Spherical Astronomy Problems And Solutions |top| Jun 2026

To solve almost any problem in this field, you need these three identities: The Sine Rule: The Four-Parts Formula: (Where are the sides—measured as angles—and are the opposite angles.) Problem 1: Converting Horizontal to Equatorial Coordinates The Challenge: An observer in London (Latitude N) observes a star at an altitude ( 40∘40 raised to the composed with power and an azimuth ( 120∘120 raised to the composed with power

cosθ=sinδ1sinδ2+cosδ1cosδ2cos(Δα)cosine theta equals sine delta sub 1 sine delta sub 2 plus cosine delta sub 1 cosine delta sub 2 cosine open paren cap delta alpha close paren

Or directly: $$\cos\sigma = \sin\phi_1\sin\phi_2 + \cos\phi_1\cos\phi_2\cos(\Delta\lambda) \tag4$$

: Circles formed by intersecting a sphere with a plane passing through its center (e.g., the celestial equator). They represent the shortest distance between two points on a sphere. spherical astronomy problems and solutions

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

$$\cos C = -\cos A \cos B + \sin A \sin B \cos c$$

$\phi = 40^\circ$ 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. To solve almost any problem in this field,

Always compute at least two functions (sin and cos) of the unknown angle and use atan2.

ϕ≥90∘−δphi is greater than or equal to 90 raised to the composed with power minus delta

(\phi = 50°N), (\delta = +20°). (\tan50 = 1.1918), (\tan20 = 0.3640) → product = 0.4336. Negate: -0.4336. (\arccos(-0.4336) = 115.7°) = 7.714 hours. Thus, star rises (H) hours before meridian transit? Wait: For rising, (H) is negative in the usual sense (east of meridian). But here (H_set = +115.7°) (since cos is symmetric). More standard: (H_rise = -\arccos(-\tan\phi\tan\delta)). Then time between rise and meridian = (|H_rise|/15) hours. $\sin a = \sin40\sin20 + \cos40\cos20\cos30 \approx 0

Unlike planar triangles, the sides of a spherical triangle are angular distances (arcs of great circles), and the sum of its angles always exceeds 180∘180 raised to the composed with power

The Sun sets (or rises) at a local hour angle of (or 138.86∘138.86 raised to the composed with power