Astronomy/Earth's rotation


Dear Philip, I know this is a very silly question, but could you please tell me if it is just the tilt of the Earth's rotation axis that makes the Sun and Moon appear to trace a semi circle in the sky? Also, if the Earth was not tilted on its axis, wouldn't the sun and moon look like they were moving in a straight line across the sky rather than going up and down? Thanks for your help!

General Declination Diagram
General Declination Di  
Hello, Liz

Actually no, there are very few 'silly questions'. Those types are the ones provided without adequate forethought and certainly do not apply to yours!

First, it isn't really the tilt of the Earth's rotation axis that causes Sun and Moon to trace out semi-circles in the sky. (What we refer to as "diurnal circles"). The point is that the circles would still be manifest even if there was no tilt - but in a different geometrical perspective and aspect.

To fix ideas, let's explore this in more detail by using something like the Sun's rising point on Dec. 21st (winter solstice) at London (latitude 51.5N) for the case where the tilt exists, and where it doesn't.

We have this relation for the Sun's azimuth (A) at sunrise on Dec. 21 for which the Sun's declination is (-23.5) and we know 23.5 degrees is the tilt of the Earth's rotation axis:

cos (A) = sin (-23.5)/ cos (51.5)

which gives approximately, 130 degrees.  We know that 180 degrees is due South so that this must be:  40 degrees SOUTH of due East. (90 + 40 = 130)

Now, if there were no tilt then the Sun's declination would be zero degrees.

In that case:

cos (A) = sin (0 )/ cos (51.5)  =   0

So that the value of A = arc cos (0) = 90 degrees

(Which you can confirm on a calculator)

Thus, in the case of no tilt, the Sun rises exactly at the due east point, there is no deviation from that cardinal compass point.

This, in fact, can be checked for any latitude on Earth you can think of, say Barbados (lat. 13 N)

cos (A) = sin (0 )/ cos (13)  =   0

Thus, diurnal circles still exist - but are now not deviated in position for different times of the year, i.e. for the observer's horizon.

In terms of the altitude of the respective circles, we'd use a diagram of the type I've appended (called a declination diagram).

From this diagram one can deduce that from London (and all other places) the Sun will always be on the celestial equator (CE = 0 degrees declination) for any point on Earth. This will then be used to determine the altitude of any "semi-circle" traced out at any point. In the case of London, with Lat. = 51.5 N, we infer from he diagram that the Sun's maximum altitude is equal 38.5 deg, at meridian transit.

In other words, the highest altitude at meridian transit (i.e. Sun at local noon) is equal to 90 - the latitude (The latitude remember is now defined by the angular distance *above* the  celestial equator - which marks the solar position).

What would the altitude of the Sun be at Barbados for the case of no axial tilt?

Barbados is at latitude = 13 N so the altitude of the Sun at meridian transit would be:

90 - 13  = 77

On the equator: we have

Latitude 0 we have: 90 - 0 = 90 degrees, so it is always passing overhead at local noon.  Which confirms our expectation.

Now, in the case of an axial tilt of 23.5 degrees how do the altitudes of the semi-circles change? Consider the case of the Sun's declination of (-23.5) on December 21st.

Then for position CE we attach the declination of (-23.5). NOw,

Declination = z + lat.  or:

z =  lat. - declination = 51.5 - (-23.5) = 51.5 + 23.5 =  75 degrees

But the altitude (from the diagram) =  90 - z = 90 deg - 75 deg = 25 degrees

So we see the semi-circle at London for the case of no tilt is going to be some (38.5 - 25) = 13.5 degrees higher than the case where there is an axial tilt.

You can use the above relations to practice and work out the quantities for a whole set of other locations and latitudes as well, and establish for yourself the new model illustrated.

In the case of the Moon, we have some added complexities but the basic principle remains the same.  Note here that the Moon's own rotation axis is  not perpendicular to its orbital plane, so the lunar equator is not in the plane of its orbit, but is inclined to it by a constant value of 6.688° (this is the 'obliquity'). Although the rotation axis of the Moon is not fixed (i.e. with respect to the stars), the angle between the ecliptic and the lunar equator is always 1.543°. This would be roughly the same angle between the lunar equator and the celestial equator if there were no inclination – i.e. the ecliptic didn’t exist.  See the detailed geometry displayed here:

In effect, you'd still get "diurnal circles" for the Moon, but again - on account of the above variations, these would now be a bit offset from what one would expect if the ecliptic didn’t exist, i.e. there was no tilt for the Earth's rotation axis.

I hope this response isn't too cumbersome or difficult - but feel free to ask any further questions if anything isn't clear!  


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Philip Stahl


I have more than forty years of experience in Astronomy, specifically solar and space physics. My specialties include the physics of solar flares, sunspots, including their effects on Earth and statistics pertaining to sunspot morphology and flare geo-effectiveness.


Astronomy: Worked at university observatory in college, doing astrographic measurements. Developed first ever astronomy curriculum for secondary schools in Caribbean. Gave workshops in astrophysics and astronomical measurements at Harry Bayley Observatory, Barbados. M.Phil. degree in Physics/Solar Physics and more than twenty years as researcher with discovery of SID flares. Developed of first ever consistent magnetic arcade model for solar flares incorporating energy dissipation and accumulation. Develop first ever loop solar flare model using double layers and incorporating cavity resonators.

American Astronomical Society (Solar Physics and Dynamical Astronomy divisions), American Mathematical Society, American Geophysical Union.

Solar Physics (journal), The Journal of the Royal Astronomical Society of Canada, The Proceedings of the Meudon Solar Flare Workshop (1986), The Proceedings of the Caribbean Physics Conference (1985). Books: 'Selected Analyses in Solar Flare Plasma Dynamics', 'Physics Notes for Advanced Level'. 'Astronomy and Astrophysics: Notes, Problems and Solutions'.

B.A. Astronomy, M. Phil. Physics

Awards and Honors
American Astronomical Society Studentship Award (1984), Barbados Government Award for Solar Research (1980), Barbados Astronomical Society Award for Service as Journal Editor (1977-91)

Past/Present Clients
Caribbean Examinations Council, Barbados Astronomical Society, Trinidad & Tobago Astronomical Society.

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