$\text{A geostationary satellite above the equator is orbiting around the earth at a fixed distance}{r}_1\text{from the center}$

$\text{of the earth.}\text{A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's}$

$\text{rotation, at a distance}{r}_2\text{from the center of the earth, such that}{\mathrm{ r}}_1=1.21r_2.$
$\text{The time period of the second satellite as measured from the geostationary satellite is}\frac{24}{p}\text{hours.}$
$\text{The value of}\mathrm{ p }\text{is}\text{\_\_\_\_\_}$

The correct answer is 2.33.

Let's break down the solution step-by-step:

Step 1: Understand the Geostationary Satellite
A geostationary satellite orbits the Earth in the equatorial plane in the same direction as the Earth's rotation (prograde direction).
The orbital time period of a geostationary satellite is:
$T_1=24\text{ hours}$
Its orbital angular velocity is:
${\omega }_1=\frac{2\pi }{T_1}$

Step 2: Find the orbital period of the second satellite using Kepler's Third Law
Kepler's Third Law of planetary motion states that the square of the orbital period is proportional to the cube of the orbital radius:
$T^2\propto r^3$
For the two satellites, we can write:
$\frac{T_1}{T_2}={\left(\frac{r_1}{r_2}\right)}^{3/2}$
We are given that:
$r_1=1.21r_2\Rightarrow \frac{r_1}{r_2}=1.21$
Substituting this value into the equation:
$\frac{T_1}{T_2}={\left(1.21\right)}^{3/2}={\left({1.1}^2\right)}^{3/2}={1.1}^3=1.331$
This gives the orbital period of the second satellite as:
$T_2=\frac{T_1}{1.331}$

Step 3: Analyze the relative angular velocity
Since the second satellite orbits in the opposite direction (retrograde direction), its angular velocity has the opposite sign:
${\omega }_2=-\frac{2\pi }{T_2}=-1.331(\frac{2\pi }{T_1})$
The relative angular velocity of the second satellite as measured by an observer on the geostationary satellite is:
${\omega }_{\text{rel}}=|{\omega }_2-{\omega }_1|$
${\omega }_{\text{rel}}=|-1.331\left(\frac{2\pi }{T_1}\right)-\frac{2\pi }{T_1}|=2.331\left(\frac{\mathrm{2<!--\; <mi-->\pi }}{T_1}\right)$

Step 4: Find the relative time period and the value of p
The time period of the second satellite as measured from the geostationary satellite is:
$T_{\text{rel}}=\frac{\mathrm{2<!--\; <mi-->\pi }}{{\omega }_{\text{rel}}}=\frac{T_1}{2.331}=\frac{24}{2.331}\text{ hours}$
Comparing this with the given format:
$T_{\text{rel}}=\frac{\mathrm{24<!--\; <mi-->p}}{}\text{ hours}$
We get:
$p=2.331\approx 2.33$