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The escape velocity of a body on the surface of the earth is 11.2" km"//s If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes |
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Answer» `22.4" KM"//s` `:. "Escape velocity " (v_(e ))=sqrt((2GM_(e ))/(R_(e ))) prop sqrt((M_(e ))/(R_(e )))` Therefore `(v_(e ))/(v._(e ))=sqrt((M_(e ))/(R_(e ))XX(0.5 R_(e ))/(2M_(e )))=sqrt((1)/(2))=(1)/(2)` `v._(e )=2v_(e )=22.4" km"//s` |
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