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A geostationary satellite is orbiting the earth at a height $5R$ above the surface of the earth , $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is
$5\,\, hr$
$10 \,\,hr$
$6\sqrt 2 \,\,hr$
$10\sqrt 2 \,\,hr$
Solution
According to Kepler's third law $T \propto {r^{3/2}}$
$\therefore \frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^{3/2}} = {\left( {\frac{{R + 2R}}{{R + 5R}}} \right)^{3/2}} = \frac{1}{{{2^{3/2}}}}$
Since ${T_1} = 24\,hours$
$So,\,\frac{{{T_2}}}{{24}} = \frac{1}{{{2^{3/2}}}}\,\,or\,\,{T_2} = \frac{{24}}{{{2^{3/2}}}} = \frac{{24}}{{2\sqrt 2 }} = 6\sqrt 2 \,hours$
Similar Questions
Match List$-I$ With List$-II$
| $(a)$ Gravitational constant $(G)$ | $(i)$ $\left[ L ^{2} T ^{-2}\right]$ |
| $(b)$ Gravitational potential energy | $(ii)$ $\left[ M ^{-1} L ^{3} T ^{-2}\right]$ |
| $(c)$ Gravitational potential | $(iii)$ $\left[ LT ^{-2}\right]$ |
| $(d)$ Gravitational intensity | $(iv)$ $\left[ ML ^{2} T ^{-2}\right]$ |
Choose the correct answer from the options given below:
