Solution

(a) Let $N$ be the north celestial pole and $d$ be our desired distance. Then, the spherical law of cosines in $\triangle N\gamma\chi$ gives

which implies $d=\arccos(\cos\delta\cos\alpha)\approx17.96^\circ$.

(b) Since $\chi$ is to his east, it has not yet culminated. Also, since he is not moving, $\chi$ achieves its minimum zenith distance as it culminates.

One way to figure out this minimum zenith distance is to consider the view from his horizon. $\chi$ moves through the small circle representing the declination line at $\delta$. Since he is located on the equator, the celestial equator from his perspective passes through the zenith. Therefore, the minimum zenith distance is the distance between the declination line at $\delta$, and the celestial equator, which is $\delta=10^\circ$.

An alternative approach is to consider his movement through the celestial sphere. Culmination occurs when his location and $\chi$ have the same right ascension. At this point, their distance is $\delta$, so this is the minimum zenith distance.

(c) $\chi$ starts in the northeastern quadrant of the sky. When $\chi$ crosses the north-south meridian, it changes from the east to west hemisphere. When $\chi$ crosses the the east-west meridian, it changes from the north to south hemisphere.

At time $t$, his location in the celestial sphere is given by $(\omega t,\omega t/4)$, so $(\alpha,\alpha/4)$ is a point in his trajectory. Since $\alpha/4<\delta$, $\chi$ culminates in his northern sky. Thus, $\chi$ passes the north-south meridian before passing the east-west meridian.

(d) $\chi$ culminates when it has the same right ascension as Andrew, so $\omega t=\alpha$. So, the zenith distance is given by

(e) Let $D$ be his position in the celestial sphere. When $\chi$ first crosses the east-west meridian, $D$ must satisfy $\angle ND\chi=90^\circ$, as shown in the diagram below. The red line in the diagram represents his trajectory in the celestial sphere (this trajectory is not a great circle, so we cannot use it in our calculations).

Now, chasing angles:

• The perpendicular condition implies that $D$ is the apex of another great circle (call $\Gamma$) that passes through both $D$ and $\chi$. $\Gamma$ intersects the equator at a point $A$, such that $DA=90^\circ$.
• As labelled in the diagram, let $B$ and $C$ be the feet of $\chi$ and $D$ from $N$, respectively (so that $\angle\chi BA=\angle DCA=90^\circ$). Since $D$ is the apex of $\Gamma$, $DA=CA=90^\circ$.
• We know $\gamma B=\alpha$ and $\gamma C=\omega t$, so $AB=90^\circ+\alpha-\omega t$.
• We also know that $\chi B=\delta$, and $DC=\omega t/4$.

Our goal is to find the zenith distance, which is $\chi D$. To find this arc, we'll need $\omega t$, which will let us solve for the right ascension of $D$. Let $\varepsilon=\angle\chi AB=\angle DAC$. The four-parts formula in both $\triangle\chi AB$ and $\triangle DAC$ gives

Substituting known values, this reduces to the equation

We can use this to numerically iterate for $\omega t$. Our initial guess does not need to be good; any value from around $10^\circ$ to $70^\circ$ converges in a few iterations, giving us $\omega t\approx46.97^\circ$.

Finally, a few more angles:

• $\angle\chi ND=\angle BNC=\omega t-\alpha$.
• $N\chi=90^\circ-\delta$.
• $ND=90^\circ-\omega t/4$.

These are all known values. The spherical law of cosines in $\triangle N\chi D$ now gives

(f) The zenith distance $\chi X$ is minimized when $\angle\chi X\gamma=90^\circ$. Let $X'$ be the foot of $X$ from $N$, so that $\angle NX'\gamma=90^\circ$. Then, if we let $\beta=\angle NX\chi$:

• $N,X,X'$ lie on the same great circle, $\angle NX\chi+\angle\chi X\gamma+\angle\gamma XX'=180^\circ$, which implies $\angle\gamma XX'=90-\beta$.
• Since the latitude of $X$ is small, and $\triangle XX'\gamma$ is right, we can say that $\angle\gamma XX'+\angle X\gamma X'\approx90^\circ$, so $\angle X\gamma X'\approx\beta$. This approximation is valid since $\gamma X'=\omega t=4X'X$, so $\triangle XX'\gamma$ can be approximated by a flat right triangle.
• Since $\angle N\gamma X'=90^\circ$, $\beta$ and $\angle N\gamma X$ add to $90^\circ$, as desired.

To compute $\beta$, we can use our approximation of $\triangle XX'\gamma$ as a flat right triangle to say that $\tan\beta\approx1/4$, so $\beta\approx14^\circ$.

(g) Consider $\triangle NX\chi$. We know the lengths of two sides, $N\chi=90^\circ-\delta$ and $NX=90^\circ-\omega t/4$, and the values of a nested and adjacent angle, $\angle N=\omega t-\alpha$ and $\angle X=\beta$. We want to find the length of the zenith angle, which is $X\chi$. Unfortunately, we can't directly solve for $X\chi$, but we may first try solving for $\omega t$ using the information we already have, which will give us enough information about the triangle to solve for $X\chi$.

Applying the four parts formula:

As in part (e), we can isolate $\omega t$ and try to solve this equation with numerical iteration:

Unfortunately, this diverges for most values. Luckily, we're given that $\omega t$ is within $(\alpha\pm2^\circ)$, which is a good enough range for us to still find a solution at $\omega t\approx16.45^\circ$.

Finally, the law of sines in $\triangle X\chi N$ gives

(h) In order for $\chi$ to be directly overhead, he needs to have the same declination and right ascension as $\chi$.

The amount of time he needs to increase his latitude to $\delta$ is bounded below by

which is when he travels straight north at all times.

On the other hand, the amount of time it takes him to have the same right ascension is bounded above by

which is when he travels straight west at all times. Since $4\alpha/3<4\delta$, he cannot get $\chi$ to his zenith when it first culminates.

(i) From part (h), he cannot get $\chi$ directly overhead the first time it culminates. Therefore, his best strategy is to try to get $\chi$ to his zenith the next time it appears above his horizon. After time $t$, the change in his right ascension due to the rotation of the earth is $\omega t$. Therefore, to optimize his distance travelled, he should travel along a great circle (defined on the surface of the Earth) that increases his longitude by $(2\pi+\alpha-\omega t)$, and increases his latitude by $\delta$.

Let $N$ be the north celestial pole, and $Y$ be his stopping point, which has coordinates $(2\pi+\alpha-\omega t,\delta)$. Note that the coordinates of $Y$ are his longitude and latitude on Earth, and not in the celestial sphere. Also, $Y\gamma=\omega t/4$, since the maximum speed his boat can travel is $\omega R_\oplus/4$.

Now, the spherical law of cosines in $\triangle NY\gamma$ gives

so

Now we can numerically iterate. Using $t=20\mathrm{hrs}$ as a first guess, we get $\omega t=300.2^\circ$, which gives $(t-20\mathrm{hrs})\approx48\mathrm s$. To account for approximation errors, a large range of values in this general vicinity are acceptable.