Density
Practice
practice problem 1
 Which is more massive on the surface of the earth?
 Which is more massive on the surface of the moon?
 Which is heavier on the surface of the earth?
 Which is heavier on the surface of the moon?
We will return to this question in the section on buoyancy. (The phrase "more massive" should be read literally as "has more mass" not "fills more space".)
solution
Answer it.
practice problem 2
solution
Start with the definition of density, replace V with s^{3} (the volume of a cube), substitute the given quantities in SI units, and solve.





practice problem 3
layer  depth range (km)  mean density (kg/m^{3})  consistency 

crust  0 ~ 20  2700  solid 
mantle  20 ~ 2890  4500  plastic 
outer core  2890 ~ 5160  ?  liquid 
inner core  5160 ~ 6370  ?  solid 
 the average density of the entire earth
 the percent of the Earth's mass located in the mantle, and
 the average density of the core.
solution
Density is the ratio of mass to volume. Use the given mass of the Earth and calculate its volume from the radius and the equation for the volume of a sphere.
ρ _{earth} = m_{earth} V_{earth} ρ _{earth} = 3m 4πr^{3}_{earth} ρ _{earth} = 3(5.97 × 10^{24} kg) 4π(6.37 × 10^{6} m)^{3} ρ _{earth} = 5510 kg/m^{3} You may want to do this is stages. Calculate the volume of the spherical shell that is the mantle.
V_{mantle} = 4 πr^{3}_{outer} − 4 πr^{3}_{inner} 3 3 V_{mantle} = 4 π ( r^{3}_{outer} − r^{3}_{inner} ) 3 V_{mantle} = 4 π [ (6,370,000 m)^{3} − (3,480,000 m)^{3} ] 3 V_{mantle} = 9.06164… × 10^{20} m^{3} Mass is density times volume. Use the density given in the data table and the mass calculated above.
m_{mantle} = ρV m_{mantle} = (4500 kg/m^{3})(9.06164… × 10^{20} m^{3}) m_{mantle} = 4.07774… × 10^{24} kg Divide the mass of the mantle by the mass of the Earth to determine the percentage of earth's mass in its most substantial layer.
m_{mantle} = 4.07774… × 10^{24} kg m_{earth} 5.97 × 10^{24} kg m_{mantle} = 0.683038… ≈ 70% m_{earth} This part could also be solved in stages, but let's try a different approach. Start from the basic definition of density, replace mass and volume with their respective fractions of the Earth's total, then multiply by the average density. Recall that the volume of a sphere is proportional to the cube of its radius.
ρ _{core} = m_{core} V_{core} ρ _{core} = (1 − 0.683038…)m_{earth} ⎛
⎝6370 km − 2890 km ⎞^{3}
⎠V_{earth} 6370 km ρ _{core} = 1.94396… ρ _{earth} ρ _{core} = (1.94396…)(5514.01… kg/m^{3}) ρ _{core} = 10,700 kg/m^{3} This number seems a bit low to me. The earth's core is homogeneous chemically, but not physically. Both layers are a mixture of iron and nickel, but one is liquid and the other solid. The liquid outer core is reported to range in density from 10,000 to 12,000 kg/m^{3} and the solid inner core from 12,300 to 12,600 kg/m^{3}. My suspicion is that the mean value is more like 10,900 kg/m^{3}, but I don't see any error in the solution presented above. I leave you with the classic excuse found in many college math textbooks, but with a slight modification. "The disproof is left to the reader."
practice problem 4
solution
Answer it.