Calculate the ratio of the mean densities of the earth and the sun from the following approximate data:

$\theta=$ angular diameter of the sun seen from the earth $=\frac{1}{2}^{\circ}$.

$l=$ length of $1^{\circ}$ of latitude on the earth's surface $=100 \mathrm{~km}$.

$t=$ one year $=3 \times 10^{7} \mathrm{~s}$.

$g=10 \mathrm{~ms}^{-2}$.
Let $r$ be the distance between the sun and the earth, $M_{e}$ and $M_{s}$ be the masses and $R_{e}$ and $R_{s}$ be the radii of the earth and the sun respectively, and $G$ be the gravitational constant. We then have

$$
\begin{aligned}
&\frac{G M_{e} M_{s}}{r^{2}}=M_{e} r \omega^{2}, \\
&\frac{2 R_{s}}{r}=\frac{1}{2} \frac{2 \pi}{360}=\frac{\pi}{360} \mathrm{rad},
\end{aligned}
$$

i.e.

$$
r=\frac{720 R_{s}}{\pi} .
$$

The above gives

$$
\frac{G M_{s}}{\left(720 R_{s} / \pi\right)^{3}}=\omega^{2}
$$

or

$$
\frac{G M_{s}}{R_{s}^{3}}=\left(\frac{720}{\pi}\right)^{3}\left(\frac{2 \pi}{3 \times 10^{7}}\right)^{2} .
$$

For a mass $m$ on the earth's surface,

$$
\frac{G m M_{e}}{R_{e}^{2}}=m g,
$$

giving

$$
\frac{G M_{e}}{R_{e}^{3}}=\frac{g}{R_{e}}=\frac{g}{\left(\frac{360 \times 100}{2 \pi}\right)}=\frac{g \pi}{18 \times 10^{3}}
$$

Hence

$$
\frac{\rho_{e}}{\rho_{s}}=\frac{g \pi}{18 \times 10^{3}}\left(\frac{720}{\pi}\right)^{-3}\left(\frac{2 \pi}{3 \times 10^{7}}\right)^{-2}=3.31
$$