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Earth's orbit is an ellipse with eccentr...

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 day. A day should be taken from noon to noon. Does this explain variation of length of the during the year ?

Text Solution

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Consider the diagram. Let `m` be the mass of the earth, `v_(p), v(a)` be the velocity of the earth at perigee and apogee respectively. Similarly, `omega_(p)` and `omega_(a)` are corresponding angular velocities.

Angular momentum and areal velocity are constant as the earth orbits the sun
At perigee, `r_(p)^(2) omega_(p) = r_(a)^(2) omega_(a)` at apogee ....(i)
If a is the semi-major axis of the earth's orbit, then `r_(p) = a(1 - e) " and" r_(a) = a (1 + e)`....(ii)
`:. (omega_(p))/(omega_(a)) = ((1 + e)/(1 - e))^(2), e = 0.0167` [from Eqs. (i) and (ii)]
`:. (omega_(p))/(omega_(a)) = 1.0691`
Let `omega` be angular speed which is geometric mean of `omega_(p)` and `omega_(a)` and corresponds to mean solar day,
`:. ((omega_(p))/(omega)) ((omega)/(omega_(a))) = 10691`
`:. (omega_(p))/(omega) = (omega)/(omega_(a)) = 1034`
If `omega` corresponds to `1^(@)` per day (mean angular speed), then `omega_(p) = 1034^(@)` per day and `omega_(a) = 0.967^(@)` per day. Since, `361^(@) = 24`, mean solar day, we get 361.034 which corresponds to `24h, 8.14''` (8.1'' longer) and `360.967^(@)` corresponds to `23h 59 min 52''` (7.9'' smaller).
This does not explain the actual variation of the length of the day during the year.
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