# Homework Help: Density of earth

1. Sep 3, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

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. Each region is taken to have a unique constant density (the average density of that region in the real Earth):

a) Use this model to predict the average density of the entire Earth (use 4 sig figs).

b) The measured radius of the Earth is 6371 km and its mass is 5.974×1024kg. Use these data to determine the actual average density of the Earth (use 4 sig figs).

3. The attempt at a solution

using the info from the chart the inner core has a radius of 1220km. The earth has a radius of 6371km so the fraction of the radius of the earth that belongs to the inner core is

(1220km/6371km)= .1915

multiply this by the density of that region

(13000)(.1915)= 2489.5

now I do the same for each other region and add the results:

outer core = 3480km-1220km = 2260km
(2260km/6371km) = .3547
(.3547)(11100) = 393.7

mantle = 6371km-3480km = 2891km
(2891km/6371km) = .4538
(.4538)(4400) = 1997

1997 + 393.7 + 2489.5 = 4880.2 which to 4 sig figs is 4880.

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2. Sep 3, 2014

### Matterwave

The Earth is a sphere, so you have to do this problem by volume, not by radius. How much of the total volume of the Earth is within the inner core? Hint: it's not 19.15%.

3. Sep 3, 2014

### toothpaste666

so i would find the volume of the inner core, then find the volume of the outer core + inner core and subtract the volume of the inner core from it. and then find the volume of the whole earth and subtract the volume of the outer core + inner core from it. then figure out what fraction each volume is of the total volume of the earth and multiply that fraction by the density of the region and add the pieces?

i will calculate all this in a second but first, i noticed the densities are in kg/m^3 and the radius of each section is in km. should i convert the km to m before starting?

4. Sep 3, 2014

### BiGyElLoWhAt

Couldn't hurt, can't add apples and kilo-apples XD

Also, I think there would be an easier way to do it. Your first steps are great, but I think you're trying to do too much with ratio's, the only ratio I would actually calculate would be the final <density>.

I think b is a pretty solid hint as to solving a. Can you make the setup for a look like the setup for b?

5. Sep 3, 2014

### toothpaste666

i could just multiply each volume by its respective density add it all together and only have to divide it by the total volume of the earth once at the end. thank you

6. Sep 4, 2014

### Orodruin

Staff Emeritus
1 keV + 1 eV = 1001 eV ...

Also, kg km^3/m^3 is a perfectly valid unit of mass ... :tongue:

Regarding the OP, I would subtract densities rather than volumes, i.e., one homogeneous sphere of density 4400 kg/m^3 and radius R = 6371 km, one homogeneous sphere of density 11100-4400 = 6700 kg/m^3 and radius 3480 km, and one homogeneous sphere of density 13000-11100 = 1900 kg/m^3 of radius 1220 km. In the end it is just bookkeeping, but for some reason I find it easier to think this way.