Question

Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 x 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{\pi}{x}$ rad h⁻¹ where x is ________.

Answer 1

3

Answer 2

1. Determine R₂ using Kepler’s Third Law:
   Since $\frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3}$, with $T_2=8$ hr, $T_1=1$ hr, and $R_1=2\times10^3\, \text{km}$:

   $$\frac{8^2}{1^2} = \frac{R_2^3}{(2\times10^3)^3} \implies R_2^3 = 64 \times (2\times10^3)^3.$$

   Taking cube roots,

   $$R_2 = 4\times (2\times10^3) = 8\times10^3\, \text{km}.$$

2. Calculate Speeds:
   For circular motion, $v=\frac{2\pi R}{T}$.
   Nearest satellite:

   $$v_1=\frac{2\pi (2\times10^3)}{1}=4\pi\times10^3\, \text{km/hr}.$$

   Farther satellite:

   $$v_2=\frac{2\pi (8\times10^3)}{8}=2\pi\times10^3\, \text{km/hr}.$$

3. Find the Relative Speed:

   $$v_{\text{rel}} = |v_2-v_1| = |2\pi\times10^3 - 4\pi\times10^3| = 2\pi\times10^3\, \text{km/hr}.$$

4. Relative Angular Speed:
   The distance between the satellites when closest is

   $$R_{\text{sep}} = R_2-R_1 = (8\times10^3 - 2\times10^3) = 6\times10^3\, \text{km}.$$

   Thus,

   $$\omega_{\text{rel}}=\frac{v_{\text{rel}}}{R_{\text{sep}}}=\frac{2\pi\times10^3}{6\times10^3}=\frac{\pi}{3}\, \text{rad/hr}.$$

   This expression is given as $\frac{\pi}{x}$ rad/hr, so $x=3$.