The relative sizes and distances of the Earth, Moon and Sun, the Sun's mean diameter, and how Eratosthenes and Aristarchus determined the sizes and distances from observations.

What this dot point is asking

Edexcel statements 3.1 to 3.4 want you to use the relative sizes and distances of the Earth, Moon and Sun, to use the Sun's mean diameter of $1.4 \times 10^6\,\text{km}$, and to understand how Eratosthenes and Aristarchus used observations of the Moon and Sun to determine successively the diameter of the Earth, the diameter of the Moon, the distance to the Moon, the distance to the Sun, and the diameter of the Sun.

Relative sizes and distances

These ratios are used in scale and standard-form calculations and explain key phenomena. A striking coincidence is that the Sun is about 400 times wider than the Moon and about 400 times further away, so the two appear almost the same size in the sky, which is why total solar eclipses just work (Topic 3 eclipses). Always keep the units consistent (kilometres) and be comfortable in standard form.

Eratosthenes and the size of the Earth

This is the classic worked calculation. From an angle of about $7.2$ degrees and a city separation of about $800\,\text{km}$, the circumference comes out near $40000\,\text{km}$, remarkably close to the true value. From the circumference the diameter follows ($C = \pi d$). The method only needs a shadow angle and one known distance, which is why it is celebrated as an early triumph of geometry.

Aristarchus and the Moon and Sun

Aristarchus's geometry was sound even though his measured angles were imprecise (he found the Sun about 19 times further, against the true 400), because the half-Moon angle is so close to $90$ degrees that a small error matters greatly. The key examinable point is the method: the right-angled triangle at half phase for the Sun's distance, and eclipse timing for the relative sizes. These were the first attempts to measure the scale of the system from observation alone.

How Edexcel examines this

This is naked-eye Paper 1 content and a prime source of calculation marks. Expect the Eratosthenes calculation: set the shadow angle over $360$ degrees equal to the city distance over the circumference, then solve, getting about $40000\,\text{km}$, and possibly convert to a diameter with $C = \pi d$. The Aristarchus statements are usually explained rather than calculated: describe the right-angled triangle at half Moon for the Sun's relative distance and the eclipse-timing method for the relative sizes. Scale and ratio questions use the Sun's mean diameter and the AU, often in standard form. A reliable synoptic link is the near-equal angular sizes of the Sun and Moon (about 400 times bigger and 400 times further), which sets up eclipses. The most common errors are swapping the two astronomers and dropping standard form, so attribute each method correctly and keep the units tidy.

Try this

Q1. State what Eratosthenes measured to find the size of the Earth. [1 mark]

• Cue. The angle of the Sun's shadow at two places a known distance apart.

Q2. State why the Sun-Moon-Earth angle is $90$ degrees at first quarter in Aristarchus's method. [1 mark]

• Cue. At half phase the Moon is exactly half lit, so the Sun's light hits it at a right angle to the Earth-Moon line.