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The radius of a planet is (1)/(4) of earth's radius and it's acceleration due to gravity is double that of the earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to it's value on the earth's surface |
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Answer» `(1)/(sqrt(2))` `(V_(S_(1)))/(V_(S_(2)))=sqrt((g_(1)R_(1))/(g_(2)R_(2)))=sqrt(((2G)(R//4))/((G)R))=(1)/(sqrt(2))` `V_(S_(1))=V_(S_(2))/(sqrt(2))` |
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