## Thursday, March 10, 2011

### More Spherical Astronomy

We left off in the last instalment with a problem to obtain Mercury's location in the sky on April 1, given by its horizontal (observer-based) coordinates, of Altitude and azimuth. To give the problem again:

The nearest planet to the Sun, Mercury, is due to make its brightest appearance on April 1. It will have a Right Ascension of 1 h 24 m and a declination of 12 deg 27 '.If an observer (located at latitude 40 deg north) wishes to find it, where would he look at about 6.45 p.m.? Give the azimuth, A and the altitude a.

First, one needs to obtain the sidereal time. From a table of sidereal times this is ~ 7h 25 m. (For an observer at 40 N).

Next, we change the Right Ascension to hour angle using: h = ST - RA

So: h = 7h 25m - 1h 24m = 6h 01m

This is then converted into degrees, using the fact that there are 15 degrees/ hr.

So: 6h 01 m ~ 90 degrees (e.g. 6h x 15 deg/ h = 90 deg)

We then find the zenith distance, z, using an astronomical triangle using sides with hour angle, declination and latitude. This implies the spherical version of the law of cosines, and by analogy with the previous solution we gave, e.g.

http://brane-space.blogspot.com/2011/03/introducing-practical-astronomy.html

we have:

cos z = sin (decl.) sin (Lat) + cos (decl.) cos h cos (Lat)

We note the following respective values:

sin (decl.) = sin 12.45 = 0.2155

sin (Lat) = sin 40 = 0.6427

cos (decl.) = cos 12.45 = 0.9764

cos h = cos 90 = 0

so, effectively:

cos z = sin (decl.) sin (Lat) = (0.2155)(0.6427) = 0.1385

Thence:

z = arc cos (0.1385) = 82 degrees

Then, a (altitude) = 90 deg - z = 90 deg - 82 deg = 8 degrees

Which shows that we need to look about 8 degrees above the horizon.

But which direction? This requires the azimuth:

The appropriate astronomic triangle yields the following equation for A, azimuth:

tan A = -cos(decl.) sin h / [sin(decl.)cos(Lat) - cos(decl) cos h sin (Lat)]

The additional values we need to what was given above are:

sin h = sin 90 = 1

cos (Lat) = cos 40 = 0.7660

Then the computation is displayed as follows:

tan A = -(0.9764)/ {(0.2155)(0.7660)}

since the 2nd term in the denominator drops out, as cos h = cos 90 = 0

then:

tan A = -(0.9764)/ (0.1650) = - 5.914

Then:

A = arc tan (-5.914) = -80.4 deg

Since the value is negative, we take:

A = 360 - 80.4 = 279.5 deg

which pins it just about at the azimuth shown in the diagram (Click on it to enlarge- note the azimuth degrees marked on periphery, i.e. below the horizon)

A similar problem for next time:

You are located in San Francisco, and the sidereal time at 11.45 p.m. local time for your location on this date, is 14 h 15 m, approximately. Saturn is visible and is at 12h 47 m Right Ascension, and Declination (- 2 deg 10m).

If your latitude is 37 deg 46 ½' north, find where Saturn will be visible in terms of its altitude and azimuth.