A geostationary satellite above the equator i | vedclass

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(D) According to Kepler's third law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.For the geostationa

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$(2.33)$

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(D) According to Kepler's third law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.For the geostationary satellite $(GSS)$ with radius $r_1$ and time period $T_1 = 24 	ext{ hours}$,and the second satellite with radius $r_2$ and time period $T_2$,we have:$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{3/2} = \left(\frac{1}{1.21}ight)^{3/2} = \left(\frac{1}{1.1^2}ight)^{3/2} = \frac{1}{1.1^3} = \frac{1}{1.331} \approx \frac{1}{1.33} = \frac{3}{4}$.Thus,$T_2 = \frac{3}{4} T_1 = \frac{3}{4} 	imes 24 = 18 	ext{ hours}$.The angular velocity of the $GSS$ is $\omega_1 = \frac{2\pi}{T_1}$ and the angular velocity of the second satellite is $\omega_2 = \frac{2\pi}{T_2}$.Since the second satellite moves in the opposite direction,their relative angular velocity is $\omega_{rel} = \omega_1 + \omega_2$.The time period $t_0$ as measured from the $GSS$ is $t_0 = \frac{2\pi}{\omega_1 + \omega_2} = \frac{2\pi}{\frac{2\pi}{T_1} + \frac{2\pi}{T_2}} = \frac{T_1 T_2}{T_1 + T_2}$.Substituting the values: $t_0 = \frac{24 	imes 18}{24 + 18} = \frac{432}{42} = \frac{72}{7} 	ext{ hours}$.Given $t_0 = \frac{24}{p}$,we have $\frac{24}{p} = \frac{72}{7}$,which gives $p = \frac{24 	imes 7}{72} = \frac{7}{3} \approx 2.33$.

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