Spherical Astronomy Problems And Solutions

To avoid quadrant ambiguity, use Cartesian vectors on unit sphere:

This yields $a$ and $A$ directly without quadrant checks.

Given: Observer latitude $\phi$, star’s altitude $a$, azimuth $A$.
Find: Declination $\delta$, hour angle $H$.

Problem: At $\phi = 35^\circ$ N, a star has $H = 45^\circ$ west, $\delta = 10^\circ$ N. Compute $a$, $A$, then verify by converting back to $H$ and $\delta$.

Step 1: $a$ from (1):
$\sin a = \sin35\sin10 + \cos35\cos10\cos45 = 0.0996 + 0.5739 = 0.6735$ → $a = 42.34^\circ$.

Step 2: $\cos A = (\sin10 - \sin35\sin42.34)/(\cos35\cos42.34) = (0.1736 - 0.4745)/(0.8192\times0.7390) = -0.3009/0.6055 = -0.4970$.
$\sin A = (\sin45 \cos10)/\cos42.34 = (0.7071\times0.9848)/0.7390 = 0.6964/0.7390 = 0.9425$.
Both sin>0, cos<0 → quadrant II → $A = 180 - \arcsin(0.9425) = 180 - 70.4 = 109.6^\circ$.

Step 3 (reverse): From (3): $\sin\delta' = \sin35\sin42.34 + \cos35\cos42.34\cos109.6 = 0.4745 + 0.6055\times(-0.3338) = 0.4745 - 0.2022 = 0.2723$ → $\delta' = 15.8^\circ$? Mismatch due to rounding? Wait, original $\delta=10^\circ$ — check: my reverse gives 15.8°, so error. Let’s recompute $\cos109.6 = -0.3338$, yes. Then product $0.6055\times(-0.3338) = -0.2021$. Add $0.4745$ → $0.2724$ → $\arcsin = 15.8^\circ$. That’s wrong; original $\delta=10^\circ$. Did I compute $\sin a$ correctly? $\sin35=0.5736, sin10=0.1736 → product 0.0995; cos35=0.8192, cos10=0.9848, cos45=0.7071 → product 0.81920.98480.7071=0.5703; sum 0.0995+0.5703=0.6698 → $a=42.07^\circ$. Then $\cos42.07=0.7417$. Then $\cos A = (0.1736 - 0.57360.6698)/(0.81920.7417) = (0.1736-0.3841)/0.6075 = -0.2105/0.6075 = -0.3465$. $\sin A = (0.70710.9848)/0.7417 = 0.6964/0.7417 = 0.9390$. A = 180-69.9=110.1°. Reverse: $\sin\delta' = 0.57360.6698 + 0.81920.7417cos110.1 = 0.3841 + 0.6075*(-0.3420) = 0.3841 - 0.2078 = 0.1763 → δ'=10.15°$. Correct. This shows sensitivity to rounding.

In the coastal town of Porto Astro, lived an elderly celestial navigator named Elara. For forty years, she had guided ships across featureless oceans using nothing but the stars. Young sailors whispered that she could “read the sky like a love letter.” But Elara knew the sky was not poetry—it was a sphere, and reading it was a matter of solving spherical triangles.

One misty evening, a frantic young captain named Marco burst into her observatory. His ship’s chronometer had broken, and his sextant’s vernier scale was jammed. He was supposed to sail to the island of Cypress Peak at dawn, but the fog would hide the horizon. “Without instruments, I’m lost,” he said. spherical astronomy problems and solutions

Elara smiled. “You’re not lost. You just don’t speak the language of the celestial sphere.” She poured two cups of tea and drew a circle on a chalkboard. “Listen. Spherical astronomy is the geometry of the sky wrapped around the Earth. Every star, every planet, every point of light sits on an imaginary sphere. Our problems are three sides and three angles—curved triangles.”

She presented the first problem:

Problem 1: Finding Latitude from Polaris
Given: Polaris is 2° away from the North Celestial Pole (due to precession). An observer measures the altitude of Polaris as 40° above the horizon. What is the observer’s latitude?
Solution: Elara explained, “On a sphere, the altitude of the celestial pole equals your latitude. But Polaris is not exactly at the pole. So we use the spherical law of sines:
[ \sin(90° - \textlat) = \sin(90° - \textalt) \cdot \sin(90° - 2°) + \cos(90° - \textalt) \cdot \cos(90° - 2°) \cdot \cos(\texthour angle) ]
But since Polaris’ hour angle is near zero when it transits, simplify: Latitude ≈ altitude – 2°·cos(hour angle). At culmination, latitude = 40° – 2°·cos(0) = 38° N.

Marco nodded slowly. “So I can find north without a compass.”

Elara nodded. “Now, your real problem: you need to find the time until sunrise without a chronometer. Let’s try a second problem.”

Problem 2: Hour Angle of Sunrise
Given: Observer at latitude 38° N. Sun’s declination = –10° (winter). Ignoring refraction, find the hour angle at sunrise (when Sun’s center is on the horizon).
Solution: On the celestial sphere, at sunrise, the zenith distance = 90°. Use spherical cosine law:
[ \cos(90°) = \sin(\textlat) \cdot \sin(\textdec) + \cos(\textlat) \cdot \cos(\textdec) \cdot \cos(H) ]
[ 0 = \sin(38°)\sin(-10°) + \cos(38°)\cos(-10°)\cos(H) ]
[ 0 = -0.1056 + 0.7660 \cdot 0.9848 \cdot \cos(H) ]
[ 0.1056 = 0.7541 \cdot \cos(H) ]
[ \cos(H) = 0.1400 \Rightarrow H = \pm 81.95° ]
Sunrise is before noon, so (H = -81.95°) (or 5.46 hours before local solar noon). She looked up: “Sunrise in 5 hr 28 min.”

Marco’s eyes widened. “But without a clock, how do I know when it’s noon?”

Elara laughed. “You measure the Sun’s shadow at its shortest—that’s noon. Now, for the real challenge: you need to sail 120 nautical miles along a great circle to Cypress Peak. But your map shows a rhumb line. The difference is a spherical problem.” To avoid quadrant ambiguity, use Cartesian vectors on

Problem 3: Great Circle Distance
Given: From (38°N, 10°W) to (32°N, 15°W). Radius of Earth = 3440 nautical miles (approx. 1 arcminute = 1 nm). Find great circle distance.
Solution: Spherical law of cosines:
[ \cos(\sigma) = \sin\phi_1\sin\phi_2 + \cos\phi_1\cos\phi_2\cos(\Delta\lambda) ]
[ \cos(\sigma) = \sin38°\sin32° + \cos38°\cos32°\cos(5°) ]
[ = 0.6157\cdot0.5299 + 0.7880\cdot0.8480\cdot0.9962 ]
[ = 0.3261 + 0.6656 = 0.9917 ]
[ \sigma = \arccos(0.9917) = 7.42° \times 60' = 445.2 \text nautical miles ]
“That’s 9% shorter than the rhumb line,” she said.

Marco spent the night solving spherical triangles by lantern light. At dawn, without chronometer or compass, he shot Polaris’ altitude, corrected for precession, found his latitude as 38° N. He watched the Sun climb, marked the shortest shadow for noon, computed the hour angle, and set sail.

Two days later, he sighted Cypress Peak exactly where the great circle track predicted.

When he returned, he brought Elara a gift—a brass armillary sphere. “For teaching me,” he said, “that the sky is not a mystery. It’s a sphere — and every problem has a solution if you know which triangle to solve.”

She placed the sphere in her window, where it caught the starlight. “Remember, Marco: spherical astronomy isn’t about memorizing formulas. It’s about understanding that you live on a curved world beneath a curved sky. The only straight line is the one you draw through the math.”

And from that day on, Porto Astro had two navigators who spoke the language of spheres.

Question: A star has a declination $\delta = -10^\circ$. At what Hour Angle ($H$) does it set for an observer at Latitude $\phi = +40^\circ$?

Concept: When a star rises or sets, its altitude $h = 0^\circ$. Therefore, $\sin h = 0$. This yields $a$ and $A$ directly without quadrant checks

Formula: From the cosine formula, setting $h=0$: $$ 0 = \sin \phi \sin \delta + \cos \phi \cos \delta \cos H $$ $$ \cos H = - \frac\sin \phi \sin \delta\cos \phi \cos \delta $$ Or simplified: $$ \cos H = - \tan \phi \tan \delta $$

Solution: $$ \cos H = - \tan(40^\circ) \tan(-10^\circ) $$

Calculate:

Substitute: $$ \cos H = - (0.839 \times -0.176) $$ $$ \cos H = - (-0.147) $$ $$ \cos H = +0.147 $$

Solve for $H$: $$ H = \arccos(0.147) \approx 81.5^\circ $$

Interpretation: The star sets at Hour Angle $H = 81.5^\circ$. Since $15^\circ = 1$ hour, the star sets $81.5 / 15 \approx 5.43$ hours after it crosses the meridian (Upper Culmination).

Answer: The Hour Angle at setting is $81.5^\circ$ (approx 5 hours 26 minutes).