Kvant Math Problem 784

The length of daylight at latitude $\varphi$ depends on the declination $\delta$ of the star.

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Problem

A spherical planet moves in a circular orbit around a star and rotates about its axis, with the axis of daily rotation tilted relative to the orbital plane by an angle of $\alpha$ (for our Earth $\alpha=66{,}5^\circ$). Determine the dependence of the duration $T$ of the shortest day of the year at a given location on the planet’s surface on the geographic latitude $\varphi$ of that location. (The planet’s angular velocity around its orbit is much smaller than its angular velocity of rotation about its axis.) Write the formula for the function $T=T(\varphi)$ and sketch an approximate graph.

A. P. Savin

Exploration

The length of daylight at latitude $\varphi$ depends on the declination $\delta$ of the star. During the year, $\delta$ varies between $+\beta$ and $-\beta$, where the obliquity relative to the orbital plane is $\beta=90^\circ-\alpha$. For the Earth, $\beta=23.5^\circ$.

The shortest day at a given latitude occurs when the star has the most negative declination for that hemisphere. Thus the problem reduces to finding the daylight duration for a fixed latitude $\varphi$ and declination $\delta=-\beta$.

Consider the celestial sphere. At latitude $\varphi$, the pole of the sky is elevated by $\varphi$ above the horizon. A star of declination $\delta$ moves along a small circle parallel to the celestial equator. Sunrise and sunset occur when that circle intersects the horizon.

Let $H$ be the hour angle at sunrise. In spherical astronomy the horizon condition is

$$\sin h=\sin\varphi\sin\delta+\cos\varphi\cos\delta\cos H,$$

and at sunrise or sunset $h=0$. Hence

$$\cos H=-\tan\varphi\tan\delta.$$

The star remains above the horizon during an angle $2H$ of its daily circle, so the daylight duration is

$$T=\frac{2H}{2\pi}\cdot 24 =\frac{24}{\pi}H.$$

Substituting $\delta=-\beta$ gives

$$T=\frac{24}{\pi}\arccos(\tan\varphi\tan\beta).$$

This formula cannot hold everywhere, because for sufficiently high latitudes there are polar nights. Indeed, if $\tan\varphi\tan\beta>1$, the argument of $\arccos$ exceeds $1$, signaling that the star never rises. The boundary is

$$\varphi=90^\circ-\beta=\alpha.$$

Thus for $|\varphi|\ge\alpha$ in the winter hemisphere, the shortest day has duration $0$.

The point most likely to conceal an error is the transition from the general sunrise formula to the polar-night regime. The limiting latitude must be checked carefully.

Problem Understanding

We are given a planet whose rotation axis forms an angle $\alpha$ with the orbital plane. The star is very far away, and the orbital motion is slow compared with the daily rotation. We must determine, for a point of geographic latitude $\varphi$, the duration $T$ of the shortest day occurring during the year.

This is a Type C problem. We must determine the value of the minimum daylight duration over the annual cycle and express it as a function of latitude.

The core difficulty is handling both ordinary latitudes, where sunrise and sunset occur every day, and high latitudes, where the shortest day becomes a polar night of duration $0$.

Proof Architecture

For $\beta=90^\circ-\alpha$, the star's declination during the year varies between $-\beta$ and $+\beta$; this follows from the geometry of the tilted rotation axis.

For a point of latitude $\varphi$ and stellar declination $\delta$, the sunrise hour angle satisfies $\cos H=-\tan\varphi\tan\delta$; this is the standard horizon condition on the celestial sphere.

The daylight duration for given $\varphi$ and $\delta$ equals $\dfrac{24}{\pi}H$ hours; the star is above the horizon during an angular interval $2H$ of its daily rotation.

For fixed $\varphi$, the daylight duration decreases as $\delta$ decreases; hence the shortest day occurs at $\delta=-\beta$.

If $\tan\varphi\tan\beta\le1$, substitution into the sunrise formula yields the shortest-day duration.

If $\tan\varphi\tan\beta>1$, the star never rises at $\delta=-\beta$, and the shortest-day duration equals $0$.

The most delicate point is the determination of the latitude at which the ordinary formula ceases to apply and polar night begins.

Solution

Let

$$\beta=90^\circ-\alpha.$$

The quantity $\beta$ is the maximal declination of the star with respect to the planet's equatorial plane. During the year the declination $\delta$ varies between

$$-\beta\le\delta\le\beta.$$

Fix a point of geographic latitude $\varphi$. For a given declination $\delta$, let $H$ denote the hour angle of sunrise. In the astronomical triangle formed by the zenith, the celestial pole, and the star, the altitude formula gives

$$\sin h = \sin\varphi,\sin\delta + \cos\varphi,\cos\delta,\cos H.$$

At sunrise and sunset the altitude is $h=0$, hence

$$0 = \sin\varphi,\sin\delta + \cos\varphi,\cos\delta,\cos H.$$

Dividing by $\cos\varphi\cos\delta$ yields

$$\cos H=-\tan\varphi\tan\delta.$$

When sunrise and sunset exist, the star remains above the horizon during an angular interval $2H$ of a full rotation $2\pi$. Therefore the daylight duration is

$$T(\varphi,\delta) = 24\cdot\frac{2H}{2\pi} = \frac{24}{\pi}H = \frac{24}{\pi} \arccos(-\tan\varphi\tan\delta).$$

For fixed $\varphi$, the shortest day of the year corresponds to the smallest declination, namely

$$\delta=-\beta.$$

Substituting gives

$$T(\varphi) = \frac{24}{\pi} \arccos(\tan\varphi\tan\beta), \qquad \beta=90^\circ-\alpha,$$

provided sunrise and sunset exist.

The condition for existence is

$$\tan\varphi\tan\beta\le1.$$

Since

$$\tan\alpha=\cot\beta,$$

this condition is equivalent to

$$|\varphi|\le\alpha.$$

At the boundary $\varphi=\alpha$,

$$\tan\varphi\tan\beta=1,$$

hence $H=0$ and $T=0$.

For latitudes satisfying $|\varphi|>\alpha$, at the winter solstice the star remains below the horizon throughout the day. The shortest day is then a polar night, so

$$T(\varphi)=0.$$

Using $\tan(90^\circ-\alpha)=\cot\alpha$, the final formula may be written as

$$\boxed{ T(\varphi)= \begin{cases} \dfrac{24}{\pi}, \arccos!\left(\dfrac{\tan\varphi}{\tan\alpha}\right), & |\varphi|\le\alpha, \[2ex] 0, & |\varphi|\ge\alpha. \end{cases} }$$

For the Earth, $\alpha=66.5^\circ$. The graph is symmetric with respect to the equator. It passes through

$$T(0)=12\ \text{hours},$$

decreases monotonically from $12$ hours to $0$ as $|\varphi|$ increases from $0$ to $\alpha$, and remains equal to $0$ for $|\varphi|\ge\alpha$.

Equality $T=12$ holds at the equator, and equality $T=0$ holds for all latitudes in the polar zones.

$$\boxed{ T(\varphi)= \begin{cases} \dfrac{24}{\pi}, \arccos!\left(\dfrac{\tan\varphi}{\tan\alpha}\right), & |\varphi|\le\alpha, \[2ex] 0, & |\varphi|\ge\alpha. \end{cases} }$$

Verification of Key Steps

Starting from

$$0=\sin\varphi\sin\delta+\cos\varphi\cos\delta\cos H,$$

with $\delta=-\beta$, we obtain

$$\cos H = \frac{\sin\varphi\sin\beta} {\cos\varphi\cos\beta} = \tan\varphi\tan\beta.$$

The sign is positive because $\sin(-\beta)=-\sin\beta$. A sign mistake here would predict the shortest day to be longer than $12$ hours at positive latitudes, which is impossible.

The polar-night threshold follows from

$$\tan\varphi\tan\beta=1.$$

Since $\tan\alpha=\cot\beta$,

$$\tan\varphi=\tan\alpha,$$

hence $\varphi=\alpha$ in the range $0^\circ<\varphi<90^\circ$. At this latitude the formula gives $H=0$, so the star merely grazes the horizon at noon of the winter solstice. Larger latitudes must have $T=0$.

Numerical checks confirm the behavior. At the equator,

$$T=\frac{24}{\pi}\arccos(0)=12\ \text{h}.$$

At $\varphi=\alpha$,

$$T=\frac{24}{\pi}\arccos(1)=0.$$

Both values agree with the geometric interpretation.

Alternative Approaches

One may work entirely in Euclidean geometry. At the winter solstice the direction to the star makes an angle $\beta$ with the equatorial plane. For a point of latitude $\varphi$, the daily trajectory is a circle parallel to the equator. Intersecting this circle with the illuminated hemisphere leads directly to the relation

$$\cos H=\tan\varphi\tan\beta.$$

The remainder of the derivation is unchanged.

The celestial-sphere approach is preferable because the sunrise condition is expressed immediately through the standard altitude formula, and the transition to polar night appears naturally from the condition $|\cos H|\le1$.