Rotational Energy: Humanity's 10^13W Impact on Day Length

[ingy489]

Homework Statement

Humanity uses energy at the rate of about 10^{13} \rm W.If we found a way to extract this energy from Earth's rotation, how long would it take before the length of the day increased by 1 minute?
Express your answer using one significant figure.

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Humanity uses energy at the rate of about 10^{13} \rm W.If we found a way to extract this energy from Earth's rotation, how long would it take before the length of the day increased by 1 minute?

Hi ingy489!

What is the rotational energy of the Earth now?

What would it be if the day was 1 minute longer?

 

[baba89]

I find this question intriguing and thought-provoking. First, it is important to note that the Earth's rotation is not a source of unlimited energy. The rotational energy of the Earth is primarily derived from its initial formation and its ongoing interaction with the Moon and other celestial bodies. Therefore, extracting energy from the Earth's rotation on a large scale would have significant consequences on the Earth's overall rotation and potentially disrupt its delicate balance.

That being said, let's assume for the sake of this question that we have found a way to extract 10^{13} \rm W of energy from the Earth's rotation without any negative effects. Using the equation E = 1/2I\omega^2, where E is energy, I is the moment of inertia, and \omega is the angular velocity, we can calculate the change in the Earth's rotation due to this energy extraction.

Assuming the Earth's moment of inertia remains constant, the change in the Earth's rotation would be directly proportional to the change in its angular velocity. Using the equation \Delta \omega = \frac{\Delta E}{I}, where \Delta \omega is the change in angular velocity and \Delta E is the change in energy, we can calculate the change in angular velocity.

\Delta \omega = \frac{10^{13} \rm W}{I}

To calculate the change in the length of the day, we can use the equation \Delta t = \frac{\Delta \theta}{\omega}, where \Delta t is the change in time and \Delta \theta is the change in angular displacement. Assuming a constant angular velocity, we can rewrite this equation as \Delta t = \frac{\Delta \theta}{\Delta \omega}.

Now, let's assume that we want to increase the length of the day by 1 minute, which is equivalent to 1/60 of a degree of angular displacement. Plugging in our calculated value for \Delta \omega, we get:

\Delta t = \frac{\frac{1}{60}^\circ}{\frac{10^{13} \rm W}{I}}

Using a moment of inertia for the Earth of approximately 8.04 \times 10^{37} \rm kg \cdot m^2, we can calculate the change in time to be approximately 8.34 \times 10^{28} \rm s. This is equivalent to 2.64 \