Quoting from the Feynman Lectures on Physics - Vol I:

The atoms are 1 or $2 \times 10^{−8}\ \rm cm$ in radius. Now $10^{−8}\ \rm cm$ is called an angstrom (just as another name), so we say they are 1 or 2 angstroms (Å) in radius. Another way to remember their size is this: if an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple.

How does this hold true?

Let us assume the radius of an average apple is about $6\ \rm cm$ ($0.06\ \rm m$). The radius of earth is about $6371\ \rm km$ ($6371000\ \rm m$). Therefore, a $\frac{6371000\ \mathrm{m}}{0.06\ \mathrm{m}} = 106183333.33$ magnification, i.e., a magnification of about $10^{-8}$ times is required to magnify an apple to the size of the earth.

If we magnify an atom of size say 1 angstrom ($10^{-10}\ \rm m$) by $106183333.33$ times, we get $0.0106\ \rm m$ or $1.06\ \rm cm$ only. The atom has not been magnified to the size of the original apple. How does the quoted statement in the book hold good?

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    $\begingroup$ I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth. $\endgroup$
    – user190081
    Oct 1 '18 at 13:11
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    $\begingroup$ 6 cm radius is an enormous apple, nearly a liter in volume. That is more the size of a cantaloupe. If you go with a 3 cm radius (more the size of a tennis ball) the ratio comes out right within a factor of of better than 2. $\endgroup$
    – Ben51
    Oct 1 '18 at 18:21
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    $\begingroup$ *we live in the Eastern U.S.A., so what we call "mountain", some people call "green hill." $\endgroup$
    – user205719
    Oct 1 '18 at 18:37
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    $\begingroup$ @Ben51 and if you choose the Van der Waals radius of a carbon atom in the apple, you get 1.7 Å and you're basically spot on. $\endgroup$
    – jkej
    Oct 1 '18 at 19:56
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    $\begingroup$ @l0b0 You're probably thinking of Fermi problem. (Often referred to as Fermi estimation or Fermi approximation.) Feynman probably used this often too, and he had better PR. ;) $\endgroup$
    – jkej
    Oct 1 '18 at 20:01

In "back of the envelope" calculations like this, all you can really do is look at orders of magnitude. As others have pointed out, not all apples have the same size, and not all atoms have the same "size". All we can then work with is orders of magnitude, so $1\ \rm cm$ and $6\ \rm cm$ (although many people in the comments are saying $3\ \rm cm$) should be considered to be the "same", since we could be dealing with apples and atoms of various sizes.

Indeed, if an apple's radius is on the order of $10^{-2}\ \rm m$, and if the Earth's radius is on the order of $10^6\ \rm m$, then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$.

If our atom radius is on the order of one Angstrom, or $10^{-10}\ \rm m$, then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^{-2}\ \rm m$, which is what we were trying to show in the first place.

You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the estimated size of an atom is on the order of $10^{-10}\ \rm m$.$^*$

Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so if you take all of this into account I would say the book is correct.

$^*$Although I have to admit, I tend to remember the size of an atom, and I can't seem to keep in my head the order of magnitude of the Earth's radius. So I might use this memory tool the other way around to remember that the size of the earth is about 8 orders of magnitude larger than an apple.

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    $\begingroup$ Speaking of mass comparisons, "a man is the geometric mean of an atom and the Sun". $\endgroup$
    – Joker_vD
    Oct 1 '18 at 15:21
  • $\begingroup$ And the book also mentions that atoms are 1-2 Angstrom, so he's already allowing for a factor of 2 error. OTOH, couldn't he have easily picked a smaller fruit like a plum or cherry? Just as familiar as an apple, but the ratio would be closer. $\endgroup$
    – Barmar
    Oct 1 '18 at 15:28
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    $\begingroup$ @Joker_vD Now we need to relate the energy gained from a man eating an apple to some sort of cosmic number... Something like "A man eating an apple is like a black hole eating ____" Although I wouldn't like to think about large scale GR while eating fruit :) $\endgroup$ Oct 1 '18 at 15:29
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    $\begingroup$ It's the internet, we need a way to relate another popular "masculine average" metric. Let's see, does the geometric mean between an AU and the Planck distance get us in the ballpark? $\endgroup$
    – Beanluc
    Oct 1 '18 at 20:55
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    $\begingroup$ @PhilipOakley how was original meter defined? ;) $\endgroup$ Oct 4 '18 at 15:04

Your question starts out with questioning whether the numbers match up very well, and then you proceed to throw away all the accuracy in your numbers to demonstrate that they don't.

While throwing away accuracy in-favor of considering just the orders of magnitude is useful for approximating (and is what the other answers are explaining is okay), it turns out if you just run the numbers through as-written, you actually get a pretty accurate correspondence.

Taking atomic radius as $1.5 \times 10^{-10}~\text{m}$ (average of range given above) and the radius of Earth as $6371000~\text{m}$ (given above), the correspondence is exact when the radius of the object is $\approx 30.91~\text{mm}$ (or diameter $61.83~\text{mm}$). The size of an apple varies by climate, variety, and harvest time, but this figure is consistent with an ordinary "175-grade" apple, albeit most apples sold in the United States are slightly larger.

  • $\begingroup$ Thank you. It makes sense now. If $ x $ is the radius of the apple, then $ \frac{6371000}{x} = \frac{x}{1.5 \times 10^{-10}} $ indeed leads to $ x = 0.03091 $ or $ 30.91 $ mm. $\endgroup$ Oct 2 '18 at 2:26

When dealing with such orders of magnitude, a factor of 6 does not greatly affect the overall picture. Furthermore, note that apples have radii that are often smaller than $6$ cm (see for instance here). If we take an apple with a radius of $3$ cm and an atom with a radius of $2$ angstrom, the "Feynman comparison" (as per your numbers) comes out to be fairly good.

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    $\begingroup$ I take issue with "a factor of 6 can easily be disregarded". An order of magnitude is purposefully vague, but in all common definitions (just the exponent in the a*10^b notation, all values within (0.5x, 5x] and so on) the order changes for most values if you multiply or divide by 6. $\endgroup$ Oct 2 '18 at 8:44
  • $\begingroup$ I don't disagree with you. In fact, the term "disregard" might be a bit too strong. I'll reword the answer so as to "weaken" the first sentence. Cheers! $\endgroup$
    – lr1985
    Oct 2 '18 at 14:16
  • $\begingroup$ I would also point out that the OP is running fast and loose with precision. That 106,183,333.33 number comes from two values precise to maybe 1.5 and 4 digits, so the most you can really say is that the magnification is about 1.1E8 (1.1 x 10 ^ 8). Suddenly the apple gets a little bigger. $\endgroup$
    – Tom Hundt
    Oct 3 '18 at 5:25

Let's test that by figuring out the size of an atom by that comparison.

Wikipedia says:

Commercial growers aim to produce an apple that is 2 3⁄4 to 3 1⁄4 in (7.0 to 8.3 cm) in diameter, due to market preference.

So we use an average radius of $3.825~\text{cm}$ or $0.03825~\text{m}$ and the radius of the earth of $6371000~\text{m}$ for the calculation.

Then the radius of an atom is calculated as: $$ r_{atom} = \frac{r_{apple}^2}{r_{earth}} = \frac{0.03825^2~\text{m}^2}{6371000~\text{m}} $$

As a result we get $2.296 \times 10^{-10}~\text{m}$ which is fairly close.


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