The Physics
Opus in profectus


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practice problem 1

Your mother gives you a kilogram of aluminum and a kilogram of lead. Both objects are solid, rectangular blocks.
  1. Which is more massive on the surface of the Earth?
  2. Which is more massive on the surface of the Moon?
  3. Which is heavier on the surface of the Earth?
  4. Which is heavier on the surface of the Moon?
Special notes:
  • 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".
A variation on this practice problem appears later in the section on buoyancy.


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".

  1. 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.
  2. 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.
  3. 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
  4. 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

A lucite cube has a mass of 142.5 g and a width of 4.9 cm. Determine its density in kg/m3.


Start with the definition of density, replace V with s3 (the volume of a cube), substitute the given quantities in SI units, and solve.

ρ =  m  =  m
V s3
ρ =  (0.1425 kg)
(0.049 m)3
ρ = 1200 kg/m3  

practice problem 3

It has been known for several thousand years that the Earth is spherical (by educated people, at least). Sometime in the 2nd century BCE the size of the Earth was determined (r = 6,370 km). By the 19th century its mass was known (m = 5.97 × 1024 kg). And in the early 20th century the structure of the Earth was deduced. The Earth has three main layers: crust, mantle, and core. The crust of the is the lightest and thinnest and, like the shell of an egg, contributes little to its overall mass. The mantle is a bit more dense, substantially thicker, and contains most of the Earth's mass. The core is the densest layer (but not the most massive) and is divided into a liquid outer core and a solid inner core. The relevant data for the interior of the Earth are summarized below.
The structure of the Earth
layer depth range (km) mean density (kg/m3) consistency
crust 0 ~ 20 2700 solid
mantle 20 ~ 2890 4500 plastic
outer core 2890 ~ 5160 ? liquid
inner core 5160 ~ 6370 ? solid
  1. the average density of the entire Earth
  2. the percent of the Earth's mass located in the mantle, and
  3. the average density of the core.


  1. 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.

    ρearth =  mearth
    ρearth =  3m
    ρearth =  3(5.97 × 1024 kg)
    4π(6.37 × 106 m)3
    ρearth = 5510 kg/m3 
  2. You may want to do this is stages. Calculate the volume of the spherical shell that is the mantle.

    Vmantle =  4  πr3outer −  4  πr3inner
    3 3
    Vmantle =  4  π ( r3outer − r3inner )
    Vmantle =  4  π [ (6,370,000 m)3 − (3,480,000 m)3 ]
    Vmantle = 9.06164… × 1020 m3  

    Mass is density times volume. Use the density given in the data table and the mass calculated above.

    mmantle =  ρV
    mmantle =  (4500 kg/m3)(9.06164… × 1020 m3)
    mmantle =  4.07774… × 1024 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.

    mmantle  =  4.07774… × 1024 kg
    mearth 5.97 × 1024 kg
    mmantle  =  0.683038… ≈ 70%
  3. 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.

    ρcore =  mcore
    ρcore =  (1 − 0.683038…)mearth

    6370 km − 2890 km 3

      6370 km
    ρcore =  1.94396… ρearth  
    ρcore =  (1.94396…)(5514.01… kg/m3)  
    ρcore = 10,700 kg/m3  

    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/m3 and the solid inner core from 12,300 to 12,600 kg/m3. My suspicion is that the mean value is more like 10,900 kg/m3, 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

Write something completely different.


Answer it.