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Calculate the kinetic energy of rotation of the earth

  1. Apr 6, 2004 #1
    Calculate the kinetic energy of rotation of the earth about its axis, and compare it with the kinetic energy of the orbital motion of the earth’s centre of mass about the sun. Assume the earth to be a homogeneous sphere of mass 6.0*10^24kg and radius 6.4*10^6m. The radius of the earth’s orbit is 1.5*10^11m.

    to get the kinetic energy you will need this formula ke=1/2mv^2 then i have calculated that the total earth's kinetic energy by spinning is 4.8x10^29 joules if we can think of a way to tap the earth's kinetic energy.
    The Earth's kinetic energy is (mv^2)/2 in its orbit about the sun.
  2. jcsd
  3. Apr 6, 2004 #2

    Doc Al

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    Is there a question in here somewhere? :smile:

    Here's a hint: To calculate the rotational KE of the Earth, you will need to know the rotational inertia of a solid sphere.
  4. Apr 6, 2004 #3
    hehe..the question was that I wanted to know if I was right with my reasoning, but I guess I wasn't :mad:
  5. Apr 7, 2004 #4


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    Not necessarily. The formula E= (1/2)mv2 is for linear motion. You can do the "kinetic energy of the earth's orbital motion" by converting "one revolution in 365.24 days" into a linear motion in meters/sec.

    To use that to calculate the "kinetic energy of rotation", you would need to imagine a small section of the earth at, say, distance "r" from the center, calculate the density of the earth times "ΔV" for mass, use r to find the linear speed of that, use (1/2)mv2 to find the kinetic energy of that little section and integrate over the volume of the earth! That's a bit difficult and is equivalent to Doc Al's "know the rotational inertia of a solid sphere."
  6. Apr 7, 2004 #5


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    I think you've got it right, and the Halls is (uncharacteristically) not reading the problem carefully enough. When you use the word spinning it sounds like you're referring to the rotation about the earths axis instead of rotation about the sun.
  7. Apr 7, 2004 #6

    Doc Al

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    Rotational versus translational energies

    The way I read the question, there are two things you have to calculate and compare:

    (1) The translational KE of the Earth in its orbit around the Sun. That will be given by KE = 1/2 M V2, where V is the speed of the earth in its orbit. Find V by considering the orbit to be circular; it takes one year to travel the circumference of that circle.

    (2) The rotational KE of the Earth due to its spinning around its axis. For this you will need the rotational KE which is given by KE = 1/2 I ω2, where I is the rotational inertia of a solid sphere (look it up!) and ω is the angular speed (radians per second) of the Earth's rotation about it's axis. Find ω by realizing that the Earth rotates 2π radians once per day.

    Make sense?
  8. Apr 7, 2004 #7
    a) velocity=(2*Pi*(1.5*10^11m))/(365days*24*60*60) why the 2pi? Can't I just calculated without the 2*pi, I mean isn't velocity = distance / time?

    b) I = 2/5 MR^2 = 2/5 ((6.0*10^24kg)(6.4*10^6m)^2)

    w = (2pi)/(24h*60min*60s) = 6.2831 / 86400s = 7.27 * 10^-5 rad/s and

    then I plug them in KE = 1/2 I ω2.

    Why in a I had to calculate the kinetic energy? I mean why not the rotational KE since the earth is in a circular motion arround the sun?
  9. Apr 7, 2004 #8

    Doc Al

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    Yes, velocity = distance/time, but what is the distance? The distance traveled is the length of the orbit, which is the circumference of the circle: the 2*pi is essential.
    I didn't check your arithmetic, but the equations look OK.
    The total KE of the earth is a combination of (1) KE due to the center of mass of the earth orbiting the sun plus (2) KE due to the earth's spinning on its axis. You are calculating each separately and comparing them. Why? Beats me, but that's what the problem asks for.
  10. Apr 7, 2004 #9
    *kiss* Doc Al, thank you, I got it!
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