question_answer

Two satellites \[{{S}_{1}}\] and \[{{S}_{2}}\] revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 h and 8 h respectively. The radius of the orbit of \[{{S}_{1}}\] is 104 km. The speed of \[{{S}_{2}}\]relative to \[{{S}_{1}}\] when they are closest, in km/h is

A)  \[{{10}^{4}}\pi \]    

B)  \[2\times {{10}^{4}}\pi \]

C)   \[{{10}^{4}}\frac{\pi }{2}\]                                         

D)  \[4\times {{10}^{4}}\pi \]

Correct Answer: A

Solution :

                 From Kepler's law, \[\frac{{{r}_{2}}}{{{r}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2/3}}\] \[\frac{{{r}_{2}}}{{{10}^{4}}}={{\left( \frac{8}{1} \right)}^{2/3}}\] \[\frac{{{r}_{2}}}{{{10}^{4}}}=4\] \[\Rightarrow \]               \[{{r}_{2}}=4\times {{10}^{4}}km\] Velocity of first satellite \[{{v}_{1}}=\frac{2\pi {{r}_{1}}}{{{T}_{1}}}\] \[=\frac{2\pi \times {{10}^{4}}}{1}=2\pi \times {{10}^{4}}km/h\] Velocity of second satellite \[{{v}_{2}}=\frac{2\pi {{r}_{2}}}{{{T}_{2}}}\] \[=\frac{2\pi \times 4\times {{10}^{4}}}{8}=\pi \times {{10}^{4}}km/h\] Velocity of\[{{S}_{2}}\]relative to \[{{S}_{1}}={{v}_{2}}-{{v}_{1}}\]                 \[=\pi \times {{10}^{4}}-2\pi \times {{10}^{4}}\]                 \[=-\pi \times {{10}^{4}}km/h\] \[\therefore \]  Relative speed\[=\pi \times {{10}^{4}}km/h\]