The Earth is not a uniform sphere, but has regions of varying ...

Question

The Earth is not a uniform sphere, but has regions of varying density. Consider a simple model of the Earth divided into three regions—inner core, outer core, and mantle. Let us assume each region has a constant density (the average density of that region in the real Earth). Use this model to predict the average density of the entire Earth.

Table displaying Earth's regions with their respective radius and density values: Inner Core, Outer Core, and Mantle.

Accepted Answer

Everyone in this problem, we're told that the core Mer Mercury can be divided into three parts. We have the inner core, the mantle and the crust. Each of the parts can be assumed to have constant densities. As given in the table below. We're asked to use the data to calculate the average density of mercury. So what we have is the inner core and we're told that it spans from the center to a radius of 1800 kilometers and has a density of 8700 kg per meter cubed. The mantle is gonna go from a radius of 1800 kilometers to 2400 kilometers with a density of 3500 kg per meter cubed. And the crust is gonna go from 2400 to 2600 kilometers with a density of 3300 kg per meter cubes. So we're getting less dense as we're getting further away from that center. So we can go ahead and draw this out. Uh We have some answer choices. We're gonna come back to just once we finish drawing this out. So we have our mercury. OK? Or the core of mercury, I should say and it has three parts. OK. We know that that inner part, this is our inner core, it has a radius of 1800 kilometers. Mm. So this is our inner core. The next section is going to be our mantle and it is gonna span from that 1800 kilometer mark out to 2400 kilometers. OK. So this is our mantle and then finally, we have the crust which goes all the way to the edge. So from the mantle out to 2600 kilometers, and so we have those three sections of our. Now our answer choices, there are four of them. They're all in kilograms per meter cubed. Option A 5200, option B 6000, option C 6400 and option D 6600. We are asked to find the average density. OK. So recall that the average density can just be written as the total mass over these three parts divided by the total volume over these three parts. OK. The density recall is the mass divided by the volume. So when we're looking for the average, we're just looking at what's the total mass, what's the total volume? OK. So if we think about the total mass, we have to think about the mass of the three different pieces we have. OK. So we're gonna have the mass of the inner core. So M IC, what's the mass of the mantle? Mm, what's the mass of the crust MC. And similarly, we have to divide by those corresponding volumes. So the volume of the inner core, what's the volume of the mantle? What's the volume of the crust? Let's think about these things. And if we think about our mass, OK. Again, we're using this equation that the density row is equal to the mass divided by the volume. So we wanna write each mass individually in terms of the density we're given in the problem and the volume which we can calculate then the mass and will just be equal to the density multiplied by the volume. OK. So we're gonna want to use that in the numerator of our equation. And then we can think about these volumes and the volume of the inner core, that's just a circle with a radius of 1800 kilometers. OK. Now when I say circle, OK, in our diagram, it's a circle. But if we're thinking about an actual planet, this shape is going to be a sphere. So we're looking for the volume of a sphere which we can recall is four thirds pi R squared. So pi multiplied by the radius of the inner core squared. Thats cute, sorry because we're dealing with volume. OK. Now we can move on to the volume of the next portion which is gonna be the volume of the mantle. And if we think about the volume of the mantle, OK. If we think about the volume of the mantle it's just going to be the area between the crust and the inner core. It's not contain that inner core. So we can't take the entire area of a sphere with radius 2400. We have to take a the area of the sphere with radius 2400 and subtract the area or the volume sorry of the sphere of radius 1800 kilometers. OK. So what we're doing is we're gonna take the volume and subtract the volume of the interval. OK. So the volume of this mantle piece is gonna be four thirds, sorry, the radius of the mantle squared. And when I say R MS cubed, sorry when I say RM cubed, I'm talking about that outer radius of the mantle. OK? And we're gonna subtract four thirds. Hi R IC Cute. Yeah. Now we're gonna do the exact same thing with the next piece. So if we're looking for the volume of the crust, we're gonna do the same thing. We're gonna have four thirds high are cute. And we're using the radius of the crust that outer radius of the crust. But then we need to subtract everything from the other components, everything in those two other sections. And so we're gonna subtract four thirds. Hi, Rm Cute. No, let's go back to our equation for our average density and substitute in what we just found, right? So what we're gonna have here. So we're gonna have that hour average density is gonna be the density of the inner core multiplied by the volume of the inner core plus the density of the mantle multiplied by the volume of the mantle plus the density of the crust multiplied by the volume of the crust all divided by the volume of the inner core plus the volume of the mantle plus the volume of the crust. OK. So let's work out the numerator in the numerator. We are gonna have the density of the inner core multiplied by the volume. So multiplied by four thirds, pi R IC cubed. And we're gonna add the density of the mantle multiplied by its volume. Four thirds pi RM cubed minus four thirds pie R IC cubed plus the density of the crust multiplied by its volume. Four thirds pr C cubed minus four thirds pi RM cut. Hey, we're gonna divide this by the total volume. OK. Now, we could have done this two ways. The total volume. If we just think about this, we think about our picture. The total volume is just gonna be the volume of a sphere with that outer radius of 2600 kilometers, right? If we were to instead do it in the pieces like we've done if we add all of those pieces up this four third pr ic cube term that's gonna cancel and the blue and the green and similar to the four third pi RF cube term will cancel as well. And we're just gonna be left with four thirds pi RCC, which is the exact same as if we had have approached it in the opposite way. And so we can write in the denominator that we have four thirds pi RC cubed. OK. So this is our big messy equation. The first thing we can see is that these four third P terms are going to cancel all the way throughout. So we can divide the entire equation by four thirds pi and those are going to go away. Yeah, what can we do next? What we're gonna do is we're gonna factor in terms of these radii. And so what we can say is that the radius of the inner cube, inner core or acute multiplied by the density of the inner core minus the density of the mantle o the radius of the mantle cute multiplied by the density of the mantle minus the density of the crust. O was the radius of the crust cubed multiplied by the density of the crust divided by the radius of the crust cubed. This is going to be our average density. OK. So we've simplified it down to all of these terms that we know these are all of the values that we were given in the table. There's nothing else we need to do but substitute them in. So let's go ahead and do that. We have 1800 kilometers. Cute multiplied by 8700 kg per meter cubed minus 3500 kg per meter cubed us 2400 kilometers cute multiplied by 3500 kg per meter cubed minus 3300 kilograms per meter cubed. We're gonna add the other term down below we're out of space. So we have plus 2600 kilometers cubed multiplied by 3300 kg per meter cut. And all of this it's divided by 2600 kilometers cubed. And you may have noticed that we did not convert our kilometers into our standard unit of meters. In this case, we have kilometers tubed in every term in the numerator as well as in the denominator. So those units are going to divide out, they're going to cancel. And so it doesn't matter what unit we're in as long as we are consistent throughout uh when you're doing problems like these, that's not always the case. So always double check that your units are correct. Um And are getting you to that correct final answer. All right. So last thing we need to put this into our calculator and work this out. And when we do, we are gonna get that the average density row up is equal to about 5182.75 kg per meter. Cute. Ok. So that is the average density of the core of mercury based on the information we were given, we're gonna round to two significant digits. And when we do that, we can see that the correct answer is going to be option a 5200 kg per meter cubed. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Density

Volume of Spheres

Weighted Average

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