Question

A geostationary satellite is rotating in circular orbit of radius $36000 \,km$ around the earth. A spy satellite which is rotating in circular orbit at a height of some hundred kilometre from earths surface, has time period approximately equal to $(R_e = 6400\, km)$:

• A. 1 hour
• B. 2 hour
• C. 24 hour
• D. 36 hour

Answer 1

Answer: (B)

From Kepler's law
$T^{2} \propto r^{3}$
$\therefore T^{2} \propto(36000)^{3}$ and $T'^{2} \propto(6400+h)^{3}$
Therefore, $\left(T'\right)^{2}=T^{2}\left(\frac{6400+h}{36000}\right)^{3}$
$>T^{2}\left(\frac{6400}{36000}\right)^{3} $
$(\because h \ll R)$
$>(24)^{2}\left(\frac{8}{45}\right)^{3}$
$\therefore T' >\frac{24 \times 8 \times \sqrt{8}}{45 \times \sqrt{45}}$
$>1.8$ hour
So, $T' \approx 2$ hour