Gravitation Part 1

Given Height of satellite from planet's surface, $h=11 R$ So, the total distance from the centre of planet $=11 R+R=12 R$ Time period of planet $=24 h$ Similarly, the total distance of second satellite from the centre of planet $=2 R+R=3 R$ Using the Kepler's law $T^{2} \propto R^{3}$ $\Rightarrow\;\; \left( \frac{T_1}{T_2} \right) = \left( \frac{R_1}{R_2} \right)^{\frac{3}{2}}$ Substituting the values in the above equation, we get $\left( \;\frac{24}{T} \right) = \left( \;\frac{12R}{3R} \right)^{\frac{3}{2}}$ $\therefore$ The time period of another satellite is $3 h$.