Gravitation Part 1

Given, the time period of the first satellite, $T_1$ = 1h The time period of the second satellite, $T_2$ = 8 h The radius of the orbit of nearer satellite, $R_1$ = 2000 kmApplying the Kepler’s third law, $T^{2} \propto R^{3}$ $\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{R_1}{R_2}\right)^3$ ⇒ $\frac{2000}{R_2} = \left(\frac{1}{8}\right)^{\frac{2}{3}}$ ⇒ $R_2$ = 8000 km As we know the relation between the angular speed and timeperiod $\omega = \frac{\theta}{T}$ So, the angular speed of the nearer satellite to the orbit, $\omega_1 = \frac{2\pi}{1}\text{ rad/h}$ and the angular speed of the farther satellite to the orbit, $\omega_2 = \frac{2\pi}{8} = \frac{\pi}{4}\text{ rad/h}$ The speed of the nearer satellite to orbit, $v_1 = \omega_1 R_1 = (2\pi) \times 2 \times 10^{3}\text{ km/h}$ The speed of the farther satellite to the orbit $V_2 = \omega_2 R_2$ $= \left(\frac{\pi}{4}\right) \times 8000$ $= \pi \times 2 \times 10^{3}\text{ km/h}$ Thus, the relative angular speed of the nearer satellite to the farthersatellite, $\omega = \frac{V_1 - V_2}{R_1 - R_2}$ $= \frac{2\pi \times 2 \times 10^{3} - \pi \times 2 \times 10^{3}}{8000 - 2000}$ $= \frac{2\pi \times 10^{3}}{6000} = \frac{\pi}{3}\text{ rad/h}$ Comparing with, $\omega = \frac{\pi}{x}$ The value of x = 3.