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Spherical astronomy I May 30, 2011

1

Math Introduction

I’m going to start with a small bit of trigonometry, both the trigonometry in the euclidean plane and spherical trigonometry. I’m sure you know enough

Figure 1: Normal triangle about basic formulas for sines and cosines, therefore I’m not going to bother you with them. I just want to mention two important theorems for both types of triangle, being law of sines and law of cosines. You might be familiar with the euclidean version of them a b c = = , c2 = a2 + b2 − 2ab cos γ. sin α sin β sin γ Generally, if the spherical triangle is small enough, you can use those formulas for a normal triangle as well. However, if you want to be precise, you have to use their following modification (see the figure with sph. triangle below) sin a sin b sin c = = , sin α sin β sin γ

cos c = cos a cos b + sin a sin b cos γ.

And this is all the math you need to know.

2 2.1

Coordinate systems Horizontal coordinate system (H)

It uses the plane of local horizon as the fundamental plane. The pole in the upper hemisphere is called zenith, the opposite one’s nadir. The first coordinate is the 1

Figure 2: Spherical trinagle altitude, denoted h. It is the angle measured between the fundamental plane (horizon) and the object. Alternatively, one can use so called zenith distance z = 90◦ − h, i.e. the angle measured between the zenith and the object. The second coordinate is azimuth A. Nowadays it is regarded as an angle measured between local meridian (the great circle going thorugh the northern point of the horizon and zenith) and the great circle going through the zenith and the object, measured from the north to the east. Not very long time ago, it was measured from the south to the west. Since it is still used in all of the literature, note that in the following text and in my problems, the azimuth will be measured from the south to the west. Basically, it is no problem to switch between these two definitions as it is equivalent to A → A+180. On the picture, azimuth is red, altitude is green.

2.2

Equatorial coordinate system (Q)

It uses the plane of the celestial equator as the fundamental plane. This plane makes an angle 90◦ − φ with the plane of local horizon (i.e. the fundamental plane of horizontal coordinates), where φ is the geographical latitude. The first coordinate is declination δ - it’s the angle measured between the plane of celestial equator and the object. The fundamental direction is taken to be that of the vernal equinox point. The second coordinate is the right-ascension α - it’s the angle which makes the great circle going through the northern celestial pole and vernal equinox point with the great circle going through the northern celestial pole and the object, measured positively to the east. See the 2

Figure 3: Horizontal coordinate system picture. You can see that equatorial coordinates of a star are approx. constant over relatively long time periods (let say years, tens of years). Basically they are constant as long as the effects of precession, abberation, paralax or proper motion are not significant for the problem. Alternatively, one can use so called hour angle t as the second coordinate. However, this coordinate varies with the period one sidereal day. It is the angle between the local meridian and the great circle determined by the northern celestial pole and the object. Let us define the hour angle of the vernal equinox point to be the local sidereal time Θ. Then we can easily relate the hour angle of an object with the right ascension as Θ = t + α (recall the definition of α and draw the picture - it’s trivial!).

2.3

Ecliptical coordinate system (E)

This coordinate system uses the plane of the ecliptics as its fundamental plane. The fundamental direction is again the direction towards the vernal equinox point. The latitudinal coordinate is the ecliptic latitude β and it is measured positively to the north (angle between the object the plane of ecliptics). The longitudinal coordinate is the ecliptic longitude λ - the angle between the great circle going through the ecliptic poles and the vernal equinox point and the graet circle going through the ecliptic poles and the object, measured positively to the east. The plane of the ecliptic makes with the plane of the celestial equator the angle ε ≈ 23.5◦ and the line of their intersection is determined by both equinox points.

2.4

Galactic coordinate system (G)

It uses the galactic plane as the fundamental plane. The longitudinal coordinate is the galactic longitude l - the angle which makes the great circle going through the galactic poles and the galactic center with the great circle going through the galactic poles and the object. The latitudinal coordinate is the

3

Figure 4: Equatorial coordinate system galactic latitude b measured positively to the north.

2.5

Supergalactic coordinate system (SG)

upergalactic coordinates are coordinates in a spherical coordinate system which was designed to have its equator aligned with the supergalactic plane, a major structure in the local universe formed by the preferential distribution of nearby galaxy clusters (such as the Virgo cluster, the Great Attractor and the PiscesPerseus supercluster) towards a (two-dimensional) plane (you don’t need to know it).

3

Transformation of coordinates

In the spherical astronomy, one very often needs to switch between the horizontal, equatorial and ecliptical coordinates (in case of galactical and supergalactical it is rather important to know what it basically is, transformations are usually not required). Therefore I’m going to derive the transformations H → Q, Q → H, let say that’s all we need. However, you SHOULDN’T memorize the equations which I shall derive now. What you SHOULD memorize are the spherical law of sines and cosines, that’s really all you need. In the following, we will use t and z - I hope that now it is simple for you to convert them to α and h.

4

3.1

Transformation Q → H

Remember, that all you need to switch between H and Q is the following triangle: Point P is the northern celestial pole, Z is zenith, H is an object. φ

is the geographical latitude, p.v. is 90 − δ (delta is declination), t is the hour angle, A is azimuth - here measured from the south. So, you can see that in this triangle are both δ, t and z, A. Moreover, there is the goegraphical latitude φ which relates them somehow. So, let me write the spherical sine theorem for this triangle sin t sin (180 − A) sin π = = sin z sin (90 − δ) sin(90 − φ) and since sin (180 − A) = sin A and sin (90 − δ) = cos δ, we have sin z sin A = sin t cos δ, Now let us write the law of cosines. We have cos z = cos (90 − δ) cos(90 − φ) + sin(90 − δ) sin(90 − φ) cos t thus cos z = sin δ sin φ + cos δ cos φ cos t. 5

And we are almost ready! If we know Q coordinates and φ, we can easily calculate z from the last equation. And if we know z, we can find out A from the previous one! However, there’s something more as you might guess, since the azimuth isn’t determined uniquely by that equation (functions arcsine and arccosine have two solutions on the interval 0 to 360). On the other hand, z is determined uniquely, because it lies in the interval -90 to 90 and here those functions have only one root. Anyway, we must write the second law of cosines for this triangle cos(90 − δ) = cos(90 − φ) cos(z) + sin(90 − φ) sin z cos(180 − A) thus (cos(180 − A) = − cos A) sin δ = sin φ cos z − cos φ sin z cos A. Let me substitute for cos z from the first law of cosines, then we have sin δ = sin2 φ sin δ + sin φ cos δ cos φ cos t − cos φ sin z cos A. Now, important thing to realize, we can put the first term from the right hand side to the left hand side: sin δ 1 − sin2 φ = sin φ cos δ cos φ cos t − cos φ sin z cos A, thus sin δ cos2 φ = sin φ cos δ cos φ cos t − cos φ sin z cos A. Now we can divide the whole equation by cos φ and obtain sin δ cos φ = sin φ cos δ cos t − sin z cos A. sin z cos A = sin φ cos δ cos t − sin δ cos φ. So, the complete transformation Q → H reads sin z sin A

=

sin t cos δ

sin z cos A

=

sin φ cos δ cos t − sin δ cos φ

cos z

=

sin δ sin φ + cos δ cos φ cos t.

So one must pay attention in which quadrant A lies - I think you must be familiar with it from math lessons. This holds once we measure the azimuth from the south. If we measure the azimuth from the north, the only thing we have to do is to change A → A + 180. And since cos(A) = − cos(180 + A) and sin A = − sin(180 + A) we would have sin z sin A

= − sin t cos δ

sin z cos A

= − sin φ cos δ cos t + sin δ cos φ

cos z

=

sin δ sin φ + cos δ cos φ cos t.

The instructions again: first of all, determine the zenith distance from the last equation of the triplet - this is unique. Then calculate the sine and cosine of the azimuth and according to that decide in which quadrant it lies. That’s all. 6

Problem 1 Determine the altitude h and azimuth A (measured from south westwards) of Thuban (α Dra) for the observer at φ = 59◦ 560 3000 at Θ = 16h 24m 33s of the local sidereal time. The equatorial coordinates of Thuban are α = 14h 01m 57s and δ = 64◦ 480 0800 . Solution:

First of all we determine the hour angle t t = Θ − α = 2m 22m 36s = 35◦ 390 0000 .

Now we can easily determine the altitude cos z = cos(90 − h) = sin h = sin δ sin φ + cos δ cos φ cos t → h = 73◦ 010 3600 . Now we can make use of remaining two transf. equations sin A =

sin t cos δ = 0.84999, cos h

sin φ cos δ cos t − sin δ cos φ = −0.52681. cos h We see that the sine is positive, whereas the cosine is negative. Therefore the solution must lie in the second quadrant: sin t cos δ A = 180 − arcsin = 121◦ 470 3000 . cos h cos A =

3.2

Transformation H → Q

This time we can use this possibility how to apply the law of cosine on that triangle (same one as before basically): cos(90 − δ) = cos(90 − φ) cos(z) + sin(90 − φ) sin z cos(180 − A) thus (cos(180 − A) = − cos A) sin δ = sin φ cos z − cos φ sin z cos A plus law of sines provides us again with sin t cos δ = sin z sin A. And we’re here again! The hour angle is not determined uniquely, hence we must introduce one more equation (again the same law of cosines as before) cos z = cos (90 − δ) cos(90 − φ) + sin(90 − δ) sin(90 − φ) cos t thus cos z = sin δ sin φ + cos δ cos φ cos t. 7

Here we can substitute from the first boxed equation in this subsection for sin δ: cos z = sin2 φ cos z − cos φ sin z cos A sin φ + cos δ cos φ cos t. Again, we do the same trick with putting the first term to the left: cos z 1 − sin2 φ = − cos φ sin z cos A sin φ + cos δ cos φ cos t. cos z cos2 φ = − cos φ sin z cos A sin φ + cos δ cos φ cos t and again, we can peacefully divide by cos φ cos z cos φ = − sin z cos A sin φ + cos δ cos t and therefore cos δ cos t = cos z cos φ + sin z cos A sin φ. Hence the complete transformation H → Q reads sin t cos δ

=

sin z sin A

cos t cos δ

=

cos z cos φ + sin z cos A sin φ

sin δ

=

sin φ cos z − cos φ sin z cos A.

Again for the azimuth measured from the north we would have sin t cos δ

= − sin z sin A

cos t cos δ

=

cos z cos φ − sin z cos A sin φ

sin δ

=

sin φ cos z + cos φ sin z cos A.

You must pay attention to the quadrant where t lies, ok? Same as the previous case: firstly calculate δ, then sine and cosine of t. And then decide. Problem 2 Determine the right ascension and declination of a star, if you know that at the place with φ = 55◦ 460 and at Θ = 11h 11m 36s it appears in the altitude h = 40◦ 440 5000 and with azimuth A = 298◦ 280 5000 (measured from the south westwards). Solution: -90 to 90)

First of all we determine the declination (unique in the interval

sin δ = sin φ cos z − cos φ sin z cos A. → δ = 19◦ 390 3000 . Then we have

sin z sin A = −0.70713, cos δ cos z cos φ + sin z cos A sin φ = 0.70709. cos t = cos δ sin t =

8

Sine is negative and cosine positive, hence it lies in the 4th quadrant and t = 315◦ . Therefore the right ascension is α = Θ − t = 14h 11m 36s . Transformation to ecliptical coordinates could be derived quite analogically, but you don’t need it really often (at least, I won’t put it in the problems). Basically, there is a more elegant way by using the unit vectors and R-matrix. However, it would be too much for the beginning.

4

Time

We have already defined one kind of time and that was local sidereal time Θ (recall the definition!). Now we define the apprent solar time Tv (tempus solare verum) to be the hour angle of the Sun. However, the Sun doesn’t move continually on the sky and hence this is not very convenient for measuring the time. Therefore we shall define the mean solar time Tm (tempus solare medium) as the hour angle of so-called mean Sun - the point which moves continually along the celestial equator with a period of one sidereal year. Tm and Tv are related by a coefficient τ where τ = Tm − Tv . τ should be given. Last but not least, we define the zone time, the time which is commonly used. As you know, the Earth is divided in 24 zones. The zone time Tz for a given zone is given by the mean solar time for the center of the zone. For instance the time GMT +1 (CR) is the mean solar time for the east longitude 15◦ . LT time is the mean solar time for the longitude 30◦ .

9

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1

Math Introduction

I’m going to start with a small bit of trigonometry, both the trigonometry in the euclidean plane and spherical trigonometry. I’m sure you know enough

Figure 1: Normal triangle about basic formulas for sines and cosines, therefore I’m not going to bother you with them. I just want to mention two important theorems for both types of triangle, being law of sines and law of cosines. You might be familiar with the euclidean version of them a b c = = , c2 = a2 + b2 − 2ab cos γ. sin α sin β sin γ Generally, if the spherical triangle is small enough, you can use those formulas for a normal triangle as well. However, if you want to be precise, you have to use their following modification (see the figure with sph. triangle below) sin a sin b sin c = = , sin α sin β sin γ

cos c = cos a cos b + sin a sin b cos γ.

And this is all the math you need to know.

2 2.1

Coordinate systems Horizontal coordinate system (H)

It uses the plane of local horizon as the fundamental plane. The pole in the upper hemisphere is called zenith, the opposite one’s nadir. The first coordinate is the 1

Figure 2: Spherical trinagle altitude, denoted h. It is the angle measured between the fundamental plane (horizon) and the object. Alternatively, one can use so called zenith distance z = 90◦ − h, i.e. the angle measured between the zenith and the object. The second coordinate is azimuth A. Nowadays it is regarded as an angle measured between local meridian (the great circle going thorugh the northern point of the horizon and zenith) and the great circle going through the zenith and the object, measured from the north to the east. Not very long time ago, it was measured from the south to the west. Since it is still used in all of the literature, note that in the following text and in my problems, the azimuth will be measured from the south to the west. Basically, it is no problem to switch between these two definitions as it is equivalent to A → A+180. On the picture, azimuth is red, altitude is green.

2.2

Equatorial coordinate system (Q)

It uses the plane of the celestial equator as the fundamental plane. This plane makes an angle 90◦ − φ with the plane of local horizon (i.e. the fundamental plane of horizontal coordinates), where φ is the geographical latitude. The first coordinate is declination δ - it’s the angle measured between the plane of celestial equator and the object. The fundamental direction is taken to be that of the vernal equinox point. The second coordinate is the right-ascension α - it’s the angle which makes the great circle going through the northern celestial pole and vernal equinox point with the great circle going through the northern celestial pole and the object, measured positively to the east. See the 2

Figure 3: Horizontal coordinate system picture. You can see that equatorial coordinates of a star are approx. constant over relatively long time periods (let say years, tens of years). Basically they are constant as long as the effects of precession, abberation, paralax or proper motion are not significant for the problem. Alternatively, one can use so called hour angle t as the second coordinate. However, this coordinate varies with the period one sidereal day. It is the angle between the local meridian and the great circle determined by the northern celestial pole and the object. Let us define the hour angle of the vernal equinox point to be the local sidereal time Θ. Then we can easily relate the hour angle of an object with the right ascension as Θ = t + α (recall the definition of α and draw the picture - it’s trivial!).

2.3

Ecliptical coordinate system (E)

This coordinate system uses the plane of the ecliptics as its fundamental plane. The fundamental direction is again the direction towards the vernal equinox point. The latitudinal coordinate is the ecliptic latitude β and it is measured positively to the north (angle between the object the plane of ecliptics). The longitudinal coordinate is the ecliptic longitude λ - the angle between the great circle going through the ecliptic poles and the vernal equinox point and the graet circle going through the ecliptic poles and the object, measured positively to the east. The plane of the ecliptic makes with the plane of the celestial equator the angle ε ≈ 23.5◦ and the line of their intersection is determined by both equinox points.

2.4

Galactic coordinate system (G)

It uses the galactic plane as the fundamental plane. The longitudinal coordinate is the galactic longitude l - the angle which makes the great circle going through the galactic poles and the galactic center with the great circle going through the galactic poles and the object. The latitudinal coordinate is the

3

Figure 4: Equatorial coordinate system galactic latitude b measured positively to the north.

2.5

Supergalactic coordinate system (SG)

upergalactic coordinates are coordinates in a spherical coordinate system which was designed to have its equator aligned with the supergalactic plane, a major structure in the local universe formed by the preferential distribution of nearby galaxy clusters (such as the Virgo cluster, the Great Attractor and the PiscesPerseus supercluster) towards a (two-dimensional) plane (you don’t need to know it).

3

Transformation of coordinates

In the spherical astronomy, one very often needs to switch between the horizontal, equatorial and ecliptical coordinates (in case of galactical and supergalactical it is rather important to know what it basically is, transformations are usually not required). Therefore I’m going to derive the transformations H → Q, Q → H, let say that’s all we need. However, you SHOULDN’T memorize the equations which I shall derive now. What you SHOULD memorize are the spherical law of sines and cosines, that’s really all you need. In the following, we will use t and z - I hope that now it is simple for you to convert them to α and h.

4

3.1

Transformation Q → H

Remember, that all you need to switch between H and Q is the following triangle: Point P is the northern celestial pole, Z is zenith, H is an object. φ

is the geographical latitude, p.v. is 90 − δ (delta is declination), t is the hour angle, A is azimuth - here measured from the south. So, you can see that in this triangle are both δ, t and z, A. Moreover, there is the goegraphical latitude φ which relates them somehow. So, let me write the spherical sine theorem for this triangle sin t sin (180 − A) sin π = = sin z sin (90 − δ) sin(90 − φ) and since sin (180 − A) = sin A and sin (90 − δ) = cos δ, we have sin z sin A = sin t cos δ, Now let us write the law of cosines. We have cos z = cos (90 − δ) cos(90 − φ) + sin(90 − δ) sin(90 − φ) cos t thus cos z = sin δ sin φ + cos δ cos φ cos t. 5

And we are almost ready! If we know Q coordinates and φ, we can easily calculate z from the last equation. And if we know z, we can find out A from the previous one! However, there’s something more as you might guess, since the azimuth isn’t determined uniquely by that equation (functions arcsine and arccosine have two solutions on the interval 0 to 360). On the other hand, z is determined uniquely, because it lies in the interval -90 to 90 and here those functions have only one root. Anyway, we must write the second law of cosines for this triangle cos(90 − δ) = cos(90 − φ) cos(z) + sin(90 − φ) sin z cos(180 − A) thus (cos(180 − A) = − cos A) sin δ = sin φ cos z − cos φ sin z cos A. Let me substitute for cos z from the first law of cosines, then we have sin δ = sin2 φ sin δ + sin φ cos δ cos φ cos t − cos φ sin z cos A. Now, important thing to realize, we can put the first term from the right hand side to the left hand side: sin δ 1 − sin2 φ = sin φ cos δ cos φ cos t − cos φ sin z cos A, thus sin δ cos2 φ = sin φ cos δ cos φ cos t − cos φ sin z cos A. Now we can divide the whole equation by cos φ and obtain sin δ cos φ = sin φ cos δ cos t − sin z cos A. sin z cos A = sin φ cos δ cos t − sin δ cos φ. So, the complete transformation Q → H reads sin z sin A

=

sin t cos δ

sin z cos A

=

sin φ cos δ cos t − sin δ cos φ

cos z

=

sin δ sin φ + cos δ cos φ cos t.

So one must pay attention in which quadrant A lies - I think you must be familiar with it from math lessons. This holds once we measure the azimuth from the south. If we measure the azimuth from the north, the only thing we have to do is to change A → A + 180. And since cos(A) = − cos(180 + A) and sin A = − sin(180 + A) we would have sin z sin A

= − sin t cos δ

sin z cos A

= − sin φ cos δ cos t + sin δ cos φ

cos z

=

sin δ sin φ + cos δ cos φ cos t.

The instructions again: first of all, determine the zenith distance from the last equation of the triplet - this is unique. Then calculate the sine and cosine of the azimuth and according to that decide in which quadrant it lies. That’s all. 6

Problem 1 Determine the altitude h and azimuth A (measured from south westwards) of Thuban (α Dra) for the observer at φ = 59◦ 560 3000 at Θ = 16h 24m 33s of the local sidereal time. The equatorial coordinates of Thuban are α = 14h 01m 57s and δ = 64◦ 480 0800 . Solution:

First of all we determine the hour angle t t = Θ − α = 2m 22m 36s = 35◦ 390 0000 .

Now we can easily determine the altitude cos z = cos(90 − h) = sin h = sin δ sin φ + cos δ cos φ cos t → h = 73◦ 010 3600 . Now we can make use of remaining two transf. equations sin A =

sin t cos δ = 0.84999, cos h

sin φ cos δ cos t − sin δ cos φ = −0.52681. cos h We see that the sine is positive, whereas the cosine is negative. Therefore the solution must lie in the second quadrant: sin t cos δ A = 180 − arcsin = 121◦ 470 3000 . cos h cos A =

3.2

Transformation H → Q

This time we can use this possibility how to apply the law of cosine on that triangle (same one as before basically): cos(90 − δ) = cos(90 − φ) cos(z) + sin(90 − φ) sin z cos(180 − A) thus (cos(180 − A) = − cos A) sin δ = sin φ cos z − cos φ sin z cos A plus law of sines provides us again with sin t cos δ = sin z sin A. And we’re here again! The hour angle is not determined uniquely, hence we must introduce one more equation (again the same law of cosines as before) cos z = cos (90 − δ) cos(90 − φ) + sin(90 − δ) sin(90 − φ) cos t thus cos z = sin δ sin φ + cos δ cos φ cos t. 7

Here we can substitute from the first boxed equation in this subsection for sin δ: cos z = sin2 φ cos z − cos φ sin z cos A sin φ + cos δ cos φ cos t. Again, we do the same trick with putting the first term to the left: cos z 1 − sin2 φ = − cos φ sin z cos A sin φ + cos δ cos φ cos t. cos z cos2 φ = − cos φ sin z cos A sin φ + cos δ cos φ cos t and again, we can peacefully divide by cos φ cos z cos φ = − sin z cos A sin φ + cos δ cos t and therefore cos δ cos t = cos z cos φ + sin z cos A sin φ. Hence the complete transformation H → Q reads sin t cos δ

=

sin z sin A

cos t cos δ

=

cos z cos φ + sin z cos A sin φ

sin δ

=

sin φ cos z − cos φ sin z cos A.

Again for the azimuth measured from the north we would have sin t cos δ

= − sin z sin A

cos t cos δ

=

cos z cos φ − sin z cos A sin φ

sin δ

=

sin φ cos z + cos φ sin z cos A.

You must pay attention to the quadrant where t lies, ok? Same as the previous case: firstly calculate δ, then sine and cosine of t. And then decide. Problem 2 Determine the right ascension and declination of a star, if you know that at the place with φ = 55◦ 460 and at Θ = 11h 11m 36s it appears in the altitude h = 40◦ 440 5000 and with azimuth A = 298◦ 280 5000 (measured from the south westwards). Solution: -90 to 90)

First of all we determine the declination (unique in the interval

sin δ = sin φ cos z − cos φ sin z cos A. → δ = 19◦ 390 3000 . Then we have

sin z sin A = −0.70713, cos δ cos z cos φ + sin z cos A sin φ = 0.70709. cos t = cos δ sin t =

8

Sine is negative and cosine positive, hence it lies in the 4th quadrant and t = 315◦ . Therefore the right ascension is α = Θ − t = 14h 11m 36s . Transformation to ecliptical coordinates could be derived quite analogically, but you don’t need it really often (at least, I won’t put it in the problems). Basically, there is a more elegant way by using the unit vectors and R-matrix. However, it would be too much for the beginning.

4

Time

We have already defined one kind of time and that was local sidereal time Θ (recall the definition!). Now we define the apprent solar time Tv (tempus solare verum) to be the hour angle of the Sun. However, the Sun doesn’t move continually on the sky and hence this is not very convenient for measuring the time. Therefore we shall define the mean solar time Tm (tempus solare medium) as the hour angle of so-called mean Sun - the point which moves continually along the celestial equator with a period of one sidereal year. Tm and Tv are related by a coefficient τ where τ = Tm − Tv . τ should be given. Last but not least, we define the zone time, the time which is commonly used. As you know, the Earth is divided in 24 zones. The zone time Tz for a given zone is given by the mean solar time for the center of the zone. For instance the time GMT +1 (CR) is the mean solar time for the east longitude 15◦ . LT time is the mean solar time for the longitude 30◦ .

9

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