question_answer

Two satellites \[{{S}_{1}}\]and \[{{S}_{2}}\]are revolving round a planet in coplanar and concentric circular orbits of radii \[{{R}_{1}}\]and \[{{R}_{2}}\]in the same direction respectively. Their respective periods of revolution are 1 hr and 8 hrs. Find the radius of the orbit of satellite \[{{S}_{1}}\]is equal to \[{{10}^{4}}km\]. Then-relative speeds when they are closest, in kmph.

Answer:

According to Kepler?s third law, \[\frac{{{T}^{2}}}{{{R}^{3}}}=\]constant \[\therefore \frac{T_{1}^{2}}{T_{2}^{2}}=\frac{R_{1}^{3}}{R_{3}^{2}}or\,\,\frac{1}{{{10}^{12}}}=\frac{64}{R_{2}^{3}}\,\,or\,\,{{R}_{2}}=4\times {{10}^{4}}km\] Distance travelled in one revolution, \[{{S}_{1}}=2\pi {{R}_{1}}=2\pi \times {{10}^{4}}\] \[{{V}_{1}}=\frac{{{S}_{1}}}{{{t}_{1}}}=\frac{2\pi \times {{10}^{4}}}{1}=2\pi \times {{10}^{4}}kmph\] \[{{V}_{2}}=\frac{{{S}_{2}}}{{{t}_{2}}}=\frac{2\pi {{R}_{2}}}{{{t}_{2}}}=\frac{2\pi \times 4\times {{10}^{4}}}{8}=\pi \times {{10}^{4}}\,kmph\] \[\therefore \]Relative velocity \[={{v}_{1}}-{{v}_{2}}=2\pi \times {{10}^{2}}-\pi \times {{10}^{4}}=\pi \times {{10}^{4}}kmph\]