What is the angular momentum of the Moon in its orbit around Earth in m^2/s? (treat the moon as a point mass)
The angular momentum of an object is given by the formula:
L = I * ω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In the case of the Moon, we can treat it as a point mass, so its moment of inertia is given by:
I = m * r^2
Where m is the mass of the Moon and r is the radius of its orbit.
The angular velocity can be calculated using the formula:
ω = (2π/T)
Where T is the period of the Moon's orbit around Earth.
Assuming a mass of 7.34 x 10^22 kg for the Moon and a radius of 3.84 x 10^8 m for its orbit, and a period of 27.3 days (which is equivalent to 2,360,952 seconds), we can calculate the angular momentum:
I = (7.34 x 10^22 kg) * (3.84 x 10^8 m)^2 = 1.083 x 10^40 kg·m^2
ω = (2π) / (2,360,952 s) = 8.39 x 10^-7 rad/s
L = (1.083 x 10^40 kg·m^2) * (8.39 x 10^-7 rad/s) = 9.08 x 10^33 kg·m^2/s
Therefore, the angular momentum of the Moon in its orbit around Earth is approximately 9.08 x 10^33 kg·m^2/s.