For satellites, the time period $ T $ is proportional to the distance to the power of $ \frac{3}{2} $:

$ T \propto r^{\frac{3}{2}} $

This gives us the ratio of the time periods for the second and first satellites:

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

Converting this to angular velocity, $ \omega $, which is inversely proportional to the time period, we have:

$ \frac{\omega_2}{\omega_1} = \left(\frac{r_1}{r_2}\right)^{\frac{3}{2}} = (1.21)^{\frac{3}{2}} $

Calculating $ (1.21)^{\frac{3}{2}} $, we find:

$ \omega_2 = \omega_1 \times 1.331 $

The combined angular displacement of both satellites is $ 2\pi $ radians in the same time period $ t_0 $:

$ (\omega_2 + \omega_1) t_0 = 2\pi $

Solving for $ t_0 $:

$ t_0 = \frac{2\pi}{\omega_2 + \omega_1} = \frac{2\pi}{\left(\frac{4}{3} + 1\right) \omega_1} = \frac{6\pi}{7\omega_1} $

Since $ T_{GSS} $, the period of the geostationary satellite, is 24 hours, we substitute:

$ t_0 = \frac{6\pi}{2\pi} \frac{T_{GSS}}{7} = \frac{3 \times 24 \text{ hours}}{7} = \frac{24}{p} \text{ hours} $

Therefore, solving for $ p $:

$ p = \frac{7}{3} = 2.33 $