Calculate the kinetic energy of rotation of the earth

In summary, we can calculate the kinetic energy of rotation of the Earth by using the formula KE = 1/2 I ω2 and the rotational inertia of a solid sphere. We can also calculate the kinetic energy of the Earth's orbital motion by using the formula KE = 1/2 M V2, where M is the mass of the Earth and V is its speed in its orbit around the sun. By comparing these two values, we can see that the Earth's kinetic energy due to its spinning is much greater than its kinetic energy due to its orbital motion.
  • #1
cristina
53
0
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.
 
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  • #2
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.
 
  • #3
hehe..the question was that I wanted to know if I was right with my reasoning, but I guess I wasn't :mad:
 
  • #4
cristina said:
hehe..the question was that I wanted to know if I was right with my reasoning, but I guess I wasn't :mad:

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."
 
  • #5
cristina said:
hehe..the question was that I wanted to know if I was right with my reasoning, but I guess I wasn't :mad:

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 Earth's axis instead of rotation about the sun.
 
  • #6
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?
 
  • #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?
 
  • #8
cristina said:
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?
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.
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.
I didn't check your arithmetic, but the equations look OK.
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?
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.
 
  • #9
*kiss* Doc Al, thank you, I got it!
 

1. How is the kinetic energy of rotation of the earth calculated?

The kinetic energy of rotation of the earth is calculated using the formula 1/2 * I * w^2, where I is the moment of inertia of the earth and w is the angular velocity of the earth.

2. What is the moment of inertia of the earth?

The moment of inertia of the earth is a measure of how difficult it is to change the rotational motion of the earth. It is calculated by integrating the mass distribution of the earth with respect to its axis of rotation.

3. How is the angular velocity of the earth determined?

The angular velocity of the earth can be determined by dividing the total angular displacement of the earth (2π radians) by the time it takes for the earth to complete one rotation (24 hours).

4. What is the unit of measurement for the kinetic energy of rotation of the earth?

The unit of measurement for the kinetic energy of rotation of the earth is joules (J), which is the standard unit for energy.

5. How does the kinetic energy of rotation of the earth impact our planet?

The kinetic energy of rotation of the earth is responsible for keeping the earth rotating on its axis and maintaining the Coriolis effect, which plays a crucial role in weather patterns and ocean currents. It also contributes to the earth's overall angular momentum and affects the length of our days.

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