Difficulty: Medium
Correct Answer: 30°
Explanation:
Introduction / Context:
This problem checks core celestial-sphere relations used in astronomical surveying. When a star crosses the local meridian it has two transits each sidereal day: the upper (higher) culmination and the lower (opposite) culmination. For circumpolar stars (those that never set), both culminations are above the horizon, and their altitudes are determined by simple formulas that involve the observer’s latitude and the star’s declination.
Given Data / Assumptions:
Concept / Approach:
For an observer at latitude φ (north positive) and a star of declination δ (north positive), the meridian transit altitudes are:
Upper culmination (on the same side of the equator as the observer): h_upper = 90° − |φ − δ|.
Lower culmination (on the opposite side around the pole for a circumpolar star): h_lower = φ + δ − 90°.
If h_lower > 0°, the star is circumpolar for that observer. If h_lower ≤ 0°, the lower culmination occurs below the horizon.
Step-by-Step Solution:
Verification / Alternative check:
Check circumpolar condition: a star is circumpolar if δ > 90° − φ. Here, 60° > 30°, so it is circumpolar. A positive h_lower (30°) is consistent with this condition.
Why Other Options Are Wrong:
Common Pitfalls:
Using h_lower = 90° − φ − δ blindly (which gives −30° here) without recognizing that for circumpolar stars the proper expression is h_lower = φ + δ − 90°. Also, mixing sign conventions for declination and latitude can lead to errors.
Final Answer:
30°
Discussion & Comments