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(B) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant.This means the area swept per unit time is co

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$3 A_{1} = 2 A_{2} = A_{3}$

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(B) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant.This means the area swept per unit time is constant:$\frac{A_{1}}{t_{1}} = \frac{A_{2}}{t_{2}} = \frac{A_{3}}{t_{3}}$Given $t_{1} = 2$ days,$t_{2} = 3$ days,and $t_{3} = 6$ days,we have:$\frac{A_{1}}{2} = \frac{A_{2}}{3} = \frac{A_{3}}{6}$To simplify this,multiply the entire equation by $6$:$6 	imes \frac{A_{1}}{2} = 6 	imes \frac{A_{2}}{3} = 6 	imes \frac{A_{3}}{6}$$3 A_{1} = 2 A_{2} = A_{3}$

$A$ planet moving around the Sun sweeps out an area $A_{1}$ in $2$ days,$A_{2}$ in $3$ days,and $A_{3}$ in $6$ days. Then,the relation between $A_{1}, A_{2}$,and $A_{3}$ is

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