Spherical Astronomy Problems And Solutions Review

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

Because of the gravitational pull of the Sun and Moon, the Earth’s axis slowly traces a circle every 26,000 years ( Precession ) and exhibits a smaller, faster "nodding" motion (

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 0.8192 0.9848 0.7071=0.5703; sum 0.0995+0.5703=0.6698 → $a=42.07^\circ$. Then $\cos42.07=0.7417$. Then $\cos A = (0.1736 - 0.5736 0.6698)/(0.8192 0.7417) = (0.1736-0.3841)/0.6075 = -0.2105/0.6075 = -0.3465$. $\sin A = (0.7071 0.9848)/0.7417 = 0.6964/0.7417 = 0.9390$. A = 180-69.9=110.1°. Reverse: $\sin\delta' = 0.5736 0.6698 + 0.8192 0.7417 cos110.1 = 0.3841 + 0.6075*(-0.3420) = 0.3841 - 0.2078 = 0.1763 → δ'=10.15°$. Correct. This shows sensitivity to rounding.