Earth's orbit is an ellipse with eccentricity $0.0167$. Thus the earth's distance from the sun and speed as it moves around the sun varies from day-to-day. This means that the length of the solar day is not constant through the year. Assume that the earth's spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year ?

Let, $m=$ mass of earth

$v_{p}=$ velocity of earth at perigee

$v_{a}=$ velocity of earth at apogee

$\omega_{p}=$ angular velocity of earth at perigee

$\omega_{a}=$ angular velocity of earth at apogee

Angular momentum and areal velocity are constant.

$\therefore r_{p}^{2} \omega_{p}=r_{a}^{2} \omega_{a}$

If $a$ is the semi-major axis of the earth's orbit, then

$r_{p}=a(1-e)$ and $r_{a}=a(1+e) \quad \ldots$ (ii)

$\therefore \frac{\omega_{p}}{\omega_{a}}=\left(\frac{1+e}{1-e}\right)^{2}$ [From equ. $(i)$ and $(ii)$]

$\therefore \frac{\omega_{p}}{\omega_{a}}=1.0691 \quad(\because e=0.0167)$

Let $\omega$ be angular speed which is geometric average of $\omega_{p}$ and $\omega_{a}$ and corresponds to average

solar day,

$\therefore\left(\frac{\omega_{p}}{\omega_{a}}\right)\left(\frac{\omega}{\omega_{a}}\right)=1.0691$

$\therefore \frac{\omega_{p}}{\omega}=\frac{\omega}{\omega_{a}}=1.034$

$\therefore \omega_{p}=1.034^{\circ}$

$\therefore \omega_{a}=0.967^{\circ}$

$361.034^{\circ}$ which corresponds to $24 \mathrm{~h}, 8.14^{\circ}$ (8.1" longer) and $360.967^{\circ}$ corresponds to $23 \mathrm{~h} 59$ min

$52^{\prime \prime}\left(7.9^{\prime \prime}\right.$ smaller).

This does not explain in actual variation of the length of the dav during the year.