Flying as fast as the sun
All of last week I had been traveling around Seattle for a mix of work and sightseeing, so this week's a short post about something that nerd-sniped me earlier in the year.
I've got a friend who occasionally writes game scenarios in Japanese that I later I pick up for translation. Normally I get involved long after drafts have been written and are being finalized for release. This time, my friend decided to ask for my opinion on what to write before the first draft was even being planned. Definitely a new experience for me.
In the discussions, my friend proposed a couple of potential plot devices he thought would be interesting. One completely nerd-sniped me (xkcd reference) and sent me down a rabbit hole. It involved using a plane to fly around the world in a way to stay within the night side of the Earth and avoid the Sun for a full day for Plot Reasons. My instinctual, perfectly normal, reaction was to ask myself whether it's possible for a plane to fly fast enough to keep up with the rotation of the earth (spoilers, yes it is possible*).
What's fun about the question is that it's not particularly hard to come to a pretty good approximation of the correct answer. If you assume the Earth's just a perfect sphere that completes one full rotation in exactly 24 hours, all you need to know is the radius of the Earth to figure out how fast a point on the equator needs to move to do a full rotation in 24 hours. If you add in a tiny bit of trigonometry, you should be able figure out the amount of distance covered per day at any latitude.
Velocity of Earth's rotation at Equator = 2 * π * 6371 km / 24 hours = 1667.92 km/h
Velocity at given latitude = cos(latitude) * Velocity of Earth's rotation at Equator
Either calculation will end up giving you the speed at which a point on the Earth's surface is traveling. The easiest way to "avoid the sun for a whole day" would simply involve starting a flight at night and head west at roughly the same speed as the Earth's surface is rotating.
But things can be even easier. Since the shadow of the Earth is ... planet-sized, you actually have a lot of room for error in terms since "night" lasts quite a few hours. You can go a bit faster, or even a bit slower, and you're quite likely to still be within the Earth's shadow for however long you need to avoid the Sun.
Slightly more complex
But of course, we used a bunch of simplifying assumptions for these calculations. For one thing, we've completely ignore the fact that the Earth is tilted, so during the height of summer or winter, one of the poles, and the whole area beyond the Arctic/Antarctic circle would be experiencing complete darkness. That would allow avoiding the sun with zero work. Maybe my friend would consider it a kind of cheating, or is thinking about the times around equinoxes.
Even if we ignore the tilt of the Earth, we know that the Earth isn't a perfect sphere but instead an oblate spheroid. The difference is that the circumference at the equator is roughly 0.3% longer than for a perfectly spherical Earth. So the real equator moves marginally faster than the idealized sphere. This detail is likely immaterial to my friend's plot but maybe he's writing something that needs atomic clock precision.
Similarly, the concept of "day" on Earth is somewhat inexact. We're used to working with "solar days", which is defined as being 24 hours long (60 * 60 * 24 seconds honestly) and time was defined so that the sun is in the same position overhead at the same time (noon). This is useful for our plane scenario because we really do want the sun to be in a certain position in the sky on the other side of us with the Earth in between.
We luckily don't have to factor in Sidereal time where the sidereal day is a few minutes shorter than the solar day. The Earth moves around the sun fast enough that, relative to distant fixed stars, we need to give a few extra minutes for the Earth to rotate in order to have sun be in the correct position at noon. Those few minutes are more or less immaterial to our main question, though it might be relevant along the edges. If somehow instead of hiding from the sun, we wanted to hide from a specific star in the sky (maybe it's gone quasar at us), then sidereal time might matter.
Either way, this problem starts sounding like real work that I don't know how to do. Luckily, by the power of the internet, someone has already asked a similar question and answered it already:
The Shape of the Earth
Due to the centrifugal force created by the Earth’s rotation, our planet is flattened at the poles and bulges at the Equator. The Earth's shape could be approximated with good accuracy by the oblate ellipsoid (ellipsoid of revolution).
Thus, the equation of the Earth's shape is:
x2/a2+y2/b2+z2/c2=1
where (for the Earth):
a = b = 6,378,137.0 meters
c = 6,356,752.3142 meters
Applying geometric methods yields the formula for the radius of the circle of latitude:
Rlat=a*cos(lat)/(1-e2sin2(lat))1/2
where:
e2 (eccentricity) = (a2-c2)/a2 = 0.00669438
Even using these more advanced approximations, it's only about a 0.3% difference in speed at the equator, where the error would be greatest. Nice to know that we don't need to really worry about it.
But what about the flying part?
Now that we know that at the equator we'd need to move at a bit over 1700 km/h relative to the ground (a.k.a. ground speed and not air speed). is it possible to even fly that fast?
Well, yes and no? Searching around some civilian aircraft sources (like this one), the fastest single engine plane, a Mooney Acclaim, can fly at an airspeed of 242 knots (448 kph). Assuming completely still air (so ground speed = air speed), you'd be able to fly as fast as the Earth rotates under you at around 75 degrees latitude. That's more north than the arctic circle which is around 66.5 degrees.
As noted before, winds can be a problem. You'll likely be facing headwinds that means you'll need a plane that goes faster to overcome the headwind and keep up with the ground. Airplane specifications are often quoted in terms of airspeed because presumably winds will very much mess with the plane's speed relative to the ground.
I have no idea where to find what the typical wind speeds are at airplane cruising altitudes cut by latitude. Best I have is this web site which takes data from the National Weather Service and generates a heat map of wind data collected for weather prediction use. If you flip through the available altitudes, they generally get stronger the higher up you go. But there's a huge variation in the actual speeds depending on the weather, ranging from close to 0 km/h to over 300 km/h.
For commercial airliners, a Boeing 737 can cruise at about 838 km/h. That would get you to somewhere around 60 degrees latitude in still air, comfortably flying over parts of Canada, Alaska. It's even further south than all of Iceland. In theory if you customized it, the plane could probably hold a ridiculous amount of fuel to support a very long flight time beyond the ~6-7 hours the longest range versions get normally.
An Airbus 350-900 runs commercial flights up to 18 hours long, cruises around 1050 km/h, and should be able to keep up with the Sun at latitudes of 52 degrees or better. That'd place it over much of Europe and North America.
Military planes can obviously fly even faster. They can even refuel in mid-air for near unlimited flight time if needed. At that point you'd just be limited by the cost of manpower and logistics.
Now if only my friend would tell me why he thought that plot device would be good...
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I’m Randy Au, Quantitative UX researcher, former data analyst, and general-purpose data and tech nerd. Counting Stuff is a weekly newsletter about the less-than-sexy aspects of data science, UX research and tech. With some excursions into other fun topics.
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