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(B) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.The energy in the first orbit $(r_1 = 2R)$ is

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$\frac{mgR}{12}$

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(B) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.The energy in the first orbit $(r_1 = 2R)$ is $E_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.The energy in the second orbit $(r_2 = 3R)$ is $E_2 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.The energy required to shift the lab is $\Delta E = E_2 - E_1$.$\Delta E = -\frac{GMm}{6R} - (-\frac{GMm}{4R}) = \frac{GMm}{4R} - \frac{GMm}{6R} = \frac{3GMm - 2GMm}{12R} = \frac{GMm}{12R}$.Since the acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting $GM = gR^2$ into the expression for $\Delta E$:$\Delta E = \frac{(gR^2)m}{12R} = \frac{mgR}{12}$.

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