Example 2.1 Trigonometric Functions in a Rectangular Spherical Triangle

Let the angle A be a right angle. When the figure is a plane triangle, the trigonometric functions of the angle B would be:

For the spherical triangle we have to use the equations in (2.7), which are now simply:

The first equation gives the sine of B:

Dividing the second equation by the third one, we get the cosine of B:

And the tangent is obtained by dividing the first equation by the second one:

The third equation is the equivalent of the Pythagorean theorem for rectangular triangles.

Example 2.2 The Coordinates of New York City

The geographic coordinates are 41° north and 74° west of Greenwich, or $ = +41°, X = -74°. In time units, the longitude would be74/15h = 4h 56min west of Greenwich. The geocentric latitude is obtained from



The geocentric latitude is 11' 26" less than the geographic latitude.

Example 2.3 The angular separation of two objects in the sky is quite different from their coordinate difference.

Suppose the coordinates of a star A are a1 = 10 h, 51 = 70° and those of another star B, a2 = 11 h, S2 = 80°.

Using the Pythagorean theorem for plane triangles, we would get







But if we use the third equation in (2.7), we get cos d = cos(«i — a2)

x sin(90c — St) sin(90c — 5-) + cos(90c — S1) cos(90c — S-) = cos(a1 — a2) cos S1 cos S2

+ sin 70c sin 80c = 0.983 , which yields d = 10.6c. The figure shows why the result obtained from the Pythagorean theorem is so far from being correct: hour circles (circles with a = constant) approach each other towards the poles and their angular separation becomes smaller, though the coordinate difference remains the same.

Example 2.4 Find the altitude and azimuth of the Moon in Helsinki at midnight at the beginning of 1996.

The right ascension is a = 2h 55min 7 s = 2.9186 h and declination S = 14c 42' = 14.70c, the sidereal time is & = 6h 19min 26 s = 6.3239 h and latitude $ = 60.16c.

The hour angle is h = & — a = 3.4053 h = 51.08c. Next we apply the equations in (2.16):

sin A cos a = sin 51.08c cos 14.70c = 0.7526 , cos A cos a = cos 51.08c cos 14.70c sin 60.16c — sin 14.70c cos 60.16c = 0.4008, sin a = cos51.08c cos 14.70c cos60.16c + sin 14.70c sin60.16c = 0.5225.

Thus the altitude is a = 31.5c. To find the azimuth we have to compute its sine and cosine:

Hence the azimuth is A = 62.0c. The Moon is in the southwest, 31.5 degrees above the horizon. Actually, this would be the direction if the Moon were infinitely distant.

Example 2.5 Find the topocentric place of the Moon in the case of the previous example.

The geocentric distance of the Moon at that time is R = 62-58 equatorial radii of the Earth. For simplicity, we can assume that the Earth is spherical.

We set up a rectangular coordinate frame in such a way that the z axis points towards the celestial pole and the observing site is in the xz plane. When the radius of the Earth is used as the unit of distance, the radius vector of the observing site is cos 0 0-4976 r0 = 0 = 0 -\sin 0/ \0-8674/

The radius vector of the Moon is

The topocentric place of the Moon is 37.53

which gives 8' = 14-00° and h' = 51-45°. Next we can calculate the altitude and azimuth as in the previous example, and we get a = 30-7°, A = 61-9°.

Another way to find the altitude is to take the scalar product of the vectors e and r0, which gives the cosine of the zenith distance:

cos z = e ■ r0 = 0-6047 x 0-4976 + 0-2420 x 0-8674 = 0-5108, whence z = 59-3° and a = 90° - z = 30-7°. We see that this is 0-8° less than the geocentric altitude; i.e. the difference is more than the apparent diameter of the Moon.

Example 2.6 The coordinates of Arcturus are a = 14 h 15-7 min, 8 = 19° 1' 1'. Find the sidereal time at the moment Arcturus rises or sets in Boston (0 = 42° 19'). Neglecting refraction, we get cos h = - tan 19° 11' tan 42° 19'

Hence, h =± 108-47° = 7 h 14 min. The more accurate result is cos h = - tan 19° 11' tan 42° 19' sin 35'

We divide this vector by its length 62.07 to get the unit vector e pointing to the direction of the Moon. This can be expressed in terms of the topocentric coordinates 5' and h':

= -0-331, whence h =± 109-35° = 7 h 17 min. The plus and minus signs correspond to setting ant rising, respectively. When Arcturus rises, the sidereal time is

0 = a + h = 14 h 16 min - 7h 17 min = 6h 59 min and when it sets, the sidereal time is

Note that the result is independent of the date: a star rises and sets at the same sidereal time every day.

Example 2.7 The proper motion of Aldebaran is ¡x = 0-20''/a and parallax n = 0-048''. The spectral line of iron at X = 440-5 nm is displaced 0-079 nm towards the

red. What are the radial and tangential velocities and the total velocity?

The radial velocity is found from


The tangential velocity is now given by (2.40), since i and n are in correct units:

The total velocity is v = ^vr2 + vt2 = V542 + 202 km/s = 58 km/s .

Example 2.8 Find the local time in Paris (longitude X = 2°) at 12:00.

Local time coincides with the zonal time along the meridian 15° east of Greenwich. Longitude difference 15°- 2° = 13° equals (13°/15°) x 60min = 52 minutes. The local time is 52 minutes less than the official time, or 11:08. This is mean solar time. To find the true solar time, we must add the equation of time. In early February, E.T. = -14 min and the true solar time is 11:08 - 14 min = 10:54. At the beginning of November, ET = +16 min and the solar time would be 11:24. Since -14 min and +16 min are the extreme values of E.T., the true solar time is in the range 10:54-11:24, the exact time depending on the day of the year. During daylight saving time, we must still subtract one hour from these times.

where T is the local solar time. This is accurate within a couple of minutes. Since the sidereal time runs about 4 minutes fast a day, the sidereal time, n days after the vernal equinox, is

At autumnal equinox culminates at 0:00 local time, and sidereal and solar times are equal.

Let us try to find the sidereal time in Paris on April 15 at 22:00, Central European standard time (= 23:00 daylight saving time). The vernal equinox occurs on the average on March 21; thus the time elapsed since the equinox is 10 +15 = 25 days. Neglecting the equation of time, the local time T is 52 minutes less than the zonal time. Hence

0 = T + 12 h + n x 4min = 21 h 8 min +12 h + 25 x 4min = 34h 48min = 10h48min .

The time of the vernal equinox can vary about one day in either direction from the average. Therefore the accuracy of the result is roughly 5 min.

Example 2.10 Find the rising time of Arcturus in Boston on January 10.

In Example 2.6 we found the sidereal time of this event, 0 = 6 h 59 min. Since we do not know the year, we use the rough method of Example 2.9. The time between January 1 and vernal equinox (March 21) is about 70 days. Thus the sidereal time on January 1 is

Example 2.9 Estimating Sidereal Time

Since the sidereal time is the hour angle of the vernal equinox V, it is 0 h when V culminates or transits the south meridian. At the moment of the vernal equinox, the Sun is in the direction of and thus culminates at the same time as V. So the sidereal time at 12:00 local solar time is 0:00, and at the time of the vernal equinox, we have

T = 0 - 7h 20 min = 6h59min-7h20min = 30 h 59 min - 7h 20min = 23 h 39 min .

The longitude of Boston is 71° W, and the Eastern standard time is (4°/15°) x 60 min = 16 minutes less, or 23:23.

Example 2.11 Find the sidereal time in Helsinki on April 15, 1982 at 20:00 UT.

The Julian date is J = 2,445,074.5 and


Next, we use (2.47) to find the sidereal time at 0 UT:

00 =- 1,506,521.0 s = -418 h 28 min 41 s = 13 h 31 min 19 s .

Since the sidereal time runs 3 min 57 s fast a day as compared to the solar time, the difference in 20 hours will be

and the sidereal time at 20 UT will be 13 h 31 min 19 s + 20 h 3 min 17 s = 33 h 34 min 36 s = 9h 34min 36 s.

At the same time (at 22:00 Finnish time, 23:00 daylight saving time) in Helsinki the sidereal time is ahead of this by the amount corresponding to the longitude of Helsinki, 25°, i. e. 1 h 40 min 00 s. Thus the sidereal time is 11 h 14 min 36 s.

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