# Centripetal? Airplane problem

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!

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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?

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

mcgooskie said:
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