Almagest Book V: Size of the Sun, Moon, and Earth

Now that we’ve worked out a distance to the sun and moon, as well as the angular diameter, we can put these together to determine the sizes which is what Ptolemy seeks to do in this chapter. Fortunately, we don’t need a new diagram and can simply make reference to the one we used in the last post:

First, we stated that $\overline{NM}$ is the radius of the earth, which we’ll take as our base unit and give a value of $1^p$ as we did in the previous post. In that post, we also showed that $\overline{\Theta H} = 0;17,33^p$ which is the radius of the moon. There, we also determined $\overline{N \Theta}$, the distance from the center of the earth to the center of the moon, was $64;10^p$.

Now consider $\triangle{N \Theta H}$. This is a similar triangle to $\triangle{NDG}$. Thus we can express the ratio of similar sides as equal to one another:

$$\frac{\overline{N \Theta}}{\overline{\Theta H}} = \frac{\overline{ND}}{\overline{DG}}$$

We just stated the value of the two pieces on the left, and also determined $\overline{ND}$ to be $1210^p$ in the last post, so we can solve for $\overline{DG}$. When doing the math, I come up with $\overline{DG} = 5;29^p$ which Ptolemy rounds off to $5 \frac{1}{2}^p$.

If we instead took the radius of the moon as $1^p$, Ptolemy states that we would then have the earth’s radius as $3 \frac{2}{5}^p$ and the sun’s as $18 \frac{4}{3}^p$. Flipping this around:

[T]he earth’s diameter is $3 \frac{2}{5}$ times the moon’s, and the sun’s diameter is $18 \frac{4}{5}$ times the moon’s, and $5 \frac{1}{2}$ time’s the earth’s.

We could easily use these values to determine the relatives volumes of each simply by cubing each of these1. Doing so, we determine:

the earth’s volume is $39 \frac{1}{4}$ [times that of the moon] and the sun’s $6644 \frac{1}{2}$ [and] the sun’s volume is about $170$ times that of the earth.

Toomer notes that this calculation of volumes isn’t actually used anywhere else, but it appears to be something that was simply traditional to do in Greek astronomy, perhaps because if everyone does so, it allows for easy comparison of results.



 

  1. The formula for the volume of a sphere is $\frac{4}{3} \pi r^3$, but the $\frac{4}{3} \pi$ is common to each and would cancel out when looking at the ratio, hence why we can only concern ourselves with the $r^3$.