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A planet moving around sun sweeps area $\mathrm{A}_{1}$ in 2 days, $\mathrm{A}_{2}$ in 3 days and $\mathrm{A}_{3}$ in 6 days. Then, the relation between $\mathrm{A}_{1}, \mathrm{~A}_{2}$ and $\mathrm{A}_{3}$ is
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The correct answer is:
$3 A_{1}=2 A_{2}=A_{3}$
By Kepler's second law of motion,
$\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}$
$\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}$
$\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}$
$3 A_{1}=2 A_{2}=A_{3}$
$\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}$
$\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}$
$\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}$
$3 A_{1}=2 A_{2}=A_{3}$
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