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Angular momentum of the earth's orbit

  1. Nov 1, 2004 #1
    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 earths 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: Nov 1, 2004
  2. jcsd
  3. Nov 1, 2004 #2

    Doc Al

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    Staff: Mentor

    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.
     
  4. Nov 1, 2004 #3
    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
     
  5. Nov 1, 2004 #4

    Doc Al

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    Staff: Mentor

    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.)
     
  6. Nov 2, 2004 #5
    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?
     
  7. Nov 2, 2004 #6

    Doc Al

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    Staff: Mentor

    Right. You've found the rotation inertias in units of kg-m^2.
     
  8. Nov 2, 2004 #7
    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: Nov 2, 2004
  9. Nov 2, 2004 #8

    Doc Al

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    Staff: Mentor

    Those are the angular velocities. Don't forget units: radians/sec.
     
  10. Nov 2, 2004 #9
    thanks. yea i just fixed it
     
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