# 2.7 Angular Size and Linear Size

Why is it so difficult to figure out the sizes and distances of celestial objects? Part of the reason is psychological. Humans have tended to regard themselves as the pinnacle of creation and the center of the universe. So there is resistance to the notion that the sky may contain objects much larger than the Earth. Throughout history, we have slowly but steadily realized that the Earth is only a tiny part of a vast universe. The other reason is practical. As the Greeks knew, it is almost impossible to measure the linear size of a distant unfamiliar object directly by eye. We can really only specify its angular size.

Linear size and linear distance are difficult to judge. People sometimes report an unfamiliar object in the sky and say something like "It looked as big as a dinner plate," but this statement conveys almost no useful information. When a bright meteor or fireball is seen in the sky, astronomers often receive reports that "It looked close; it landed just over the hill." These reports are almost always wrong. Fireballs are typically in the upper atmosphere, 60 to 100 miles from the observers. To use angular measurement and say the fireball looked ½° across, the same size as the Moon" would be correct; but to unconsciously convert to linear measurement and say "It was as big as an airplane" is likely to be wrong if the object itself is unfamiliar.

Schematic for calculating the parallax of a star. Click here for original source URL.

The ancient Greeks invented trigonometry, which allows quantitative relationships between angles, linear sizes, and linear distances. Through simple mathematical relations, involving basic trigonometry, we can calculate distances of remote objects whose sizes are known (or sizes, if the distances are known).

Angles and linear measures can be combined in an extremely useful and simple equation called the small-angle equation. This equation involves the angular size of an object, its linear size, and its distance. If any two of these quantities are known, the third can be calculated. Let us refer to the angular size with the symbol a, expressed in seconds of arc. Let us refer to the diameter of the object as d and its distance as D. Then the small-angle equation is

a / 206,265 = d / D

A chart for the conversion between degrees and radians, along with the signs of the major trigonometric functions in each quadrant. Click here for original source URL.

The number 206,265 is called a constant of proportionality; it stays the same in all applications of the equation. The number 206,265 is actually the number of seconds of arc in an angle of 57.3°, which is a special angle called a radian. A radian is defined as an angle subtending one radius of a circle, laid along the circumference of the circle. Since the circumference of a circle is 2πr, a radian is 360° / 2 π =57.3° or about a sixth of a full circle. It is an important angle with many applications in geometry.

Here is an example of the use of the small angle equation. Suppose a friend who is 2 meters tall is standing across a field from you, where he or she subtends an angle of ½°, or 1800". How far away are they? We want to solve the equation for D. Rearranging the equation, we have

D = 206,265 d / a

Using metric units, we would have

D = (2.1 x 10^{5} x 2) / 1.8 x 10^{3} = 2.3 x 10^{2} meters = 230 meters

If your friend is 2 meters tall and subtends an angle of ½° (or 1800 arc seconds), their distance D is 230 meters. In other words, your friend is about one third of a mile away. Notice we are rounding off all our estimates to two significant figures because the angle measurement is not likely to be very accurate.

As the Greeks realized, the small-angle equation can be used to investigate astronomical distances. They could not measure the Moon's diameter accurately, but they knew its angular size a, which is also roughly ½°, or 1800". If we use the modern knowledge that the Moon is about 3500 kilometers in diameter, we can estimate its distance just as we did for the friend's distance above. In metric units, d would be 3.5 × 10^{6} meters. The equation would read

D = (2.1 × 10^{5} × 3.5 × 10^{6}) / (1.8 × 10^{3}) ? 4 x 10^{8} meters ? 4 x 10^{5} kilometers

This is about 400,000 kilometers. Notice once again the symbol "?" meaning "approximately equal to." It is useful whenever approximate values such as 1° are involved. In other words the measurement of angular size is approximate, so the resulting estimate of distance must also be approximate.

Here is another example of the use of the small angle equation: How big are the smallest craters that we can see on the Moon with a backyard telescope? To solve this we start with the information that a good backyard telescope can resolve or make out angular detail as small as one second of arc, or 1". The moon is 384,000 kilometers away. So we are asking how big is an object that subtends 1" at that distance. Rearranging the small angle equation again

d = aD / 206,265

Plugging in the Moon's distance and the angle of one second of arc

d = (1 x 3.8 x 10^{5}) / (2.1 x 10^{5}) = 1.8 kilometers

We can see features as small as a mile across on the Moon with a telescope. The eye can only make out angular scales of about 3 minutes of arc. So when you stare at the Moon, the smallest features you can see are 3 × 60 × 1.8 ? 320 kilometers or about 200 miles across.

Finally, if the Sun has a diameter of about 1.4 million kilometers and is 150 million kilometers away (also called one Astronomical Unit or 1 A.U.), what angle does the Sun subtend in the sky? We need to solve the small-angle equation for a, so rearranging it one last time

a = (d / D) x 206,265

Substituting the values for diameter d and distance D, we get

a = (1.4 x 10^{6} / 1.5 x 10^{8}) x (2.1 x 10^{5}) = 1960"

Since there are 3600" in one degree, this angle is 1960 / 3600 = 0.54°. The Sun and the Moon differ greatly in size and in distance from the Earth. It is because of the coincidence that they both subtend the same angle of ½° that eclipses are possible.