Question

If the distance between the earth and the sun were reduced to half its present value, then the number of days in one year would have been
A) $65$
B) $129$
C) $183$
D) $730$

Answer 1

The number of days in a year or as we commonly call it, one calendar year, is the number of days of the earth’s orbital period. So if we can find the change in the time which the earth takes to complete one revolution around the earth, we can find the new duration of a calendar year.

Complete step by step solution:
The laws explaining the orbiting of planets, asteroids and comets around are known as Kepler’s laws.
Kepler’s laws are a compilation of three laws which are as follows:
Every planet’s orbit is an ellipse with the sun at a focus
A line joining the sun and a planet sweeps out equal areas in equal times
The square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit.
Since we are concerned with the orbital period of the earth, we will only deal with the third law.
Earth’s orbit has an eccentricity of less than $0.02$. Due to this fact, we can approximately consider the earth’s orbit to be circular. Reviewing the third law with this approximation, we can state that
${{T}^{2}}={{r}^{3}}$ where $T$ is the period of revolution of the earth and $r$ is the radius of the earth’s orbit.
Considering two cases, where the radius in the second case is half of the radius in the first case, we can say that
$\dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{r_{1}^{3}}{r_{2}^{3}}$ where ${{T}_{1}}$ and ${{T}_{2}}$ are the time periods of revolution of the earth in the two corresponding cases.
Substituting the values in the above expression, we get

& \dfrac{{{(365)}^{2}}}{T_{2}^{2}}=\dfrac{r_{1}^{3}}{{{(0.5{{r}_{1}})}^{3}}} \\\ & \Rightarrow \dfrac{{{(365)}^{2}}}{T_{2}^{2}}={{(2)}^{3}} \\\ & \Rightarrow T_{2}^{2}=\dfrac{{{(365)}^{2}}}{8} \\\ & \Rightarrow {{T}_{2}}=\dfrac{365}{2\sqrt{2}}=129.06 \\\ & \Rightarrow {{T}_{2}}\simeq 129 \\\ \end{aligned}$$ **Hence the new number of days in the year would be $$129$$ . Thus the correct option is (B).** **Note:** Kepler’s laws are used to plot and time the positions of comets and asteroids as they orbit the sun, plot the orbit of moons or man-made space satellites and also used to plot a course to send a rocket into space. Kepler’s laws encapsulate the principle of conservation of angular momentum for planetary systems.