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(D) The gravitational force exerted by a satellite of mass $m_s$ at a distance $r$ from the center of the earth is $F = \frac{G M_E m_s}{r^2}$.For the

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$1: 5$

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(D) The gravitational force exerted by a satellite of mass $m_s$ at a distance $r$ from the center of the earth is $F = \frac{G M_E m_s}{r^2}$.For the first satellite: mass $m_1 = m$,height $h_1 = R_E$,so distance $r_1 = R_E + R_E = 2 R_E$. Force $F_1 = \frac{G M_E m}{(2 R_E)^2} = \frac{G M_E m}{4 R_E^2}$.For the second satellite: mass $m_2 = 1.5 m$,height $h_2 = 2 R_E$,so distance $r_2 = R_E + 2 R_E = 3 R_E$. Force $F_2 = \frac{G M_E (1.5 m)}{(3 R_E)^2} = \frac{1.5 G M_E m}{9 R_E^2} = \frac{G M_E m}{6 R_E^2}$.Maximum force $F_{\max} = F_1 + F_2 = \frac{G M_E m}{R_E^2} (\frac{1}{4} + \frac{1}{6}) = \frac{G M_E m}{R_E^2} (\frac{3+2}{12}) = \frac{5}{12} \frac{G M_E m}{R_E^2}$.Minimum force $F_{\min} = F_1 - F_2 = \frac{G M_E m}{R_E^2} (\frac{1}{4} - \frac{1}{6}) = \frac{G M_E m}{R_E^2} (\frac{3-2}{12}) = \frac{1}{12} \frac{G M_E m}{R_E^2}$.Ratio $\frac{F_{\min}}{F_{\max}} = \frac{1/12}{5/12} = \frac{1}{5} = 1: 5$.

Two satellites of masses $m$ and $1.5 m$ are revolving around the earth in two circular orbits of heights $R_E$ and $2 R_E$ respectively,where $R_E$ is the radius of the earth. The ratio of the minimum and maximum gravitational forces exerted by the two satellites on the earth is

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