Question

Assuming that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. The length of the martian year in days is:
A.\[{\left( {1.52} \right)^{2/3}} \times 365\]
B.\[{\left( {1.52} \right)^{3/2}} \times 365\]
C.\[{\left( {1.52} \right)^2} \times 365\]
D.\[{\left( {1.52} \right)^3} \times 365\]

Answer 1

Hint: Use the expression for Kepler’s third law of planetary motion. This expression gives the relation between the time period of the planet around the sun and the radius of orbit of the planet around the sun. Rewrite this equation for the earth and mars. Substitute the given value of orbital radius of mars and the earth and the period of the earth around the sun in days in these equations and solve it.

Formula used:
The expression for Kepler’s third law of planetary motion is given by
\[{T^2} = k{a^3}\] …… (1)
Here, \[T\] is the time period of motion of the planet around the sun, \[k\] is the constant and \[a\] is the semi-major axis or radius of the orbit of the planet around the sun.

Complete step by step answer:
We have given that the earth and mars are moving around the sun in their orbits.
Let \[{T_E}\] be the time period of the earth around the sun and \[{T_M}\] be the time period of mars around the sun.
The time period of the earth around the sun is 365 days.
\[{T_E} = 365\,{\text{days}}\]
Let \[{a_E}\] be the radius of the earth around the sun and \[{a_M}\] be the radius of mars around the sun.
We have given that the radius of the orbit of mars around the sun is 1.52 times the radius of orbit of the earth around the sun.
\[{a_M} = 1.52{a_E}\]
Rewrite equation (1) for the earth.
\[T_E^2 = ka_E^3\] …… (2)
Rewrite equation (1) for mars.
\[T_M^2 = ka_M^3\] …… (3)
Divide equation (3) by equation (2).
\[\dfrac{{T_M^2}}{{T_E^2}} = \dfrac{{ka_M^3}}{{ka_E^3}}\]
\[ \Rightarrow \dfrac{{T_M^2}}{{T_E^2}} = \dfrac{{a_M^3}}{{a_E^3}}\]
Take square root on both sides of the above equation.
\[ \Rightarrow \dfrac{{{T_M}}}{{{T_E}}} = {\left( {\dfrac{{a_M^3}}{{a_E^3}}} \right)^{3/2}}\]
Rearrange the above equation for \[{T_M}\].
\[ \Rightarrow {T_M} = {\left( {\dfrac{{a_M^3}}{{a_E^3}}} \right)^{3/2}}{T_E}\]
Substitute \[1.52{a_E}\] for \[{a_M}\] and \[365\,{\text{days}}\] for \[{T_E}\] in the above equation.
\[ \Rightarrow {T_M} = {\left( {\dfrac{{1.52{a_E}}}{{{a_E}}}} \right)^{3/2}}\left( {365\,{\text{days}}} \right)\]
\[ \therefore {T_M} = {\left( {1.52} \right)^{3/2}} \times 365\,{\text{days}}\]

Therefore, the length of the martian year in days is \[{\left( {1.52} \right)^{3/2}} \times 365\].Hence, the correct option is B.

Note:The students should keep in mind that we have asked the length of the martian year in days. Hence, the time period of the earth around the sun should also be taken in days. If this time period is taken in a year and not in days the final result we get will be correct but other than the options given.