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# Average distance of the earth from the sun is ${L_1}$. If one year of the earth di equal to $D$ days, one year of another planet whose average distance from the sun is ${L_2}$ will beA) $D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{1}{2}}}$ daysB) $D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{3}{2}}}$ daysC) $D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{2}{3}}}$ daysD) $D\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)$ days

Last updated date: 04th Aug 2024
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Hint: In this question, the concept of Kepler's third law will be used that is, It states that the squares of the orbital periods of the planets are directly proportional to the cubes of the size of its orbit.
Complete step by step solution:

As we know that Kepler’s laws describe the planets orbit around the sun. The third law describes that any planet’s orbital period is proportional to the size of its orbit. It states that the squares of the orbital periods of the planets are directly proportional to the cubes of the size of its orbit.
So, here it is given that the average distance of the earth from the sun is ${L_1}$ and one year at earth is $D$ days.
Average distance of another planet from the sun is ${L_2}$.
As we know from Kepler’s third law that, ${T^2} \propto {a^3}$ [$T$ is the orbital period and$a$is the size of its orbit].
Let us assume one year of another planet is ${D_1}$ days.
Hence, for earth ${D^2} \propto {\left( {{L_1}} \right)^3}$ and for another planet ${D_1}^2 \propto {L_2}^3$ now we get,
$\Rightarrow {\left( {\dfrac{{{D_1}}}{D}} \right)^2} = {\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^3}$
Now, we will simplify the above equation and obtain,
$\therefore {D_1} = D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{3}{2}}}$
Thus, one year of another planet whose average distance from the sun is ${L_2}$ will be $D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{3}{2}}}$ days.

Hence, the correct option is (B).
Note:As we know that Kepler's third law implies that if the distance from the sun to a planet increases then the time for a planet to orbit the sun increases rapidly. Thus, it takes less days for Mercury to orbit the sun than earth.