How can we draw the Solar System to scale?

Draw a Sun 10 cm across, then add Earth. How much space should separate them: a handspan, a page or a whole room? If sizes and distances really shrink by the same factor, Earth has a diameter of just 0.916 mm, and its orbital reference lies about 10.75 m from the Sun’s centre. Neptune needs roughly 323 m.

One scale means one factor for every length: body diameters, distances between centres and orbit sizes. Enlarging only the planet circles so that we can see them produces a useful schematic, but it no longer shares one scale for sizes and distances. NASA/JPL: building a model with a shared scale.

Choose one ruler first

Start with our 10 cm Sun. For the calculation, use a solar diameter of 1,391,400 km: twice the IAU nominal solar radius. Here, “nominal” means an agreed exact conversion standard close to the Sun’s physical size; it does not mean observations have determined its diameter with no uncertainty. Prša and colleagues: the reasoning behind IAU 2015 Resolution B3.

The real Sun must shrink by about 13.9 billion times. Divide every other length by that same number. Earth’s diameter then becomes a dot less than a millimetre across, while the enormous Earth–Sun gap becomes a little over ten metres. The ratio of distance to Earth’s diameter stays the same: about 11,740 Earth diameters for the 1 au reference.

The astronomical unit, au, is exactly 149,597,870,700 m, or 149,597,870.7 km. It is close to the characteristic scale of Earth’s orbit. It is a unit of length, not a promise that Earth always stays the same distance from the Sun. JPL: definition of au.

The table below uses the same kind of size for every planet: the diameter of a sphere with the planet’s volume. We obtain it by doubling the volumetric mean radius in JPL’s table. This consistently compares both nearly spherical Earth and the noticeably flattened giants. Saturn’s rings are excluded from its diameter. JPL: planetary sizes and column definitions.

A small Sun needs a long path

We can now plan a walk. Place the model Sun at the start of a straight measuring line. Measure the planets’ orbital scales from this origin: roughly 4.16 m for Mercury, 10.75 m for Earth and 55.94 m for Jupiter. Neptune lies at 323.30 m, although its little sphere is only 3.54 mm across.

Distances are measured from the Sun’s centre, rather than starting again at each preceding planet. Our orbital reference is the rounded semimajor axis: half the longest axis of the orbital ellipse. It describes orbit size, rather than the current distance to a planet. This is a ruler comparing orbit sizes, not a photograph of the Solar System: the planets do not stand along this path at one instant. JPL: orbital elements.

A 10 cm Sun: one model for every length

The bars are rulers of orbital scale, not planet disks. Each starts at the Sun’s centre; full width represents Neptune.

Planet	Diameter, mm	Orbital scale, m
Mercury	0.35	4.16
Venus	0.87	7.77
Earth	0.92	10.75
Mars	0.49	16.39
Jupiter	10.05	55.94
Saturn	8.37	102.54
Uranus	3.65	206.31
Neptune	3.54	323.30

Diameters and numerical distances use one physical scale. Orbital anchors are rounded semimajor axes, not positions on a date. Diameters use volumetric mean radii; rings are excluded. The Sun uses the IAU nominal diameter of 1,391,400 km. Sources: JPL · r, JPL · a, IAU · 2015 B3.

The table shows why the inner planets appear “crowded” on an honest linear scale: even the entire route to Mars takes up only about five percent of the route to Neptune. Much of the model’s length falls in the gaps between outer orbital references. This follows from their ratios; choosing another shared scale will not make those fractions equal.

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Getting 10.75 m without losing zeros

Let $k$ be the ratio of model length to real length. Both lengths must use the same units when calculating this ratio. One kilometre contains 100,000 cm, so the solar diameter is $1.3914\times10^{11}$ cm.

\begin{aligned} k&=\frac{10}{1.3914\times10^{11}},\\ k&\approx7.187\times10^{-11}. \end{aligned}

Centimetres have cancelled. For any length $L$ , the corresponding model length is $l=kL$ . One au equals $1.495978707\times10^{13}$ cm. For Earth’s orbital reference:

\begin{aligned} l&=kL,\\ l&\approx1075.16\ \text{cm},\\ l&\approx10.75\ \text{m}. \end{aligned}

Divide centimetres by 100 to get metres. For Neptune, multiply the unrounded model length of one au by our adopted 30.070 au, giving about 323.30 m. The same $k$ applies to Earth’s diameter; a shared scale has no separate “planet factor”.

On a page, the planets nearly disappear

Try fitting the Sun-to-Neptune-reference distance onto 30 cm of paper. Our previous model must shrink by another factor of roughly 1,078. The Sun becomes a dot about 0.093 mm across, and Earth just 0.85 micrometres, or 0.00085 mm. Earth will not look like a noticeable circle in that drawing.

Keeping visible disks and convenient equal gaps instead can communicate planetary order or appearance well. But that drawing can no longer be used to measure real length ratios. If distances are linear while planets are enlarged, the caption needs to say so immediately. Uneven compression of distances or a logarithmic scale also needs an explicit label.

That is why our table has no large circles at the ends of the bars. The bars show only orbital scale; diameters appear as numbers. On a physical walking model, a large name sign can help you find a tiny planet. The sign serves as a pointer, rather than an enlarged body presented as part of the exact scale.

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Why an orbital reference is not a planet’s address

In a simple Keplerian description, an orbit is an ellipse with the Sun at one focus. The semimajor axis is half the longest axis of that ellipse. It describes its size without specifying the Sun–planet distance at every instant. In a circular idealization it would equal the radius; real planetary orbits must not silently be replaced with circles. JPL: geometric definition of the semimajor axis.

Our ruler uses rounded semimajor-axis values from the initial coefficients in JPL’s table for epoch J2000. For instance, Jupiter uses 5.203 au and Neptune 30.070 au. The source’s Earth row represents the Earth–Moon centre of mass; here it is rounded to the teaching reference of 1 au. That is adequate for our coarse scale, but not for precise telescope pointing. JPL: table and limits of approximate elements.

Drawing positions for a particular date also requires the phase of motion, the orbit’s orientation and inclination, and a consistent coordinate system. A straight path does not reproduce these. It honestly answers a narrower question: how do body sizes and characteristic orbit sizes compare when every length shrinks equally?

A good scale is easy to check

Before drawing, decide what needs to fit: a Sun of a chosen diameter, or the farthest orbital reference within a chosen length. Use that anchor to find one factor. Apply it to every diameter and distance; only then decide how to mark objects that are too small. A number or a callout preserves information where a true disk cannot be seen.

Check your understanding: we increase the Sun from 10 to 20 cm, preserving one shared scale. What happens to Earth and the route to Neptune’s orbital reference? Earth’s diameter doubles to about 1.83 mm, and the route to 647 m. Doubling only the planets would break the shared scale.

We can therefore draw the Solar System to scale by shrinking every length equally and identifying which orbital quantities we show. The difficulty is that a compact drawing leaves bodies nearly invisible, while visible bodies require a long model. Return to what belongs to the Solar System and try identifying which familiar illustrations show order, which show sizes, and which really preserve distances.