Question
A geostationary satellite is orbiting the earth at a height of $$5R$$ above that surface of the earth, $$R$$ being the radius of the earth. The time period of another satellite in hour at a height of $$2R$$ from the surface of the earth is
A.
$$5$$
B.
$$10$$
C.
$$6\sqrt 2 $$
D.
$$\frac{6}{{\sqrt 2 }}$$
Answer :
$$6\sqrt 2 $$
Solution :
From Kepler's third law $${T^2} \propto {r^3}$$
\[\left[ {\begin{array}{*{20}{c}}
{{\rm{where,}}\,T = {\rm{time \,period \,of \,satellite}}}\\
{r = {\rm{radius\, of \,elliptical\, orbit}}\left( {{\rm{semi \,major \,axis}}} \right)}
\end{array}} \right]\]
Hence, $$T_1^2 \propto r_1^3\,{\text{and}}\,\,T_2^2 \propto r_2^3$$
$$\eqalign{
& {\text{So,}}\,\,\frac{{T_2^2}}{{T_1^2}} = \frac{{r_2^3}}{{r_1^3}} = \frac{{{{\left( {3R} \right)}^3}}}{{{{\left( {6R} \right)}^3}}} \cr
& {\text{or}}\,\,\frac{{T_2^2}}{{T_1^2}} = \frac{1}{8} \cr
& T_2^2 = \frac{1}{8}T_1^2 \Rightarrow {T_2} = \frac{{24}}{{2\sqrt 2 }} = 6\sqrt 2 \,h \cr} $$