Difficulty: Medium
Correct Answer: Horizon and equator
Explanation:
Introduction / Context:
At meridian transit, simple relations connect latitude (φ), declination (δ), and zenith distance (z). Depending on where the star lies (relative to the observer’s zenith, the celestial equator, and the pole), different sign conventions apply. Recognizing the correct geometric case is key in astronomical surveying.
Given Data / Assumptions:
Concept / Approach:
General meridian relations: z = |φ − δ| when the body and the zenith are on the same side of the celestial equator; z = |φ + δ| when they are on opposite sides. Rearranging gives φ = z ± δ, with the sign depending on the body’s position. The case φ = z − δ corresponds to the star lying such that δ reduces the numerical value of z to reach the latitude—geometrically consistent with the star positioned between the horizon and the equator as seen from the observer’s latitude.
Step-by-Step Solution:
Start from z = φ + (−δ) in the opposite-side configuration.Rearrange to φ = z − δ.This matches the situation where the star lies lower in altitude on the equatorial side of the sky (between horizon and equator).
Verification / Alternative check:
Textbook sign charts show that when the body is on the opposite side of the equator from the zenith, declination algebraically subtracts from z in solving for latitude.
Why Other Options Are Wrong:
Zenith and pole / Equator and zenith / Pole and horizon: correspond to other standard cases leading to φ = δ − z, φ = z + δ, or φ = 180° − (z + δ).
Common Pitfalls:
Mixing up which angle is subtracted or added; always sketch the celestial meridian with equator, pole, zenith, and the body to select the correct case.
Final Answer:
Horizon and equator.
Discussion & Comments