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:

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

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2 A_{1} = 3 A_{2} = 6 A_{3}

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3 A_{1} = 2 A_{2} = 6 A_{3}

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6 A_{1} = 3 A_{2} = 2 A_{3}

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Solution

The correct option is D $3 A_{1} = 2 A_{2} = A_{3}$
According to Kepler's 2nd law of planetary motion, equal area is swept in equal time interval i.e. $\frac{Δ A}{Δ t} =$ constant
$∴$ $\frac{A_{1}}{t_{1}} =$ $\frac{A_{2}}{t_{2}} =$ $\frac{A_{3}}{t_{3}}$
Or $\frac{A_{1}}{2} =$ $\frac{A_{2}}{3} =$ $\frac{A_{3}}{6}$
$⟹$ $3 A_{1} = 2 A_{2} = A_{3}$

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A planet moving around sun sweeps area A1 in 2 days, A2 in 3 days and A3 in 6 days. Then the relation between A1,A2 and A3 is:

The correct option is D 3A1=2A2=A3According to Kepler's 2nd law of planetary motion, equal area is swept in equal time interval i.e. ΔAΔt= constant∴ A1t1= A2t2= A3t3Or A12= A23= A36 ⟹ 3A1=2A2=A3

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: