What is a proportion, and how does scale work?

A tabletop display has small models of Earth and the Moon. Someone decides that Earth is hard to see and doubles its diameter, leaving the Moon unchanged. Each ball still resembles a celestial body, but does the pair still show their relative sizes correctly?

It does not: the Moon is now half as large relative to Earth. To enlarge the whole model while preserving the size relationship, we must multiply both diameters by the same number. A proportion is an equality between two ratios. It lets us state exactly what should remain unchanged when the sizes themselves change. OpenStax: definition of a proportion.

Compare two defined lengths, not just two numbers

We will use diameters: lengths through the centre of a sphere from one side to the other. Earth and the Moon are not perfect spheres, so for this problem we will use spheres with their volumetric mean radii. These are the radii of spheres with the same volumes as the respective bodies. NASA’s table gives 6,371.0 km for Earth and 1,737.4 km for the Moon. Doubling each radius gives diameters of 12,742 km and 3,474.8 km. NASA: Moon–Earth comparison.

What fraction of Earth’s diameter is the Moon’s diameter? Divide the Moon’s diameter by Earth’s. The answer is approximately 0.273, or 27.3%. In this comparison, the Moon is slightly wider than one quarter of Earth. This is a diameter ratio; it does not say that the Moon has 27.3% of Earth’s mass or volume.

Order matters. “Moon / Earth” gives a number smaller than one. Reverse the order and “Earth / Moon” gives about 3.67: Earth’s diameter is approximately 3.67 times the Moon’s.

For this ratio, express both lengths in the same units. You can divide kilometres by kilometres or centimetres by centimetres. Expressing the same length in metres makes its numerical value a thousand times greater without making the body grow. Comparing bare numbers without matching the units would lose that meaning. OpenStax: ratios and units.

A common multiplier changes sizes but preserves the ratio

For the first model, let one centimetre represent 2,000 kilometres. Earth’s diameter will be approximately 6.37 cm and the Moon’s 1.74 cm. We have reduced each real length using the same rule.

Now double both diameters. They become approximately 12.74 cm and 3.47 cm, and one model centimetre now represents 1,000 km. We have chosen a different scale for the whole pair. Moon / Earth remains approximately 0.273 because both numerator and denominator changed by the same factor.

Compare the second and third pairs in the figure. In the third, Earth is as large as in the second, but the Moon is left over from the first. That pair no longer has a common scale.

Two scales and one mistake

Compare diameters. Both disks are doubled in the second pair; only Earth is doubled in the third.

Original model

1 model cm ↔ 2,000 km

Earth’s diameter: 6.37 cm
Moon’s diameter: 1.74 cm
Moon / Earth: ≈ 0.273

Original ratio

Both diameters ×2

1 model cm ↔ 1,000 km

Earth’s diameter: 12.74 cm
Moon’s diameter: 3.47 cm
Moon / Earth: ≈ 0.273

Ratio preserved

Only Earth’s diameter ×2

No common scale

Earth’s diameter: 12.74 cm
Moon’s diameter: 1.74 cm
Moon / Earth: ≈ 0.136

Ratio halved

Spherical diameters use NASA’s volumetric mean radii. The centimetres describe hypothetical models, not measurements of your screen. Labels are rounded; ratios are calculated before rounding. Gaps are arbitrary; no orbit is shown. Source: NASA · Moon Fact Sheet.

Labels are rounded for readability; ratios are calculated from the original numbers. Dividing just the rounded values 1.74 by 6.37 gives slightly different final digits. This is a rounding effect, not a change in the model.

We are comparing diameters only. The gap between disks does not represent the distance to the Moon. A spatial model would also need to reduce the centre-to-centre distance using the same rule; moving the balls arbitrarily closer does not preserve the scale of the whole system.

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Why the multiplier disappears from the ratio

Let $a$ be the model Moon’s diameter, $b$ the model Earth’s diameter, and $q>0$ the common enlargement factor. For positive diameters:

\frac{qa}{qb}=\frac{a}{b}.

The factor $q$ cancels. This equality is a proportion: the ratio after enlargement equals the ratio before it. Adding the same length behaves differently: $2/6=1/3$ , but adding 2 to both numbers gives $4/8=1/2$ .

Find the model Moon’s diameter $d$ when the model Earth is 10 cm across. Moon / Earth must match the ratio for our chosen spheres:

\begin{aligned} \frac{d}{10\ \text{cm}}&=\frac{3\,474.8}{12\,742},\\ d&=10\ \text{cm}\times\frac{3\,474.8}{12\,742},\\ d&\approx2.73\ \text{cm}. \end{aligned}

The numerator and denominator of the fraction on the right are diameters in kilometres; the kilometres cancel. The result $d$ has units of centimetres, like the given Earth diameter. We preserved the ratio first, then used it to find the required length.

Direct proportionality needs a fixed condition

Within a single scale, a larger real length always gets a larger model length using the same multiplier. Twice as long becomes twice as long; three times as long becomes three times as long. This relationship is called direct proportionality.

We write it as $y=kx$ , where $x$ is real length, $y$ is model length, and $k$ is a constant coefficient. For the first model, with $x$ in kilometres and $y$ in centimetres, $k$ is 0.0005 cm/km. Multiplying each length in kilometres by this coefficient gives centimetres. The second model uses a different $k$ , shared once again by both bodies. OpenStax: direct variation.

Saying “both quantities increase” does not establish direct proportionality. Add two centimetres to each diameter and both models become larger, but the small Moon changes more in relative terms. Direct proportionality keeps the ratio fixed, rather than the difference.

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“Increase by” and “increase to” give different instructions

Suppose a ball’s initial diameter is 6 cm. Increase it by 2 cm: add two to get 8 cm. Increase it to 8 cm: this states the final length directly. Double it: multiply by two to get 12 cm.

Percentages also need a starting quantity. Increase 6 cm by 50% of its initial value: add 3 cm to get 9 cm. Increase it to 150% of its initial value gives the same 9 cm. But to 50% of its initial value means 3 cm, a reduction.

So a phrase such as “it became 200% larger” deserves careful reading. Adding 200% of the initial value produces 300% of the initial value, or a tripling. Reaching 200% means doubling. A model needs an unambiguous operation as well as a number.

The diameter doubles; area and volume change differently

When we say “enlarge the model by a factor of two”, we must specify that this applies to each corresponding length. In similar bodies, every length changes by the same factor: diameter, radius and circumference. But area involves two dimensions of length, and volume involves three.

A square and a cube make this easy to see. A square with a 1 cm side covers 1 cm². A square with a 2 cm side contains two rows of two such little squares: 4 cm² altogether. A cube with a 2 cm side contains two layers of four unit cubes: 8 cm³ altogether. OpenStax: areas, volumes and geometric calculations.

The same principle applies to spheres: doubling the diameter multiplies the entire surface area by four and the volume by eight. This is a geometric result about similar shapes. It does not mean we can stretch a real planet while preserving its density, internal structure and other physical properties.

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Where the factors four and eight come from

For a sphere of radius $r$ , the entire surface area is $S=4\pi r^2$ , and the volume is $V=\frac{4}{3}\pi r^3$ . Write the new radius as $r'=qr$ , where $q>0$ . Then:

\begin{aligned} S'&=4\pi(qr)^2,\\ &=q^2S,\\ V'&=\frac{4}{3}\pi(qr)^3,\\ &=q^3V. \end{aligned}

For $q=2$ , we obtain $q^2=4$ and $q^3=8$ . For $q=1/2$ , lengths halve, area falls to a quarter and volume to one eighth of the original. The prime mark denotes the new quantity here. Both spheres have the same shape, and all their lengths have changed uniformly.

Check what changed and what stayed the same

In a correct Earth–Moon pair, divide both diameters by three. Does Moon / Earth become three times smaller? No: it stays the same, approximately 0.273. Both lengths and the common scale change, but the common factor cancels from the ratio.

What if we double only the Moon’s diameter, leaving Earth unchanged? The ratio doubles to approximately 0.545. The pair no longer preserves the real ratio of our chosen diameters. To check this, identify the numerator, the denominator, and which quantity changed.

For your own drawing, use an Earth disk 10 cm across and a Moon disk approximately 2.73 cm across. Then imagine a copy with both lengths halved. Using the original unrounded ratio, the correct diameters are 5 cm and approximately 1.36 cm, rather than 5 cm and an unchanged 2.73 cm. Halving the already rounded 2.73 cm could change the final digit of the result. Choose a convenient gap between disks, but remember that such a drawing compares body sizes without representing their separation to scale.

A proportion helps preserve a specified ratio. Before calculating, name the quantities, match the units, and decide what stays constant. Enlarging a model then becomes an understandable transformation instead of an arbitrary change in numbers. Next in the course, we will explore why squared numbers appear so often in physics.