# Angular momentum of the earth's orbit

i need help on this.

how much greater is the angular momentum of the earth orbiting about the sun than the moon orbiting about the earth? (using a ratio of angular momenta)

angular momentum = rotational inertia x rotational velocity

mass of earth 5.98x10^24
mass of moon 7.36x10^22

thanks

Last edited:

Doc Al
Mentor
aneima6 said:
angular momentum = rotational inertia x rotational velocity
Nothing wrong with that. (I assume by "rotational velocity" you mean the angular velocity.) But what is the rotational inertia of a mass m which is at a distance r from the axis of rotation? (Look up the definition of rotational inertia, if you need to.) And you'll have to figure out the angular velocity of each.

ok i get Rotational Inertia: I=(2/5)(m)(r^2)

Earth
I=(.4)(5.98x10^24)((6.37x10^6)^2)
I=9.71x10^37

Moon
I=(.4)(7.36x10^34)((1.74x10^6)^2)
I=8.91x10^34

Doc Al
Mentor
aneima6 said:
ok i get Rotational Inertia: I=(2/5)(m)(r^2)
That formula gives you the rotational inertia of a solid ball about an axis. But that's not what you need here, since you are not asked to calculate the angular momentum of the Earth or Moon rotating on their axes. Instead you need to find the rotational inertia of Earth as it orbits the Sun, and the Moon as it orbits the Earth.

I assume you may treat the Earth and Moon as point masses (thus ignoring their rotation). Hint: The rotational inertia of a mass m which is at a distance r from the axis of rotation is $I = mr^2$. (The axis of rotation of the Earth orbiting the Sun is the Sun; the distance from the axis is the Earth's distance from the Sun; the mass is the mass of the Earth.)

Earth Sun:
(5.98x10^24)((1.5x10^11)^2)
= 1.35x10^47

Earth Moon:
(7.36x10^24)((3.84x10^8)^2)
=1.09x10^40

so far so good?

Doc Al
Mentor
Right. You've found the rotation inertias in units of kg-m^2.

thanks for the help.

is this rotational velocity?

earth sun
365 days or 31536000 seconds

earth moon