0
$\begingroup$

Today there was a seminar regarding modern air crafts held in my school. At the starting of the lecture, we were asked a question which I found quite intriguing.

"Why does a helicopter or a fixed wing aircraft or anything stationary in mid air need to move to get to a certain place? Why cant it just stay mid air and wait for the destination to arrive to itself?"

(We were asked to answer it at the end of the lecture. Throughout the lecture I was thinking about the answer. I know that on ground due to friction, the relative velocity between the earth and us is zero hence we don't observe any such effect. Also according to what I can imagine, at lower altitudes, the air closer to earth would also be in motion and that might be the reason due to which the helicopter stays in one place with respect to the ground.)

But what about high altitudes where the air density is low. I know my reasoning might be incorrect. I looked through the net but did not find a satisfying answer.

(Edited) Add-ons to make the question clear (1) Initially when an airplane is on the ground, it moves with the Earth.

(2) Assume this airplane takes off in a direction opposite to the Earth's direction of rotation. There will be a point, when the plane acquires equal speed of the Earth but in opposite direction.

(3) So as viewed from the space, the airplane would appear stationary.

(4) Assume that the engine of the airplane is self sufficient and it does not require external air as oxidizer , and that the viscous effects are neglected (though theoretically , without viscosity, the plane would not fly, but this is just an assumption) .

(5) Now in such a case where the plane is mid air and is stationary when viewed from space, would my proposition hold true?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ see curious.astro.cornell.edu/disclaimer/134-physics/… $\endgroup$ – anna v Oct 29 '18 at 13:30
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/16390/2451 , physics.stackexchange.com/q/1193/2451 and links therein. $\endgroup$ – Qmechanic♦ Oct 29 '18 at 13:45
  • $\begingroup$ @anna v, the question you pointed out is regarding the time and its different from what i am asking actually.. $\endgroup$ – Mike Victor Oct 29 '18 at 14:23
  • $\begingroup$ Anyways i had a genuine doubt. duplicate or not, idc $\endgroup$ – Mike Victor Oct 29 '18 at 14:23
  • $\begingroup$ @MikeVictor motion has inherently the concept of time. We are sitting in front of the computer and moving with 1000miles an hour with respect to the rest system of the earth. With respect to the local rest system we are stationary. $\endgroup$ – anna v Oct 29 '18 at 15:16
3
$\begingroup$

For an aircraft or a helicopter, "stationary in mid air" means stationary with respect to the air, not to the ground.

The thing that you are missing is that, like almost all real fluids, air is viscous. Its viscosity is obviously much lower than a liquid like oil, but it isn't zero. Because of its viscosity, if we ignore "small scale" disturbances like weather patterns, all the atmosphere, even at very high altitudes, is rotating at the same speed as the earth. Because the viscosity is small, you might imagine it would take a very long time to get to that situation, but the earth has been rotating and had an atmosphere for billions of years, which is "long enough"!

In fact the previous paragraph is over-simplified, because there are actually stable patterns in the air flow at different latitudes and at different altitudes. The east-to-west component of the air velocity is not quite the same as the earth's surface velocity, which results in steady wind patterns such as the "trade winds" over the oceans (which were very important in the days of sailing ships) and the "jet stream" at high altitudes (important for air travel). But an explanation of those effects is beyond the scope of an answer to your question.

In theory, you could launch a balloon into such a wind pattern and it could "wait for the destination to arrive" - but the balloon would take far longer than 24 hours to circumnavigate the earth, which is probably what your lecturer meant by "waiting for the destination to arrive".

| cite | improve this answer | |
$\endgroup$

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