A body of mass m rises to a height h = R/5 fr | vedclass

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(B) The gravitational potential energy at the surface of the earth is $PE_1 = -\frac{GMm}{R}$.At a height $h = R/5$,the potential energy is $PE_2 = -\

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$\frac{5}{6} mgh$

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(B) The gravitational potential energy at the surface of the earth is $PE_1 = -\frac{GMm}{R}$.At a height $h = R/5$,the potential energy is $PE_2 = -\frac{GMm}{R+h} = -\frac{GMm}{R + R/5} = -\frac{GMm}{6R/5} = -\frac{5GMm}{6R}$.The increase in potential energy is $\Delta PE = PE_2 - PE_1 = -\frac{5GMm}{6R} - (-\frac{GMm}{R}) = \frac{GMm}{R} (1 - 5/6) = \frac{GMm}{6R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the expression,$\Delta PE = \frac{(gR^2)m}{6R} = \frac{mgR}{6}$.Given $h = R/5$,then $R = 5h$.Substituting $R = 5h$ into the expression,$\Delta PE = \frac{mg(5h)}{6} = \frac{5}{6} mgh$.

$A$ body of mass $m$ rises to a height $h = R/5$ from the earth's surface,where $R$ is the earth's radius. If $g$ is the acceleration due to gravity at the earth's surface,the increase in potential energy is:

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