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?
 The phrase "more massive" should be read literally as "has more mass" not "fills more space".
 The phrase "heavier" should be read as "is pulled down more strongly by gravity" not "is more dense".
solution
This is a question about how people sometimes confuse the meanings of words. The answer to all four parts of this problem are, "They are the same".
 People sometimes confuse mass and volume when using the word "massive". A kilogram of aluminum takes up more volume than a kilogram of lead because it is less dense, but this is a question about mass, not volume. All identical masses are identical no matter what they are made of. A kilogram of aluminum on Earth has the same mass as a kilogram of lead on Earth.
 Mass is an invariant quantity — it changes for nothing. All identical masses are identical no matter where they are. A kilogram of aluminum on the moon has the same mass as a kilogram of lead on the moon.
 People sometimes confuse density and weight when using words like "heavier" or "lighter". Lead may be said to be "heavier" as a material, but this is a question about weight, not density. Weight is affected by mass and gravity so two identical massed objects in the same gravitational field have the same weight. A kilogram of aluminum on Earth has the same weight as a kilogram of lead on Earth
 Weight varies with location, so objects weigh less on the moon than on the Earth. But objects with the same mass have the same weight when located in the same place. It doesn't matter where that place is. A kilogram of aluminum on the moon has the same weight as a kilogram of lead on the moon.
The future version of this question is about how the presence of an atmosphere affects the precise measurement of weight.
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} = 5,510 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}_{3}πr^{3}_{outer} − ^{4}_{3}πr^{3}_{inner}
V_{mantle} = ^{4}_{3}π(r^{3}_{outer} − r^{3}_{inner})
V_{mantle} = ^{4}_{3}π[(6,370,000 m)^{3} − (3,480,000 m)^{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} = (4,500 kg/m^{3})(9.06164… × 10^{20} m^{3})
m_{mantle} = 4.07774… × 10^{24} kgDivide 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} ⎛
⎜
⎝6,370 km − 2,890 km ⎞^{3}
⎟
⎠V_{earth} 6,370 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.