Sun, Earth, Moon

• 16 July 2026
• celestial

Celestial sphere

I saw this picture on Wikipedia the other day.

It struck me that the ecliptic is almost vertical in this picture, super close to the zenith. I've always lived in the 51°N-53°N range where the ecliptic never moves up so high off the horizon.

Everyone knows the Earth orbits the Sun, the Moon orbits the Earth, and the Earth's axis is tilted. But it's surprisingly easy to know those facts without having a mental model of what they imply. This post is an attempt to build that model.

So I'll put myself on the line a bit and try to do a digest of the obvious, i.e. the Sun - Earth - Moon system. I obviously won't cover everything but I'll try to do an overview of the important stuff. And yeah, some of this is obvious - though hopefully still fun to think about. By the end we'll have enough machinery to work out roughly when the picture above was taken - without peeking at the metadata.

As a bonus, I've been enjoying Three.js based visualizations I can get Gemini to cook up. They're almost good as-is and need very few tweaks. Incidentally, I find these three.js visualizations the best way to actually do 3D diagrams.

Initially I wanted to also cover the Moon's nodal precession here but I'll have to do that later.

The celestial geometry

Before we get to the ecliptic, we need one useful abstraction: the celestial sphere. Imagine the Earth at the centre of an enormous sphere with the stars, Sun and Moon projected onto its inside surface. The stars are, of course, nowhere near the same distance from us in reality, but they are so far away compared with the Earth-Moon system that treating them as if they all lie on one giant sphere turns the geometry of the sky into ordinary geometry on a sphere.

With that picture in mind, we can talk about paths, poles and great circles on the sky without worrying about how far away anything actually is.

Let's start with the ecliptic: the plane of the Earth's orbit around the Sun - equivalently, the plane the Sun appears to move in when watched from the Earth. The Earth orbits in that plane but crucially, its axis of rotation is tilted with respect to it by about $ϵ={23.4}^{\circ }$. This is where we get the seasons and days of varying length. If the Earth's axis were perpendicular to the ecliptic (i.e. the equator lined up with the ecliptic plane) we'd have nice 12h days and 12h nights, and every place on Earth would experience them the same way.

So: we have the ecliptic plane; the Earth orbits in that plane, and its axis is tilted with respect to it. The Moon then orbits around the Earth but to make it more complicated, its orbital plane is tilted another ~5.1 degrees with respect to the ecliptic. We end up with something like this.

I added two nodes on the Moon's orbit: those are the points where the Moon's orbit crosses the ecliptic plane. One is the ascending node (since the Moon is crossing on the way UP), the other the descending node.

If you keep watching you might notice the nodes themselves are slowly rotating too - that's the precession of the lunar nodes, an 18.6-year cycle.

Equinoxes and solstices

Due to the axial tilt the Sun will have a different declination throughout the year. When the North Pole side of Earth is facing the Sun we see it at its highest declination $ϵ={23.4}^{\circ }$. When the North Pole side is facing the opposite way the Sun's at its lowest declination - $-ϵ=-{23.4}^{\circ }$. The two in-between points are the equinoxes, where the Sun crosses the celestial equator. Those are the 2 days a year where we see 12h days and nights (or close enough to it due to atmospheric refraction and the fact that we need just the top of the Sun to be visible to count as a sunrise).

Note some consequences of this fact:

• on the equinoxes the Sun will rise and set due east and west, respectively^[1]
• during the Northern hemisphere summer the Sun will rise and set north of east, west respectively; and south of them during the winter
• during the summer solstice the Sun will be directly overhead at noon at the Tropic of Cancer
• during the winter solstice the Sun will be directly overhead at noon at the Tropic of Capricorn

Let's call the Sun's declination $\delta$ and our latitude $\varphi$. At noon the Sun sits at an altitude of ${90}^{\circ }-|\varphi -\delta |$: if we're at latitude $\varphi =\delta$ the Sun passes directly overhead, and every degree of latitude away from that costs us one degree of altitude.

This also tells us when the Sun never sets: it has to stay above the horizon even at its lowest point, on the opposite side of the sky from noon, which happens when $\delta \ge {90}^{\circ }-\varphi$. Since the Sun's declination maxes out at $ϵ={23.4}^{\circ }$, the midnight sun is only possible above $\varphi \approx {90}^{\circ }-{23.4}^{\circ }={66.6}^{\circ }$ - and right at that latitude, only on the summer solstice itself. Same (or the opposite) with the Southern hemisphere during its summer. These are the polar circles.

Sun vs Horizon

The Sun will peak at noon but what about a few hours either side of noon? Let's calculate the Sun's altitude throughout the day. To do that, we'll need to use the spherical law of cosines.

Spherical Law of Cosines

This follows the derivation presented in here. I've LaTeX'd the equations, redone the illustrations and streamlined the normalisation step (since I assume a unit sphere to begin with).

Let's take $V$ to be the center of any sphere of radius $R$ and let's draw 3 points $A$, $B$, $C$ on the surface of the sphere.

We have $|VA|=|VB|=|VC|=R$. For simplicity we'll take $R=1$ but this won't matter since one can normalise the vectors first.

The arclength of each side is $a$, $b$ and $c$.

We want to find the angle $A$ between the planes BVA and CVA. We can find the normals to each plane and then calculate the dot product between them. That will give us the cosine of the angle we want. $\overrightarrow{x}$ is the normal to the BVA plane, $\overrightarrow{y}$ is the normal to the CVA plane.

$$\begin{array}{r}\overrightarrow{x}=\overrightarrow{VA}\times \overrightarrow{VB}\\ \overrightarrow{y}=\overrightarrow{VA}\times \overrightarrow{VC}\end{array}$$

Since the arc $AB$ has length $c$ and the arc $AC$ has length $b$, the magnitudes of the two normals are

$$|\overrightarrow{x}|=|\overrightarrow{VA}\times \overrightarrow{VB}|=\sin (c),|\overrightarrow{y}|=|\overrightarrow{VA}\times \overrightarrow{VC}|=\sin (b)$$

so the dot product gives us the cosine of our angle: $$(\overrightarrow{VA}\times \overrightarrow{VB})\cdot (\overrightarrow{VA}\times \overrightarrow{VC})=\sin (c)\sin (b)\cos (A).$$

Now using the cross product identity:

$$(\overrightarrow{VA}\times \overrightarrow{VB})\cdot (\overrightarrow{VA}\times \overrightarrow{VC})=\overrightarrow{VA}\cdot [\overrightarrow{VB}\times (\overrightarrow{VA}\times \overrightarrow{VC})]$$

$$\overrightarrow{VA}\cdot [\overrightarrow{VB}\times (\overrightarrow{VA}\times \overrightarrow{VC})]=\overrightarrow{VA}\cdot [\overrightarrow{VA}(\overrightarrow{VB}\cdot \overrightarrow{VC})-\overrightarrow{VC}(\overrightarrow{VA}\cdot \overrightarrow{VB})]$$

$$\overrightarrow{VA}\cdot [\overrightarrow{VA}(\overrightarrow{VB}\cdot \overrightarrow{VC})-\overrightarrow{VC}(\overrightarrow{VA}\cdot \overrightarrow{VB})]=(\overrightarrow{VB}\cdot \overrightarrow{VC})-(\overrightarrow{VA}\cdot \overrightarrow{VC})(\overrightarrow{VA}\cdot \overrightarrow{VB})$$

$$(\overrightarrow{VB}\cdot \overrightarrow{VC})-(\overrightarrow{VA}\cdot \overrightarrow{VC})(\overrightarrow{VA}\cdot \overrightarrow{VB})=\cos (a)-\cos (b)\cos (c)$$

And we end up with: $$\cos (a)-\cos (b)\cos (c)=\sin (b)\sin (c)\cos (A)$$

and so

$$\cos (a)=\cos (b)\cos (c)+\sin (b)\sin (c)\cos (A)$$

We take $A$ to be the north celestial pole, $B$ to be our zenith and $C$ to be the position of the Sun. Matching the diagram, we refer to $A$ as $P$, $B$ as $Z$ and $C$ as $S$. The sides of this triangle are:

• $a=ZS$, the angle from our zenith to the Sun - the one we are looking for
• $b=PS$, the angle between the celestial pole and the Sun; since declination is measured from the celestial equator this is ${90}^{\circ }-\delta$
• $c=PZ$, the angle between the pole and the zenith: ${90}^{\circ }-\varphi$ where $\varphi$ is our latitude

The angle at the pole (the $A$ in the law) is the hour angle $H$: how far the Sun is from our meridian, advancing ${15}^{\circ }$ per hour from local solar noon.

Plugging into the law and using $\cos ({90}^{\circ }-\varphi )=\sin (\varphi )$ and $\cos ({90}^{\circ }-\delta )=\sin (\delta )$ (and the same for the sines) we get:

$$\cos (a)=\sin (\varphi )\sin (\delta )+\cos (\varphi )\cos (\delta )\cos (H)$$

Note that $a$ is measured from the zenith and we usually want the opposite - the altitude above the horizon $h={90}^{\circ }-a$, which satisfies

$$\sin (h)=\sin (\varphi )\sin (\delta )+\cos (\varphi )\cos (\delta )\cos (H)$$

As a sanity check: Manchester ($\varphi \approx {53.5}^{\circ }$) at noon on the June solstice ($\delta ={23.4}^{\circ }$, $H=0$) gives $h={90}^{\circ }-({53.5}^{\circ }-{23.4}^{\circ })\approx {60}^{\circ }$.

Some interesting things to look at:

• the poles - they get 6 month days, 6 month nights
• the polar circles around 66-67 degrees north and south
• the tropics around 23 degrees north and south
• some of the preset cities I put on there

Overhead Ecliptic

The Sun is one thing - what about the ecliptic in general? You'll notice that in the picture the ecliptic is pretty much exactly overhead - at night, possibly close to but definitely before sunrise.

The angle that the ecliptic makes with the horizon will differ throughout the day: the tilt between the celestial equator (i.e. Earth's equator projected out onto the sky) and the ecliptic causes a "wobble" as the Earth rotates. Let's see how we can calculate it based on time of day; then we can work out approximately when the picture was taken.

Also, this was very confusing to me at first. To take a very specific example, let's pretend we're in Manchester, near ${53}^{\circ }$N. Our peak ecliptic angle will be about ${60.4}^{\circ }$ (${90}^{\circ }-({53}^{\circ }-{23.4}^{\circ })$). 6 months later, we'll get the same peak angle, it'll just happen at night. In other words - we go through the full cycle every single day, and the time of year only decides how that cycle lines up with the Sun being on the "day" side.

Here are two visualizations. The first one is an Earth view at different days of the year and different times of the day. There are two different perspectives: the first one locks the observer in place (zenith always points up) and rotates the rest accordingly. The second one locks the ecliptic and rotates the observer.

Personally I find the fixed ecliptic perspective a bit easier to reason about than the fixed observer.

The second is a graph of this ecliptic vs horizon angle throughout the year.

Back to the picture

First, why is a picture like this possible at all? For the ecliptic to pass overhead, the zenith has to lie on the ecliptic - i.e. the zenith-to-NEP distance $d$ has to hit exactly ${90}^{\circ }$. Setting $\cos H^{\prime }=\pm 1$ in our formula shows that over a day $\cos d$ swings between $\sin (L-ϵ)$ and $\sin (L+ϵ)$, so $d={90}^{\circ }$ is only reachable when $|L|\le ϵ$: in the tropics. Our photographer sits at $|L|\approx ϵ$, right on the boundary, where the ecliptic just grazes the zenith once a day.

Now for the when. Take $L=-ϵ$ and set $d={90}^{\circ }$ in

$$\cos d=\sin L\sin {\delta }_{\text{NEP}}+\cos L\cos {\delta }_{\text{NEP}}\cos H^{\prime }$$

Plugging in $\sin {\delta }_{\text{NEP}}=\cos ϵ$ and $\cos {\delta }_{\text{NEP}}=\sin ϵ$:

$$0=-\sin ϵ\cos ϵ+\cos ϵ\sin ϵ\cos H^{\prime }⟹\cos H^{\prime }=1⟹H^{\prime }=0$$

So the ecliptic is overhead exactly when the NEP crosses our meridian, i.e. when $\text{LST}={\alpha }_{\text{NEP}}={270}^{\circ }$.

The picture looks like deep pre-dawn - call it 3:30 am. From our LST formula:

$${\alpha }_{\odot }=\text{LST}-(t-12)\cdot {15}^{\circ }={270}^{\circ }-(3.5-12)\cdot {15}^{\circ }={397.5}^{\circ }\equiv {37.5}^{\circ }$$

Inverting $\tan {\alpha }_{\odot }=\sin {\lambda }_{\odot }\cos ϵ,/\cos {\lambda }_{\odot }$ gives $\tan {\lambda }_{\odot }=\tan {\alpha }_{\odot }/\cos ϵ$, so ${\lambda }_{\odot }\approx {40}^{\circ }$. And finally the day of year:

$$N=80+{40}^{\circ }\cdot \frac{365.25}{{360}^{\circ }}\approx 121$$

Day 121 is the very start of May.

Time to check our deduction against the metadata: the picture shows the Eta Aquariids, a meteor shower that peaks around the 5th of May.

Wrapping up

So that's the digest of the "obvious": the tilt $ϵ$ gives the Sun its yearly declination swing, which sets the seasons, the sunrise directions, the tropics and the polar circles. One application of the spherical law of cosines gives the Sun's altitude at any hour; another one - with the Sun swapped for the ecliptic pole - gives the angle the whole ecliptic makes with the horizon.

The Moon is another topic for another day.