Watching Airplanes Fly Toward You

Recently, I was enjoying the stars on a cool summer night and saw an airplane flying toward me from the distance. I observed its path as it started out as a small dot in the distance until it flew directly overhead. This made me wonder, is it possible to determine whether an airplane is ascending or descending or maintaining level flight as it is also flying toward you? For example, looking at an airplane flying toward you from the distance looks very similar to something flying directly upward (like a rocket). Is it possible to determine whether the plane is ascending or descending by using just the angle of sight between your eyes and the plane, a timer, and knowledge that the plane is flying at a constant velocity? Also let's assume we cannot actually resolve any surfaces of the plane. Let's just assume it looks like a point light in the night sky.

My first guess was to calculate dθ/dt and to see if d^2θ/dt^2 could somehow be used to determine whether the plane was ascending or not, but I wasn't sure if this was possible.

• Surely, but this question is basic mathematics math.stackexchange.com, not physics. May 21, 2016 at 19:43
• May 21, 2016 at 19:52
• How do you know that the plane is flying at a constant velocity? You don't. May 21, 2016 at 20:05
• @CuriousOne Without making some assumptions I don't think it would be possible to determine whether the plane is ascending/descending/maintaining level flight. I think it is reasonable to assume a plane at cruising conditions would maintain constant velocity. Even so, I think it may not be possible to make a determination without knowing the actual velocity of the plane (i.e, knowing that velocity is just constant may not be enough). May 21, 2016 at 21:11

From your diagram you have $x = y \tan \theta$. Taking the time derivative, and using the speed of the plane is $\dot{x} = v$ (taken as constant) you have $v = y \, \sec^2 \theta \, \dot{\theta}$ assuming that the height of the plane, $y$, is constant. So, for constant height (and taking the speed constant) you have $\dot{\theta} = C \cos^2 \theta$ where $C=\frac{v}{y}$. If this relationship isn't obeyed, and you are keeping $v$ constant it must be because $y$ is changing.