Time Period of Counter Orbiting Equatorial Satellite

The time period of a geostationary satellite orbiting the Earth is equal to the rotational period of the Earth: $T_1 = 24 \text{ hours}$

According to Kepler's Third Law, the time period of a satellite in a circular orbit around the Earth is related to its orbital radius $r$ by: $T^2 \propto r^3 \implies T \propto r^{3/2}$

Therefore, the ratio of the time periods of the second satellite ($T_2$) and the geostationary satellite ($T_1$) is: $\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$

Given that $r_1 = 1.21 \, r_2$, we have: $\frac{r_2}{r_1} = \frac{1}{1.21} = \frac{1}{(1.1)^2}$

Substituting this into the ratio: $\frac{T_2}{T_1} = \left(\frac{1}{(1.1)^2}\right)^{3/2} = \frac{1}{(1.1)^3} = \frac{1}{1.331}$

So, the time period of the second satellite with respect to the Earth's center is: $T_2 = \frac{24}{1.331} \text{ hours}$

The angular speeds of the two satellites with respect to the Earth's center are: $\omega_1 = \frac{2\pi}{T_1}$ $\omega_2 = \frac{2\pi}{T_2}$

Since the second satellite orbits in the equatorial plane in the direction opposite to the geostationary satellite, their relative angular velocity as seen from the geostationary satellite is the sum of their individual angular velocities: $\omega_{\text{rel}} = \omega_1 + \omega_2 = \frac{2\pi}{T_1} + \frac{2\pi}{T_2}$

The time period of the second satellite as measured from the geostationary satellite ($T_{\text{rel}}$) is given by: $T_{\text{rel}} = \frac{2\pi}{\omega_{\text{rel}}} = \frac{1}{\frac{1}{T_1} + \frac{1}{T_2}}$

Given $T_{\text{rel}} = \frac{24}{p}$ hours, we set up the equation: $\frac{24}{p} = \frac{1}{\frac{1}{24} + \frac{1.331}{24}} = \frac{24}{1 + 1.331} = \frac{24}{2.331}$

Equating the denominators: $p = 1 + 1.331 = 2.331$

Rounding off to two decimal places gives: $p \approx 2.33$