Question

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?

Answer 1

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.