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The Moon has a mass of 7.35×10²² kg and a radius of 1.74×10⁶ m. It orbits around the Earth at a distance of 3.84×10⁸ m, completing a full rotation every 27.3 days. It also spins on its axis at a rate such that the same side of the moon is always facing the Earth.

What is the angular momentum, in newton seconds, of the Moon in its orbit around Earth?

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Final answer:

The angular momentum of the Moon in its orbit around Earth can be calculated using the formula: L = I * ω, where I is the moment of inertia and ω is the angular velocity. By substituting the given values and solving the equation, we can determine the angular momentum of the Moon.

Step-by-step explanation:

The angular momentum of an object in orbit is equal to the product of its moment of inertia and angular velocity.

The moment of inertia of a spherical object can be calculated using the formula:

I = (2/5) * m * r^2

where m is the mass of the object and r is its radius.

In this case, the mass of the Moon is 7.35×10²² kg and its radius is 1.74×10⁶ m. Using these values, we can calculate the moment of inertia of the Moon:

I = (2/5) * (7.35×10²²) * (1.74×10⁶)^2

Once we have the moment of inertia, we can calculate the angular momentum using the formula:

L = I * ω

where ω is the angular velocity of the Moon. Since the Moon completes a full rotation every 27.3 days, we can convert this to radians per second:

ω = (2π) / (27.3 * 24 * 60 * 60)

Substituting the values into the equation, we find:

L = (2/5) * (7.35×10²²) * (1.74×10⁶)^2 * (2π) / (27.3 * 24 * 60 * 60)

Now we can calculate the angular momentum of the Moon in its orbit around the Earth.

User Matthew Pateman
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