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Average distance of the earth from the sun is L 1 . If one year of the earth is equal to D days, then one year of another planet whose average distance from the sun is L 2 will be:
$
A:D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{1}{2}}}days \\
B:D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{3}{2}}}days \\
C:D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{2}{3}}}days \\
D:D\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)days \\
$
Answer
456.9k+ views
Hint:In the given question, the concept of Kepler’s third law can be used. Kepler's third law states that the square of the orbital period of the planet is proportional to the cube of semi- major axis of the orbit. In simpler terms, the square of the 'year' of each planet divided by the cube of its distance from the Sun, gives you the same number for all the planets.
Complete step by step answer:: We know that according to Kepler’s third law, square of the orbital periods of planets is directly proportional to the cube of size of its orbit i.e. ${T^2}\alpha {R^3}$(where T = orbital period and R = size of its orbit).
In the question, we are provided with the following information:
Average distance of the earth from the sun = L 1 (Given)
One year of the earth = D days (Given)
Average distance of another planet from the sun = L 2 (Given)
One year of another planet = X days (Assumption)
According to Kepler’s third law:
For Earth:
${D^2}\alpha {L_1}^3$
Similarly for another planet:
${X^2}\alpha {L_2}^3$
Thus, we can say:
${\left( {\dfrac{X}{D}} \right)^2} = {\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^3}$
On solving, we get:
$X = D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{3}{2}}}$days
Thus, If one year of the earth is equal to D days, then 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 answer is Option B.
Note:Mercury takes less time period or days to orbit the sun in comparison to Earth. It can be justified with Kepler's third law according to which the time period for a planet to orbit the sun increases with the increase in the distance of the planet from the sun.
Complete step by step answer:: We know that according to Kepler’s third law, square of the orbital periods of planets is directly proportional to the cube of size of its orbit i.e. ${T^2}\alpha {R^3}$(where T = orbital period and R = size of its orbit).
In the question, we are provided with the following information:
Average distance of the earth from the sun = L 1 (Given)
One year of the earth = D days (Given)
Average distance of another planet from the sun = L 2 (Given)
One year of another planet = X days (Assumption)
According to Kepler’s third law:
For Earth:
${D^2}\alpha {L_1}^3$
Similarly for another planet:
${X^2}\alpha {L_2}^3$
Thus, we can say:
${\left( {\dfrac{X}{D}} \right)^2} = {\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^3}$
On solving, we get:
$X = D{\left( {\dfrac{{{L_2}}}{{{L_1}}}} \right)^{\dfrac{3}{2}}}$days
Thus, If one year of the earth is equal to D days, then 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 answer is Option B.
Note:Mercury takes less time period or days to orbit the sun in comparison to Earth. It can be justified with Kepler's third law according to which the time period for a planet to orbit the sun increases with the increase in the distance of the planet from the sun.
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