Angular Momentum of an airplane

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SUMMARY

The discussion focuses on calculating the angular momentum of an airplane with a mass of 12,000 kg flying at a constant speed of 175 m/s at an altitude of 10 km. The formula for angular momentum, L = r × mv, is applied, where 'r' is the distance from the observer to the airplane. The initial calculation suggests a value of 21,000,000 kg·m²/s, but participants question the accuracy and seek clarification on the relationship between linear velocity and angular momentum, particularly in the context of a spherical Earth.

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An airplane of mass 12000kg flies level to the ground at an altitude of 10. km with a constant speed of 175 m/s relative to the earth. (a) what is the magnitude of the airplane's angular momentum relative to a ground observer directly below the airplane? (b) Does this value change as the airplane continues its motion along a straight line?

desperate...was out with the flu now I am lost :confused:
so L= r X Mv...
L=r??X (12000)(175)
r=10??
so 10X(12000)(175)
21000000. this doesn't look correct...
 
Last edited:
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What equations do you know that relate linear velocity with angluar momentum (assuming the Earth is spherical)
 
Hootenanny said:
What equations do you know that relate linear velocity with angluar momentum (assuming the Earth is spherical)

just edited first post. i guess maybe
I=(2/5)Mr^2 might come in handy?
 
L = Mvr would be easier to use with regards to the plane, where r is the radius of orbit and v is linear velocity.
 

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