A geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance $r_2$ from the center of the earth, such that $r_1 = 1.21\,r_2$. The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is _____.

Step 1: Identify the given parameters.

The geostationary satellite has a time period $T_1 = 24\text{ h}$ and orbital radius $r_1$, while the second satellite has orbital radius $r_2$.

$$r_1 = 1.21\,r_2$$

Step 2: Calculate the orbital period of the second satellite.

Apply Kepler's Third Law to relate the orbital periods and radii.

$$T_2 = T_1\left(\frac{r_2}{r_1}\right)^{3/2}$$ $$T_2 = 24\left(\frac{1}{1.21}\right)^{3/2}$$ $$T_2 = \frac{24}{1.331} \approx 18.03\text{ h}$$

Step 3: Formulate the relative angular speed.

Since the satellites orbit in opposite directions, their relative angular speed is the sum of their individual angular speeds.

$$\omega_{\text{rel}} = \omega_1 + \omega_2$$ $$\omega_{\text{rel}} = \frac{2\pi}{T_1} + \frac{2\pi}{T_2}$$

Step 4: Calculate the relative time period.

The relative time period $T_{\text{rel}}$ is the time period of the second satellite as measured from the geostationary satellite.

$$T_{\text{rel}} = \frac{2\pi}{\omega_{\text{rel}}} = \frac{T_1 T_2}{T_1 + T_2}$$ $$T_{\text{rel}} = \frac{24 \times 18.03}{24 + 18.03} \approx 10.30\text{ h}$$

Step 5: Solve for the parameter $p$.

Equate the calculated relative period to the given expression $\frac{24}{p}$.

$$\frac{24}{p} = 10.30$$ $$p = \frac{24}{10.30} \approx 2.33$$

Alternatively, expressing $p$ analytically:

$$p = 1 + \left(\frac{r_1}{r_2}\right)^{3/2}$$ $$p = 1 + 1.21^{3/2} = 2.331$$

Answer: $2.33$ (or $2.331$)