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Assuming one had access to all modern technology but couldn't go into space, would it be possible for the protagonist to permanently stay in sunlight? While it would be trivial near the poles during summer and winter, I presume there's no aircraft capable of staying within sunlight around the time of the equinox.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Monica Cellio Oct 18 '16 at 3:31

13 Answers 13

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With current aerial refueling technology, which is capable of refueling some aircraft in the air, one could "easily" keep the sun over the horizon, even near the equator, in a sufficiently fast aircraft.

The circumference of the Earth is about 40000 km, so you'd need to go about $\frac{40\,000}{24} = 1667$ km/h to stay ahead of nightfall. That's just over the speed of sound at 1236 km/h, but we have plenty of supersonic aircraft capable of the task, many of which can be refueled in the air.

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    $\begingroup$ @Crowley that's valid at the equator... Can't you just move north/south, thus making the distance -and thus need speed -less? Equator is worse case $\endgroup$ – WernerCD Oct 15 '16 at 23:44
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    $\begingroup$ @Crowley I think you would need to have 10 000 km altitude to be forced to move almost thrice as fast as ground-level - and that's pretty far out of range for commercial jets. $\endgroup$ – Cthulhu Oct 15 '16 at 23:49
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    $\begingroup$ @Crowley you calculations are simply wrong radius of earth if near 6370 km, equator length in this case is 40003km, at 10km altitude, 6380km radius, length will be 40066km, just 63km longer, so speed have to be 1.001585 faster then at altitude 0.0km $\endgroup$ – MolbOrg Oct 16 '16 at 0:02
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    $\begingroup$ He doesn’t mean the circle is longer, but rather the speed of sound is lower at thinner air. This affects the functioning of the aerodynamics so the real Mach number is important, not just a choice of scale. $\endgroup$ – JDługosz Oct 16 '16 at 1:14
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    $\begingroup$ @Bohemian more than 95% of aircraft wear and tear is caused during take off and landing. You can stay up an awfully long time before requiring maintenance. $\endgroup$ – Roger Willcocks Oct 17 '16 at 2:15
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Yes, it is possible, but for a normal travel airplane very hard to accomplish. In a former answer I errornously used an estimate of 24 hours, but as WhatRoughBeast pointed out, the real time is 12 hours (at equinox every latitude day and night are equally long). That is a real challenge.

You start at the poles during the equinox in a plane, either with an Boeing 777-200LR (range 17,395km) or an Airbus A340-500 (range 16,670km). As you will see you need more range so pull out the seats and fill the empty room with fuel tanks. This allows us to reach our destination without air refueling (which would be the only other option, because we have no spare time).

Your plane stays always exactly after the terminator shadow to reach 12 h flight time until no longer possible. Before I assumed an 45° angle, now we need to get as close to the equator as possible.

Cruise speed is 900 km/h.

At what latitude following the sun is not possible anymore ?

1667 km/h * cos (latitude) = 900 km/h =>
latitude = arccos(0.54) = 57,5°

After that we still have 12 h light if we would fly straight southwards and stay on the same longitude...this distance would be 115° x 111.3 km/h = 12 800 km. But we can extend the time by flying eastwards after the sun to widen the distance between us and the terminator line.

Now give your plane the beans, you must have an airspeed of more than 850 km/h, else you will be caught by the terminator line before you reach the safe -57.5° line. The faster, the better !

This is the flight plan for 900 km/h:

Terminator lines are at 0 and +- 180° when we enter the latitude.

Lat: 57,500000 Long: 0,000000
Lat: 55,000000 Long: 1,005882
Lat: 52,500000 Long: 2,009119
Lat: 50,000000 Long: 3,031186
Lat: 47,500000 Long: 4,056425
Lat: 45,000000 Long: 5,040777
Lat: 42,500000 Long: 6,109931
Lat: 40,000000 Long: 7,121132
Lat: 37,500000 Long: 8,113210
Lat: 35,000000 Long: 9,163885
Lat: 32,500000 Long: 10,088179
Lat: 30,000000 Long: 11,122173
Lat: 27,500000 Long: 12,294797
Lat: 25,000000 Long: 13,231579
Lat: 22,500000 Long: 14,349810
Lat: 20,000000 Long: 15,346429
Lat: 17,500000 Long: 16,193566
Lat: 15,000000 Long: 17,093546
Lat: 12,500000 Long: 18,037084
Lat: 10,000000 Long: 19,019831
Lat: 7,500000 Long: 20,221031
Lat: 5,000000 Long: 21,123519
Lat: 2,500000 Long: 22,210219
Lat: 0,000000 Long: 23,235670
Lat: -2,500000 Long: 24,551104
Lat: -5,000000 Long: 25,594283
Lat: -7,500000 Long: 26,447472
Lat: -10,000000 Long: 27,615605
Lat: -12,500000 Long: 28,637297
Lat: -15,000000 Long: 29,308408
Lat: -17,500000 Long: 30,056890
Lat: -20,000000 Long: 31,585313
Lat: -22,500000 Long: 32,427503
Lat: -25,000000 Long: 33,287608
Lat: -27,500000 Long: 34,620467
Lat: -30,000000 Long: 35,282857
Lat: -32,500000 Long: 36,352323
Lat: -35,000000 Long: 37,561653
Lat: -37,500000 Long: 38,595209
Lat: -40,000000 Long: 39,178347
Lat: -42,500000 Long: 40,832200
Lat: -45,000000 Long: 41,588653
Lat: -47,500000 Long: 42,181960
Lat: -50,000000 Long: 43,299990
Lat: -52,500000 Long: 44,240993
Lat: -55,000000 Long: 45,488449
Lat: -57,500000 Long: 46,147859

Distance : 13 534 km
Flight time : 15.037863 h
Terminator line : 45,567938

The values are calculated by a numerical routine I had scraped together in short time. I think the optimization is still off (check it later), but the values indicate that a 1° increase for every 2,5° is a good estimate. All necessarily assumed that we have still air (in contrast to jetstreams, westerlies and the howling sixties in the south).

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  • $\begingroup$ best aircraft answer) $\endgroup$ – MolbOrg Oct 16 '16 at 0:06
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    $\begingroup$ The direct north/south path must take less than 12 hours, or it will be carried into the earth's shadow. So the path must have an eastward component, and it may not be possible at the specified speed. $\endgroup$ – WhatRoughBeast Oct 16 '16 at 0:10
  • $\begingroup$ @WhatRoughBeast Correct, I forgot that being on the edge of the terminator line does not increase the time to 24 hours :(. I will optimize the flight path this evening. $\endgroup$ – Thorsten S. Oct 17 '16 at 14:03
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Kim Stanley Robinson in the Mars trilogy (Red Mars, Green Mars, Blue Mars) has a train built around the equator of Mercury such that the expansion and contraction of the rails as they enter and exit sunlight perpetually propels a train to stay always in shadow. He worked out the details of the system pretty well... and if you put a train 180 degrees around, nearly the same system could keep a train always in sunlight.

A vast series of parabolic mirrors could also do the trick.

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  • $\begingroup$ I was about to post the same, but wasn't it a city on rails, rather than just a train? Or am I thinking of some other, similar, story? $\endgroup$ – flith Oct 15 '16 at 18:11
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    $\begingroup$ @filth What is a city on rails if not a giant train? :P $\endgroup$ – Trotski94 Oct 15 '16 at 19:12
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    $\begingroup$ In his book 2312 it is a city not a train. I don't remember about the Mars Trilogy. $\endgroup$ – BSteinhurst Oct 16 '16 at 1:09
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    $\begingroup$ It's a city in both books. (And of course also a train.) $\endgroup$ – Nathaniel Oct 16 '16 at 2:22
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    $\begingroup$ Mercury is smaller than the earth and rotates about 58 times more slowly, so this isn't applicable to earth. And going there counts as going into space, until you start talking about native born citizens of the train city. $\endgroup$ – Chris Stratton Oct 16 '16 at 3:47
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Much depends on your timescale. Just exactly how long do need to be in sunlight? If it's 8 or 10 hours, Thorsten S has the right answer.

However, no aircraft can maintain flight continuously, and supersonic aircraft in particular will start to get very unhappy after 10 hours or so in the air. As a result, you'd need to create a Pony Express system, with multiple aircraft spaced at, let's say, 8-hour intervals, with the payload being transferred to each in turn. So you'd need a minimum of 3 aircraft, plus a fleet of tankers, and perhaps 2 or 3 times as many to deal with larger maintenance issues.

Note that the B2 has flown missions of 70 hours continuous operation, but a) this was considered a quite remarkable performance, and b) the B2 is subsonic and incapable of keeping up with the sun.

Of course, this assumes an equatorial flight, and this is not necessary. By flying along the solar equivalent of the Arctic Circle, and staying constantly in twilight, a ground distance of about 10,000 miles is adequate. This implies a ground speed of about 400 miles per hour, so subsonic CAN do the job. In this case, you might get away with only 2 aircraft, with each aircraft operating for 24 hours in turn. Eventually this will require more aircraft, since major overhauls typically take more than 24 hours, and sooner or later something major will need replacing.

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    $\begingroup$ +1 for noting that airplanes don't last forever (or even very long!) without maintenance $\endgroup$ – Steve V. Oct 16 '16 at 13:58
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Of course!

Both Arctic and Antarctic midnight suns stay in the sky for 6 months, alternating:

https://www.youtube.com/watch?v=ndlQNicOeso

Quote from Wikipedia:

Around the summer solstice (approximately 21 June in the Northern Hemisphere and 22 December in the Southern Hemisphere) the sun is visible for the full 24 hours, given fair weather. The number of days per year with potential midnight sun increases the farther towards either pole one goes. Although approximately defined by the polar circles, in practice the midnight sun can be seen as much as 55 miles (90 km) outside the polar circle, as described below, and the exact latitudes of the farthest reaches of midnight sun depend on topography and vary slightly year-to-year.

So stay in one place for 6 months, and then take a (fairly) quick flight to the other pole.

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    $\begingroup$ But the change between midnight sun in the north and the south isn't instantaneous, so you'd still have to fly around the earth for a couple of weeks. $\endgroup$ – JonathanReez Oct 15 '16 at 21:40
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    $\begingroup$ @JonathanReez No: assuming neither pole is in some sort of depression, there is always sun at one of the two poles. And, given that the sun has size, there will even be a period at the equinoxes where there is sun at both poles. $\endgroup$ – Daniel Griscom Oct 17 '16 at 1:59
  • $\begingroup$ @JonathanReez I believe Daniel to be correct on his assessment. In any case, even if there's a depression, we can use a prosaic baloon (or, if we feel particularly industrious, a tower.) $\endgroup$ – OnoSendai Oct 17 '16 at 17:25
  • $\begingroup$ There are planes that fast. But the three planes that fast that can land on the poles? $\endgroup$ – paparazzo Oct 17 '16 at 17:48
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One answer that hasn't been mentioned yet is an orbiting heliostat.

By positioning mirrors in orbit and aiming them at a specific point on Earth, it is possible to stay on the earth and yet maintain constant sunlight irradiation by simply directing the space-based mirrors.

This paper describes a number of possible modes of such space-based 24-hour irradiation mirrors.

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  • $\begingroup$ It was not mentioned because the OP stipulates «but couldn't go into space» $\endgroup$ – JDługosz Oct 16 '16 at 1:16
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    $\begingroup$ @JDługosz I don't see why not, from my reading of the question, the restriction only applies for the protagonist. $\endgroup$ – March Ho Oct 16 '16 at 2:02
  • $\begingroup$ You ought to add that quallification to the body of the Answer. $\endgroup$ – JDługosz Oct 16 '16 at 15:59
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Well, if you can stay in the air, you just need to pick a pole, and stay above cloud level.

At that altitude, axial tilt is not sufficient to hide the sun from you, so you are safe.

Stationkeeping, resupply, repairs, etc on the other hand may be an issue.

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  • $\begingroup$ "At that altitude, axial tilt is not sufficient to hide the sun from you, so you are safe." Citation needed. $\endgroup$ – Taemyr Oct 17 '16 at 9:29
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    $\begingroup$ Bah, missed a decimal point. You'd need to be ~125km up, not 12. :( Given the angles 11.25, 90 and 78.75 and sides of 6371 (radius of earth), 6371 + B, and C, solve for B. C/6371 = TAND(11.25), C = 1267, 1267^2 + 6371^2 = (6371 + B)^2, 1267^2 + 6371^2 = 6371^2 + B^2 + 6371*2*B, 1267^2 = B^2 + 12742*B, B = SQRT(42194930) - 6371, B = 6496 - 6371, B = 125 $\endgroup$ – Roger Willcocks Oct 18 '16 at 1:40
  • $\begingroup$ But, if you move closer to the mid point between the orbital pole and the axial pole, the height decreases while the distance per day to cover increases, so there would be an option there somewhere. $\endgroup$ – Roger Willcocks Oct 18 '16 at 1:42
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You could live on the north pole for 6 months and south pole for 6 months, and travel with an aircraft at slightly less than the speed of sound between the two. If the planet has no tilt then on the same pole.

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You already gave the answer. Stay at one pole during its light time, which is half the year, then take a super fast jet to the other pole at just the switching time.

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    $\begingroup$ This is the 5th answer of "stay at the poles". At least have an original twist on it... $\endgroup$ – JohnP Oct 16 '16 at 16:47
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    $\begingroup$ @JohnP: It's actually the third such answer. $\endgroup$ – MichaelS Oct 16 '16 at 23:34
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Couldn't you just make camp at one of the poles and move with the tilt of the earth to remain at the pole year round? If the earth's circumference is 40,000km and earth tilt is 23.5 degrees that would mean the pole moves 2611km/year or 7.15km/day. Surely you could walk that distance everyday.

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  • $\begingroup$ This would only work if Earth was not rotating. As it is, you would have to go some more to compensate for that rotation. A lot more. $\endgroup$ – Daerdemandt Oct 16 '16 at 16:51
  • $\begingroup$ I get a different than you $\endgroup$ – paparazzo Oct 17 '16 at 17:49
  • $\begingroup$ Oh ya, now I see my error. The tilt means the axis of the rotation is not perpendicular to the sun. $\endgroup$ – Peter Quiring Oct 18 '16 at 1:26
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SR-71 Blackbird

Since you have access to all technology (I assume including discontinued tech), set up a network of SR-71s and in-flight refueling planes, so you can zip around with enough time to land and switch planes when needed.

It is far from simple though, and can be terrifying (and very expensive), here's a link:

Blackbird Pilot Interview

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Position several large mirrors in orbit, relaying the sunlight to your position at all times. No travelling necessary ;)

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Travel from pole to pole twice a year at 45 degrees. Need to travel at a 45 west to keep up with the sun. Pole to pole in 24 hours at a 45 degrees is about 1465 mph. Problem is landing and taking off on the poles. And it is cold - want to live on a pole even if have sunshine.

The Artic Circle is at 63 degrees. Assume you live there 1/2 the year. One day a year (June 21) is all daylight. After that you merely need to travel toward the pole. Each day add about 75 miles. June 22 75 miles, June 23 150 miles, ... On the worst day if you are splitting poles by land 6712. miles. In the air more like 6000 miles with is only 125 mph. But you probably don't want to travel until is starts to get dark so need to do it in 12 hours for 500 mph. Can do that with a LearJet. On average you could spend 18 hours a day on the ground. On your commute days twice a year to the other circle at a 45 you would need to travel 1075 mph. 1075 mph is exotic / military). In the south you would need to be on Antartica.

If wanted to stay subsonic (90 mph) then you could only commute to 55 latitude. UK and and south end of South America.

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