# Why doesn't an airplane simply go up in the air and stay there?

To get where a plane wants to go for example if I wanted to go from California to Japan, why can't an airplane simply stay up in the sky and let the earth rotate underneath it?

Is it the gravitational force between the earth and the plane so the plane actually has to overcome this velocity??? I thought the earth was spinning crazy fast.

Maybe I am getting off topic here but: aren't we in its frame of reference so we wouldn't have to accelerate into another?

• Um...no. The plane is moving with the earth - losing contact with the earth doesn't mean the earth rotates underneath it, just as a ball you throw in a uniformly moving car doesn't smash into your rear window when you let go of it. – ACuriousMind Mar 30 '15 at 3:08
• @ACuriousMind You should type that into the answer box, because that is a pretty good answer. – Jimmy360 Mar 30 '15 at 3:43

Let's forget about air momentarily. What will happen is that the moment our air craft leaves contact with the ground, its initial tangential velocity will be the same tangential velocity of earths rotation, given by $v_{T i}=R\omega_{\text{Earth}}$. Now let's assume that in a very short time we achieve the desired elevation $h$, for the earth example our tangential speed is much less than the speed required to maintain orbit, so we will start to fall, but our vehicle will propel itself against gravity to maintain the elevation $h$.
So now we are orbiting earth with tangential velocity $R\omega_{Earth}$ (and our angular velocity is there for $R\omega_{\text{Earth}}/(R+h)$), however to stay above our launch point we need $(R+h)\omega_{Earth}$. So if $h\ll R$, you will basically stay above the launch point as @ACuriousMind succinctly remarked. However for large enough $h$ you will be moving opposite to earth's rotation (or due west), with a $\textbf{ground speed}$ of $$R\omega_{\text{Earth}}\left( 1 - \frac{R}{R+h} \right) = \omega_{\text{Earth}}\frac{Rh}{R+h} \approx \omega_{\text{Earth}}\left( h - R\left(\tfrac{h}{R}\right)^2 + \ldots\right)$$ So you need to be about $13$ million feet above see level to reach a ground speed of $300$ meter per second, which is approximately the speed of commercial air planes (however they only need to be at about $40,000$ feet).
If you are still not convinced of the inefficiency of such vehicle, remember that we must also take into account the atmosphere. On average the atmosphere rotates with the same speed as the earth, so if you just hover, air drag will tend to push you eastwards because you will be experiencing $670$ mph winds, with no propulsion against it. So this again will push you back east so that your ground speed will decay quickly to zero. And all the distance that you would have managed to cross will be very small indeed.