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An airplane of mass 12,939 kg flies level to the ground at an altitude of 19 km with a constant speed of 172 m/s relative to the Earth. What is the magnitude of the airplane's angular momentum relative to a ground observer directly below the airplane in kg·m²/s?

Answer :

Final answer:

The magnitude of the airplane's angular momentum relative to a ground observer directly below is calculated using L = mvr. With the given values, the angular momentum is found to be 4218475800 kg·m²/s.

Explanation:

The magnitude of an airplane's angular momentum relative to a ground observer directly below can be calculated using the formula L = mvr, where L is the angular momentum, m is the mass of the airplane, v is the tangential velocity, and r is the radius vector from the point of rotation (observer) to the airplane.

Given the mass (m) of the airplane as 12939 kg, its constant speed (v) as 172 m/s, and its altitude (which can be taken as the radius r) as 19,000 meters, we can calculate the angular momentum as:

L = mvr = 12939 kg* 172 m/s *19,000 m

L = 12939 *172 * 19000 = 4218475800 kg·m2/s

Therefore, the magnitude of the airplane's angular momentum relative to a ground observer directly below the airplane is 4218475800 kg·m2/s.

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