AP EAPCET 18 May 2024 Shift 1 Solved Paper

Let the mass of the Earth be $M$.For the first satellite of mass $m$ at height $R_E$, the distance from Earth's center is $r_1=R_E+R_E=2R_E$.The magnitude of its gravitational pull on the Earth is $F_1=\frac{G M m}{(2R_E)^2}=\frac{G M m}{4R_E^2}$.For the second satellite of mass $1.5m$ at height $2R_E$, the distance from Earth's center is $r_2=R_E+2R_E=3R_E$.The magnitude of its gravitational pull on the Earth is $F_2=\frac{G M(1.5m)}{(3R_E)^2}=\frac{1.5G M m}{9R_E^2}=\frac{G M m}{6R_E^2}$.The net gravitational force on the Earth due to the two satellites is maximum when the two pulls are in the same direction:$F_{\max}=F_1+F_2=\frac{G M m}{4R_E^2}+\frac{G M m}{6R_E^2}=\frac{5G M m}{12R_E^2}$.The net gravitational force is minimum when the two pulls are opposite in direction:$F_{\min}=|F_1-F_2|=\left|\frac{G M m}{4R_E^2}-\frac{G M m}{6R_E^2}\right|=\frac{G M m}{12R_E^2}$.Therefore, the required ratio is:$\frac{F_{\min}}{F_{\max}}=\frac{\frac{G M m}{12R_E^2}}{\frac{5G M m}{12R_E^2}}=\frac{1}{5}$.Thus, $F_{\min}:F_{\max}=1:5$, which matches option D.