Explanation

(a) According to Kepler’s law of periods
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto a^{3}\lbrack a = semi$-major axis]
Here, in case I $a\ is\ 7R$ as satellite is 6R above the earth and for a geostationary satellite T = 24 h
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (24)^{2} \propto (7R)^{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Similarly for case II
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto (3.5R)^{3}\ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{(24)^{2}}{T^{2}} = \frac{(7R)^{3}}{(3.5R)^{3}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} = \frac{(24)^{2}}{8}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = 6\sqrt{2}\ h$