Spherical Astronomy Problems And Solutions [upd] Page

The Sun sets at approximately 7:28 PM Local Apparent Time. Problem 3: Finding the Angular Distance Between Two Stars The Challenge: Star A is at RA 5h5 to the h-th power +10∘positive 10 raised to the composed with power . Star B is at RA 7h7 to the h-th power +25∘positive 25 raised to the composed with power . What is the angular separation ( ) between them? The Solution: Step 1: Calculate the difference in Right Ascension (

$\phi = 40^\circ$ 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.

cos(H)=sin(a)−sin(ϕ)sin(δ)cos(ϕ)cos(δ)cosine open paren cap H close paren equals the fraction with numerator sine a minus sine open paren phi close paren sine open paren delta close paren and denominator cosine open paren phi close paren cosine open paren delta close paren end-fraction is greater than or less than -1negative 1

) must be treated as negative numbers in trigonometric functions. Functions like arcsinarc sine arccosarc cosine spherical astronomy problems and solutions

) during the summer solstice, when the Sun's Declination reaches

cos(90∘−δ)=cos(90∘−ϕ)cos(90∘−a)+sin(90∘−ϕ)sin(90∘−a)cos(360∘−A)cosine open paren 90 raised to the composed with power minus delta close paren equals cosine open paren 90 raised to the composed with power minus phi close paren cosine open paren 90 raised to the composed with power minus a close paren plus sine open paren 90 raised to the composed with power minus phi close paren sine open paren 90 raised to the composed with power minus a close paren cosine open paren 360 raised to the composed with power minus cap A close paren

θ=arccos(0.8824)≈28.07∘theta equals arc cosine 0.8824 is approximately equal to 28.07 raised to the composed with power The angular separation between Star A and Star B is 28.07∘28.07 raised to the composed with power (or Problem 3: Finding the Duration of Daylight The Sun sets at approximately 7:28 PM Local Apparent Time

At $\phi = 35^\circ$ N, a star has $H = 45^\circ$ west, $\delta = 10^\circ$ N. Compute $a$, $A$, then verify by converting back to $H$ and $\delta$.

These two stars will have the same hour angle only if they are the same star at different altitudes, which would be impossible. Instead, we have two different stars with different declinations.

cosθ=sin(-12.5∘)sin(-11.17∘)+cos(-12.5∘)cos(-11.17∘)cos(12.5∘)cosine theta equals sine open paren negative 12.5 raised to the composed with power close paren sine open paren negative 11.17 raised to the composed with power close paren plus cosine open paren negative 12.5 raised to the composed with power close paren cosine open paren negative 11.17 raised to the composed with power close paren cosine open paren 12.5 raised to the composed with power close paren What is the angular separation ( ) between them

): The angular distance measured eastward along the horizon, typically starting from true North ( 0∘0 raised to the composed with power ) or true South ( 180∘180 raised to the composed with power Equatorial System

cosH=−tan(52∘)tan(35∘)cosine cap H equals negative tangent open paren 52 raised to the composed with power close paren tangent open paren 35 raised to the composed with power close paren Compute the tangents: Multiply the values: Calculate the inverse cosine to find

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

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