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(C) The gravitational force on a body of mass $m$ at the surface of the earth is $F_0 = \frac{G M_e m}{R^2}$.At a height $h$,the force is $F_h = \frac

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$\frac{8 \pi \rho Gmh}{3}$

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(C) The gravitational force on a body of mass $m$ at the surface of the earth is $F_0 = \frac{G M_e m}{R^2}$.At a height $h$,the force is $F_h = \frac{G M_e m}{(R+h)^2}$.The error in weighing is the difference in forces: $\Delta F = F_0 - F_h = \frac{G M_e m}{R^2} - \frac{G M_e m}{(R+h)^2}$.$\Delta F = \frac{G M_e m}{R^2} \left[ 1 - (1 + \frac{h}{R})^{-2} \right]$.Using binomial expansion $(1+x)^n \approx 1+nx$ for $h \ll R$,we get $(1 + \frac{h}{R})^{-2} \approx 1 - \frac{2h}{R}$.So,$\Delta F \approx \frac{G M_e m}{R^2} \left[ 1 - (1 - \frac{2h}{R}) \right] = \frac{G M_e m}{R^2} \left( \frac{2h}{R} \right) = \frac{2 G M_e m h}{R^3}$.Since the density $\rho = \frac{M_e}{\frac{4}{3} \pi R^3}$,we have $M_e = \frac{4}{3} \pi R^3 \rho$.Substituting $M_e$ into the expression: $\Delta F = \frac{2 G m h}{R^3} \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{8}{3} \pi \rho G m h$.

Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$,the error in weighing is

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