Question

The moon's mass is 7.4×1022kg and it orbits 3.8×108m from the earth with a period of 27.3 days. part a what is the angular momentum of the moon around the earth?

Answer 1

To find the angular momentum of the moon around the earth, we need to use this equation: Angular momentum = mass x velocity x radius of the orbit

The mass and the radius of the orbit is already given, so we need to find its velocity.
We know that the moon’s angular velocity, which is: ω = 2π / T = 2π / 2.3606e6 s

Then we need to find the orbital velocity of moon

(2pi x 380,000,000)/ (2.36/cdot 10x^6 s) = 1011m/s

So, angular momentum = (7.4x10^22 kg) x 1011 m/s x (380,000,000 m)

Angular momentum = 2.84 x 10^34 km2s

From here, we can conclude that the orbital angular momentum of the moon is about four times that of the rotational angular momentum of the earth and that is 7.1 x 10^33 kgm^2/s.

Answer 2

The angular momentum of the Moon around the Earth is given by:

`L=mvr`
where m is the Moon's mass, v its speed and r the radius of the orbit.

We already have the mass, m and the radius of the orbit r, so we need to find the velocity first. Assuming the orbit of the Moon is like a circle, the velocity is the perimeter divided by the time to complete one orbit:

`v= (2 \pi r)/(t)`
The time t is (keeping in mind that 1 day=86400 s):

`t=27.3 days = 2.36 \cdot 10^6 s`

So the velocity is:

`v=(2 \pi r)/(t)= ((2 \pi)(3.8 \cdot 10^8 m))/((2.36 \cdot 10^6 s))=1011 m/s`

And so we can finally calculate the angular momentum of the Moon around the Earth:

`L=mvr=(7.4 \cdot 10^(22)kg)(1011 m/s)(3.8 \cdot 10^8 m)=2.84 \cdot 10^(34)kgm^2 s`