Step-by-step Solution:

1. Understand the properties of the geostationary satellite (S1). A geostationary satellite orbits the Earth in the equatorial plane in the same direction as the Earth's rotation. Its time period is equal to the Earth's rotational period, which is 24 hours.

   • Time period of S1, $T_1 = 24$ hours.
   • Orbital radius of S1 is $r_1$.

2. Understand the properties of the second satellite (S2). The second satellite orbits in the equatorial plane in the opposite direction to the Earth's rotation.

   • Let its time period be $T_2$.
   • Its orbital radius is $r_2$.
   • We are given the relation between the radii: $r_1 = 1.21 r_2$.

3. Apply Kepler's Third Law of Planetary Motion. Kepler's third law states that the square of the orbital period of a planet (or satellite) is directly proportional to the cube of the semi-major axis of its orbit. For circular orbits, this means $T^2 \propto r^3$. Applying this law to both satellites: $\left( \frac{T_2}{T_1} \right)^2 = \left( \frac{r_2}{r_1} \right)^3$

4. Calculate the time period of the second satellite (T2). Substitute the given relation $r_1 = 1.21 r_2$ into the equation from Step 3: $\left( \frac{T_2}{T_1} \right)^2 = \left( \frac{r_2}{1.21 r_2} \right)^3 = \left( \frac{1}{1.21} \right)^3$ Taking the square root of both sides: $\frac{T_2}{T_1} = \left( \frac{1}{1.21} \right)^{3/2} = \left( \frac{1}{1.1^2} \right)^{3/2} = \frac{1}{(1.1)^3} = \frac{1}{1.331}$ Since $T_1 = 24$ hours, the time period of the second satellite is: $T_2 = \frac{T_1}{1.331} = \frac{24}{1.331} \text{ hours}$

5. Determine the relative angular velocity. The angular velocity of a satellite is given by $\omega = \frac{2\pi}{T}$.

   • Angular velocity of S1: $\omega_1 = \frac{2\pi}{T_1}$.
   • Angular velocity of S2: $\omega_2 = \frac{2\pi}{T_2}$.

   The satellites are orbiting in opposite directions. To find the time period of S2 as measured from S1, we need their relative angular velocity. Since they move in opposite directions, their angular velocities add up. $\omega_{rel} = \omega_1 + \omega_2$

6. Calculate the relative time period (T_rel). The relative time period is the time taken for the relative angle between the satellites to change by $2\pi$. It is given by: $T_{rel} = \frac{2\pi}{\omega_{rel}} = \frac{2\pi}{\omega_1 + \omega_2} = \frac{2\pi}{\frac{2\pi}{T_1} + \frac{2\pi}{T_2}}$ $T_{rel} = \frac{1}{\frac{1}{T_1} + \frac{1}{T_2}} = \frac{T_1 T_2}{T_1 + T_2}$

7. Substitute the values to find T_rel. Now, substitute $T_1 = 24$ hours and $T_2 = \frac{24}{1.331}$ hours into the expression for $T_{rel}$: $T_{rel} = \frac{24 \times \frac{24}{1.331}}{24 + \frac{24}{1.331}}$ Factor out 24 from the denominator: $T_{rel} = \frac{24 \times \frac{24}{1.331}}{24 \left(1 + \frac{1}{1.331}\right)} = \frac{\frac{24}{1.331}}{1 + \frac{1}{1.331}}$ $T_{rel} = \frac{\frac{24}{1.331}}{\frac{1.331 + 1}{1.331}} = \frac{24}{1.331 + 1} = \frac{24}{2.331} \text{ hours}$

8. Determine the value of p. The problem states that the time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. Comparing our result with the given expression: $\frac{24}{2.331} = \frac{24}{p}$ Therefore, $p = 2.331$.

9. Final Answer. The value of $p$ is 2.331. This value is within the specified range of 2.3 to 2.4.