Angular momentum of the earth's orbit

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SUMMARY

The discussion focuses on calculating the angular momentum of the Earth in its orbit around the Sun compared to the Moon in its orbit around the Earth. The relevant formulas for angular momentum and rotational inertia are established, with the Earth’s rotational inertia calculated as 1.35x10^47 kg-m² and the Moon’s as 1.09x10^40 kg-m². Angular velocities are also determined, with the Earth’s angular velocity at 1.99x10^-7 rad/sec and the Moon’s at 2.66x10^-6 rad/sec. The final goal is to find the ratio of angular momenta between the two celestial bodies.

PREREQUISITES
  • Understanding of angular momentum and its formula: angular momentum = rotational inertia x angular velocity.
  • Knowledge of rotational inertia and its calculation for point masses.
  • Familiarity with the concept of angular velocity in radians per second.
  • Basic understanding of celestial mechanics and the orbits of the Earth and Moon.
NEXT STEPS
  • Calculate the angular momentum of the Earth and Moon using the derived rotational inertia and angular velocity values.
  • Research the implications of angular momentum in celestial mechanics and its conservation.
  • Explore the effects of gravitational interactions on the angular momentum of orbiting bodies.
  • Learn about the differences in orbital dynamics between point masses and solid bodies.
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Astronomy students, physics enthusiasts, and educators seeking to understand the principles of angular momentum in celestial orbits will benefit from this discussion.

aneima6
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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

radius of Earth (equatorial) 6.37x10^6
radius of Earth's orbit 1.5x10^11
radius of moon (average) 1.74x10^6
radius of moon orbit 3.84x10^8

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

thanks
 
Last edited:
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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
 
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?
 
Right. You've found the rotation inertias in units of kg-m^2.
 
thanks for the help.

is this rotational velocity?

earth sun
360 degrees = 6.28318531 radians
365 days or 31536000 seconds
6.28318531 radians/31536000 seconds
1.99x10^-7 rad/sec

earth moon
360 degrees = 6.28318531 radians
27.3 days or 2358720 seconds
6.28318531 radians/2358720 seconds
2.66x10^-6 rad/sec
 
Last edited:
Those are the angular velocities. Don't forget units: radians/sec.
 
thanks. yea i just fixed it
 

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