# Airplane executing horizontal loop

The Problem:

An aircraft executes a horizontal loop at a speed of $$720$$ kmph (or $$200$$ m/s) with its wings banked at an angle of $$15^{\circ}$$. What is the radius of the loop.

My Confusion: Unlike the case of cars on banked roads executing a loop (which have the aid of friction and normal forces to provide the centripetal force), there seems to be no force (except air resistance) that aids the plane to execute a loop. So how does an aircraft maintain a loop horizontally?

• What about he horizontal component of the lift? – Farcher Feb 9 at 7:14
• Well, obviously, a force which is perpendicular to the wings is present. This is the same force which makes the plane fly. – Photon Feb 9 at 7:15
• @farcher I see, that can be a viable reason! But my book tells me that the only relation needed to find the radius of the loop is $\tan \theta = v^2/rg$. How is there no component of the lift in it as well? – Apekshik Panigrahi Feb 9 at 7:23
• I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. – Ben Crowell Feb 9 at 21:50
• @ja72 That is exactly what I found surprising. Two completely different scenarios (well not completely, but you get the point), but the exact same equation works! Physics is truly marvelous! – Apekshik Panigrahi Feb 10 at 7:32

$$\tan \left( \beta \right) =\dfrac {f_{z}}{f_g}=\dfrac {m\dfrac {v^{2}}{r}}{m\cdot g}\tag 1$$
Where $$f_z$$ Is the centrifugal forces ,$$f_g$$ Is the weight forces and $$r$$ is the radius of the loop.
\begin{aligned}r=\dfrac {v^{2}}{\tan \left( \beta \right) \cdot g}= \dfrac {200^2}{10\cdot \tan \left( \dfrac {15\cdot \pi }{180}\right) }\simeq 14.9 \end{aligned} \quad [km]