Question

If the semi-major axis of earth's elliptical orbit near sun becomes twice then the length of a year approximately in days-
A. $730$
B. $\left( {\sqrt 2 } \right)730$
C. $\left( {\sqrt 2 } \right)365$
D. $183$

Answer 1

Kepler's third law concept tells us that the earth's orbital time period is directly proportional to the cube of its semi-major axis. We will use this law equation to determine the new orbital time period when the semi-major axis is increased to twice its original length.

Complete step by step answer:
Assume:
The initial semi-major elliptical axis of the elliptical orbit of earth near the sun is ${r_1}$.
The final semi-major elliptical axis of the elliptical orbit of earth near the sun is ${r_2}$.
The number of days in a year at the semi-major axis $r{}_1$ is ${T_1}$.
The number of days in a year at the semi-major axis ${r_2}$ is ${T_2}$.

From the concept of the third law of Kepler, we can write:
${T^2} = \dfrac{{4{\pi ^2}{r^3}}}{{GM}}$
Here, T is the time period of the orbital, r is the elliptical semi-major axis, G is the constant of gravity, and M is the sun's mass.

Let us write the above expression for the time period of the orbital when the semi-major axis is ${r_1}$.
$T_1^2 = \dfrac{{4{\pi ^2}r_1^3}}{{GM}}$......(1)

The expression for the orbital time period when the semi-major axis is ${r_2}$ is:
$T_2^2 = \dfrac{{4{\pi ^2}r_2^3}}{{GM}}$......(2)

Divide equation (1) and equation (2).

\dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{\dfrac{{4{\pi ^2}r_1^3}}{{GM}}}}{{\dfrac{{4{\pi ^2}r_2^3}}{{GM}}}}\\\ \dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{r_1^3}}{{r_2^3}} \end{array}$$ On further simplifying the above equation, we can write: $$\dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^{\dfrac{3}{2}}}$$……(3) It is given that the semi-major axis is increased to double its initial value. $${r_2} = 2{r_1}$$ Substitute $$2{r_1}$$ for $${r_2}$$ in equation (3). $$\begin{array}{c} \dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{{{r_1}}}{{2{r_1}}}} \right)^{\dfrac{3}{2}}}\\\ = {\left( {\dfrac{1}{2}} \right)^{\dfrac{3}{2}}} \end{array}$$……(4) We know that the orbital time period of the earth is $${\rm{365 days}}$$ when its semi-major axis is $${r_1}$$. $${T_1} = 365{\rm{ days}}$$ Substitute $$365{\rm{ days}}$$ for $${T_1}$$ in equation (4). $$\begin{array}{c} \dfrac{{365{\rm{ days}}}}{{{T_2}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{3}{2}}}\\\ {T_2} = \sqrt 8 \cdot 365{\rm{ days}}\\\ {\rm{ = }}\left( {\sqrt 2 } \right){\rm{730 days}} \end{array}$$ Therefore, the length of a year will become $$\left( {\sqrt 2 } \right){\rm{730 days}}$$ when the semi-major axis of earth's elliptical orbit near the sun becomes twice its initial length **So, the correct answer is “Option B”.** **Note:** It would be an added advantage to remember that the number of days in a year is $$365$$ when the semi-elliptical orbit is not increased. Take extra care while rearranging the equations to get the value of the orbital time period $${T_2}$$.