A Guide to Spherical Astronomy: Olympiad Edition

Great Circle and Small Circle

A great circle is a circle on the surface of a sphere that has:

  • The same radius as the sphere itself.
  • A center that coincides with the center of the sphere (meaning the plane defining the circle passes through the sphere’s midpoint).

A small circle is a circle on the surface of a sphere that has:

  • A radius smaller than the sphere’s radius.
  • A center that does not coincide with the center of the sphere. 

Spherical Triangle

A spherical triangle is formed by the intersection of three great circles on the surface of a sphere. The sides of these triangles are arcs of great circles, and the angles between these arcs are the angles of the spherical triangle. Unlike planar triangles, the sum of the angles of a spherical triangle is greater than 180 degrees and less than 540 degrees. 

Spherical Triangle Formulas

Law of cosines

Cos(a) = Cos(b) Cos(c) + Sin(b) Sin(c) Cos(A)

 

Law of Sine

 

The Spherical Law of Sines states that in a spherical triangle, the ratio of the sine of an angle to the sine of the opposite side is constant.

   

sin(A)/sin(a) = sin(B)/sin(b) = sin(C)/sin(c)


Four Parts Formula 

 

Cos(A) Cos(c) = Sin(c) Cot(b) – Sin(A) Cot(B)

Practice Problems

  1. Angle A=75°, Side b=40°, Side c=50°. Find side a
  2. Angle B=65°, Side a=30°, Angle C=40°. Find side b
  3. Side a=35°, Side b=50°, Side c=65°. Find angle A
  4. Side b=50°, Side c=80°, Angle A=60°. Find angle B

Geographical Coordinate

Longitude and Latitude are used in distance calculations between two points on Earth.

  • Equator – the great circle equidistant from the poles
  • North Pole and South Pole – fixed rotation points of Earth
  • Latitude (): Angle north/south of the equator (-90° to +90°).
  • Longitude (λ): Angle east/west of the prime meridian (0° to 360° or -180° to +180°). [Prime meridian=Reference for longitude (passes through Greenwich, UK)]

Longitude Rules

 

  1. If both longitudes are East (E) or both are West (W): Subtract the smaller from the larger.
  2. If one is East and the other is West: Add them. 
  3. If one is East and the other is West and the result is >180° then subtract it from 360°. 

How to Calculate the Distance Between Two Places (in km)

1st step 

Use Cosine law to find angular separation between two coordinates. 

2nd step 

Distance = Angular separation x 2 π R/360 

R= Radius of Earth 

Practice Problems

  1. What is the angular separation between Jakarta (θ=6° S, λ=106°49′ E) and Tashkent (θ=41° N, λ=69° E)?
  2. What is the distance between Slovenia (θ=46° N, λ=15°32′ E) and Brazil (θ=23° S, λ=43° W)? What angle from the North would you need to take to travel along the shortest path? Earth Radius=6400 km.
  3. If we fly from Alberton (θ=26° S, λ=27°54′ E) in a westward direction, bearing 9.67°, and landing after 8171 km, what are the final coordinates of the landing site?
  4. Alex moves from point New York (θ=40° N, λ=74°54′ W) to Melbourne (θ=37° S, λ=144° E) using the shortest path (great circle route). Calculate the highest latitude reached on that route.

Celestial Coordinate System

The Celestial Coordinate System is used to locate stars, planets, and other celestial objects in the sky, similar to how latitude and longitude work on Earth. There are two primary systems:


  1. Equatorial Coordinate System (Earth-based, fixed relative to stars)
  2. Horizontal Coordinate System also known as alt-azimuth system (Observer-based, changes with time/location)

Horizontal Coordinate System

Horizon is the apparent line where the sky meets the Earth, separating what is visible above from what is hidden below an observer’s line of sight.

 

Zenith is the point in the sky directly overhead an observer (at a 90° angle from the horizon).

 

Altitude is the angle between an object in the sky and the observer’s horizon, measured upward from 0° at the horizon to 90° at the zenith.

 

Azimuth is the angle measured clockwise from the north point on the horizon to the point directly below a celestial object, ranging from 0° to 360°.

 

The horizontal coordinate system is fixed to the Earth, not to the stars. Therefore, the altitude and azimuth of our star in the sky changes with time, as the star appears to move across the sky. 

 

The main disadvantage of the alt-az system is that it is a local coordinate system – i.e. two observers at different points on the Earth’s surface will measure different altitudes and azimuths for the same star at the same time. 

 

So, we must notice our position on the horizon and the time. Maybe we want to later show our star to somebody who is far away from our place. Then, we could not use the horizontal system.

 

But the horizontal coordinates are still very useful for determining the rise and set times of an object in the sky. When an object’s altitude is 0°, it is on the horizon. If at that moment its altitude is increasing, it is rising, but if its altitude is decreasing, it is setting.


Equatorial Coordinate System

The equatorial coordinate system is basically the projection of the latitude and longitude coordinate system we use here on Earth, onto the celestial sphere. By direct analogy, lines of latitude become lines of declination (Dec; measured in degrees, arcminutes and arcseconds) and indicate how far north or south of the celestial equator (defined by projecting the Earth’s equator onto the celestial sphere) the object lies. Lines of longitude have their equivalent in lines of right ascension (RA), but whereas longitude is measured in degrees, minutes and seconds east the Greenwich meridian, RA is measured in hours, minutes and seconds east from where the celestial equator intersects the ecliptic (the vernal equinox).
 
 

Declination(δ)

The declination of a star is its angular distance in degrees measured from the celestial equator along the meridian through the star. It is measured from north and south of the celestial equator and ranges from 0 degree at the celestial equator to 90 degree at the celestial poles, being taken to be positive when north of the celestial equator and negative when south. The declination of a star does not change with time. Example: North pole has a declination of +90 degree. 

Hour Angle 

Hour angle of a star is measured from the observer’s meridian westwards to the meridian through the star (from 0° to 360°). Because of the rotation of the Earth, the hour angle increases uniformly with time, going from 0° to 360° in 24 hours. Hour angle of a particular object is therefore a measure of the time since it crossed the observer’s meridian. Example: If HA = 0h then the object is right on your meridian (highest in the sky).

Right Ascension

Right Ascension is measured from 0h to 24h along the celestial equator eastwards from the first point of Aries, that is, in the opposite direction to that in which the hour angle is measured. Right ascension is typically expressed in units of time (hours, minutes, seconds) rather than degrees, which helps astronomers locate celestial objects more easily. 

1 Hour (RA) = 15°

To convert Right Ascension into degrees, you multiply the RA in hours by 15. 

Vernal Equinox 

It is the point in the sky where the ecliptic (the Sun’s apparent yearly path) intersects the celestial equator. 

 


Horizontal-Equatorial system transformation

To convert between the horizontal and equatorial coordinates for an object S (in our case a star), we use a spherical triangle often called the Astronomical Triangle: SPZ, where Z is the zenith, P is the North Celestial Pole, and S is the object. 

The sides of the triangle are:

  • PZ is the observer’s co-latitude, PZ = 90° —θ. [ =Latitude]
  • ZS is the zenith distance of S, ZS= 90° – h. [h=altitude of star]
  •  PS is the North Polar Distance of S, PS= 90° —δ. [=declination]

The angles of the triangle are:

  •  The angle at P is H, the local Hour Angle of S.
  •  The angle at Z is 180° – A, where A is the azimuth of S.

Using cosine law in triangle of fig 2

cos (90-h) = cos(90-θ) cos(90-δ) + sin(90-θ) sin(90-δ) cos(H)

sin(h) = sin(θ) sin(δ) + cos(θ) cos(δ) cos(H)              Equation 1

You can use Cosine Law and Sine law in this triangle to derive other quantities

Practice Problem

  1. An observer at latitude 22°21′ N observes a star positioned at an altitude of 58°10′ and an azimuth of 310°15′ (measured from North through East). Assuming the observation is made under ideal conditions (no atmospheric refraction or precession effects), calculate the Hour Angle (HA) and Declination (δ) of the star at the time of observation. (Hint: use spherical triangle diagram from fig 2 and apply cosine law to find value)
  2. An observer located at latitude 22°21′ N observes a star with an hour angle of 4ʰ 15ᵐ and a declination of +37°40′. Assuming ideal observational conditions, calculate the altitude of the star above the horizon and its azimuth, measured from North through East, as seen by the observer.
  3. An observer at 35°45′ south latitude sees a star with a declination of +18°20′. Assuming clear skies and a flat horizon (no atmospheric effects), at what azimuth will the star rise?
  4. An observer located at latitude 28°10′ N observes a star as it sets on the western horizon. The star’s declination is −12°40′. Determine the azimuth at which the star sets.  (neglect atmospheric refraction)


Ecliptic Coordinate System

The Ecliptic coordinate system is a celestial coordinate system commonly used for representing the positions and orbits of Solar System objects. Because most planets (except Mercury), and many small Solar System bodies have orbits with small inclinations to the ecliptic, using it as the fundamental plane is convenient. The system’s origin can be either the center of the Sun or the center of the Earth, its primary direction is towards the vernal equinox, and it has a right-handed convention.

Earth-centered ecliptic coordinates as seen from outside the celestial sphere. The ecliptic coordinate system has two poles: the North Ecliptic Pole and the South Ecliptic Pole, located perpendicular to the plane of the ecliptic.

 

Ecliptic longitude (red) is measured along the ecliptic from the vernal equinox. 

 

Ecliptic latitude (yellow) is measured perpendicular to the ecliptic. 

 

Obliquity is the angle between an object’s rotational axis and its orbital axis. It is also called obliquity of the ecliptic. The obliquity of the ecliptic is the angle between the plane of the earth’s orbit and that of the celestial equator, equal to approximately 23°27′ at present. 

All the objects considered so far have been “fixed stars”, which keep almost constant values of Right Ascension and declination. But bodies within the Solar System change their celestial positions. The most important one to consider is the Sun. The Sun’s declination can be found by measuring its altitude when it’s on the meridian (at midday). The Sun’s Right Ascension can be found by measuring the Local Sidereal Time of meridian transit. We find that the Sun’s RA increases by approximately 4 minutes a day, and its declination varies between +23°27′ and -23°27′. This path apparently followed by the Sun is called the ecliptic.

The reason the Sun behaves this way is that the Earth’s axis is tilted to its orbital plane. The angle of tilt is +23°27′, which is called the obliquity of the ecliptic (symbol ε).

Any two great circles intersect at two nodes. The node where the Sun crosses the equator from south to north (the ascending node) is called the vernal (or spring) equinox.


The Sun passes through this point around March 21st each year. This is the point from which R.A. is measured, so here RA = 0h. At RA = 12h, the descending node is called the autumnal equinox; the Sun passes through this point around September 23rd each year. At both these points, the Sun is on the equator and spends 12 hours above horizon and 12 hours below. (“Equinox” means “equal night”: night equals day.)

Right Ascension and Declination of the Sun on Key Dates of the Year

21 March – Vernal Equinox

  • RA: 0ʰ

  • Declination (δ): 0°

23 September – Autumnal Equinox

  • RA: 12ʰ

  • Declination (δ): 0°

22 June – Summer Solstice

  • RA: 6ʰ

  • Declination (δ): +23.5°

22 December – Winter Solstice

  • RA: 18ʰ

  • Declination (δ): −23.5°

Formula to calculate Equatorial coordinate of Sun

Ecliptic Longitude of Sun

Conversion to Equatorial Coordinates

Right Ascension of Sun

Declination of Sun

n = number of days since the Vernal Equinox

T=365.25 = number of days in a year

ε=23.5° = obliquity of the ecliptic



Equatorial-Ecliptic Coordinate System transformation

In the ecliptic system of coordinates, the fundamental great circle is the ecliptic. The zero-point is still the vernal equinox. Take K as the northern pole of the ecliptic, K’ as the southern one.

 

To fix the ecliptic coordinates of an object X on the celestial sphere, draw the great circle from K to K’ through X. The ecliptic (or celestial) latitude of X (symbol β) is the angular distance from the ecliptic to X, measured from -90° at K’ to +90° at K. Any point on the ecliptic has ecliptic latitude 0°. 

 

The ecliptic (or celestial) longitude of X (symbol λ) is the angular distance along the ecliptic from the vernal equinox to the great circle through X. It is measured eastwards (like R.A.), but in degrees, 0°-360°. To convert between ecliptic and equatorial coordinates, use the spherical triangle KPX.

Example of using Cosine law in spherical triangle KPX

Cos(90-δ) = Cos(ε) Cos(90-β) + Sin(ε) Sin(90-β) Cos(90-λ)

Sin(δ) = Cos(ε) Sin(β) + Sin(ε) Cos(β) Sin(λ)     Equation 2

You can use Cosine Law and Sine law in this triangle to derive other quantities. 


Practice Problems

  1. Vega, a bright star in the constellation Lyra, has equatorial coordinates: Right Ascension (RA): 18ʰ 36ᵐ, Declination (δ): +38°47′. Assuming the obliquity of the ecliptic is ε = 23°27′, calculate the ecliptic longitude (λ) and ecliptic latitude (β) of Vega. 
  2. Show that, for any object on the ecliptic, tan(δ) = sin(α) tan(ε), where (α, δ) are the object’s Right Ascension and declination, and ε is the obliquity of the ecliptic.
  3. An asteroid located in the constellation Perseus has the following ecliptic coordinates: Ecliptic Longitude (λ): 247°10′ Ecliptic Latitude (β): +36°15′. Assuming the obliquity of the ecliptic is ε = 23°27′, calculate the equatorial coordinates of the asteroids. 
  4. Show that the declination of the Sun can be expressed in terms of its ecliptic longitude (λ​) as: sin⁡δ=sin⁡ε⋅sin⁡λ. 

Galactic Coordinate System

The galactic coordinate system is a celestial coordinate system used in astronomy, centered on the Sun and aligned with the center of the Milky Way Galaxy. It uses spherical coordinates, where galactic latitude (b) measures the angle north or south of the galactic plane, and galactic longitude (l) measures the angle around the galactic equator from the galactic center. The inclination of the galactic equator to Earth’s equator is 63⁰.

 

Galactic Longitude (l) is Measured eastward from the Galactic Center (located at right ascension 17ʰ46ᵐ, declination −29°00′ in the constellation Sagittarius) along the galactic equator. It ranges from 0° to 360°.

 

Galactic Latitude (𝑏) is measured perpendicular to the galactic plane, ranging from +90° at the North Galactic Pole to −90° at the South Galactic Pole, with 0° corresponding to the galactic plane. The North Galactic Pole lies at right ascension 12ʰ51.4ᵐ, declination +27.13° in the constellation Coma Berenices.


Equatorial-Galactic coordinate System transformation

δ’ = 27.13° & a’=12ʰ51.4ᵐ are constant. Position angle=123°

Example of Using Cosine law in triangle RPG

Cos(90-δ) = Cos(90-δ’) Cos(90-b) + Sin(90-δ’) Sin(90-b) Cos(123-λ)

Sin(δ) = Sin(δ’) Sin(b) + Cos(δ’) Cos(b) Cos(123-λ)                 Equation 3

Practice Problems

  1. The brightest star Sirius has coordinates δ’ = −16° and right ascension = 6 hour 43 minutes. Calculate its galactic coordinates.  
  2. A star has Galactic coordinates l = 202° and b = -12°. Calculate its equatorial coordinate (a, δ). 
  3. Draw a celestial sphere to show the position of a star with Galactic coordinates l =150 and b = +35°. 

Concept of Time

Solar Time

Solar time is a measurement of time based on the position of the Sun in the sky, calculated from Earth’s rotation relative to the Sun. It is fundamentally linked to the synodic rotation period, which defines the length of a solar day. Solar time can vary due to factors such as the elliptical shape of Earth’s orbit and its axial tilt, which affect the apparent motion of the Sun.  

There are two types of Solar time:

Apparent Solar Time

  • It is based on the actual position of the real Sun in the sky.
  • Noon = when the Sun crosses the local meridian (upper culmination).
  • Varies throughout the year due to Earth’s elliptical orbit and axial tilt.
Apparent Solar Time = local solar time (LST). If you are using the real Sun to define it.
 
 
 

Mean Solar Time

  • It is a way of measuring time based on an imaginary Sun that moves at a constant speed across the sky.
  • It is the kind of time we use every day (Standard time), like the time on your phone or in your time zone (for example, UTC).

Local Solar Time

𝐿𝑇=𝐻𝐴⊙+12 hour
LT = Local Solar Time in Hour
𝐻𝐴⊙ = Hour Angle of Sun

Position of Sun, Hour Angle, Local Time

Sun on the meridian (solar noon)

  • HA⊙: 0h

  • Local Time: 12:00 (noon)

Sun at bottom culmination (midnight position)

  • HA⊙: 12h

  • Local Time: 0:00 (midnight)

Sun setting (western horizon)

  • HA⊙: 6h

  • Local Time: 18:00 (6:00 PM)

   Sun setting (western horizon)

  • HA⊙: -6h

  • Local Time: 18:00 (6:00 PM)

Equation of Time

Equation of time (EoT) is the difference between mean solar time and apparent solar time. 

EoT = Apparent Solar Time – Mean Solar Time

EOT = (HA apparent Sun + 12 hour) – (HA mean Sun + 12 hour)

Hence, 

EoT = HA apparent Sun − HA mean Sun                 (Equation 4)

Relation between right ascension and EOT


EoT = (LST- RA of apparent Sun) – (LST – RA of mean Sun) 

EoT= RA of mean Sun – RA of apparent Sun


Formula to calculate right ascension of mean Sun

N is the number of days that have passed since January 1st, starting from n = 0 on January 1st.

T = Number of days in a year i.e 365.25

24 hour = Total RA range in one full circle 

Formula to calculate right ascension of Apparent Sun

(Taken from formula to calculate Equatorial coordinate of Sun)

Practice Problem

  1. On December 10 at a certain location, the local sidereal time (LST) is 8h 30m. The Sun’s apparent right ascension is 7h 45m, and the mean right ascension is 7h 30m. Calculate the hour angles of the apparent Sun and the mean Sun at that moment.Using these hour angles, determine the Equation of Time (EoT) in minutes.
  2. Assume the Earth moves in a circular orbit. Calculate the apparent right ascension of the Sun on October 5. Using this value, determine the Equation of Time (EoT) for that date.
  3. On July 21, the Equation of Time (EoT) is +10 minutes, and the mean right ascension of the Sun is 18h 15m. Assuming the Earth moves in a circular orbit, calculate the apparent right ascension of the Sun on that date.


Sidereal Time

Sidereal time is a system of timekeeping used primarily by astronomers, measuring the rotation of the Earth relative to the stars rather than the Sun. It is defined as the right ascension of any star when that star is at its highest point in the sky. Instead of tracking the position of Sun, it uses the Vernal Equinox (♈︎) as a reference point.

Calculating Local Sidereal Time

LST = Local Sidereal Time 

HA​(♈︎) = Hour Angle of Vernal Equinox

Special Case

LST = RA of culminating object​​

This relation is true for an object on the meridian (highest point in the sky). If a star or any celestial object is culminating (crossing the observer’s meridian), its RA = LST at that exact time.

Local Sidereal time (LST) = Hour Angle (HA) + Right Ascension (RA)    [Fundamental Equation]


Calculating Local Sidereal Time from Solar Time

HA⊙ = Hour Angle of Sun 

RA⊙ = Right Ascension of Sun 

You can calculate the hour angle of sun and the right ascension of sun using date and time. 

From date you can calculate RA⊙ using this formula


Practice Problems

  1. The Local Sidereal Time (LST) at 14:20 local time on March 10 is observed to be 5h 48m. Determine the calendar date and local time.
  2. Calculate the Local Sidereal Time (LST) on May 12 at 16:20 local time.
  3. The Local Sidereal Time (LST) is calculated to be 11h 52m at a location with coordinates 40.71°N, 74.01°W (New York City) on September 14 at 21:15 EDT (UTC -4). Determine the calendar date and local time at which this LST occurs.


Rising and Setting Time

From Equation 1 

sin(h) = sin(θ) sin(δ) + cos(θ) cos(δ) cos(HA)

The object rises and sets when its altitude (h) = 0°. (neglect atmospheric refraction)

0 = sin(θ) sin(δ) + cos(θ) cos(δ) cos(HA)

– cos(θ) cos(δ) cos(HA) = sin(θ) sin(δ)

Cos(HA) = -tan(δ) tan(θ)                         Equation 5

At Rising

Local Sideral Time (LST) = Right ascension of Celestial object (RA) – Hour angle at rising (HA)

When the star rises, it is east of the meridian →hour angle (HA) is negative. 

 

At Setting

Local Sideral Time (LST) = Right ascension of Celestial object (RA) – Hour angle at rising (HA)

When the star sets, it is west of the meridian → hour angle (HA) is positive

 


Practice Problems

  1. A star has celestial coordinates of Right Ascension (α) = 6h and Declination (δ) = +20°. An observer is located at a latitude of 40° N. Calculate the hour angle and determine the Local Sidereal Time (LST). 

Upper and lower culmination

Upper culmination and lower culmination are the moments when a circumpolar object reaches its highest and lowest points in the sky, respectively. At upper culmination, the object is at its greatest altitude and has an hour angle of 0h. At lower culmination the object is at its lowest visible point, passing between the pole and the horizon, with an hour angle of 12h. For non-circumpolar objects, lower culmination happens below the horizon, so it cannot be seen.

Upper Culmination

From Equation 1 

sin(h) = sin(θ) sin(δ) + cos(θ) cos(δ) cos(HA)

[At upper culmination H=0]

sin(h) = sin(θ) sin(δ) + cos(θ) cos(δ)

Sin(h) = Cos(θ-δ)                                     [using, Cos(A-B) = Cos(A) Cos(B) + Sin(A) Sin(B)]

Sin(h) = Sin(90 – (θ-δ))                           [using, cos(x)=sin(90-x)]

This gives two possible values of h.

 h up = 90 – θ + δ

If the star culminates south of zenith because δ<θ

h up = 90 + θ – δ

If the star culminates north of zenith because δ>θ

Lower Culmination

From Equation 1 

sin(h) = sin(θ) sin(δ) + cos(θ) cos(δ) cos(HA)

[At lower culmination H=12h= 180]

sin(h) = sin(θ) sin(δ) – cos(θ) cos(δ)

sin(h) = -(cos(θ) cos(δ) – sin(θ) sin(δ))

Sin(h) = -Cos(θ+δ)          [using, Cos(A+B) = Cos(A) Cos(B) – Sin(A) Sin(B)]

Sin(h) = Sin(θ + δ – 90)

h low = θ + δ – 90

Condition: Circumpolar

  • Visibility: Always above the horizon

  • Declination (δ) > 90° − Latitude

Condition: Never visible

  • Visibility: Always below the horizon

  • Declination (δ) < 90° − Latitude

Condition: Rises and sets

  • Visibility: Crosses the horizon daily

  • θ – 90< δ < 90 – θ

Adding the value of upper and lower culmination

h up + h low = (90 – θ + δ) + (θ + δ – 90)

h up + h low = 2δ

Subtracting the Value of Upper and Lower culmination

h up – h low = (90 – θ + δ) – (θ + δ – 90)

h up + h low = 180 – 2θ 


Julian Date

A Julian Date (JD) is a continuous count of days and fractions of a day starting from noon Universal Time (UT) on January 1, 4713 BCE in the Julian calendar.

The starting point is called the Julian Epoch:

JD=0 at 12.00 UTC on January 1, 4713 BC (Julian Calendar) 

K is the year — but subtract 1 if the month is January or February. 

 

Example: If July 25 (2025) then K = 2025 

If February 19 (2025) then k = year – 1 so k = 2024

 

M is the month, but January = 13 and February = 14 

 

Example: If June 5 (2025) then M = 6

If February 14th (2024) then M = 14

 

I is the day of the month.

 

Example: If April 3rd (2025) then I = 3 

If December 29th (2025) then I = 29

 

UT = Universal Time at the moment you’re calculating, in decimal hours.

 

Modified Julian Date (MJD)

The Modified Julian Date (MJD) is a simplified version of the Julian Date, commonly used in astronomy and space science to avoid large numbers.

MJD=JD−2400000.5

Practice Problems

  1. Find the Julian Date for March 4 (2025) at 21:30 UT.
  2. Find the Julian Date for January 28, 2023, at 02:15 UT and calculate MJD

Precession

Precession is the slow movement of Earth’s rotational axis as it revolves around the ecliptic poles.

So far, we have looked at how we assign coordinates to points in the sky and how different effects can change where things appear. But there is a bigger challenge — the celestial equator and the ecliptic don’t stay fixed in space.

The Earth’s axis is tilted to its orbital plane. The gravitational pull of Sun and Moon on the Earth’s equatorial bulge tends to pull it back towards the plane of the ecliptic. The North Celestial Pole traces out a precessional circle around the pole of the ecliptic, and this means that the equinoxes precess backwards around the ecliptic, at the rate of 50.35 arc-seconds per year (around 26,000 years for a complete cycle).

Formula to Calculate ecliptic longitude (Δλ)

t = time span (in years)

T= precession period ( 26,000 years)
λ′=λ+Δλ (new ecliptic longitude)
β′=β (ecliptic latitude remains unchanged)

For small time intervals, approximate changes in equatorial coordinates due to precession can be calculated as:

Δδ = Δλ⋅sinε⋅cosα

Δα = Δλ (cos⁡ε + sin⁡ε⋅tanδ⋅sin⁡α)

ε = obliquity of the ecliptic
α = right ascension
δ = declination

Practice Problems

  1.  Betelgeuse is located at α = 05ʰ55ᵐ, δ = +07°24′, calculate its equatorial coordinates 6000 years from now due to precession.
  2. Charles is an observer at 45° latitude. Over a span of 2000 years, 270 stars have remained circumpolar. Originally, how many circumpolar stars could Charles see 2000 years ago?

Proper Motion

Proper motion is the astrometric measure of changes in the apparent positions of stars or other celestial objects as they move relative to the center of mass of the Solar System. This motion is observed from Earth and measured in arcseconds per year.


The motion has two components:

 
● Change in Right Ascension (RA): μa cos δ

 
Here, the factor cos δ appears because motion in RA must be scaled by this factor to reflect actual sky motion at declination δ.
μa= Proper motion in RA, measured in arcseconds/year.

 
● Change in Declination (Dec): μδ

Total Proper Motion (Angular Speed)

Total Proper Motion is the star’s actual angular speed across the sky, combining its motion in both Right Ascension and Declination.

μ = Total angular motion (arcsec/year)

 

Transverse Velocity

Transverse velocity is the linear speed of an object moving perpendicular to our line of sight, derived from its proper motion and distance.

D: Distance to the star measured in parsecs (pc) or light-years (ly)

 

Radial Velocity

The component of motion toward or away from us is called radial velocity, and it’s determined using the Doppler effect:

Radial Velocity (Vr)= Speed of light (c) x Redshift (Z)

Space Velocity

Space velocity is the star’s total three-dimensional speed relative to the Sun, combining both its radial and transverse velocities. 

Space Velocity (V)=Transverse Velocity (Vt) + Radial Velocity (Vr)