Question

A geostationary satellite is rotating in circular orbit of radius $36000km$ 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 $(R_e=6400km)$:

• 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 $T^{\prime 2}\propto (6400+h{)}^3$
Therefore, ${\left(T^{\prime }\right)}^2=T^2{\left(\frac{6400+h}{36000}\right)}^3$
$>T^2{\left(\frac{6400}{36000}\right)}^3$
$(\because h\ll R)$
$>(24{)}^2{\left(\frac{8}{45}\right)}^3$
$\therefore T^{\prime }>\frac{24\times 8\times \sqrt{8}}{45\times \sqrt{45}}$
$>1.8$ hour
So, $T^{\prime }\approx 2$ hour