• "الفلك الموضعي"

• مقدمة
• الكرة الأرضية
• هندسة المثلثات الكروية
• ملاحظات حول الهندسة الكروية
• إحداثيات أفقية "alt-az"
• إحداثيات استوائية "HA-dec"
• إحداثيات استوائية "RA-dec"
• الزمن النجمي
• الانتقال بين أفقي و استوائي
• إحداثيات مجرية
• إحداثيات بروجية
• الانتقال بين بروجي و استوائي
• حركة الشمس و ضبط الوقت
• القمر
• الإنكسار الجوي
• شروق و غروب الشمس، الشفق
• اختلاف المنظر المركزي اليومي
• اختلاف المنظر الموسمي
• الزيغ
• المبادرة و الترنح :اضطراب محور الأرض
• تقاويم
• الامتحان الأخير
• الإحداثيات الفلكية، خضر الأحمد
• الحركة الظاهرية للنجوم (تفاعلي)
• الأقطاب السماوية (تفاعلي)

اختلاف المنظر الموسمي (السنوي)

English المواضيع

{Note: If your browser does not distinguish between "a,b" and "α, β" (the Greek letters "alpha, beta") then I am afraid you will not be able to make much sense of the equations on this page.}

Geocentric or diurnal parallax varies with the daily spinning of the Earth around its axis.
Annual parallax is caused by the Earth's yearly orbit around the Sun.

The Earth shifts by 2a from side to side,
where a is the radius of the Earth's orbit (assumed circular) = 1 Astronomical Unit.

For the star S₁, the maximum shift occurs
as the Earth moves from position E₁ to E₂.
If the distance between the Sun and the star S₁ is r,
then we define the annual parallax as Π, where tan(Π) = a/r.
And since a/r is always extremely small,
we may write Π = a/r (in radians).

If the star is not at S₁, but at some other arbitrary position S₂,
then the shift in position as Earth moves from E₁ to E₂ will appear less.
Let the direction from the Sun to the star make an angle θ
with the line E₁E₂ .
But the star appears at angle θ' from Earth at E₁.

By plane trigonometry,
     sin(θ-θ')/a = sin(θ')/r
so
     sin(θ-θ') = sin(θ') a/r
                  = sin(θ') sin(Π)

Since (θ-θ') is a very small angle,
we can replace θ' by θ, and write
     θ-θ' = Π sin(θ)

The apparent shift is towards the Sun,
and it alters the star's ecliptic longitude λ
(so this is another occasion for using ecliptic coordinates).
If the star is not in the plane of the ecliptic,
there is a shift in ecliptic latitude β too.

The star X at (λ, β) is shifted to X' at (λ+Δλ, β+Δβ)
along a great circle arc towards the position of the Sun S.
XX' is the parallactic shift Π sin(θ).
We need to find the shifts Δλ and Δβ.

UX is the arc of a small circle centred on the ecliptic pole K,
passing through the star X.
The length of arc UX is Δλ cos(β) – the shift in longitude.
The length of arc UX' is -Δβ – the shift in latitude.

Consider the tiny triangle UXX' as a plane right-angled triangle,
and denote the angle at X by the arbitrary symbol ψ:
UX = XX' cos(ψ) = Π sin(θ) cos(ψ)
UX' = XX' sin(ψ) = Π sin(θ) sin(ψ)

In other words,
Δλ cos(β) = Π sin(θ) cos(ψ)
Δβ = -Π sin(θ) sin(ψ)                                    (equations 1)

To eliminate θ and ψ from these two equations,
we use the spherical triangle KXS.

First, by the sine rule:
sin(90°+ψ/sin(90°) = sin(λ_S-λ)/sin(θ)
i.e. sin(θ) cos(ψ) = sin(λ_S-λ)                                         (expression 2)
where λ_S is the ecliptic longitude of the Sun.

Then, by the cosine rule:
cos(90°) = cos(θ) cos(90°-β) + sin(θ) sin(90°-β) cos(90°+ψ)
i.e. 0 = cos(θ) sin(β) - sin(θ) cos(β) sin(ψ)
So: sin(θ) sin(ψ) = cos(θ) sin(β)/cos(β)

To get rid of the cos(θ) on the right-hand side of this expression,
we use the cosine rule again:
cos(θ) = cos(90°-β) cos(90°) + sin(90°-β) sin(90°) cos(λ_S-λ)
i.e. cos(θ) = cos(β) cos(λ_S-λ)

Substituting this expression for cos(θ) in the previous equation:
sin(θ) sin(ψ) = cos(β) cos(λ_S-λ) sin(β) /cos(β)
i.e. sin(θ) sin(ψ) = cos(λ_S-λ) sin(β)                                        (expression 3)

Now we can substitute these expressions (2) and (3) in equations (1), to get:

Δλ cos(β) = Π sin(λ_S-λ)
Δβ = Π cos(λ_S-λ) sin(β)

This is actually the formula for an ellipse, of the form:
x = a cos(θ), y = b sin(θ)
where x is the shift parallel to the ecliptic [ Δλ cos(β) ]
y is the shift perpendicular to the ecliptic [ Δβ ]
and θ is temporary shorthand for [ 90° - (λ_S-λ) ].
So we have a = Π and b = Π sin(β)

In other words, this parallactic ellipse has
semi-major axis Π, parallel to the ecliptic,
and semi-minor axis Π sin(β), perpendicular to the ecliptic.

So, during the year,
the star appears to trace out a parallactic ellipse,
which is a reflection of the Earth's orbit.
For a star on the ecliptic (β = 0°) it reduces to a straight line;
for a star at the pole of the ecliptic (β = 90°) it becomes a circle.

The size of a star's parallactic ellipse yields its distance,
in units of parsecs (parallax-seconds):
r (in parsecs) = 1 / Π (in arc-seconds),
so a star at 1 parsec would have parallax Π = 1 arc-second.
(In fact, no star is this close.)