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

Hint: 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).
\[\begin{array}{l}
\dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{\dfrac{{4{\pi ^2}r_1^3}}{{GM}}}}{{\dfrac{{4{\pi ^2}r_2^3}}{{GM}}}}\\
\dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{r_1^3}}{{r_2^3}}
\end{array}\]
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 \[{T_2}\].