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InterviewSolution
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An object is dropped from height `h=2 R` on the surface of earth. Find the speed with which it will collide with ground by neglecting effect of air. (Where R is radius of earth, take mass of earth M) |
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Answer» The initial potential energy of object is `U_(i)=-(GMm)/(3R)` Final potential energy `U_(f)=-(GMm)/(R)` By law of conservation of energy, `Delta KE = - Delta PE` `rArr (1)/(2)mv^(2)=-(U_(f)-U_(i))=U_(i)-U_(f)` `rArr (1)/(2)mv^(2)=-(GMm)/(3R)+(GMm)/(R)` `rArr (1)/(2)v^(2)=(2GM)/(3R)` `rArr v=sqrt((4GM)/(3R))=2sqrt((GM)/(3R))`. |
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