Airplane with mass of 24113 kg?

Question

flies level to the ground at an altitude of 12 km with a constant speed of 186 m/s relative to the Earth.

What is the magnitude of the plane's angular momentum relative to a ground observer directly below the plane in kg . m^2/s?

Does this value change as the plane continues in motion along a straight line (increase, change in random pattern, decrease, remain constant)?

What is the magnitude & how does it change?

Answer 1

Hmmm...
The question is a bit weird and perhaps that makes it a bit interesting.

An angular momentum is defined as
Ma=RxMl where
Ma – angular momentum
R – distance from the center of mass of the object to the point about which it rotates
Ml – linear momentum or mass times the velocity (mV) of the object
x - represents a vector (cross) product

So if we assume that a this particular instant of time a plane ‘rotates’ about the point where observer is located then we have
(We can compute only absolute value (magnitude) since the direction of the plane motion is unknown.)

|Ma|= R Ml=
|Ma|= R m V= 12,000m (24113kg) 186 m/s=
|Ma|=53,820,216,000 or~5.382 E9 kg m^2/s

The value of the angular momentum will increase with the R value or the distance to the observer.

Answer 2

yes