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

(N/A) Let $m$ be the mass of the Earth.
Let $\omega_p$ and $\omega_a$ be the angular velocities of the Earth around the Sun at perihelion and aphelion,respectively.
According to Kepler's second law,the areal velocity is constant,which implies $r_p^2 \omega_p = r_a^2 \omega_a$.
Given $r_p = a(1-e)$ and $r_a = a(1+e)$,we have $\frac{\omega_p}{\omega_a} = \left(\frac{1+e}{1-e}\right)^2$.
Substituting $e = 0.0167$,we get $\frac{\omega_p}{\omega_a} = \left(\frac{1.0167}{0.9833}\right)^2 \approx 1.0691$.
Let $\omega$ be the average angular velocity corresponding to the mean solar day. Then $\omega^2 = \omega_p \omega_a$.
Thus,$\frac{\omega_p}{\omega} = \frac{\omega}{\omega_a} = \sqrt{1.0691} \approx 1.034$.
In one day,the Earth rotates $360^\circ$ relative to the stars,plus the angle $\theta$ it moves in its orbit. The solar day is $T = \frac{360^\circ + \theta}{\omega_{spin}}$.
Since $\theta \propto \omega_{orbit}$,the variation in the solar day is $\Delta T \approx T_{mean} \times \frac{\Delta \omega}{\omega_{spin}}$.
Calculating the extremes,the variation is approximately $\pm 8 \text{ s}$.
This calculation shows that the variation due to orbital eccentricity is small and does not fully explain the observed variations in the length of the day,which are also significantly affected by the tilt of the Earth's axis (obliquity).