Solution 2: Spherical vs Cylindrical Earth
1. Spherical Case:
\[ g_s = \frac{GM}{r^2} \propto \frac{R^3}{r^2} \implies v_s = \sqrt{r g_s} \propto \sqrt{\frac{R^3}{r}} \]
\[ g_s = \frac{GM}{r^2} \propto \frac{R^3}{r^2} \implies v_s = \sqrt{r g_s} \propto \sqrt{\frac{R^3}{r}} \]
2. Cylindrical Case:
\[ g_c = \frac{2G\lambda}{r} = \frac{2G(\pi R^2 \rho)}{r} \implies v_c = \sqrt{r g_c} \propto \sqrt{R^2} \] Velocity is constant with distance for the cylinder.
\[ g_c = \frac{2G\lambda}{r} = \frac{2G(\pi R^2 \rho)}{r} \implies v_c = \sqrt{r g_c} \propto \sqrt{R^2} \] Velocity is constant with distance for the cylinder.
3. Ratio:
\[ \frac{v_c}{v_s} = \sqrt{\frac{3r}{2R}} \] Since Moon’s distance $r \approx 60R$, the ratio is $\approx \sqrt{90} \approx 9.5$. The speed increases significantly.
\[ \frac{v_c}{v_s} = \sqrt{\frac{3r}{2R}} \] Since Moon’s distance $r \approx 60R$, the ratio is $\approx \sqrt{90} \approx 9.5$. The speed increases significantly.
Answer: (d) It will increase several times.