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 \[={{v}_{1}}-{{v}_{2}}=2\pi \times {{10}^{2}}-\pi \times {{10}^{4}}=\pi \times {{10}^{4}}kmph\]
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