Solution 1: Relative Time Period
1. Relative Angular Velocity:
Both Earth and Satellite revolve West to East. \[ \omega_{rel} = \omega_s – \omega_E = \frac{2\pi}{T_s} – \frac{2\pi}{T_E} \]
Both Earth and Satellite revolve West to East. \[ \omega_{rel} = \omega_s – \omega_E = \frac{2\pi}{T_s} – \frac{2\pi}{T_E} \]
2. Time Calculation:
The time $t$ to meet again overhead (cover $2\pi$ relative angle): \[ t = \frac{2\pi}{\omega_{rel}} = \frac{1}{\frac{1}{T_s} – \frac{1}{T_E}} = \frac{T_E T_s}{T_E – T_s} \]
The time $t$ to meet again overhead (cover $2\pi$ relative angle): \[ t = \frac{2\pi}{\omega_{rel}} = \frac{1}{\frac{1}{T_s} – \frac{1}{T_E}} = \frac{T_E T_s}{T_E – T_s} \]
3. Final Value:
\[ t = \frac{24 \times 8}{24 – 8} = \frac{192}{16} = 12 \text{ h} \]
\[ t = \frac{24 \times 8}{24 – 8} = \frac{192}{16} = 12 \text{ h} \]
Answer: (b) 12 h