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Mass of earth’s oceans

Estimate the total mass of the oceans on Earth.

Strategy

We know the density of water is about 10 3 kg/m 3 , so we start with the advice to “get masses from densities and volumes.” Thus, we need to estimate the volume of the planet’s oceans. Using the advice to “get areas and volumes from lengths,” we can estimate the volume of the oceans as surface area times average depth, or V = AD . We know the diameter of Earth from [link] and we know that most of Earth’s surface is covered in water, so we can estimate the surface area of the oceans as being roughly equal to the surface area of the planet. By following the advice to “get areas and volumes from lengths” again, we can approximate Earth as a sphere and use the formula for the surface area of a sphere of diameter d —that is, A = π d 2 , to estimate the surface area of the oceans. Now we just need to estimate the average depth of the oceans. For this, we use the advice: “If all else fails, bound it.” We happen to know the deepest points in the ocean are around 10 km and that it is not uncommon for the ocean to be deeper than 1 km, so we take the average depth to be around ( 10 3 × 10 4 ) 0.5 3 × 10 3 m. Now we just need to put it all together, heeding the advice that “one ‘sig. fig.’ is fine.”

Solution

We estimate the surface area of Earth (and hence the surface area of Earth’s oceans) to be roughly

A = π d 2 = π ( 10 7 m ) 2 3 × 10 14 m 2 .

Next, using our average depth estimate of D = 3 × 10 3 m, which was obtained by bounding, we estimate the volume of Earth’s oceans to be

V = A D = ( 3 × 10 14 m 2 ) ( 3 × 10 3 m ) = 9 × 10 17 m 3 .

Last, we estimate the mass of the world’s oceans to be

M = ρ V = ( 10 3 kg/m 3 ) ( 9 × 10 17 m 3 ) = 9 × 10 20 kg .

Thus, we estimate that the order of magnitude of the mass of the planet’s oceans is 10 21 kg.

Significance

To verify our answer to the best of our ability, we first need to answer the question: Does this make any sense? From [link] , we see the mass of Earth’s atmosphere is on the order of 10 19 kg and the mass of Earth is on the order of 10 25 kg. It is reassuring that our estimate of 10 21 kg for the mass of Earth’s oceans falls somewhere between these two. So, yes, it does seem to make sense. It just so happens that we did a search on the Web for “mass of oceans” and the top search results all said 1.4 × 10 21 kg, which is the same order of magnitude as our estimate. Now, rather than having to trust blindly whoever first put that number up on a website (most of the other sites probably just copied it from them, after all), we can have a little more confidence in it.

Check Your Understanding [link] says the mass of the atmosphere is 10 19 kg. Assuming the density of the atmosphere is 1 kg/m 3 , estimate the height of Earth’s atmosphere. Do you think your answer is an underestimate or an overestimate? Explain why.

3 × 10 4 m or 30 km. It is probably an underestimate because the density of the atmosphere decreases with altitude. (In fact, 30 km does not even get us out of the stratosphere.)

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How many piano tuners are there in New York City? How many leaves are on that tree? If you are studying photosynthesis or thinking of writing a smartphone app for piano tuners, then the answers to these questions might be of great interest to you. Otherwise, you probably couldn’t care less what the answers are. However, these are exactly the sorts of estimation problems that people in various tech industries have been asking potential employees to evaluate their quantitative reasoning skills. If building physical intuition and evaluating quantitative claims do not seem like sufficient reasons for you to practice estimation problems, how about the fact that being good at them just might land you a high-paying job?

For practice estimating relative lengths, areas, and volumes, check out this PhET simulation, titled “Estimation.”

Summary

  • An estimate is a rough educated guess at the value of a physical quantity based on prior experience and sound physical reasoning. Some strategies that may help when making an estimate are as follows:
    • Get big lengths from smaller lengths.
    • Get areas and volumes from lengths.
    • Get masses from volumes and densities.
    • If all else fails, bound it.
    • One “sig. fig.” is fine.
    • Ask yourself: Does this make any sense?

Problems

Assuming the human body is made primarily of water, estimate the volume of a person.

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Assuming the human body is primarily made of water, estimate the number of molecules in it. (Note that water has a molecular mass of 18 g/mol and there are roughly 10 24 atoms in a mole.)

10 28 atoms

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Estimate the mass of air in a classroom.

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Estimate the number of molecules that make up Earth, assuming an average molecular mass of 30 g/mol. (Note there are on the order of 10 24 objects per mole.)

10 51 molecules

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Estimate the surface area of a person.

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Roughly how many solar systems would it take to tile the disk of the Milky Way?

10 16 solar systems

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(a) Estimate the density of the Moon. (b) Estimate the diameter of the Moon. (c) Given that the Moon subtends at an angle of about half a degree in the sky, estimate its distance from Earth.

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The average density of the Sun is on the order 10 3 kg/m 3 . (a) Estimate the diameter of the Sun. (b) Given that the Sun subtends at an angle of about half a degree in the sky, estimate its distance from Earth.

a. Volume = 10 27 m 3 , diameter is 10 9 m.; b. 10 11 m

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Estimate the mass of a virus.

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A floating-point operation is a single arithmetic operation such as addition, subtraction, multiplication, or division. (a) Estimate the maximum number of floating-point operations a human being could possibly perform in a lifetime. (b) How long would it take a supercomputer to perform that many floating-point operations?

a. A reasonable estimate might be one operation per second for a total of 10 9 in a lifetime.; b. about (10 9 )(10 –17 s) = 10 –8 s, or about 10 ns

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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