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The period of a satellite of the earth very near to the surface of the earth is $T_{0} .$ What is the approximate period of the geostationary satellite in terms of $T_{0} ?$ (Height of geostationary satellite above the surface of earth is 36000km)

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$T_{0} / \sqrt{7}$

$\sqrt{7} T_{0}$

$7 T_{0}$

$7 \sqrt{7} T_{0}$

answer is D.

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Detailed Solution

$T^{2} α r^{3}, R = 6400 km$

$∴ r = (6400 + 36000) km = 42, 400 km$

$∴ \frac{T}{T_{0}} = {(\frac{r}{R})}^{3 / 2} = {(\frac{42400}{6400})}^{3 / 2} \Rightarrow T \approx 7 \sqrt{7} T_{0}$

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