xx10=110 km//s`. |
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