The Moon is currently moving away from the Earth at about 4 cm per year. a. At this rate, how ma

The Moon is currently moving away from the Earth at about 4...

Key Concepts

Secular Orbital Evolution

Secular orbital evolution refers to the long-term, gradual changes in the parameters of an orbit. In the context of the Moon’s recession, the cumulative effect of tidal friction over millions of years has led to significant alterations in the Moon’s orbit. Although the current recession rate might seem small on a yearly basis, its effects over geologic timescales are substantial.

Distance Dependence of Tidal Interactions

The strength of tidal forces is highly dependent on the distance between the interacting bodies, generally decreasing as the separation increases. In the early history of the Earth-Moon system, when the two bodies were closer together, the tidal forces were stronger, leading to a more rapid transfer of energy and angular momentum. Therefore, the tidal effects that drive the recession of the Moon would have been larger in the past compared to the present day.

Tidal Forces and Tidal Friction

Tidal forces arise due to the gravitational interaction between two bodies, such as Earth and the Moon. These forces distort the shape of the bodies, creating tidal bulges. In the Earth-Moon system, the misalignment of these bulges relative to the line connecting the centers of mass leads to friction, which dissipates energy. This process, known as tidal friction, is a key driver of the Moon’s gradual recession from Earth as energy is transferred from Earth’s rotation to the Moon’s orbital motion.

Angular Momentum Transfer

Angular momentum in a two-body system like the Earth and Moon is conserved. The tidal interactions and associated friction cause a transfer of angular momentum from the Earth’s rotation to the Moon’s orbit. As a consequence, Earth’s rotation slows down while the Moon moves to a higher orbit. This conservation and transfer of angular momentum is fundamental to understanding why the Moon is gradually moving away from the Earth.

Transcript

0:00 All right.

00:01 So in this question, we're dealing with the recession of the moon.

00:05 So we have measurements that say that the moon is getting further and further away from the earth every year at a rate of 3 .0 centimeters per year further away from us.

00:21 And this problem asks us to use this information and the information in the back of the textbook about like the size of the moon and how it's radius to figure out what is the change in time of its angular velocity and whether it's increasing or decreasing we're supposed to assume that no angular momentum is transferred from the moon to the earth or vice versa in this equation and so yeah let's go ahead and get started so it gives us the hint that if l is constant then the change in angular momentum is obviously zero, but that also means that dldt is equal to zero.

01:10 So that tells us that the change in angular momentum doesn't is, well, there is no change in angular momentum from one point in time to another.

01:21 So that means that angular momentum at one point is equal to angular momentum at another point.

01:27 So we could use this to set up an equation saying that the initial i omega is the same as the final i omega but it's actually a little bit easier to use this to do this from a calculus point of view and using uh differentiation so um the first thing i would like to do is i would like to go ahead and say that this 3 .0 centimeters per year will be our d r d t uh the change in radius over time um and so we'll go ahead and rewrite this as 0 .030 meters per year because that's how we'll end up using it later.

02:10 So this is the change in the distance of the moon over time, and this will come in handy with our equation later.

02:18 So the way that we can do this is since dldt is equal to zero, if we rewrite l as being equal to i omega, then we can say that d, d, dt, the change in i omega over time is also equal to zero.

02:38 And then we can go ahead and expand this out a little bit and kind of isolate the factors that are changing over time.

02:45 So omega, we can't really expand any further, but i, we can go ahead and say that i is equal to some constant times the mass of the moon times the radius squared.

02:57 The radius being how far it is from the earth.

03:01 So then we can rewrite this equation as being d over dt of c, m r squared times omega, and then that's equal to zero.

03:14 So now let's take a closer look at this.

03:16 C is a constant and it's just some number, like two -fifths or one -half or one, you know, it is just some number.

03:26 It's a pure number and it doesn't change over time.

03:29 So if we're trying to differentiate it over time, then the c will just come out to the front.

03:34 It'll come out of the differentiation.

03:37 And now let's take a look at m, the mass of the moon.

03:39 Well, the mass of the moon isn't changing.

03:42 That's also constant.

03:43 So that's also going to come out of the differentiation.

03:46 So we can then rewrite the equation as being cm times d, dt of r squared omega.

03:56 And then that is equal to zero.

03:58 We know that r is going to change because we're told that r changes over time in the very beginning.

04:05 And we should assume that omega is also changing because we're asked to find what the change in omega is.

04:14 That's the goal of the problem.

04:15 But also, i mean, if we think about the fact that angular momentum is constant and we know that r is changing, no other factor and we know that the mass isn't changing, then no other factor can change besides its angular velocity.

04:27 So omega also has to change over time.

04:31 So we can't just pull either r squared or omega out of this integration or of this differentiation.

04:38 But we can go ahead and cancel out c and m because if we divide both sides by c and m, well zero divided by cm is just going to be zero.

04:47 So then our equation just becomes d d t of r squared omega is equal to zero.

04:57 Now if you've taken some calculus classes you should know how to do this using the product rule.

05:04 If you don't know how to do this, i'll just walk you through it.

05:09 And you can kind of, if you haven't done calculus before, hopefully we'll do a calculus class pretty soon because that will be very helpful with future physics courses.

05:18 If this is just a one -off course for you, then just follow me as i go through this.

05:25 So when we do the product, when we use the product rule for differentiation, if we do d d t of x and y and y and y are both functions of t then it's going to be equal to d x d t times y plus x times d y d t so what that means is we take the derivative of the first of the two the first of the two in this case the r squared and then just leave the second one as it is and then we add to that leaving the first one alone and taking the derivative of the second with respect to whatever.

06:10 So what that means is in order to do this ddt of r squared times omega, first we'll want to do ddt of r squared and leave omega alone.

06:23 So ddt of r squared will be 2r times drdt because of chain rule.

06:30 We know that r is a function of time.

06:32 So the derivative of r squared is going to be 2r.

06:36 We bring the exponent down and then have it multiply as a constant out of the front.

06:43 And then we lower the exponent by 1.

06:45 So we multiply it by the r by 2.

06:48 And then for the exponent, the 2 becomes a 1.

06:51 So it becomes 2r.

06:52 And then for chain rule, we multiply it by the derivative of r.

06:58 So deriving r squared becomes 2.

07:01 R times d r d t and then we want to multiply it by omega and just leave omega alone.

07:08 Then we add to it the second term, which will be the opposite, where we leave r squared alone and then derive omega in terms of time.

07:19 So we leave r squared alone out at the front and then, well, the time derivative of omega is just the time derivative of omega.

07:28 So we can just say that we derive, we do ddt on omega, which is just d omega d .t.

07:37 And then that is equal to zero.

07:42 So that is where kind of the real calculusy bit happens.

07:47 At this point, we just need to know that drdt is the change in the radius over time of the moon's orbit, and domega dt is what we're solving for, the change in the angular velocity over time...

Please give Ace some feedback