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Consider the Earth-Moon system. Construct a problem in which you calculate the total angular momentum of the system including the spins of the Earth and the Moon on their axes and the orbital angular momentum of the Earth-Moon system in its nearly monthly rotation. Calculate what happensto the Moon's orbital radius if the Earth's rotation decreases due to tidal drag. Among the things to be considered are the amount by which the Eart''s rotation slows and the fact that the Moon will continue to have one side always facing the Earth.

$- L _ { \text {total } } = 3.20 \times 10 ^ { 31 } \mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \mathrm { s }$- The moon's orhit radins increnses.

Physics 101 Mechanics

Chapter 10

Rotational Motion and Angular Momentum

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

Hope College

University of Sheffield

Lectures

04:16

In mathematics, a proof is…

04:48

In mathematics, algebra is…

04:12

The Moon orbits the Earth …

05:19

07:29

(a) What is the angular mo…

03:26

04:04

Determine the angular mome…

05:55

Determine the mass of the …

10:02

(a) Calculate the angular …

02:55

04:07

The Moon and Earth rotate …

03:41

01:09

A moon has an elliptical o…

02:34

00:36

Earth's rotational sp…

05:26

Assume that the moon orbit…

11:05

03:56

Phases of the Moon As the …

09:18

Model the Earth as a unifo…

00:53

Assume that the Moon'…

so we know that first, um, the rate, the rate at which the moon has been moving away from the earth is four centimeters per year for 179 years. This is giving us 7.36 meters. And so we can say that then the angular momentum of the moon would be equaling the massive of the moon times the velocity of the moon times the orbital radius of the moon. No, we can say that the velocity of the moon, the linear velocity of the moon, would be calculated by the square root of the orbital radius of the moon times the gravitational force between the Earth and the moon divided by the mass of the moon. The gravitational force between the Earth and the moon would be the gravitational, constant times the mass of the moon times the mass of the earth divided by the orbital radius of the So we know that first, um, the rate the rate at which the moon has been moving away from the earth is four centimeters per year for 179 years. This is giving us 7.36 meters and so we can say that. Then the angular momentum of the moon would be equaling the mass of the moon squared. We can then say that substituting this equation in for the velocity equation we find the velocity of the moon is equaling the square root of the orbital radius of the moon times a gravitational constant of the moon multiplied by the mass of the moon Times the mass of the earth divided by the orbital radius of the moon Squared moon times the velocity of the moon times the orbital radius of the moon No, we can say that the velocity of the moon the linear velocity of the moon would be calculated by the square root of the orbital radius of the moon times the gravitational force between the Earth and the moon divided by the mass of the moon. The gravitational force between the Earth and the moon would be the gravitational, constant times the mass of the moon times the mass of the earth divided by the orbital radius of the And then this would be divided by the mass of the moon. And so the mass of the moon cancels out, Um, essentially this second power cancels out with one of these essentially one horrible radius of the moon. And so we have that this is equal in the square root of the gravitational, constant times the mass of the earth divided by the orbital radius of the moon. And so we can plug this into the angular moon squared. We can then say that substituting this equation in for the velocity equation we find the velocity of the moon is equaling the square root of the orbital radius of the moon times a gravitational constant of the moon multiplied by the mass of the moon times the mass of the earth divided by the orbital radius of the moon squared momentum equation. And then we find that the angular momentum of the moon is equaling the mass of the moon times we can say the square root I the gravitational constant times the mass of the earth times the orbital radius of the moon where we can see that the 1/2 but rather the orbital radius of the moon to the negative. And then this would be divided by the mass of the moon. And so the mass of the moon cancels out um essentially this second power cancels out with one of these essentially one horrible radius of the moon. And so we have that this is equaling the square root of the gravitational, constant times the mass of the earth divided by the orbital radius of the moon. And so we can plug this into the angular 1/2 power multiplied by the radius of the moon to the first power would essentially give us a positive 1/2 power for the rate for the orbital radius of the moon. And so we can then say that the change in the angry momentum of the moon with respect to the orbital radius of the moon is gonna be equaling the mass of the moon over too, multiplied by the square root of the gravitational, constant times the mass of the earth divided by the orbital radius momentum equation. And then we find that the angular momentum of the moon is equaling the mass of the moon times we can say the square root, Uh, the gravitational, constant times, the mass of the earth times, the orbital radius of the moon, where we can see that the 1/2 but rather the orbital radius of the moon to the negative one of the moon. And so we can say that the change and the the change in the angular momentum of the moon would be equaling the mass of the moon over too, multiplied by the square root of the gravitational, constant times the mass of the earth divided by the orbital radius of the moon multiplied by the differential orbital radius of the moon. And so essentially, we're going to then solved on half power multiplied by the radius of the moon to the first power would essentially give us a positive 1/2 power for the rate for the orbital radius of the moon. And so we can then say that the change in the angry momentum of the moon with respect to the orbital radius of the moon is gonna be equaling the mass of the moon over too, multiplied by the square root of the gravitational, constant times, the mass of the earth divided by the orbital radius of the And we can say that then the orbital, the change in the linear momentum of the moon of angular momentum of the moon would be equaling the mass of the moon. Seven. Rather, we can say 1/2 times the mass of the moon. 7.35 times 10. So the 22nd kilograms this would be multiplied by the square root of the gravitational constant 6.67 times 10 mood. And so we can say that the change and the, uh the change in the angular momentum of the moon would be equaling the mass of the moon over two multiplied by the square root of the gravitational, constant times the mass of the earth divided by the orbital radius of the moon, multiplied by the differential orbital radius of the moon. And so, essentially, we're going to then solve to the negative 11th Newton meters squared per kilogram squared, multiplied by the mass of the earth. 5.98 or other 5.97 times 10 to the 24th kilograms extend the square root. This would be divided. This would be divided by the orbital radius of the moon's 3.84 times, 10 to the A. And we can say that then the orbital, the change in the linear momentum of the moon of angular momentum of the moon would be equaling the mass of the moon. Seven. Rather, we can say 1/2 times the mass of the moon, 7.35 times 10 to the 22nd kilograms. This would be multiplied by the square root of the gravitational constant 6.67 times 10 leaders. And then this would be all be multiplied by the differential radius differential orbital radius of the moon D R C M, which we found to be 7.36 meters. And so we confined then that the change in the linea in the angular momentum of the moon is going to be equaling two positive 2.75 So it increases by 2.75 times 10 to the 26th to the negative. 11th Newton meters squared per kilogram squared multiplied by the mass of the earth. 5.98 or other 5.97 times 10 to the 24th kilograms extend the square root. This would be divided. This would be divided by the orbital radius of the moon's 3.84 times, 10 to the eighth and this would be kilograms meter squared per second. This would be our final answer, and we can say that the title interaction, the title interaction essentially sat meters. And then this would be all be multiplied by the differential radius differential orbital radius of the moon D R C M, which we found to be 7.36 meters. And so we confined, then, that the change in the linea in the angular momentum of the moon is going to be equaling. Two positive 2.75 So it increases by 2.75 times 10 to the 26th. Small momentum from the horrible orbiting body and transfers it to the primary rotation. So it, you could say, saps momentum from the orbiting body and transfers it Houthi Primary rotation. So essentially here it's It's the angular and this would be kilograms meter squared per second. This would be our final answer, and we can say that the title interaction the title interaction essentially saps momentum from the horrible orbiting body and transfers it to the primary went um, from the From the moon, essentially orbiting around the earth is decreased, and the actual rotation of the moon is the mo mentum of the rotation of the moon about its own axis is increased. And so this will. Essentially, we could say, lowering the orbits. Multitude your rotation. So it, you could say, saps momentum from the orbiting body and transfers it to the primary rotation. So essentially here it's It's the angular moment until frictional until friction Essentially, um, till friction uh, becomes in effect. That is the end of the solution. Thank you. Some from the from the moon, essentially orbiting around the earth is decreased. And the actual rotation of the moon is the mo mentum of the rotation of the moon. About its own axis is increased. And so this will essentially, we could say, lowering the orbits multitude.

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