Castaway

If days are longer than nights in the southern hemisphere, the Sun is south of the celestial equator, $\delta <0$. It then rises south of east and sets south of west, symmetrically about the meridian. The 120° is the angle across the south, so each point is $(180°-120°)/2=30°$ south of east or west. The azimuth of sunrise is $A=120°$.

(a) With $\cos A=\sin \delta /\cos \phi$,

The Sun's declination can be anywhere from 0 to −23.44°, so $\cos \phi \le 2\sin 23.44°=0.796$, which gives $|\phi |\ge 37.3°$. The island can be anywhere from 37.3° S to very close to the South Pole.

(b) The angle is growing, so sunrise is moving back towards the east and $|\delta |$ is shrinking. The Sun is therefore between the December solstice and the March equinox, at an ecliptic longitude ${\lambda }_0$ between 270° and 360°. In 40 days it moves $40\times 360°/365.24=39.43°$ along the ecliptic. The new angle of 163° puts sunrise 8.5° south of east.

The Sun's declination follows from its longitude by $\sin \delta =\sin \epsilon \sin \lambda$ (Section 1.6), with $\epsilon =23.44°$. The two observations give

Dividing one by the other removes both $\epsilon$ and $\phi$, which leaves

Expanding the sine of the sum gives

so $\cot {\lambda }_0=-0.751$ and ${\lambda }_0=306.9°$. The Sun's declination on the first day was therefore ${\delta }_0=-18.55°$, and from step 2

so $\phi =50.5°$ S.