How Does Banking Angle Affect the Radius of an Airplane's Turn?

[mcgooskie]

In this problem I need to figure out how far I am from the airport.

The pilot tells us that we have to circle the aiport before we can land. We will maintain a speed of 450mph at an altitude of 20,000ft. while traveling in a horizontal circle around the airport. I notice that the pilot banks the plane so that the wings are oriented at 10deg to the horizontal. An article in the in-flight magazine says that planes can fly because the air exerts a force "lift" on the wings which is perpendicular to the wing surface.

So:
V=660 ft/s
h=20,000 ft
bank = 10deg to the horizontal

So I drew a FBD with my normal force perpendicular to my wings, which were banked to the horizontal.

Sum Fx=Wx=ma where a=v^2/r
sin10 (mg)=mv^2/r (mass drops out)
sin10 (32.2ft/s^2)=(660ft/s)/r
solving for r=118.03ft

add that to the current altitude of 20,000ft gives a distance of 20,118ft from the airport?

It doesn't seem like this problem could be this "easy". I feel like I am totally missing a point here.

Thanks!

 

[mcgooskie]

oops! I forgot to square my velocity. That being said, I have a new r=77,904ft.

If you figure 77,904 as the x value of a right triangle, and 20,000 as the y value, then the hypotenuse or distance from the airport would be 80,431ft or approx 15miles. This sounds more reasonable?

 

[remcook]

Why have you taken "mg" as your lift force?
Take another look!

 

[e(ho0n3]

Sum Fx=Wx=ma where a=v^2/r
sin10 (mg)=mv^2/r (mass drops out)
sin10 (32.2ft/s^2)=(660ft/s)/r
solving for r=118.03ft

It seems from what you wrote above, the force F that is giving the plane lift has magnitude of mg. This is clearly not true.
Using F = ma in the vertical direction gives

0 = Fcos(\theta) - mg \rightarrow F = mg/cos(\theta)​

where \theta is the bank angle.
Plug this F back into what you have above, solve for r, use the pythagorean theorem and you're done.

e(ho0n3