Airplanes Landing & Altitude Problem

[GilGiy]

First I'm going to say that I was supposed to select a prefix for this thread... I'm not sure what that's for so my apologies if I did it wrong.

So here's a problem that I was presented with a bit ago (by a flat earther[emoji849][emoji19][emoji44]) and I'm sort of confused. It's been about a year since I took physics but I don't think that's a good excuse.

Let's say an airplane takes off from New York and flies to an Airport directly south, both runways run North and South. Also, the Earth is spinning directly East.

As the plane takes off, it will be traveling East at the same speed of the Earth. However, as it climbs to 10,000 feet it must start traveling much faster to the east to stay caught up with the earth, correct?

Say it is then flying at 10,000' but must land. It is traveling East at the rate of the surface of the Earth plus X m/s to stay traveling in a straight line above the Earth's surface. As it descends to land, how does it stay lined up with the runway? Will it not still be traveling at X m/s relative to the surface of the Earth when it reaches the runway? What does the pilot do with that extra sideways velocity?

If the question itself is confusing that's cause I'm confused... Here's the video that got my wheels turning:

Thanks in advance!
Gilbert

 

[berkeman]

Welcome to the PF. .

To a first approximation, at normal plane altitudes, the atmosphere moves with the surface of the Earth (before you account for whatever direction the wind is blowing that day).

Just stay on your compass heading toward the destination airport, and you should be fine.

 

[jbriggs444]

However, as it climbs to 10,000 feet it must start traveling much faster to the east to stay caught up with the earth, correct?

The Earth does not spin very rapidly. Just a tad more than one rotation in 24 hours. At the equator, that amounts to a respectable 1000 miles per hour eastward. 10,000 feet (about two miles) above the equator, would need an extra 2 miles radius times 2π = 6 miles circumference traveled eastward in 24 hours to keep up. About ½ of a mile per hour faster than the surface of the Earth below. Slower than a walking pace.

Edit: Corrected ¼ to ½

 

[GilGiy]

The Earth does not spin very rapidly. Just a tad more than one rotation in 24 hours. At the equator, that amounts to a respectable 1000 miles per hour eastward. 10,000 feet (about two miles) above the equator, that amounts to an extra 2 miles times 2π = 6 miles circumference traveled in 24 hours -- approximately an extra ¼ of a mile per hour eastward. That's slower than a walking pace.

Awesome that makes perfect sense, thanks!