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(B) The minimum energy required to escape the gravitational field of a planet is equal to the magnitude of the gravitational potential energy of the o

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$\frac{E}{16}$

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(B) The minimum energy required to escape the gravitational field of a planet is equal to the magnitude of the gravitational potential energy of the object at the surface: $E = \frac{GMm}{R}$.Given that the densities $\rho$ of the Earth and the Moon are equal,we have $M = \rho V = \rho \cdot \frac{4}{3}\pi R^3$.Since $V_e = 64 V_m$,we have $R_e^3 = 64 R_m^3$,which implies $R_e = 4 R_m$.Also,$M_e = \rho V_e = 64 \rho V_m = 64 M_m$.The energy required is $E = \frac{GMm}{R}$.Therefore,$\frac{E_m}{E_e} = \frac{M_m}{M_e} \cdot \frac{R_e}{R_m} = \frac{1}{64} \cdot \frac{4}{1} = \frac{1}{16}$.Thus,$E_m = \frac{E}{16}$.

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