What is the declination of the sun ? ?

Question

i thought it should be 0 this should be its position on the equinox but it isn't its position on google sky

Answer 1

Due to tilt in the earths axis,we see the Sun travelling to north and south by 23.5degrees and it will be zero on the two equinoxes.

The north it is 23.5degree  and south it is minus 23.5degree.

It is zero degrees on the equinoxes on many other sky simulation planetarium programmes.

Answer 2

The declination of the Sun is the angle between the rays of the sun and the plane of the earth's equator. Since the angle between the earth axis and the plane of the earth orbit is nearly constant due to Earth's low eccentricity,  declination values vary with the seasons and its period is one year.

When the projection of the earth axis on the plane of the earth orbit is on the same line linking the earth and the sun, the angle between the rays of the sun and the plane of the earth equator is maximum and its value is 23degrees 27minutes. This happens at the solstices. Therefore declination = +23degrees 27minutes at the northern hemisphere summer solstice and declination = -23degrees 27minutes at the northern hemisphere winter solstice. Due to the changes in the tilt of the Earth's axis, the angle between the rays of the sun and the plane of the earth equator is slightly decreasing.

When the projection of the earth axis on the plane of the earth orbit is perpendicular to the line linking the earth and the sun, the angle between the rays of the sun and the plane of the earth equator is null. This happens at the equinoxes. Therefore declination is 0degrees at the equinoxes.

Sun's declination is equal to inverse sine of the product of sine of Sun's maximum declination and sine of Sun's tropical longitude at any given moment. This is measured daily at 0:00 Greenwich or UT.

Declination measures the longitudinal coordinate. To find the exact location of an orb at any point in time, Right Ascention would be used to measure its distance from the ecliptic.

Answer 3

I might as well have some fun with this. You want to know the declination of the sun at any moment of time, yes? Not just for today. Well, here's how to figure it out.

t = the time of observation expressed in Julian Date format

To convert a calendar date to a Julian date, go to this page:

http://wwwmacho.mcmaster.ca/JAVA/JD.html

a = 1.00000261

e = 0.01671123

m = (0.0172020316 radians/day) (t - 2454468.5833)

Adjust m to the interval [0, 2 pi).

U1 = m

REPEAT...

. U0 = U1

. F0 = U0 - e sin U0 - m

. F1 = 1 - e cos U0

. F2 = e sin U0

. F3 = e cos U0

. D1 = -F0 / F1

. D2 = -F0 / [ F1 + D1 F2 / 2 ]

. D3 = -F0 / [ F1 + D1 F2 / 2 + (D2)^2 F3 / 6 ]

. U1 = U0 + D3

UNTIL |U1-U0| < 1E-15

u = U1

Note: m and u are both angles, and they must be both expressed in radians within the repeat-loop shown just above.

x'' = a (cos u - e)

y'' = a (sin u) sqrt(1 - e^2)

w = 102.93768 degrees

x' = x'' cos w - y'' sin w

y' = x'' sin w + y'' cos w

phi = 23.439282 - (3.563E-7)( t - 2451543.5 )

Note: phi is an angle expressed in degrees by the above equation. If you need it in radians, you should multiply it by (pi/180).

x = -x'

y = -y' cos(phi)

z = -y' sin(phi)

r = sqrt ( x^2 + y^2 + z^2 )

Although you didn't ask for it, the RIGHT ASCENSION of the sun at time t is found from

if x = 0 and y > 0 then RA = 6 hours

if x = 0 and y < 0 then RA = 18 hours

if x = 0 and y = 0 then RA = 0 hours

if x > 0 and y > 0 then RA = Arctan( y / x ) / 15

if x < 0 then RA = Arctan( y / x ) / 15 + 12 hours

if x > 0 and y< 0 then RA = Arctan( y / x ) / 15 + 24 hours

Where the Arctan function is assumed to give its result in degrees.

The DECLINATION of the sun at time t is found from:

dec = Arcsin ( z / r )

WORKED EXAMPLE.

t = 12h UT on 30 August 2008 = JD 2454709.0

m = 4.135655671 radians

u = 4.121775338 radians

x'' = -0.57358355

y'' = -0.83048530

x' = +0.93782299

y' = -0.37308437

x = -0.93782299

y = +0.3423012

z = +0.1483976

r = 1.0093086 AU

RA = 10.66321 hours = 10h 39m 47.6s

dec = +8.45479 degrees = 8d 27' 17"

Let me check my (hastily written) program against Bikenbee's result for 20h UT on this same date. I get dec=8d 20' 3". Oh no! The sun's dec is shrinking! (Heh.)

A table showing the average declination of the sun for each day of the year can be found at

http://www.wsanford.com/~wsanford/exo/su...

Answer 4

+63.87°

Answer 5

It depends on the season.  At the equinoxes, it's zero.  It ranges to its maxima at the solstices: + or - 23.4 degrees.

This is because RA and DEC coordinates are based on the earth's polar axis, (the equatorial), which is inclined to the path of the earth's orbit around the sun, (the ecliptic).  The equatorial, or celestial equator is the zero degree line.  The ecliptic makes a nice sine wave across its length.

Edit: Right now it's about +9 deg 13'.

Edit: Bikenbee's figure is probably better because mine is taken from a table of mean values and doesn't specify time of day.

Answer 6

I had a look on Google Sky and moved the pointer on to the image of the Sun and the display gives the declination as about 8º 40' which is correct for the present time (20h UT on August 30th). It's declination will be 0º on the equinox, but that's not until September 22nd.

Answer 7

I'll give you a very compact formula:

From June 21 count the number of days to your target date, call it 'n'.

Let D = Declination,

Tan D = Tan (23.45d) Cos n/[365.2475/360]

D = Arc Tan{0.433775 Cos n/[365.2475/360]}.

Since Cos (-A) = Cos A & argument of Cos in degrees.

you can count 'n' backwards from June 21 and plug into the formula.

It takes care of Earth's speed variation due to Aphelion/Perihelion.