Posted Aug 25, 2023 by Ray Patrick

As I’ve written before, SI absolutely dunks on the 11th-century measurement scheme that Americans have chosen to hobble ourselves with.

To drive this point home, let’s estimate the mass of the atmosphere. We only need to make use of three facts: the earth’s mean radius (6.37 Mm), the mean atmospheric pressure (101 kPa), and the mean acceleration due to gravity (9.81 m/s^{2}.) Our method is as follows:

- Compute the surface area of the earth.
- Multiply the mean atmospheric pressure by the earth’s surface area to get the atmosphere’s weight.
- Divide the atmosphere’s weight by the mean acceleration due to gravity to get the atmosphere’s mass.

Well, the surface area of a sphere is given by 4 π r^{2}. Expressing the earth’s radius in megameters as I did above will give the answer in square megameters, resulting in a comfortably-sized number. This turns out to be 510 Mm^{2}.

Now we need to multiply this area by the atmospheric pressure to get the atmosphere’s weight in newtons. The mean atmospheric pressure is 101 kPa, which is 101 kN/m^{2}. Multiplying these two figures together will give us the correct *numerical* result, but we can’t be sure of the *magnitude* unless we take care to keep the units consistent. Therefore, we need to rearrange the expression for atmospheric pressure so it has square megameters in the denominator.

This is easier to do than it might sound. There are 1000 meters in a kilometer, so a square kilometer contains 10^{3} m × 10^{3} m = 10^{6} m^{2}. Similarly, a square megameter contains 10^{6} km^{2}. Putting these facts together, we see that a square megameter is equivalent to 10^{12} m^{2}, so the atmospheric pressure is therefore equal to 101 × 10^{12} kN/Mm^{2}. Well, 10^{12} kN is just 10^{12} × 10^{3} N = 10^{15} N. The metric prefix for 10^{15} is peta, making the atmospheric pressure equal to 101 PN/Mm^{2}.

Therefore, the weight of the atmosphere is about 510 Mm^{2} × 101 PN/Mm^{2} = 51 510 PN, or about 51.5 EN.

Now, to obtain the mass of the atmosphere, we just divide 51.5 EN by 9.81 m/s^{2}. We have to be careful here, too. As before, we will get the correct *numerical* result if we simply divide these two quantities right now. However, to ensure we have the correct magnitude, we’ll have to rearrange the expression for acceleration due to gravity.

The definition of the newton is: N = kg m/s^{2}. Therefore, 1 EN = 10^{18} kg m/s^{2}. Notice that, by the rules of multiplication, the 10^{18} term can apply to *any* term within the definition of the newton. For instance, we can multiply it by the first term to get (10^{18} kg) m/s^{2} = (10^{21} g) m/s^{2} = Zg m/s^{2}.

Notice also that this expression includes the term m/s^{2}. Dividing by the acceleration due to gravity will cancel the m/s^{2} term, leaving only zettagrams. The mass of the atmosphere is therefore 51.5/9.81 = 5.25 Zg.

There’s our answer: the mass of the atmosphere is 5.25 Zg. (WolframAlpha gives a result of 5.1441 Zg, a difference of only 2.06%.)

Think about this for a moment. The power of SI reduced the formidable problem of *weighing the earth’s atmosphere* to an exercise in simple arithmetic! Furthermore, the elegant metric prefix system ensured that we could perform this computation with none but comfortably-sized numbers.

How would we even do this in pounds? Well, the average atmospheric pressure is 14.7 pounds per square inch (psi). In order to find the weight of the atmosphere, we will have to multiply this value by the earth’s surface area, expressed in square inches.

The radius of earth is 3959 miles. To find the surface area of the earth in square inches, we must first convert the earth’s radius to inches:

3959 miles × 5280 feet/mile × 12 inches/feet = 250,842,240 inches

Now we can use the spherical surface area formula:

4 × π × (250,842,240 inches)^{2} = 790,298,176,864,813,100 square inches.

Now we have to multiply 14.7 psi by 790,298,176,864,813,100 square inches. That brings us to 11,617,383,199,912,750,000 pounds, or, if we use the largest possible unit for weight, 5,808,691,599,956,376 tons. Not too easy on the eyes. And of course there’s no way to abbreviate this quantity without going back to the old “Sagan Standby” of saying “5.8 quadrillion tons.” Come on, now.

Of course, the other way to do it would be to keep the earth’s radius in miles. That would make the surface area “only” 196,861,433 square miles. Now we convert the atmospheric pressure to something with square miles in the denominator:

1 ft^{2} = 144 in^{2} 1 mi^{2} = (5280 × 5280) = 27,878,400 ft^{2} = 4,014,489,600 in^{2} 14 lb = 14/2000 ton 14 psi = (14/2000)/(4014489600) = 0.00000000000174 tons per square mile …

Okay, I’m not even going to take that any further. You get the idea, anyway. There are two ways to do this problem in medieval English units: the way that gives you eighteen-digit numbers or the one that gives you a decimal point followed by eleven zeros. Charming!

I think I’d rather use SI. Wouldn’t you?

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Topics: metric