Spherical Astronomy Problems And Solutions Guide

Problem: Determine the semi-major axis of a planet's orbit with an eccentricity of 0.5 and a perihelion distance of 1.5 AU.

Solution:

where e is the eccentricity, r_a is the aphelion distance, and r_p is the perihelion distance.

r_a ≈ 1.5 * (1 + 0.5) / (1 - 0.5) ≈ 4.5 AU a ≈ (4.5 + 1.5) / 2 ≈ 3 AU

The semi-major axis of the planet's orbit is approximately 3 AU.

These problems and solutions demonstrate some of the fundamental concepts in spherical astronomy, including celestial coordinates, time and date, parallax and distance, and orbital elements.

Additional Resources

For more practice problems and a deeper understanding of spherical astronomy, I recommend:

By mastering spherical astronomy, you'll gain a deeper understanding of the techniques used to study celestial objects and events, which is essential for a wide range of astronomical applications.

Spherical astronomy is the branch of astronomy that deals with the celestial sphere—a projection of celestial objects onto an imaginary sphere centered on the observer. It is the foundation for determining positions, timekeeping, and navigation. spherical astronomy problems and solutions

This guide covers the essential concepts, formulas, and worked solutions to typical problems.

A celestial body rises when $a = 0^\circ$ (ignoring refraction). From equation (1) with $a=0$:

$$0 = \sin\phi \sin\delta + \cos\phi \cos\delta \cos H$$

$$\cos H = -\tan\phi \tan\delta \tag5$$

Solution exists only if $|\tan\phi \tan\delta| \le 1$.

Hour angle at rising: $H_r = \arccos(-\tan\phi \tan\delta)$ (positive for setting after meridian crossing).
Set $H_s = -H_r$ (for rising before meridian).
Duration above horizon: $2H_r$ in hour angle (convert to hours: $H_r/15$ hours).

Special cases:

Given: Two points on Earth (or celestial sphere) with coordinates $(\phi_1, \lambda_1)$ and $(\phi_2, \lambda_2)$ (latitude/longitude).
Find: Angular distance $\sigma$ (great circle arc) and initial azimuth $\alpha_1$.

The astronomical triangle connects:

Sides:
$PZ = 90^\circ - \phi$ (co-latitude)
$PX = 90^\circ - \delta$ (polar distance)
$ZX = 90^\circ - a$ (zenith distance)

Angle at $P$ = hour angle $H$ (for upper culmination).
Angle at $Z$ = $360^\circ - A$ if azimuth measured from north westward, but conventionally we use $A$ measured from north eastward. We adopt: Angle at Z = $A$ (azimuth) only after careful quadrant check.

Using law of cosines for angle $A$ (at Z):

$$\cos(90^\circ - \delta) = \cos(90^\circ - \phi)\cos(90^\circ - a) + \sin(90^\circ - \phi)\sin(90^\circ - a)\cos A$$

$$\sin \delta = \sin \phi \sin a + \cos \phi \cos a \cos A \tag2$$

Solve for $\cos A$: $$\cos A = \frac\sin \delta - \sin \phi \sin a\cos \phi \cos a$$

Quadrant ambiguity: $\sin A$ from law of sines: $$\frac\sin H\sin(90^\circ - a) = \frac\sin A\sin(90^\circ - \delta) \implies \sin A = \frac\sin H \cos \delta\cos a$$

Use both $\sin A$ and $\cos A$ to determine $A$ in $[0^\circ, 360^\circ)$.

Example: $\phi = 40^\circ$ N, $\delta = 20^\circ$ N, $H = 30^\circ$ (west).
$\sin a = \sin40\sin20 + \cos40\cos20\cos30 \approx 0.2198 + 0.6634 = 0.8832$ → $a \approx 62.0^\circ$.
$\cos A = (\sin20 - \sin40\sin62)/(\cos40\cos62) \approx (0.3420 - 0.588)/0.359 \approx -0.685$, $\sin A = (\sin30 \cos20)/\cos62 \approx (0.5*0.9397)/0.4695 \approx 1.001$ → $A \approx 135^\circ$ (southeast? Wait, sin positive, cos negative → quadrant II → $A \approx 180-43=137^\circ$). So azimuth ≈ 137° from north. Note: These coordinates change constantly as the Earth

Problem: On 2024-10-15 at 4h UT, an observer at (\phi = 35^\circ N), longitude (= 75^\circ W) observes a star with (\alpha = 6h 45m 12s), (\delta = +16^\circ 20'). Find the star’s altitude and azimuth at that moment.

Step 1: Convert UT to LST.
J2000.0 = Jan 1, 2000, 12h UT. Days from J2000.0 to Oct 15, 2024 ≈ 9060 days.
GMST0 = 100.46 + 0.9856479060 = 100.46 + 8929.4 = 9029.86° → mod 360 = 9029.86 – 25360 = 9029.86 – 9000 = 29.86°.
UT = 4h = 60°.
GMST = 29.86° + 60°*1.0027379 ≈ 29.86 + 60.164 = 90.024°.
LST = GMST – longitude (75°W = –75°) = 90.024 – (-75) = 165.024° (or mod 360 = 165.024°).
Star’s RA: 6h45m12s = 6.7533h = 101.3°.
Hour angle H = LST – RA = 165.024° – 101.3° = 63.724°.

Step 2: Compute altitude.
(\phi = 35°), (\delta = 16.333°), (H=63.724°).
(\sin h = \sin35 \sin16.333 + \cos35 \cos16.333 \cos63.724)
= (0.5736)(0.2813) + (0.8192)(0.9596)(0.4423)
= 0.1613 + (0.81920.95960.4423) = 0.1613 + (0.7859*0.4423) = 0.1613 + 0.3476 = 0.5089.
(h = \arcsin(0.5089) = 30.58^\circ).

Step 3: Compute azimuth.
(\sin A = (\cos\delta \sin H) / \cos h = (0.9596 * 0.8960) / 0.8608 = 0.8598 / 0.8608 \approx 0.9988) → (A \approx 86.9^\circ) or 93.1°?
(\cos A = (\sin\delta - \sin\phi \sin h) / (\cos\phi \cos h) = (0.2813 - 0.57360.5089) / (0.81920.8608))
Numerator: 0.2813 – 0.2918 = -0.0105. Denominator: 0.7054.
(\cos A = -0.0149) → (A \approx 90.85^\circ) (since cos slightly negative, sin near 1).
Thus Azimuth ≈ 91° (just east of north? Wait – 91° from north = just west of north? No, 0°=N, 90°=E, 180°=S. 91° is slightly east of north? Mist: 91° is 1° past east? No: 90° = east, so 91° is 1° past east = east-southeast? Let’s check: quadrant – sin positive, cos negative → angle in second quadrant (90–180°), so A = 180 – 89.15 = 90.85°? Actually atan2(0.9988, -0.0149) = 180 – 0.854°? No – atan2 positive y, negative x returns >90 and <180. Value: tan^-1(0.9988/0.0149)=89.15°, so angle = 180-89.15=90.85°. Correct. Thus azimuth = 90.85° from north = just east of north? That’s nearly east. Fine.)

Azimuth ≈ 90.9° (east-north).

From equation (2) rearranged for $\sin \delta$: $$\sin \delta = \sin \phi \sin a + \cos \phi \cos a \cos A \tag3$$

Then from equation (1) rearranged: $$\cos H = \frac\sin a - \sin \phi \sin \delta\cos \phi \cos \delta$$

And $\sin H = \frac\sin A \cos a\cos \delta$ (from law of sines).

Quadrant check for $H$ (0–360°, east negative, west positive in hour angle convention). Problem: Determine the semi-major axis of a planet's