18. Polar motion

The Earth rotates around an axis, an imaginary straight line that all points not on the line move around in circles. The axis passes through the Earth’s North Pole and the South Pole. So the positions of the two Poles are defined by the position of the rotation axis.

Earth rotation and poles

The Earth’s North and South Poles are defined as the points where the axis of rotation passes through the surface of the planet. (Earth photo is a public domain image from NASA.)

Interestingly, the Earth’s rotation axis is not fixed – it moves slightly. This means that the Earth’s poles move.

The positions of the Earth’s poles can be determined by looking at the motions of the stars. As we’ve already seen, if you observe the positions of stars throughout a night, you will see that they rotate in the sky about a central point. The point on the Earth’s surface directly underneath the centre of rotation of the stars is one of the poles of the Earth.

Star trails in the northern hemisphere

Star trails above Little Hawk Lake in Canada. The northern hemisphere stars rotate around the North Celestial Pole (the point directly above the Earth’s North Pole). The bright spot in the centre is Polaris, the pole star. The circles are somewhat distorted in the upper corners of the photo because of the wide angle lens used. (Creative Commons Attribution 2.0 image by Dave Doe.)

Through the 19th century, astronomers were improving the precision of astronomical observations to the point where the movement of the Earth’s rotational poles needed to be accounted for in the positions of celestial objects. The motion of the poles was also beginning to affect navigation, because as the poles move, so does the grid system of latitude and longitude that ships rely on to reach their destinations and avoid navigational hazards. In 1899 the International Geodetic Association established a branch known as the International Latitude Service.

The fledgling International Latitude Service established a network of six observatories, all located close to latitude 39° 08’ north, spread around the world. The initial observatories were located in Gaithersburg, Maryland, USA; Cincinatti, Ohio, USA; Ukiah, California, USA; Mizusawa, Japan; Charjui, Turkestan; and Carloforte, Italy. The station in Charjui closed due to economic problems caused by war, but a new station opened in Kitab, Uzbekistan after World War I. Each observatory engaged in a program of observing the positions of 144 selected reference stars, and the data from each station were cross referenced to provide accurate measurements of the location of the North Pole.

International Latitude Service station in Ukiah

International Latitude Service station in Ukiah, California. (Public domain image from Wikimedia Commons.)

In 1962, the International Time Bureau founded the International Polar Motion Service, which incorporated the International Latitude Service observations and additional astronomical observations to provide a reference of higher accuracy, suitable for both navigation and defining time relative to Earth’s rotation. Finally in 1987, the the International Astronomical Union and the International Union of Geodesy and Geophysics established the International Earth Rotation Service (IERS), which took over from the International Polar Motion Service. The IERS is the current authority responsible for timekeeping and Earth-based coordinate systems, including the definitions of time units, the introduction of leap seconds to keep clocks in synch with the Earth’s rotation, and definitions of latitude and longitude, as well as measurements of the motion of the Earth’s poles, which are necessary for accurate use of navigation systems such as GPS and Galileo.

The motion of Earth’s poles can be broken down into three components:

1. An annual elliptical wobble. Over the period of a year, the Earth’s poles move around in an ellipse, with the long axis of the ellipse about 6 metres in length. In March, the North Pole is about 6 metres from where it is in September (though see below). This motion is generally agreed by scientists to be caused by the annual shift in air pressure between winter and summer over the northern and southern hemispheres. In particular there is an imbalance between the Northern Atlantic ocean and Asia, with higher air pressure over the ocean in the northern winter, but higher air pressure over the Asian continent in summer. This change in the mass distribution of the atmosphere is enough to cause the observed wobble.

Annual wobble of North Pole

Annual elliptical wobble of the Earth’s North Pole. Deviation is given in milliarcseconds of axial tilt; 100 milliarcseconds corresponds to a bit over 3 metres at ground level. (Figure adapted from [1].)

2. Superimposed on the annual elliptical wobble is another, circular, wobble, with a period of around 433 days. This is called the Chandler wobble, named after its discoverer, American astronomer Seth Carlo Chandler, who found it in 1891. The Chandler wobble occurs because the Earth is not a perfect sphere. The Earth is slightly elliptical, with the radius at the equator about 20 kilometres larger than the polar radius. When elliptical objects spin, they experience a slight wobble in the rotation known as free nutation. This is the sort of wobble seen in a spinning rugby ball or American football in flight (where the effect is exaggerated by the ball’s exaggerated elliptical shape). This wobble would die away over time, but is driven by changes in the mass distribution of cold and warm water in the oceans and high and low pressure systems in the atmosphere. The Chandler wobble has a diameter of about 9 metres at the poles.

The combined effect of the annual wobble and the Chandler wobble is that the North and South Poles move in a spiralling pattern, sometimes circling with a diameter up to 15 metres, then reducing down to about 3 metres, before increasing again. This beat pattern occurs over a period of about 7 years.

Annual _ Chandler wobble of North Pole

Graph showing the movement of the North Pole over a period of 4500 days (12.3 years), with time on the vertical axis and the spiralling motion mapped in the x and y axes. The motion tickmarks are 0.1 arcsecond in rotation angle of the axis apart, corresponding to about 3 metres of motion along the ground at the Pole. (Public domain image from Wikimedia Commons.)

3. The third and final motion of the Earth’s poles is a systematic drift, of about 200 millimetres per year. Since 1900, the central point of the spiral wobbles of the North Pole has drifted by about 20 metres. This drift is caused by changes in the mass distribution of Earth due to shifts in its structure: movement of molten rock in the mantle, isostatic rebound of crust following the last glacial period, and more recently the melting of the Greenland ice sheet. The melting of the Greenland ice sheet in the last few decades has shifted the direction of polar drift dramatically; one of the serious indications of secondary changes to the Earth caused by human-induced climate change. Changes in Earth’s mass distribution alter its rotational moment of inertia, and the rotational axis adjusts to conserve angular momentum.

Motion of North Pole since 1900

Plot of motion of the North Pole since 1900. The actual position of the Pole from 2008 to 2014 is shown with blue crosses, showing the annual and Chandler wobbles. The mean position (i.e. the centre of the wobbles) is shown for 1900 to 2014 as the green line. The pole has mostly drifted towards the 80° west meridian, but has changed direction dramatically since 2000. (Figure reproduced from [2].)

Each of the three components of Earth’s polar motion are: (a) observable with 19th century technology, (b) accurately measurable using current technology, and (c) understandable and quantitatively explainable using the fact that the Earth is a rotating spheroid and our knowledge of its structure.

If the Earth were flat, it would not be possible to reconcile the changes in position of the North and South Poles with the known shifts in mass distribution of the Earth. The Chandler wobble would not even have any reason to exist at close to its observed period unless the Earth was an almost spherical ellipsoid.

References:

[1] Höpfner, J. “Polar motion at seasonal frequencies”. Journal of Geodynamics, 22, p. 51-61, 1996. https://doi.org/10.1016/0264-3707(96)00012-9

[2] Dick, W., Thaller, D. IERS Annual Report 2013. International Earth Rotation Service, 2014. https://www.iers.org/IERS/EN/Publications/AnnualReports/AnnualReport2013.html

17. Light time corrections

In the 16th century, the naval powers of Europe were engaged in a race to explore and colonise lands previously unknown to Europeans (though many were of course already inhabited), and reap the rewards of the new found resources. They were limited by the accuracy of navigation at sea. Determining latitude was a relatively simple matter of sighting the angle of a star or the sun through a sextant. But because of the daily rotation of the Earth, determining the longitude by sighting a celestial object required knowing the time of day. Mechanical clocks of the era were rendered useless by the rocking of a ship, making this a major problem.

Solving the problem would give such an advantage to the country holding the secret that in 1598 King Philip III of Spain offered a prize of 6000 ducats plus an annual pension of 2000 ducats for life to whoever could devise a means of measuring longitude at sea. In 1610 the prize was still unclaimed, and in that year Galileo Galilei trained his first telescope on Jupiter, becoming the first person to observe the planet’s largest four moons. He studied their movements, and a couple of years later had produced orbital tables that allowed their positions to be calculated months or years in advance. These tables included the times when a moon would slip into Jupiter’s shadow, and be eclipsed, disappearing from view because it no longer reflected sunlight.

Galileo wrote to King Philip in 1616, proposing a method of telling the time at sea by observing the eclipses of Jupiter’s moons. One could pinpoint the time by observing an eclipse, and then use an observation of a star to calculate the longitude. Although the method could work in principle, observing an eclipse of a barely visible object through the narrow field of view of a telescope while standing on a rocking ship was practically impossible, and it never worked in practice.

Jupiter and Io

Jupiter and its innermost large moon, Io, as seen by NASA’s Cassini space probe. (Galileo’s view was nowhere near as good as this!) (Public Domain image by NASA.)

By the 1660s, Giovanni Cassini had developed Galileo’s method as a way of measuring precise longitudes on land, as an aid to calculating distances and making accurate maps. In 1671 Cassini moved to take up directorship of the Royal Observatory in Paris. He dispatched his assistant Jean Picard to Uraniborg, the former observatory of Tycho Brahe, near Copenhagen, partly to make measurements of eclipses of Jupiter’s moon Io, to accurately calculate the longitude difference between the two observatories. Picard himself employed the assistance of a young Dane named Ole Rømer.

Ole Rømer

Portrait of Ole Rømer by Jacob Coning. (Public domain image from Wikimedia Commons.)

The moon Io orbits Jupiter every 42.5 hours and is close enough to be eclipsed on each orbit, so an eclipse is visible every few days, weather and daylight hours permitting. After observing well over 100 eclipses, Rømer moved to Paris to assist Cassini himself, and continued recording eclipses of Io over the next few years. In these observations Cassini noticed some odd discrepancies. In particular, the time between successive eclipses got shorter when the Earth was approaching Jupiter in its orbit, and longer several months later when the Earth was moving away from Jupiter. Cassini realised that this could be explained if the light from Io did not arrive at Earth instantaneously, but rather took time to travel the intervening distance. When the Earth is closer to Jupiter, the light has less distance to cover, so the eclipse appears to occur earlier, and vice versa: when the Earth is further away the eclipse appears to be later because the light takes longer to reach Earth. Cassini made an announcement to this effect to the French Academy of Sciences in 1676.

Ole Rømer's eclipse notes

Ole Rømer’s notebook showing recordings of the dates and times of eclipses of Io from 1667 to 1677. “Imm” means immersion into Jupiter’s shadow, and “Emer” means emergence from Jupiter’s shadow. (Public domain image from Wikimedia Commons.)

However, it was common wisdom at the time that light travelled instantaneously, and Cassini later retreated from his suggestion and did not pursue it further. Rømer, on the other hand, was intrigued and continued to investigate. In 1678 he published his findings. He argued that as the Earth moved in its orbit away from Jupiter, successive eclipses would each occur with the Earth roughly 200 Earth-diameters further away from Jupiter than the previous one. Using the geometry of the orbit and his observations, Rømer calculated that it must take light approximately 11 minutes to cross a distance equal to the diameter of the Earth’s orbit. This is a little low—it actually takes about 16 and a half minutes—but it’s the right order of magnitude. So for the first time, we had some idea how fast light travels. And as we’ve just seen, the finite speed of light can have a significant effect on the observed timing of astronomical observations.

Ole Rømer's figure

Figure 2 from Rømer’s paper, illustrating the difference in distance between Earth and Jupiter between successive eclipses as Earth recedes from Jupiter (LK) and approaches Jupiter (FG). Reproduced from [1].

The finite speed of light means that astronomical events don’t occur when we see them. We only see the event after enough time has elapsed for the light to travel to Earth. This is important for events with precisely measurable times, such as eclipses, occultations, the brightness variations of variable stars, and the radio pulses of remote pulsars.

Not only do you need to correct for the time it takes light to reach Earth, but the correction is different depending on where you are on Earth. An observer observing an object that is directly overhead is closer to it than an observer seeing the same object on the horizon. The observer seeing the object on the horizon is further away by the radius of the Earth. The radius of the Earth is 6370 km, and it takes light a little over 21 milliseconds to travel this distance. So astronomical events observed on the horizon appear to occur 21 milliseconds later than they do to someone observing the same event overhead. This effect is significant enough to be mentioned explicitly in a paper discussing the timing of variable stars:

“More disturbing effects become significant which require more conventions and more complex reduction procedures. By far the biggest effect is the topocentric light-time correction (up to 20 msec).”[2]

Topocentric refers to measuring from a specific point on the surface of the Earth. Depending where on Earth you are, the timing of observed astronomical events can appear to vary by up to 20 ms.

Not only does the light travel time affect the observed time of astronomical events, it also affects the observed position of some astronomical objects, most importantly solar system objects that move noticeably over the few hours that light takes to travel to Earth from them. When we observe an object such as a planet or an asteroid, we see it in the position that it was when the light left it, not where it is at the time that we see it. So for such objects, a corrected position needs to be calculated. The correction in observed position of a moving astronomical object due to the finite speed of light is, somewhat confusingly, also known as light time correction.

Light time correction of observed position is critical in determining the orbits of bodies such as asteroids and comets with accuracy. A paper describing general methods for determining orbital parameters from observations notes that Earth-based observations are necessarily topocentric, and states in the description of the method that:

“In the case of asteroid or comet orbits, the light-time correction has been computed.”[3]

Finally, a recent paper on determining the orbital parameters of near-Earth objects (which pose a potential threat of catastrophic collision with Earth) points out, where ρ is the topocentric distance:

“Note that we include a light-time correction by subtracting ρ/c from the observed epochs for any propagation computation with c as speed of light.”[4]

All of these corrections, which must be applied to astronomical observations where either (a) timings must be known to less than a second or (b) positions must be known accurately to determine orbits, are different by a light travel time of 21 ms for observers looking at objects directly overhead versus observers looking towards the horizon. And in between the light time corrections are 21 ms × (1 minus the sine of the observation zenith angle).

light time corrections on a spherical Earth

Diagram of light time corrections. Observation points where an astronomical event are on the horizon are 6370 km further away than observation points where the event is directly overhead.

This implies that places on Earth where an astronomical object appears near the horizon are a bit over 6000 km further away from the object than the location where the object is directly overhead. This is true no matter which object is observed, meaning it is independent of which position on Earth is directly under it. This cannot be so if the Earth is flat.

light time corrections on a flat Earth

Geometry of light time corrections on a flat Earth.

Observation points on Earth where an astronomical event is overhead and on the horizon are separated by 10,000 km. If the Earth is flat, then the geometry must be something like that shown in the diagram above. The astronomical event is a distance x above the flat Earth, such that the distance from the event to a point 10,000 km along the surface is x plus the measured light travel time distance of 6370 km. Applying Pythagoras’s theorem:

(6370 + x)2 = 100002 + x2

Solving for x gives 4660 km. So measurements of light time correction imply that all astronomical events are 4660 km above the flat Earth. This means the elevation angle of the event seen from 10,000 km away is arctan(4660/10,000) = 25°, well above the horizon, which is inconsistent with observation (and the trigonometry of all the intermediate angles doesn’t work either). It’s also easy to show by other observations that astronomical objects are not all at the same distance – some are thousands, millions, or more times further away than others, and they are all much further away than 4660 km.

So the measurement of light time corrections imply that observers on Earth are positioned on the surface of a sphere. In other words, that the Earth is spherical in shape.

References:

[1] Rømer, O. (“A Demonstration Concerning the Motion of Light”.) Philosophical Transactions of the Royal Society, 12, p. 893-94, 1678. (Originally published in French as “Demonstration touchant le mouvement de la lumiere trouvé”. Journal des Sçavans, p. 276-279, 1677.) https://www.jstor.org/stable/101779

[2] Bastian, U. “The Time Coordinate Used in the Variable-star Community”. Information Bulletin on Variable Stars, No. 4822, #1, 2000. https://ui.adsabs.harvard.edu/abs/2000IBVS.4822….1B/abstract

[3] Dumoulin, C. “Unified Iterative Methods in Orbit Determination”. Celestial Mechanics and Dynamical Astronomy, 59, 1, p. 73-89, 1994. https://doi.org/10.1007/BF00691971

[4] Frühauf, M., Micheli, M., Santana-Ros, T., Jehn, R., Koschny, D., Torralba, O. R. “A systematic ranging technique for follow-ups of NEOs detected with the Flyeye telescope”. Proceedings of the 1st NEO and Debris Detection Conference, Darmstadt, 2019. https://ui.adsabs.harvard.edu/abs/2019arXiv190308419F/abstract

16. Lunar eclipses

Lunar eclipses occur when the Earth is positioned between the sun and the moon, so that the Earth blocks some or all of the sunlight from directly reaching the moon. Because of the relative sizes of the sun, Earth, and moon, and their distances from one another, the Earth’s shadow is large enough to completely cover the moon.

To talk about eclipses, we need to define some terms. The sun is a large, extended source of light, not a point source, so the shadows that objects cast in sunlight have two components: the umbra, where light from the sun is totally blocked, and the penumbra, where light from the sun is partially blocked.

Umbra and penumbra

Diagram showing the umbra and penumbra cast by the Earth. Not to scale. Public domain image from Wikimedia Commons.

When the moon passes entirely inside the Earth’s umbra, that is a total lunar eclipse. Although no sunlight reaches the moon directly, the moon is not completely dark, because some sunlight refracts (bends) through the Earth’s atmosphere and reaches the moon. This light is red for the same reason that sunsets on Earth tend to be red: the atmosphere scatters blue light more easily than red, so red light penetrates large distances of air more easily. This is why during a total lunar eclipse the moon is a reddish colour. Although a totally eclipsed moon looks bright enough to our eyes, it’s actually very dark compared to a normal full moon. Our eyes are very good at compensating for the different light levels without us being aware of it.

Total lunar eclipse

Total lunar eclipse of 28 August 2007 (photographed by me). 1 second exposure at ISO 800 and aperture f/2.8.

The amount of refracted light reaching the moon depends on the cleanliness of the Earth’s atmosphere. If there have been recent major volcanic eruptions, then significantly less light passes through to reach the moon. The brightness of the moon during a total lunar eclipse can be measured using the Danjon scale, ranging from 0 for very dark eclipses, to 4 for the brightest ones. After the eruption of Mount Pinatubo in the Philippines in 1991, the next few lunar eclipses were extremely dark, with the eclipse of December 1992 rating a 0 on the Danjon scale.

When the moon is only partly inside the Earth’s umbra, that is a partial lunar eclipse. A partial phase occurs on either side of a total lunar eclipse, as the moon passes through the Earth’s shadow, and it can also occur as the maximal phase of an eclipse if the moon’s orbit isn’t aligned to carry it fully within the umbra. During a partial eclipse phase, you can see the edge of the Earth’s umbral shadow on the moon.

Partial lunar eclipse phase

Partial phase of the same lunar eclipse of 28 August 2007. 1/60 second exposure at ISO 100 and f/8, which is 1/3840 the exposure of the totality photo above. If this photo was 3840 times as bright, the dark part at the bottom would look as bright as the totality photo (and the bright part would be completely washed out).

Lunar eclipses can only occur at the full moon – those times when the sun and moon are on opposite sides of the Earth. The moon orbits the Earth roughly once every 29.5 days and so full moons occur every 29.5 days. However, lunar eclipses occur only two to five times per year, because the moon’s orbit is tilted by 5.1° relative to the plane of the Earth’s orbit around the sun. This means that sometimes when the moon is full it is above or below the Earth’s shadow, rather than inside it.

Okay, so what can lunar eclipses tell us about the shape of the Earth? A lunar eclipse is a unique opportunity to see the shape of the Earth via its shadow. A shadow is the same shape as a cross-section of the object casting the shadow. Let’s have another look at the shape of Earth’s shadow on the moon, in a series of photos taken during a lunar eclipse:

Lunar eclipse montage

Montage of photos taken over 83 minutes during the lunar eclipse of 28 August 2007. Again, the bottom row of photos have 3840 times the exposure of the top row, so the eclipsed moon is nearly 4000 times dimmer than the full moon.

As you can see, the edge of the Earth’s shadow is curved. The fact that the moon’s surface is curved doesn’t affect this, because we are looking from the same direction as the Earth, so we see the same cross-section of the moon. (Your own shadow looks the shape of a person to you, even if it falls on an irregular surface where it looks distorted to someone else.) So from this observation we can conclude that the edge of the Earth is rounded.

Many shapes can cause a rounded shadow. However, if you observe multiple lunar eclipses, you will see that the Earth’s shadow is always round, and what’s more, it always has the same radius of curvature. And different lunar eclipses occur at any given location on Earth with the moon at different points in the sky, including sometimes when the moon is not in the sky (because the location is facing away from the moon). This means that different lunar eclipses occur when different parts of the Earth are facing the moon, which means that different parts of the Earth’s edge are casting the shadow edge on the moon. So from these observation, we can see that the shape of the shadow does not depend on the orientation of the Earth to the moon.

There is only one solid shape for which the shape of its shadow doesn’t depend on the object’s orientation. A sphere. So observations of lunar eclipses show that the Earth is a globe.

Addendum: A common rebuttal by Flat Earthers is that lunar eclipses are not caused by the Earth’s shadow, but by some other mechanism entirely – usually another celestial object getting between the sun and moon and blocking the light. But any such object is apparently the same colour as the sky, making it mysteriously otherwise completely undetectable, and does not have the simple elegance of explanation (and the supporting evidence from numerous other observations) of the moon moving around the Earth and entering its shadow.

Colour naming experiment

Firstly, sorry for the delay in getting a new proof written. I’ve been travelling, and then got sick on the flight home and was mostly incapacitated for two weeks. And I have deadlines for other stuff that then got in the way.

But one of those deadlines also involves science, and it’s pretty cool so I thought I’d share it with you. I do volunteer work with CSIRO’s STEM Professional in Schools program. As a professional scientist, I am partnered with a primary school and visit the school several times a year to talk to and engage the students with science topics. In past years I’ve mostly done presentations and Q&A sessions, but this year the school science coordinator suggested running a science club with some of the keenest science students from each year.

My Science Club is made of 13 students from years 2 to 6 (so ages 7 to 11). I’m running several experiments with them throughout the year. One of them is actually Eratosthenes’ method of measuring the size of the Earth, modified slightly. I’m getting the kids to measure the length of a vertical stick’s shadow every day at noon. At the end of the year I’ll help them plot the length versus day of the year, and we’ll fit a sine curve and extract the parameters to let us calculate the size of the Earth.

This Monday, I have another Science Club meeting, and I’ve been preparing a different experiment, on colour perception and naming. This is a cool topic that I’ve been interested in ever since I attended an imaging conference and saw some talks about the psychophysics and cultural psychology of colour perception. What I’ve done is to visit a local hardware store and raid their set of house paint sample brochures. Then I cut them up:

Cutting up paint brochures

I had way too many colours, many of which were very similar to others, so I selected a representative subset to try and span as much of the colour space as I could. Then I arranged them and used double sided tape to stick them into manila folders:

Sticking samples into folders

A couple of hours later, and I had 13 folders with identically laid out colour swatches inside:

Colour swatch folders

I used a marker to label all the swatches in each folder with a number. There are 35 swatches:

The 35 colour swatches

Now, here’s the experiment: On Monday in Science Club I’ll give each of the students one of the folders. I’ll also give them a potential list of colour names, with over 100 possible colour names on it:

List of colour names

Their task is to look at each colour, decide which name is the best name for it, and write the colour number on the sheet next to that name. Repeat for all 35 colours. So a lot of the names are going to be left unused. And I’ve included a few write-in slots for any cases where a student is positive that a certain colour really must be called “nasty bruise” or whatever. I’ve been careful to pick names that young children can relate to, and avoid weird things like “heliotrope” and “malachite” that they’ve probably never heard.

The science behind this experiment is that we’re all pretty good and consistent at naming very basic colours like red, and yellow, and blue, but when it comes to naming more subtle shades we are actually highly inconsistent. Is that particular shade of red: rose red, or raspberry, or cherry, or something else? Ask a lot of people and you’ll get a lot of different answers. There are classic studies showing this. (And yes, Randall Munroe of xkcd did a similar thing online a while back and published the results.)

There’s also a study showing that people are inconsistent with themselves, if given exactly the same task a few weeks later. Nearly everyone changes their mind on what certain shades should be called. So this is my experiment with my Science Club! I’m not going to tell the kids that we’ll be repeating this task later in the year. It’ll be interesting to see how closely they can reproduce their own results then, and also how closely their answers align with one another.

Basically, I’m doing something that happens with all good science. I’m replicating an experiment to see if I can reproduce the results. And now that I have this experiment ready to go, I’ll get on to writing up another proof that the Earth is a globe… hopefully within the next few days.

15. Trilateration

Trilateration is the method of locating points in space based on measuring the distances from known reference locations. It is used in surveying and navigation, similarly to the related method of triangulation, which technically uses the measurement of angles, not distances. For this entry we’re going to get practical and attempt to do some trilateration, using distances between some major cities in the world. To do this, I’ll need some equipment:

Equipment used

I acquired graph paper, a ruler, a tape measure, a pen, a pair of compasses, and a couple of large polystyrene balls.

I began my first scale drawing on a piece of graph paper. I’ve picked Auckland, New Zealand, as one of my cities of interest. Since nothing is on the paper yet, I can place Auckland wherever I want to. So I draw a cross indicating the position of Auckland and label it AKL (the city’s international airport code).

Auckland's position

For my second city, I’ve chosen Tokyo, Japan. According to a flight distance reference website, the travelling distance between Auckland and Tokyo, or more specifically between Auckland Airport and Tokyo’s Narita Airport, is 8806 kilometres. My graph paper has 2 mm squares, and (for reasons that will become clear in a minute) I’m using a scale of 86.1 km/mm. So I take a pair of compasses and set the distance from the metal point to the pen tip to be 102.3 mm as best I can. That’s 51 and a bit grid squares. I place the point in the centre of the AKL cross and mark a point on the paper 102.3 mm away with the pen tip. I enlarge the point to a cross and label it NRT (for Narita Airport). It doesn’t matter which direction I choose to place Tokyo from Auckland, because at this point there are no other constraints.

Tokyo's position

For my third city, I choose Los Angeles, USA. Los Angeles Airport, LAX, is 10467 km from Auckland, and 8773 km from Tokyo Narita. To locate LAX on my scale drawing, I first set my compasses with a distance of 10467 / 86.1 = 121.6 mm. With this distance setting, I draw an arc centred on AKL.

Los Angeles' position from Auckland

All of the points on this arc are the correct distance from Auckland to be Los Angeles. But we have another constraint – Los Angeles also has to be the correct distance from Tokyo. So I set my compasses to 8773 / 86.1 = 101.9 mm, and draw an arc centred at NRT.

Los Angeles' position nailed down

The intersection of these two arcs is the point that is both the correct scale distance from Auckland and Tokyo, so I label the intersection point LAX. So far, so good. We have three world cities with their relative positions accurately plotted to scale. Let’s add a fourth city! For the fourth city, I’ll choose something somewhere in the middle of the first three: Honolulu, USA. For starters, Honolulu is 7063 km from Auckland. So I draw an arc with radius 7063 / 86.1 = 82.0 mm centred on AKL.

Honolulu's position from Auckland

Honolulu is 6146 km from Tokyo. So I draw an arc with radius 6146 / 86.1 = 71.4 mm centred on NRT.

Honolulu's position from Auckland and Tokyo

Now in theory this is enough to give us the location of Honululu. It must be on both the arc centred at Auckland and on the arc centred at Tokyo – so it has to be at the intersection of those two arcs. But wait! We have more information than that. We also know that Honolulu is 4113 km from Los Angeles. So I draw an arc with radius 4113 / 86.1 = 47.8 mm centred on LAX.

Honolulu's position from Auckland, Tokyo, and Los Angeles

For the flight distances to be correct, Honolulu Airport (HNL) must be on all three arcs that I’ve drawn. But the arcs don’t all intersect at the same point. So where is Honolulu? According to the rules of geometry, anywhere we put it results in at least one of the distances being wrong. In the worst case, the the AKL-LAX intersection is 10 mm on the drawing from the NRT-LAX intersection, an error of 861 kilometres, which is 300 km longer than the entire chain of populated Hawaiian Islands from Niihau to Hawaii. Obviously a navigation error this large when trying to find Honolulu in the midst of the Pacific Ocean would be disastrous.

What’s gone wrong? Well, I’ve attempted to draw these distances to scale on a flat piece of paper. The error shows the distortion caused by trying to map the shape of the Earth onto a flat surface. The distances are all correct, but in reality they don’t lie in the same plane. So let’s try another approach. I’m going to map the distances onto a scale model of the Earth as a sphere.

To do this, I got a polystyrene sphere from an art supply shop. I measured the circumference using a tape measure to be 465 mm. Dividing the average circumference of the Earth by this gives me a scale of 86.1 km/mm (which is where I got the scale that I used for the drawing above). Now I just need to repeat the steps above, but plot the points and arcs on the surface of the sphere. But there’s one small wrinkle: flight distances are measured along the surface of the Earth, but the compasses step off the distance in a straight line, as measured through the Earth. To get the correct scale distance to set the compasses, we need to do a little geometry:

Geometry figure: surface distance versus straight line distance

The distance along the surface of the Earth is d, the distance through the Earth is x, and the radius of the Earth is r. In radians, the angle θ is d/r. Now according to the cosine rule of trigonometry:

x2 = r2 + r2 – 2r2 cos θ

x2 = 2r2(1 – cos(d/r))

So plugging in d and r we can find the distance x to set the compasses to (at the correct scale). Here’s a summary table:

Cities Distance (km) Scale distance (mm) Compasses distance (mm)
AKL-NRT 8806 102.3 94.3
AKL-LAX 10467 121.6 108.4
NRT-LAX 8773 102.0 94.0
AKL-HNL 7063 82.1 77.9
NRT-HNL 6146 71.4 68.6
LAX-HNL 4113 47.8 46.9

Using the distances in the Compasses column on my polystyrene sphere, and following the same steps as above for the graph paper, produced this:

Honolulu's position on a sphere

The arcs drawn with the correct scale distances of Honolulu from Auckland, Tokyo, and Los Angeles all intersect at exactly the same point on the surface of the sphere. We’ve found Honolulu!

So by experiment, trilateration of points on the Earth’s surface does not work if you use a flat surface to map the points. It only works if you use a sphere.

Addendum: I bought two spheres because I was prepared for the first attempt to be a little bit out due to any small inaccuracies or mistakes in my setting the correct compasses distances. But as it turned out I only needed the one. I was pleasantly surprised when it worked so well the first time.

14. Map projections

Cartography is the science and art of producing maps – most commonly of the Earth (although there are also maps of astronomical bodies and fictional worlds). The best known problem in cartography is that of representing the Earth on a flat map with minimal distortion.

If the Earth were itself flat, then this problem would not exist. A flat map would simply be a constant scale drawing of the flat Earth, and it woud be accurate and distortion-free at all points. But it is well known that such a map cannot be made. The reason, of course, is that the Earth is spherical, and the surface of a sphere cannot be projected onto a flat plane without some sort of distortion.

There are numerous different map projections, which render areas of the Earth onto a flat map with varying types and amounts of distortion. In all of these projections, some trade off must be made between the different goals of preservation of distances (i.e. a constant distance scale), preservation of directions (e.g. north is always up, east is always to the right), preservation of shapes (geographical regions look the same shape as they do when viewed from the air or a satellite), and preservation of areas (geographical regions of the same area appear the same area on the map). The familiar Mercator projection preserves directions at the expense of all the others, and is infamous for its large distortions of area between the polar and equatorial regions.

Mercator projection

Mercator projection map of Earth, showing gross area distortions. (Public domain image from Wikimedia Commons.)

The area distortion is apparent when you consider that Africa has an area of 30.4 million square kilometres, while North America, including Central American, the Caribbean islands, the northern Canadian islands, and Greenland, is only 24.7 million square kilometres. On a Mercator map, Greenland all by itself looks larger than Africa, but it is in reality less than a third the size of Australia.

There are projections which give a better impression of the relative areas, but these necessarily distort shapes and distances. The Gall-Peters projection also maps lines of latitude and longitude to straight lines like Mercator, but preserves areas.

Gall-Peters projection

Gall-Peters projection map of Earth, showing gross shape distortions. (Public domain image from Wikimedia Commons.)

Both of these projections have the disadvantage that distortion becomes extreme at the poles. In the Gall-Peters projection, the North and South Poles are mapped to horizontal lines spanning the width of the map, rather than to points. The Mercator projection cannot even show the poles at all, because the projection puts them at infinity.

A map specifically designed to compromise between all the various distortions is the Winkel tripel projection. This projection was adopted by the National Geographic Society as its standard world map projection in 1998 (replacing the Robinson projection, a similar compromise projection), and many textbooks and educational materials now use it.

Winkel tripel projection

Winkel tripel projection map of Earth, showing compromised distortions. (Public domain image from Wikimedia Commons.)

In the Winkel tripel projection, the lines of latitude and longitude are both curved, indicating that directions and shapes are not preserved faithfully. Areas are also distorted somewhat – Greenland looks almost the same size as Australia, even though it is less than a third the area. But all of the different distortions are moderate compared to the more extreme distortions visible in some of these features in other projections.

The Hammer projection goes further in rectifying the distance distortion issues with the polar regions, by mapping the North and South Poles to single points, as they are on Earth. However, this distorts the shapes of areas near the poles and away from the central meridian even more.

Hammer projection

Hammer projection map of Earth, showing poles mapped to points, but large shape distortions. (Public domain image from Wikimedia Commons.)

If you want to minimise distortions in a sort of T-shaped area encompassing, say, the Old World continents of Europe, Asia, and Africa, then you can do a bit better by adopting the Bonne projection. This maps a chosen so-called “standard” parallel of latitude to a circular arc, which reduces distortion along that parallel (since in reality parallels of latitude are circles, not straight lines).

Bonne projection

Bonne projection map of Earth with standard parallel 45°N, showing poles mapped to points and small distortions in Africa, Europe, Asia, but large distortions in South America, Australia, Antarctica. (Public domain image from Wikimedia Commons.)

Minimising distortions along the straight central meridian and a parallel in the northern hemisphere naturally increases distortions in the southern hemisphere, making this a good choice for Old World maps, which it has been used for extensively, but pretty bad for South America and Oceania.

Oddly enough, we’re now not so far from the standard map advocated by most Flat Earthers. If you map all the parallels of latitude to complete circles (rather than partial circular arcs as in the Bonne projection), increasing in radius by a constant amount per degree of latitude, you end up with an azimuthal equidistant projection, centred on the North Pole.

Azimuthal equidistant projection

Azimuthal equidistant projection map of Earth centred at the North Pole. (Public domain image from Wikimedia Commons.)

The result is that the northern hemisphere is moderately distorted, but the distortion grows extreme in the south, and the South Pole is mapped to a large circle encircling the whole map. This map projection is good for showing directions relative to the central point (so a variant centred on Mecca is useful for Muslims who wish to know which direction to face during prayer). It’s not a good projection for much else though, because of the severe distance, shape, direction, and area distortions in the southern hemisphere. If you were plotting a trip from Australia to South America, it would be utterly useless.

If the Earth were in fact flat, then it would be possible—indeed trivial—to construct flat maps which accurately show the shapes and distances over large areas of the Earth’s surface at a constant scale. No such maps exist. And the fact that cartographers have struggled for centuries to make flat maps of the world, trading off various compromises with arguable degrees of success, is evidence that it’s not possible, and that the Earth is a globe.

Admin update: travelling 11-25 May

Sorry there hasn’t been a new post for a few days. I’ve been dealing with some urgent deadlines for stuff that needs to be done before I travel next week. I’m going to Portugal for 2 weeks, from 11-25 May. There won’t be any new posts while I’m away, but I hope to get one or two more done before I leave – after I deal with more urgent stuff this weekend.

13. Hydrostatic equilibrium

The theory of gravity is wildly successful in explaining and predicting the behaviours of masses. Isaac Newton’s formulation of gravity (published in his Principia Mathematica in 1686) is a simple formula that works very well for most circumstances of interest to people. When the gravitational potential energy or the velocity of a mass is very large, Albert Einstein’s general relativity (published 1915) is required to correctly determine behaviour. Newton’s gravity is in fact an approximation of general relativity that gives almost exactly the correct answer when the gravitational energy per unit mass is small compared to the speed of light squared, and the velocity is much smaller than the speed of light. For almost all calculation purposes, Newton’s law is sufficiently accurate to be used without worrying about the difference.

Newton’s law says that the force of gravitational attraction F between two bodies equals the universal gravitational constant G, multiplied by the masses of the two bodies m1 and m2, divided by the square of the distance r between them: F = G m1 m2/(r2).

Newton's law of gravitation

Newton’s law of gravitation describes the force F between two bodies m1 and m2 separated by a distance r between their centres of mass.

Newton himself had no idea why this simple formula worked. Although he showed that it was accurate to the limits of the measurements available to him, he was deeply concerned about its philosophical implications. In particular, he couldn’t imagine how such a force could occur between two bodies separated by any appreciable distance or the vacuum of space. He wrote in a letter to Richard Bentley in 1692:

“That one body may act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action and force may be conveyed from one another, is to me so great an absurdity that, I believe, no man who has in philosophic matters a competent faculty of thinking could ever fall into it.”

Newton was so concerned about this that he added an appendix to the second edition of the Principia – an essay titled the General Scholium. In this he wrote about the distinction between observational, experimental science, and the interpretation of observations (translated from the original Latin):

“I have not as yet been able to discover the reason for these properties of gravity from phenomena, and I do not feign hypotheses. For whatever is not deduced from the phenomena must be called a hypothesis; and hypotheses, whether metaphysical or physical, or based on occult qualities, or mechanical, have no place in experimental philosophy.”

In other words, Newton was being led by his observations to deduce physical laws and how the universe behaves. He refused to countenance speculation unsupported by evidence, and he accepted that the world behaved as observed, even if he didn’t like it. Commenting on Newton’s words in 1840, the philosopher William Whewell wrote:

“What is requisite is, that the hypotheses should be close to the facts, and not connected with them by other arbitrary and untried facts; and that the philosopher should be ready to resign it as soon as the facts refuse to confirm it.”

This affirms the position of a scientist as one who observes nature and tries to describe it as it is. Any hypothesis formed about how things are or why they behave the way they do must conform to all the known facts, and if any future observation contradicts the hypothesis, then the hypothesis must be abandoned (perhaps to be replaced with a different hypothesis). This is the scientific method in a nutshell, and guides our understanding of the shape of the Earth in these pages.

The universal gravitational constant G is a rather small number in familiar units: 6.674×10−11 m3 kg−1 s−2. This means that the force of gravity between two everyday objects is so small as to be unnoticeable. For example, even large objects such as two 1-tonne cars a metre apart experience a gravitational force between them of only 6.674×10−5 newtons – far too small to move the cars against rolling friction, even with the brakes off. Also, the distance between the masses in Newton’s formula is the distance between the centres of mass of the objects, not the closest surfaces. The centres of mass of two cars can’t be brought closer together than about 2 metres in practice, even with the cars touching each other (unless you crush the cars).

Gravity really only starts to significantly affect things when you gather millions of tonnes of mass together. On Earth, the mass of the Earth (5.9722×1024 kilograms) itself dominates our experience with gravity. Removing mass 2 from Newton’s formula, we can calculate the acceleration a towards the centre of mass of the Earth, caused by the Earth’s gravity, as experienced at the surface of the planet (r = 6370 kilometres): a = G m1/(r2) = 6.674×10−11 × 5.9722×1024 / (6370×103)2 = 9.82 m/s2. This number matches experimental observations we can make of the gravity on the surface of the Earth (for example, using a pendulum: see also Airy’s coal pit experiment).

So large object like planets or other astronomical bodies experience a significant gravitational force on parts of themselves. Think about a tall mountain, such as Mount Everest. Let’s estimate the mass of Mount Everest – just roughly will do for our purposes. It is 8848 metres tall, above sea level. Let’s imagine it’s roughly a cone, with sides sloping at 45°. That makes the radius of the base 8848 metres, and its volume is π × 88483 / 3 = 7.25×1011 cubic metres. The density of granite is 2.75 tonnes per cubic metre, so the mass of Mount Everest is roughly 2×1015 kg. It experiences a gravitational force of approximately 2×1016 newtons, pulling it down towards the rest of the Earth.

Newton's law of gravitation

Approximating Mount Everest as a cone of rock to calculate the pressure on the base.

Obviously Mount Everest is strong enough to withstand this enormous force without collapsing. But how much higher could a mountain be without collapsing under its own mass? The taller a mountain gets, the more force pulls it down, but the structural strength of the rock making up the mountain does not increase. At some point there is a limit. Our conical Mount Everest model spreads that mass over an area of π × 88482 square metres. This means the pressure of the rock above on this area is 2×1016 / (π × 88482) = 8×107 pascals, or 80 megapascals (Mpa). Now, the compressive strength of granite is about 200 MPa. We’re pretty close already! Not to mention that rock can also shear and deform plastically, so we probably don’t even need to get as high as 200 MPa before something bad (or spectacular, depending on your point of view!) happens. A mountain twice as high as Everest would almost certainly be unstable and collapse very quickly.

As mountains get pushed up by tectonic activity, their bases spread out under the pressure of the rock above, so that they can’t exceed the limit of the tallest possible mountain. In practice, it turns out that glaciation also has a significant effect on the maximum height of mountains on Earth, limiting them to something not much higher than Everest [1].

Now, compared to the size of the Earth, even a mountain as tall as Everest is pretty insignificant. It is barely a thousandth of the radius of the planet. It’s often said that if shrunk down to the same size, the Earth would be smoother than a billiard ball. In a sense, this is actually true! Billiard and snooker balls are specified to be 52.5 mm in diameter, with a tolerance of 0.05 mm [2]. That is just under a 500th of the radius, so it would be acceptable to have billiard balls for professional play that are twice as rough as the Earth – although in practice I suspect that billiard balls are manufactured smoother than the quoted tolerance.

So, there is a physical limit to the strength of rock that means that Earth can’t have any protruding lumps of any significant size compared to its radius. Similarly, any deep trenches can’t be too deep either, or else they’ll collapse and fill in due to the gravitational stress on the rock pulling it together. The Earth is spherical in shape (more or less) because of the inevitable interaction of gravity and the structural strength of rock. Any astronomical body above a certain size will also necessarily be close to spherical in shape. The size may vary depending on the materials making up the body: rock is stronger than ice, so icy worlds will necessarily be spherical at smaller sizes than rocky ones.

The phenomenon of large bodies assuming a spherical shape is known as hydrostatic equilibrium, referring to the fact that this is the shape assumed by any body with no resistance to shear forces, in other words fluids. For ice and rock, the resistance to shear force is overcome by gravity for objects of size a few hundred to a thousand or so kilometres in diameter. The asteroid Ceres is a hydrostatic spherical shape, with a diameter of 945 km. On the other hand, Saturn’s moon Iapetus is the largest known object to deviate significantly from hydrostatic equilibrium, with a diameter of 1470 km. Iapetus is almost spherical, but has an unusual ridge of mountains running around its equator, with a height around 20 km – about 1/36 of the moon’s radius.

Iapetus

Iapetus, one of the moons of Saturn, photographed by NASA’s Cassini space probe. (Public domain image by NASA.)

It’s safe to say, however, that any planetary sized object has to be very close to spherical – or spheroidal if rotating rapidly, causing a slight bulge around the equator due to centrifugal force. This is because of Newton’s law of gravity, and the structural strength of rock. Our Earth, naturally, is such a sphere.

Flat Earth models must either conveniently ignore this conclusion of physics, or posit some otherwise unknown force that maintains the mass of the Earth in a flat, non-spherical shape. By doing so, they violate Newton’s principle that one must be guided by observation, and discard any hypothesis that does not fit the observed facts.

References:

[1] Mitchell, S. G., Humphries, E. E., “Glacial cirques and the relationship between equilibrium line altitudes and mountain range height”. Geology, 43, p. 35-38, 2015. https://doi.org/10.1130/G36180.1

[2] Archived from worldsnooker.com on archive.org: https://web.archive.org/web/20080801105033/http://www.worldsnooker.com/equipment.htm

12. The sun

Possibly the most obvious property of our sun is that it is visible from Earth during daylight hours, but not at night. The visibility of the sun is in fact what defines “day time” and “night time”. At any given time, the half of the Earth facing the sun has daylight, while the other half is in the shadow of the Earth itself, blocking the sun from view. It’s trivial to verify that parts of the Earth are in daylight at the same time as other parts are in night, by communicating with people around the world.

The first physical property of the sun to be measured was how far away it is. In the 3rd century BC, the ancient Greek Aristarchus of Samos (who we met briefly in 2. Eratosthenes’ measurement) developed a method to measure the distance to the sun in terms of the size of the Earth, using the geometry of the relative positions of the sun and moon. Firstly, when the moon appears exactly half-illuminated from a point on Earth, it means that the angle formed by the sun-moon-Earth is 90°. If you observe the angle between the sun and the moon at this time, you can determine the distance to the sun as a multiple of the distance to the moon.

Aristarchus's method 1

Geometry of the sun, moon, and Earth when the moon appears half-illuminated.

In the figure, if you measure the angle θ, then the ratio of the distance to the sun S divided by the distance to the moon M is the reciprocal of the cosine of θ. Aristarchus then used the size of the shadow of Earth on the moon during a lunar eclipse to obtain further equations relating the distances to the sun and moon and the size of the Earth.

Aristarchus's method 2

A medieval copy of Aristarchus’s drawing of the geometry of the sun-Earth-moon system during a lunar eclipse. (Public domain image.)

By combining these results, you can calculate the distances to both the sun and the moon in terms of the radius of the Earth. Aristarchus got the wrong answer, estimating that the sun was only about 19 times further away than the moon, because of the limited precision of his naked eye angle measurements (it’s actually 390 times further away). But Eratosthenes later made more accurate measurements (which were again discussed in Eratosthenes’ measurement), most likely using the same method.

The first rigorous measurement of the absolute distance to the sun was made by Giovanni Cassini in 1672. By this time, observations of all the known celestial bodies in our solar system and some geometry had well and truly established the relative distances of all the orbits. For example, it was known that the orbital radius of Venus was 0.72 times that of Earth, while the orbit of Mars was 1.52 times that of Earth. To measure the absolute distance to the sun, Cassini used a two-step method, the first step of which was measuring the distance to the planet Mars. This is actually a lot easier to do than measuring the distance to the sun, because Mars can be seen at night, against the background of the stars.

Cassini dispatched his colleague Jean Richer to Cayenne in French Guiana, South America, and the two of them arranged to make observations of Mars from there and Paris at the same time. By measuring the angles between Mars and nearby stars, they determined the parallax angle subtended by Mars across the distance between Paris and Cayenne. Simple geometry than gave the distance to Mars in conventional distance units. Then applying this to the relative distances to Mars and the sun gave the absolute distance from the Earth to the sun.

Since 1961, we’ve had a much more direct means of measuring solar system distances. By bouncing radar beams off the moon, Venus, or Mars and measuring the time taken for the signal to return at the speed of light, we can measure the distances to these bodies to high precision (a few hundred metres, although the distances to the planets change rapidly because of orbital motions) [1].

The Earth orbits the sun at a distance of approximately 150 million kilometres. Once we know this, we can work out the size of the sun. The angular size of the sun as seen from Earth can be measured accurately, and is 0.53°. Doing the mathematics, 0.53°×(π/180°)×150 = 1.4, so the sun is about 1.4 million kilometres in diameter, some 109 times the diameter of the Earth. This is the diameter of the visible surface – the sun has a vast “atmosphere” that we cannot see in visible light. Because of its vast distance compared to the size of the Earth, the sun’s angular size does not change appreciably as seen from different parts of the Earth. The difference in angular size between the sun directly overhead and on the horizon (roughly the Earth’s radius, 6370 km, further away) is only about 6370/150000000×(180°/π) = 0.002°.

Our sun is, in fact, a star – a huge sphere composed mostly of hydrogen and helium. It produces energy from mass through well-understood processes of nuclear fusion, and conforms to the observed properties of stars of similar size. The sun appears much larger and brighter than stars, and heats the Earth a lot more than stars, because the other stars are all so much further away.

The Sun

Our sun, observed in the ultraviolet as a false colour image by NASA’s Solar Dynamics Observatory satellite. (Public domain image by NASA.)

Like all normal stars, the sun radiates energy uniformly in all directions. This is expected from the models of its structure, and can be inferred from the uniformity of illumination across its visible disc. The fact that the sun’s polar regions are just as bright as the equatorial edges implies that the radiation we see in the ecliptic plane (the plane of Earth’s orbit) is reproduced in all directions out of the plane as well.

NASA’s Ulysses solar observation spacecraft was launched in 1990 and used a gravity slingshot assist from Jupiter to put it into a solar orbit inclined at about 80° to the ecliptic plane. This allowed it to directly observe the sun’s polar regions.

Ulysses' orbit

Polar orbit of Ulysses around the sun, giving it views of both the sun’s north and south poles. (Public Domain image by NASA.)

Now, I tried to find scientific papers using data from Ulysses to confirm that the sun indeed radiates electromagnetic energy (visible light, ultraviolet, etc.) uniformly in all directions. However, it seems that no researchers were willing to dedicate space in a paper to discussing whether the sun radiates in all directions or not. It’s a bit like looking for a research paper that provides data on whether apples fall to the ground or not. What I did find are papers that use data from Ulyssessolar wind particle flux detectors to measure if the energy emitted by the sun as high energy particles varies with direction.

Solar proton flux versus latitude

Proton flux density observed by Ulysses at various heliographic (sun-centred) latitudes. -90 is directly south of the sun, 0 would be in the ecliptic plane. The track shows Ulysses’ orbit, changing in distance and latitude as it passes under the sun’s south polar regions. Figure reproduced from [2].

Solar wind energy flux versus latitude

Various solar wind plasma component energy fluxes observed by Ulysses at various heliographic latitudes. Figure reproduced from [3].

As these figures show, the energy emitted by the sun as solar wind particles is pretty constant in all directions, from equatorial to polar. Interestingly, there is a variation in the solar wind energy flux with latitude: the solar wind is slower and less energetic close to the plane of the ecliptic than at higher latitudes. The solar wind, unlike the electromagnetic radiation from the sun, is affected by the structure of the interplanetary medium. The denser interplanetary medium in the plane of the ecliptic slows the wind. The amount of slowing provides important constraints on the physics of how the solar wind particles are accelerated in the first place.

Anyway, given there are papers on the variation of solar wind with direction, you can bet your bottom dollar that there would be hundreds of papers about the variation of electromagnetic radiation with direction, if it had been observed, because it goes completely counter to our understanding of how the sun works. The fact that the sun radiates uniformly in all directions is such a straightforward consequence of our knowledge of physics that it’s not even worth writing a paper confirming it.

Now, in our spherical Earth model, all of the above observations are both consistent and easily explicable. In a Flat Earth model, however, these observations are less easily explained.

Why is the sun visible in the sky from part of the Earth (during daylight hours), while in other parts of the Earth at the same time it is not visible (and is night time)?

The most frequently proposed solution for this is that the sun moves in a circular path above the disc of the Flat Earth, shining downwards with a sort of spotlight effect, so that it only illuminates part of the disc. Although there is a straight line view from areas of night towards the position of the sun in the sky, the sun does not shine in that direction.

Given that we know the sun radiates uniformly in all directions, we know this cannot be so. Furthermore, if the sun were a directional spotlight, how would such a thing even come to be? Directional light sources do occur in nature. They are produced by synchrotron radiation from a rapidly rotating object: for example, a pulsar. But pulsars rotate and sweep their directional beams through space on a timescale of approximately one second. If our sun were producing synchrotron radiation, its spotlight beam would be oscillating many times per minute – something which is not observed.

Even furthermore, if the sun is directional and always above the plane of the Flat Earth, it should be visible in the night sky, as an obscuration passing in front of the stars. This prediction of the Flat Earth model is not seen – it is easy to show that no object the size of the sun obscures any stars at night.

And yet furthermore, if the sun is directional, there are substantial difficulties in having it illuminate the moon. Some Flat Earth models acknowledge this and posit that the moon is self-luminous, and changes in phase are caused by the moon itself, not reflection of sunlight. This can easily be observed not to be the case, since (a) there are dark shadows on the moon caused by the light coming from the location of the sun in space, and (b) the moon darkens dramatically during lunar eclipses, when it is not illuminated by the sun.

In addition to the directional spotlight effect, typical Flat Earth models state that the distance to the sun is significantly less than 150 million kilometres. Flat Earth proponent Wilbur Glenn Voliva used geometry to calculate that the sun must be approximately 3000 miles above the surface of the Earth to reproduce the zenith angles of the sun seen in the sky from the equator and latitudes 45° north and south.

Voliva's distance to the sun calculation

Wilbur Glenn Voliva’s calculation that the sun is 3000 miles above the Flat Earth. Reproduced from Modern Mechanics, October 1931, p. 73.

Aside from the fact that Voliva’s distance does not give the correct zenith angles for any other latitudes, it also implies that the sun is only about 32 miles in diameter, given the angular size seen when it is overhead, and that the angular size of the sun should vary significantly, becoming only 0.53°/sqrt(2) = 0.37° when at a zenith angle of 45°. If the sun is this small, there are no known mechanisms than can supply the energy output it produces. And the prediction that the sun would change in angular size is easily disproved by observation.

The simplest and most consistent way of explaining the physical properties of our sun is in a model in which the Earth is a globe.

References:

[1] Muhleman, D. O., Holdridge, D. B., Block, N. “The astronomical unit determined by radar reflections from Venus”. The Astrophysical Journal, 67, p. 191-203, 1962. https://doi.org/10.1086/108693

[2] Barnes, A., Gazis, P. R., Phillips, J. L. “Constraints on solar wind acceleration mechanisms from Ulysses plasma observations: The first polar pass”. Geophysical Research Letters, 22, p. 3309-3311, 1995. https://doi.org/10.1029/95GL03532

[3] Phillips, J. L., Bame, S. J., Barnes, A., Barraclough, B. L., Feldman, W. C., Goldstein, B. E., Gosling, J. T., Hoogeveen, G. W., McComas, D. J., Neugebauer, M., Suess, S. T. “Ulysses solar wind plasma observations from pole to pole”. Geophysical Research Letters, 22, p. 3301-3304, 1995. https://doi.org/10.1029/95GL03094