Those long summer days...

the sun cause an angular shift in one day I 'll calculate it then I'll compare the maximum with the minimum value of it. the sun's Mass we can calculate it using Kepler's third Law $\frac{T^2}{a^3}=\frac{4\pi^2}{GM}$ then from the mechanical Energy equation for planets motion $E=-\frac{GMm}{2a}$ here $E=\frac12 mv^2 -\frac{GMm}{D}$ so $v=\sqrt{GM\frac{2a-D}{aD}}$ we all know "according to kepler's laws" that the velocity at the nearest point is the biggest and at the furthest point is the smallest so V is the biggest and v is the smallest then

$V= \sqrt{GM\frac{2a-D'}{aD'}}$ here $D'=1.47*10^8 km$

and we have $v=\sqrt{GM\frac{2a-d}{ad}}$ here $d=1.521*10^8 km$ the angular shift for each case is

$\delta\theta'=\frac{TV}{D'}$
where $T=24*3600 sec$

$\delta\theta=\frac{Tv}{d}$ so the shift in seconds in each case is $\Delta\tau'=T\frac{\delta\theta'}{2\pi}$

$\Delta\tau=T\frac{\delta\theta}{2\pi}$
$\Delta\tau'-\Delta\tau=16.13 sec$

The Earth goes around the Sun in $T=365.25$ days and rotates around its axis in $t=1$ day. These two motions are in opposite directions. Since the angular momentum of the earth around the sun is a constant of motion and $L \sim r^2 \omega$ , $\omega \sim r^{-2}$ and the different angular velocities when the earth is close or far from the Sun is $\Delta \omega = \frac{2\pi}{T} (d_f^2-d_s^2)/d_0^2$ where $d_0$ is the average of $d_f$ and $d_s$ .

As the earth spins it also goes around the sun, which means that if the period of the earth's spin is t, the total angle seen during daylight is $\pi+\beta,\beta=-\omega t$ . The length of daylight $t_d$ is therefore

$t_d=\frac{\pi + \beta}{2\pi}t$ .

The difference in $t_d$ during the longest and shortest day of the year is then

$t_l-t_s=\frac{\Delta \omega}{2\pi}t^2$ ,

which works out to be 16.13 seconds.