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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 throughout the year. Assume that 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 the variation of the length of the day during the year?

Answers (1)

Perihelion distance for an ellipse of eccentricity e and semi-major axis a, $r_{p}=a(1-e)$

Aphelion distance $r_{a}=a(1+e)$

$\frac{dA}{dt}=\frac{L}{2m}=const$ (Kepler's law)

$L_{a}=L_{p}$

$r_{a}\times p_{a}=r_{p}\times p_{p}$

$r_{a}\times mv_{a}=r_{p}\times mv_{p}$

$r_{a}\times m\omega _{a}r_{a}=r_{p}\times m\omega _{p}r_{p}$

$\omega _{a}r_{a}^{2}=\omega _{p}r_{p}^{2}$

$\frac{\omega _{p}}{\omega _{a}}=\left ( \frac{r_{a}}{r_{p}} \right )^{2}$

$\frac{\omega _{p}}{\omega _{a}}=\left ( \frac{a(1+e)}{a(1-e)} \right )^{2}$

$e=0.0167(given)$

$\frac{\omega _{p}}{\omega _{a}}=\frac{(1+0.0167)^{2}}{(1-0.0167)^{2}}=\left ( \frac{1.0167}{0.9833} \right )^{2}=1.0691$

$\frac{\omega _{p}}{\omega }\frac{\omega }{\omega _{a}}=1.0691$ where $\omega$ is the meanangular velocity of earth

$\frac{\omega _{p}}{\omega }=\frac{\omega }{\omega _{a}}=\sqrt{1.0691}=1.034$

If $\omega$ is $1^{o}$ it corresponds to 1 day i.e., avg angular speed $=1.034^{0}$ per day

and $\omega _{a}0.9833^{0},\omega _{p}=1.0167^{0}$

Since $361^{0}$ = 24 hrs mean solar day. We get $361.034^{0}$ which corresponds to 24 hrs 8.14'

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

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