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\[=\pi \times {{10}^{4}}km/h\]You need to login to perform this action.
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