[Physics] How fast would one have to travel in an airplane in order to experience a continuous sunset

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For the sake of simplifying this problem and removing any guess work on how high the plane is flying, we'll say that our airplane is at 30,000 ft. as this is the average altitude for commercial airplanes.

I have a feeling this should also depend on the latitude where the plane is flying. Again, in an effort to simplify the problem, let's say that we're on the equator flying due west into the sunset.

We needn't set a parameter on how close the sun needs to be above the horizon to qualify as a sunset as the main idea here is to keep up with the sun. That is, the sun should appear stationary to a passenger on the plane.

How fast do we need to move in order to keep up with the sun so that it appears stationary.

My first guess at answering this question is not very technical, but I want a more technical answer. Essentially, the question is to have "time" be unchanging. That is, the time of the day stays the same. We achieve this by moving across time zones. The earth is divided into 24 time zones each. if we're traveling at an altitude of 30,000 ft. Then the width of each time zone at our altitude should be approximately (not all time zones are equally spaced) should be $(1/24)*(2*\pi (r_{earth} +30,000ft))\approx 1039$ miles. Additionally, we need to cover each time zone in one hours time. Thus, $v_{aircraft} = 1039$ $mi/hr$. A quick google search turned up that the fastest aircraft (the Lockheed SR-71 Blackbird) has achieved a speed of 2193 mi/hr. So, assuming my approximation is somewhere near the real answer, this seems doable.

So, is my approximation in the right ballpark? Can someone take a more rigorous approach to this problem and find a better approximation than mine?

Edit: There appears to be aircrafts that can achieve much faster speeds than the aircraft I noted above.

Best Answer

good computation ...

let's assume that the plane flies on the equator , the earth circumference D= 40075 km ( wiki earth ) , its mean radius R= 6371 km and the altitude a= 10000 ft = 9.144 km. The day length d = 24h ( and not 23.9344 h in this case )

  • wiki : Earth orbits the Sun at an average distance of about 150 million kilometers every 365.2564 mean solar days, or one sidereal year. This gives an apparent movement of the Sun eastward with respect to the stars at a rate of about 1°/day, which is one apparent Sun or Moon diameter every 12 hours. Due to this motion, on average it takes 24 hours—a solar day—for Earth to complete a full rotation about its axis so that the Sun returns to the meridian.

The upper circumference C = $D * (R+a) / R $ and the speed to be Sun oriented stationary is $C / d = 40075 * (( 6371+9.144) / 6371 ) / 24 = 1672.2 km/h = 1039.1 miles/h$

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