The continents on Earth drift relative to each other at...

The continents on Earth drift relative to each other at rates of up to a few centimeters per year; this drift is caused by convective motions in the mantle. (a) Assuming typical bulk motions in the mantle of Earth are $0.01 \mathrm{~m} / \mathrm{yr}$, calculate the total energy associated with mantle convection. (b) Calculate the kinetic energy associated with the Earth's rotation. (c) Calculate the kinetic energy associated with the Earth's orbital motion about the Sun.

The continents on Earth drift relative to each other at rates of up to a few centimeters per year; this drift is caused by convective motions in the mantle.
(a) Assuming typical bulk motions in the mantle of Earth are $0.01 \mathrm{~m} / \mathrm{yr}$, calculate the total energy associated with mantle convection.
(b) Calculate the kinetic energy associated with the Earth's rotation.
(c) Calculate the kinetic energy associated with the Earth's orbital motion about the Sun.

Key Concepts

Kinetic Energy Calculation

Kinetic energy is calculated using the formula KE = ½mv², where m is mass and v is velocity. This fundamental concept in mechanics is used to quantify the energy associated with any form of motion, whether it be bulk fluid motions in the mantle, rotational movement, or orbital motion around a star.

Mantle Convection

Mantle convection refers to the slow, thermally driven movement of Earth's mantle material. This process, driven by temperature gradients and buoyancy, is responsible for plate tectonics and is characterized by comparatively low velocities. The associated kinetic energy of these convective motions provides insight into the energy budget of Earth's interior dynamics.

Rotational Dynamics

Rotational dynamics deals with the behavior of rotating bodies and involves calculating rotational kinetic energy using the object's moment of inertia and angular velocity. For a rotating planet like Earth, the energy due to its spin is computed by ½I?², where I is the moment of inertia and ? is the angular velocity. This concept is crucial for understanding the energy stored in the rotation of celestial bodies.

Orbital Dynamics

Orbital dynamics encompasses the study of objects in orbit, such as the Earth revolving around the Sun. The kinetic energy in orbital motion is calculated using the standard kinetic energy formula, taking into account the orbital velocity of the Earth and its mass. This concept is fundamental in celestial mechanics and helps in comparing the energy scales of different dynamic systems in space.

Transcript

00:01 In question a, we have to calculate the rotational kinetic energy of the earth.

00:07 And in order to do that, i'm going to model the earth as a solid sphere that has a moment of inertia of 5 fifths of m are squared, where m is the mass of the earth, or 6 times 10 to the 24 kilograms, and r is the radius of the earth, which is 6 .37 times 10, to the 6 .37 ,000, to the 6 .6 .7.

00:31 6 meters.

00:38 And remember that the rotational kinetic energy is equal to the moment of inertia i times omega squared over 2, where omega is the angular speed.

00:48 So all i need to do is to substitute the numbers i have.

00:51 So i is 2 5ths of m r squared.

00:58 Omega can be calculated as 2 pi over the period of rotation, and then we have to divide it all by 2.

01:06 So the 2 is here can about and we're left with the mass, actually 4 pi squared times the mass r squared, divided by 5 times the period of rotation.

01:22 So plugging in the numbers, we have 4 pi squared times 6 times 10 to the 24 kilograms times r squared, that's 6 .37 times 10 to the 6 meters squared, divided by 5 times the period of rotation, that's 24 hours.

01:46 Actually, the period of rotation should be squared.

01:49 That's 24 hours, which if we want to find the period of rotation in seconds, we have to multiply by 1 ,000 ,600 seconds per hour.

02:03 And the result is 2 .58 times 10 to the 29.

02:14 Joules.

02:16 This is the rotational kinetic energy of the earth.

02:22 In question b, we have the information that each year, the period of rotation of the earth increases by 10 microseconds.

02:33 So delta t y represents the increase in the period of rotation of the earth.

02:41 And with this information, our goal in question b is to calculate the variation...

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