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Question
Earth’s orbit is an ellipse with eccentricity 0.0167. Thus, 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 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?
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Solution
From the geometry of the ellipse of eccentricity e and semi-major axis a, the aphelion and perihelion distances are:
Angular momentum and areal velocity are constant as the earth orbits the sun
At perigee `r_p^2ω_p = r_a^2ω_a` at apogee.
If 'a' is the semi-major axis of earth's orbit, then `r_p = a(1 - e)` and `r_a = a(l + e)`
∴ `ω_p/ω_a = ((1 + e)/(1 - e))^2, e = 0.0167`
∴ `ω_p/ω_a = 1.0691`
Let ω be angular speed which is the geometric mean of ωp and ωa and corresponds to mean solar day,
∴ `(ω_p/ω) (ω/ω_a) = 1.0691`
∴ `ω_p/ω = ω/ω_a = 1.034`
If ω corresponds to 1° per day (mean angular speed), then w, = 1.034° per day and ωa = 0.967 per day. Since 361° = 14hrs: mean solar day, we get 361.034° which corresponds to 24 hrs 8.14" (8.1" longer) and 360.967° corresponds to 23 hrs 59 min 52" (7.9" smaller).
This does not explain the actual variation in the length of the day during the year.
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