Question

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 ___

Hint:

Kepler’s third law relates orbital period and radius by \(T \propto r^{3/2}\). For satellites orbiting in opposite directions, relative angular velocity sums, affecting observed time period.

Answer 1

Correct Answer: 2.5

Step 1: The time period \(T_1\) of the geostationary satellite is 24 hours.

Step 2: Given, \(r_1 = 1.21 r_2\). The time period \(T\) is proportional to \(r^{3/2}\) (Kepler’s third law), so \[ \frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = (1.21)^{3/2} \approx 1.5 \]
Step 3: The second satellite moves in the opposite direction, so the relative angular velocity is the sum of their angular velocities. The time period observed from the first satellite is \[ \frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} \implies T = \frac{T_1 T_2}{T_1 + T_2} \]
Step 4: Substitute \(T_2 = \frac{T_1}{1.5} = 16\) hours, then \[ T = \frac{24 \times 16}{24 + 16} = \frac{384}{40} = 9.6 \text{ hours} \] Step 5: Since the measured time period is \(\frac{24}{p} = 9.6\), \[ p = \frac{24}{9.6} = 2.5 \]