Gravitation

Mass of first satellite, $m_1=m$Mass of second satellite, $m_2=1.5 m$Radius of Earth, $R_E$Calculating Distances:For the first satellite:Distance from Earth's center, $r_1=R_E+R_E=2 R_E$For the second satellite:Distance from Earth's center, $r_2=R_E+2 R_E=3 R_E$Gravitational Forces:Gravitational force by the first satellite:$F_1 = \frac{G M \cdot m}{(2 R_E)^2} = \frac{G M m}{4 R_E^2}$Gravitational force by the second satellite:$F_2 = \frac{1.5 \times G M m}{(3 R_E)^2} = \frac{1.5 G M m}{9 R_E^2}$Since $r_2 > r_1$, it follows that $F_1 > F_2$.Ratio of Forces:$\frac{F_2}{F_1} = \frac{ \frac{1.5 \cdot G M m}{9 R_E^2} }{ \frac{G M m}{4 R_E^2} } = \frac{1.5}{9} \times \frac{4}{1} = \frac{6}{9} = \frac{2}{3}$Thus, the ratio $F_2 : F_1 = 2 : 3$.