Question

A geostationary satellite is orbiting the earth at a height of $5\,R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2\,R$ from the surface of the earth is

• A. 5
• B. 10
• C. $6\sqrt2$
• D. $\frac {6}{\sqrt2}$

Answer 1

Answer: (C)

$6\sqrt2$

Answer 2

According to Kepler's third law, $T ^{2} \propto r ^{3}$
$\frac{ T _{2}}{ T _{1}}= \left(\frac{ r _{2}}{ r _{1}}\right)^{3 / 2}$
Given $T _{1}=24\, h$
$r _{1}= R + h _{1}= R +5 R =6 R$
$r _{2}= R + h _{2}= R +2 R =3 R$
On solving we get
$T _{2}=6 \sqrt{2}\, h$