1.

A body released at the distance r(r gt R) from the centre of the earth. What is the velocity of the body when it strikes the surface of the earth?

Answer»

Solution :TOTAL Energy of the body `=KE+PE=0+[-(GMM)/(r^2)]= -(mgR^2)/(r )`
Let v be velocity acquired by body on reaching the surface of earth.
Total Energy on the surface `=(1)/(2)MV^(2)+[-(mgR^2)/(R )]= (1)/(2)mv^(2)-mgR`
According to law of conservatives of energy
`(1)/(2)mv^(2)-mgR= (mgR^2)/(r )`
`v^(2)= 2gR - (2gR^2)/(r )= 2gR^2[(1)/(R )-(1)/(r )]`
`implies v= Rsqrt(2G((1)/(R )-(1)/(r )))`.


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