Angular momentum of the earth's orbit

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Homework Help Overview

The discussion revolves around calculating the angular momentum of the Earth as it orbits the Sun compared to the Moon's orbit around the Earth, using a ratio of their angular momenta. The subject area includes concepts of rotational inertia and angular velocity in the context of celestial mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for angular momentum and the appropriate definitions of rotational inertia and angular velocity. There are attempts to calculate the rotational inertia for both the Earth and the Moon, with some questioning the assumptions made regarding their treatment as point masses.

Discussion Status

The discussion is progressing with participants sharing calculations for rotational inertia and angular velocities. Guidance has been offered regarding the correct interpretation of rotational inertia in the context of orbital motion, and participants are verifying their calculations.

Contextual Notes

Participants are working under the constraints of homework rules, focusing on the mathematical relationships without providing complete solutions. There is an emphasis on understanding the definitions and applications of the relevant physical concepts.

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 [itex]I = mr^2[/itex]. (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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