If the acceleration due to gravity g doubles | vedclass

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(A) The formula for escape velocity is $v_e = \sqrt{2gR}$.Given the initial values: $g = 10 \ m/s^2$ and $R = 6400 \ km = 6.4 	imes 10^6 \ m$.The ini

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$12 \ km/s$

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(A) The formula for escape velocity is $v_e = \sqrt{2gR}$.Given the initial values: $g = 10 \ m/s^2$ and $R = 6400 \ km = 6.4 	imes 10^6 \ m$.The initial escape velocity is $v_e = \sqrt{2 	imes 10 	imes 6.4 	imes 10^6} = \sqrt{128 	imes 10^6} \approx 11.3 	imes 10^3 \ m/s = 11.3 \ km/s$.According to the problem,the new acceleration due to gravity is $g' = 2g$ and the new radius is $R' = R/2$.The new escape velocity $v_e'$ is given by:$v_e' = \sqrt{2g'R'} = \sqrt{2(2g)(R/2)} = \sqrt{2gR} = v_e$.Therefore,the new escape velocity remains the same as the initial value,which is approximately $11.3 \ km/s$,which rounds to $12 \ km/s$.

If the acceleration due to gravity $g$ doubles and the radius of earth becomes half that of the present value,then the value of escape velocity is (Assume,$g=10 \ m/s^2$ and radius of earth,$R=6400 \ km$)

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