1. Given data

• Period of nearer satellite: $T_1 = 1\,\text{h}$
• Period of farther satellite: $T_2 = 8\,\text{h}$
• Radius of nearer satellite: $r_1 = 2\times 10^3\,\text{km}$
• Both move in the same plane, in anticlockwise circular motion.

We need the angular speed of the farther satellite as observed from the nearer satellite at the instant when they are closest.

2. Find the radius of the farther satellite using Kepler's third law

For circular orbits around the same planet, $T^2 \propto r^3$ So, $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$ Substitute $T_2=8$ and $T_1=1$: $8^2 = \left(\frac{r_2}{r_1}\right)^3$ $64 = \left(\frac{r_2}{r_1}\right)^3$ $\frac{r_2}{r_1} = 4$ Hence, $r_2 = 4r_1 = 4(2\times 10^3) = 8\times 10^3\,\text{km}$

3. Angular speeds of the satellites

For uniform circular motion, $\omega = \frac{2\pi}{T}$ Thus, $\omega_1 = \frac{2\pi}{1} = 2\pi\,\text{rad h}^{-1}$ $\omega_2 = \frac{2\pi}{8} = \frac{\pi}{4}\,\text{rad h}^{-1}$

4. Configuration when they are closest

The satellites are closest when they lie on the same radial line from the planet and on the same side. At that instant, let both be on the positive $x$-axis.

Then:

• nearer satellite position: $(r_1,0)$
• farther satellite position: $(r_2,0)$

Relative position of farther w.r.t. nearer: $\vec R = (r_2-r_1,0)$ Initially, this line of sight is along the $x$-axis.

5. Relative velocity of farther satellite with respect to nearer satellite

At that instant, both velocities are perpendicular to the radius vector, i.e. along positive $y$-direction.

So, $v_1 = r_1\omega_1, \qquad v_2 = r_2\omega_2$ Compute: $v_1 = (2\times 10^3)(2\pi)=4000\pi\,\text{km h}^{-1}$ $v_2 = (8\times 10^3)\left(\frac{\pi}{4}\right)=2000\pi\,\text{km h}^{-1}$

Therefore relative velocity of farther with respect to nearer is $v_{rel} = v_2-v_1 = 2000\pi-4000\pi = -2000\pi\,\text{km h}^{-1}$ Magnitude: $|v_{rel}|=2000\pi\,\text{km h}^{-1}$

6. Angular speed as observed from the nearer satellite

The observed angular speed is the angular speed of the relative position vector $\vec R$ about the nearer satellite.

At the closest instant, $\vec R$ has magnitude $|\vec R| = r_2-r_1 = 8000-2000=6000\,\text{km}$ Since the relative velocity is perpendicular to $\vec R$ at that instant, $\omega_{obs} = \frac{|v_{rel}|}{|\vec R|}$ So, $\omega_{obs} = \frac{2000\pi}{6000} = \frac{\pi}{3}\,\text{rad h}^{-1}$

Thus, $\frac{\pi}{x} = \frac{\pi}{3} \implies x=3$

7. Final answer

$\boxed{3}$

The derived answer matches the stored correct answer.