Question

A planet moving around sun sweeps 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

• A. $3 A_1 \,=\,2A_2 \,=\, A_3$
• B. $2 A_1 \,=\,3A_2 \,=\, 6A_3$
• C. $3 A_1 \,=\,2A_2 \,=\, 6A_3$
• D. $6 A_1 \,=\,3A_2 \,=\, 2A_3$

Answer 1

The Correct Option is A

By Kepler's second law of motion,
$\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}\,\,\,\,\,\dots(i)$
$\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}$
or $\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}$
Or $3 A_{1}=2 A_{2}=A_{3}$