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 ______.

The geostationary satellite revolves in the same sense as the earth’s rotation, so its period in an inertial frame is the sidereal day:

$$T_1 = 24 \text{ h}$$ (to the accuracy required in the problem)

According to Kepler’s third law, the time period is proportional to $$r^{3/2}$$:

$$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$$

Given $$r_1 = 1.21\,r_2$$, we obtain

$$T_2 = T_1 \left(\frac{1}{1.21}\right)^{3/2}$$

Compute the numerical factor:
$$1.21^{3/2} = (1.21)^{1.5} \approx 1.331$$
$$\left(\frac{1}{1.21}\right)^{3/2} \approx \frac{1}{1.331} \approx 0.752$$

Hence

$$T_2 = 24 \times 0.752 \text{ h} \approx 18.0 \text{ h}$$

The second satellite moves in the opposite (retrograde) direction, so its angular velocity is opposite in sign to that of the geostationary satellite. In magnitude:

$$\omega_1 = \frac{2\pi}{T_1}, \quad \omega_2 = -\,\frac{2\pi}{T_2}$$

Their relative angular speed is

$$\omega_{\text{rel}} = \omega_1 - \omega_2 = \frac{2\pi}{T_1} + \frac{2\pi}{T_2} = 2\pi\left(\frac{1}{T_1} + \frac{1}{T_2}\right)$$

The time period with which the second satellite is seen to pass the geostationary satellite (as measured from the geostationary frame) is therefore

$$T_{\text{rel}} = \frac{2\pi}{\omega_{\text{rel}}} = \frac{1}{\dfrac{1}{T_1} + \dfrac{1}{T_2}} = \frac{T_1\,T_2}{T_1 + T_2}$$

Substituting $$T_1 = 24\text{ h}$$ and $$T_2 \approx 18.0\text{ h}$$:

$$T_{\text{rel}} = \frac{24 \times 18.0}{24 + 18.0} \text{ h} = \frac{432}{42} \text{ h} \approx 10.3 \text{ h}$$

This observed period is given in the problem as $$\dfrac{24}{p} \text{ h}$$, so

$$\frac{24}{p} = 10.3 \Longrightarrow p = \frac{24}{10.3} \approx 2.33$$

Thus the required value is

$$\boxed{p \approx 2.3 - 2.4}$$