question_answer

For a satellite orbiting close to the surface of the earth the period of revolution is 84 min. Find the time period of another satellite orbiting at a distance three times the radius of the earth from its surface.

Answer:

Using, \[{{T}^{2}}\propto {{R}^{3}}\Rightarrow {{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}\] \[\Rightarrow {{\left( \frac{{{T}_{2}}}{84} \right)}^{2}}={{\left( \frac{3{{R}_{1}}}{R} \right)}^{3}}\,\,(or)\] \[\left( \frac{{{T}_{2}}}{84} \right)=27\Rightarrow T_{2}^{2}=27\times {{84}^{2}}\] \[(or)\,\,{{T}_{2}}=84\times 3\sqrt{3}\min \]