1
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ My back-of-the envelope got 975 mph, so I think you've done the right thing. It will certainly depend on the latitude, with higher latitudes requiring lower speeds (until you get to the arctic circle, where you can stay stationary and see the Sun all day). Also, notice that you aren't actually flying around the equator - you want to fly around the elliptic or the Sun would move side to side). $\endgroup$
    – levitopher
    Commented Jun 4, 2015 at 19:26
  • 2
    $\begingroup$ It is obviously doable. The (now retired) Concorde took about 3.5 hours from Paris to New York. These cities are 76 degrees apart, which translates to 5.07 hours. So, during that trip, you actually caught up with the sun -- given the right timing, the sun would RISE from the west. $\endgroup$
    – safkan
    Commented Jun 4, 2015 at 22:22
  • $\begingroup$ On our sister Aviation site: Is it possible to fly around the world to stay in daylight in a fighter jet? $\endgroup$ Commented Feb 5 at 14:07

2 Answers 2

3
$\begingroup$

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 $C / d = 40075 * (( 6371+9.144) / 6371 ) / 24 = 1672.2 km/h = 1039.1 miles/h$

$\endgroup$
3
  • 1
    $\begingroup$ If you take the speed of sound as 1225km/h, this works out to Mach 1.365. It will vary, but is definitely supersonic. $\endgroup$
    – safkan
    Commented Jun 4, 2015 at 22:09
  • 1
    $\begingroup$ Wow! i'm surprised that my estimation came out so close to your answer. This was the exact kind of answer i was looking for. +1 $\endgroup$ Commented Jun 4, 2015 at 23:03
  • $\begingroup$ @safkan at 30,000 ft the speed of sound is different from sea level. It would be more like Mach 1.532, which indeed is still supersonic. $\endgroup$
    – fibonatic
    Commented Jun 5, 2015 at 0:40
0
$\begingroup$

The earth spin around itself with a specific or a given velocity. If we assume that earth is a perfect sphere (more likely not the case but it is a safe assumption to understand the situation) and the plane is flying on the great circle (largest possible circle that can be drawn around a sphere), the velocity of the plane should approximate the angular velocity of the earth. Literally consider the plane as fixed point that belongs to the earth. Exactly as the upper comment showed.

$\endgroup$
1
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Feb 4 at 21:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.