Module 3
Module 3
Module 3
Celestial sphere - Astronomical terms and definitions - Motion of sun and stars -
Apparent altitude and corrections - Celestial co-ordinate systems - Different time
systems –Use of Nautical almanac - Star constellations - calculations for azimuth of a
line.
PART - A
TWO MARK QUESTIONS AND ANSWERS
1. Define Celestial Sphere.
The millions of stars that we see in the sky on a clear cloudless night are all at
varying distances from us. Since we are concerned with their relative distance rather
than their actual distance from the observer. It is exceedingly convenient to picture
the stars as distributed over the surface of an imaginary spherical sky having its
center at the position of the observer. This imaginary sphere on which the star
appears to lie or to be studded is known as the celestial sphere. The radius of the
celestial sphere may be of any value – from a few thousand meters to a few
thousand kilometers. Since the stars are very distant from us, the center of the earth
may be taken as the center of the celestial sphere.
8. Write a short note on: Co-altitude or Zenith Distance (z) and azimuth (A).
It is the angular distance of heavenly body from the zenith. It is the
complement or the altitude, i.e z = (90° - α).
The azimuth of a heavenly body is the angle between the observer’s meridian
and the vertical circle passing through the body.
9. Write a short note on: Declination (δ) and Co-declination or Polar Distance
(p).
The declination of a celestial body is angular distance from the plane of the
equator, measured along the star’s meridian generally called the declination circle,
(i.e., great circle passing through the heavenly body and the celestial pole).
Declination varies from 0° to 90°, and is marked + or – according as the body is north
or south of the equator.
It is the angular distance of the heavenly body from the near pole. It is the
complement of the declination. i.e., p = 90° - δ.
10. Define the terms: Hour Circle, Hour Angle and Right ascension (R.A).
Hour circles are great circles passing though the north and south celestial
poles. The declination circle of a heavenly body is thus its hour circle.
The hour angle of a heavenly body is the angle between the observer’s
meridian and the declination circle passing through the body. The hour angle is
always measured westwards.
Right ascension (R.A): It is the equatorial angular distance measured
eastward from the First Point of Aries to the hour circle through the heavenly body.
The spherical excess of a spherical triangle is the value by which the sum of
three angles of the triangle exceeds 180°.
Thus, spherical excess E = (A + B + C - 180°)
A spherical triangle is that triangle which is formed upon the surface of the
sphere by intersection of three arcs of great circles and the angles formed by the arcs
at the vertices of the triangle are called the spherical angles of the triangle.
The latitude is north or positive when measured above the equator, and is south or negative
when measured below the equator. The latitude of a point upon the equator is thus 0°, while at
the North and South Poles, it is 90° N and 90° S latitude respectively. The co-latitude is the
complement of the latitude, and is the distance between the point and pole measured along the
meridian.
The longitude (φ) of a place is the angle made by its meridian plane with some fixed
meridian plane arbitrarily chosen, and is measured by the arc of equator intercepted between
these two meridians. The prime meridian universally adopted is that of Greenwich. The
longitude of any place varies between 0° to 180°, and is reckoned as φ° east or west of
Greenwich. All the points on meridian have the same longitude.
2. What is a Spherical Triangle? Briefly explain its properties.
A spherical triangle is that triangle which is formed upon the surface of the sphere by
intersection of three arcs of great circles and the angles formed by the arcs at the vertices of
the triangle are called the spherical angles of the triangle.
AB, BC and CA are the three arcs of great circles and intersect each other at A, B and
C. It is usual to denote the angles by A, B and C and the sides respectively opposite to them,
as a, b and c. The sides of spherical triangle are proportional to the angle subtended by them
at the centre of the sphere and are, therefore, expressed in angular measure. Thus, by sin b we
mean the sine of the angle subtended at the centre by the arc AC. A spherical angle is an angle
between two great circles, and is defined by the plane angle between the tangents to the
circles at their point of intersection. Thus, the spherical angle at A is measured by the plane
angle A1AA2 between the tangents AA1 and AA2 to the great circles AB and AC.
The spherical excess of a spherical triangle is the value by which the sum of three
angles of the triangle exceeds 180°.
Thus, spherical excess E = ( A + B + C - 180° )
1 1 1 1 1
Also, tan²
E = tan s tan (s − a) tan (s − b) tan (s − c)
2 2 2 2 2
In geodetic work the spherical triangles on the earth’s surface are comparatively small
and the spherical excess seldom exceeds more than a few seconds of arc. The spherical
excess, in such case, can be expressed by the approximate formula
∆
E =
seconds
R 2 sin 1"
where R is the radius of the earth and ∆ is the area of triangle expressed in the same linear
units as R.
3. Derive the relationship between co-ordinates?
1. The Relation between Altitude of the Pole and Latitude of the Observer.
In the sketch, H-H is the horizon plane and E-E is the equatorial plane. O is the centre
of the earth. ZO is perpendicular to HH while OP is perpendicular to EE.
Now latitude of place = θ = ∠EOZ
And altitude of pole = α = ∠HOP
∠EOP = 90° = ∠EOZ + ∠ZOP
= θ + ∠ZOP …. (i)
∠HOZ = 90° = ∠HOP + ∠POZ
= α + ∠POZ …. (ii)
Equating the two, we get
θ + ∠ZOP = α + ∠POZ or θ=α
Hence the altitude of the pole is always equal to the latitude of the observer.
2. The Relation between Latitude of Observer and the Declination and Altitude of a
Point on the Meridian.
If the star is north of the zenith but above the pole, as at M2, we have
ZP = Z M2 + M2 P
or ( 90° - θ ) = ( 90° - α ) + p, where p = polar distance = M2 P
or θ=α–p …. (2)
Similarly, if the star is north of the zenith but below the pole, as at M3, we have
ZM3 = ZP + PM3
( 90° - α ) = (90° - θ) + p, where p = polar distance = M3 P
θ=α+p …. (3)
The above relations form the basis for the usual observation for latitude.
3. The Relation between Right Ascension and Hour Angle.
Fig 1.22. shows the plan of the stellar sphere on the plane of the equator. M is
the position of the star and ∠SPM is its westerly hour angle. HM. Y is the position of
the First Point of Aries and angle SPY is its westerly hour angle. ∠YPM is the rit
ascension of the star. Evidently, we have
∴ Hour angle of Equinox = Hour angle of star + R.A. of star.
4. Find the difference of longitude between two places A and B from their following
longitudes : ]
(1) Longitude of A = 40° W
Longitude of B = 73° W
(2) Long. Of A = 20° E
Long. Of B = 150° E
(3) Longitude of A = 20° W
Longitude of B = 50° W
Solution.
(1) The difference of longitude between A and B = 73° - 40° = 33°
(2) The difference of longitude between A and B = 150° - 20° = 130°
(3) The difference of longitude between A and B = 20° - (- 50°) = 70°
(4) The difference of longitude between A and B = 40° - ( - 150°) = 190°
Since it is greater than 180°, it represents the obtuse angular difference. The acute
angular difference of longitude between A and B, therefore, is equal to
360° - 190° = 170°.
5. Calculate the distance in kilometers between two points a and B along the parallel of latitude,
given that
(1) Lat. Of A, 28° 42’ N : longitude of A, 31° 12’ W
Lat. Of B, 28° 42’ N : longitude of B, 47° 24’ W
(2) Lat. Of A, 12° 36’ S : longitude of A, 115° 6’ W
Lat. Of B, 12° 36’ S : longitude of B, 150° 24’ E.
Solution.
The distance in nautical miles between A and B along the parallel of latitude =
difference of longitude in minutes x cos latitude.
6. Find the shortest distance between two places A and B, given that the longitudes of A and B are
15° 0’ N and 12° 6’ N and their longitudes are 50° 12’ E and 54° 0’ E respectively. Find also
the direction of B on the great circle route.
Radius of earth = 6370 km.
Solution.
6370 x 4 °.7 x
Now, arc ≈ radius x central angle π = 522 .54 km .
=
180 °
Hence distance AB = 522.54 km.
Direction of A from B :
The direction of A from B is the angle B, and the direction of B from A is the angle A.
Angles A and B can be found by the tangent semi-sum and semi-difference formulae
1
cos (a −b)
2 1
Thus tan ½ ( A + B) 1 cot p
2
= (a+b)
cos
2
1
sin (a
1
And tan ½ ( A – B ) = −b) cot 2 p
2
1
sin (a +b)
2
1
∴ tan (A +B ) cos 1°27 ' cot 1°54
=
2 cos 76 °27 '
A +
From = 89 ° 35 ' …. (i)
B
which,
2
sin 1 °27
'
1
and tan (A−B)= cot 1 °54 '
2 sin 76 °27
'
A −
From which, B = 38 °6' …. (ii)
2
∴ Direction of B from A = angle A = 89°35’ + 38°6’ = 127°41’ = S 52°19’ E
∴ Direction of A from B = angle B = 89°35’ + 38°6’ = 51°29’ = N 51°29’ W.
7. Determine the hour angle and declination of a star from the following data :
(i) Altitude of the star = 22° 36’
(ii) Azimuth of the star = 42° W
(iii) Latitude of the place of observation = 40° N.
Solution.
Since the azimuth of the star is 42° W, the star is in the western hemi-sphere.
In the astronomical ∆ PZM, we have
PZ = co-latitude = 90° - 40° = 50° ;
ZM = co-altitude = 90° - 22° 36° = 67 24’ ;
angle A = 42°
Knowing the two sides and the included angle, the third side can be calculated from
the cosine formula
Thus, cos PM = cos PZ . cos ZM + sin PZ. Sin ZM. cos A
= cos 50°. cos 67° 24’ + sin 50°. sin 67° 24’. cos 42°
= 0.24702 + 0.52556 = 0.77258
∴ PM = 39° 25’
∴ Declination of the star = δ = 90° - PM = 90° - 39° 25’ = 50° 35’ N.
Similarly, knowing all the three sides, the hour angle H can be calculated from Eq.
1.2
cos H = cos ZM − cos PZ . cos PM = cos 67 °24 '− cos 50 °. cos 39 °25 '
sin PZ . sin PM sin 50 °. sin 39 °25 '
0.38430 −
0.49659
= = = −0.23086
0.48640
∴ cos ( 180° - H ) = 0.23086 ∴ 180° - H = 76° 39’
H = 103° 21’.
8. Explain the astronomical parameters of the earth and the sun. The
Earth:
The Earth is considered approximately spherical in shape. But actually it is very
approximately an oblate spheroid. Oblate spheroid is the figure formed by revolving an ellipse
about its minor axis. The earth is flattened at poles – its diameter along the polar axis being
lesser than its diameter at the equator. The equatorial radius a of the earth, according to
Hayford’s spheroid is 6378.388 km and the polar radius b of the earth is 6356.912 km. The
The sun:
The sun is at a distance of 93,005,000 miles from the earth. The distance is only about
1
of that of the nearest star. The diameter of the sun is about 109 times the diameter of
250,000
the earth, and subtends and angle of 31’ 59” at the centre of the earth. The mass of the sun is
about 332,000 times that of the earth. The temperature at the centre of the sun is computed to
be about 20 million degrees.
The sun has twp apparent motions, one with respect to the earth from east to west, and
the other with respect to the fixed stars in the celestial sphere. The former apparent path of the
sun is in the plane which passes through the centre of the celestial sphere and intersects it in a
great circle called the ecliptic. The apparent motion of the sun is along this great circle. The
angle between the plane of equator and the ecliptic is called the Obliquity of Ecliptic, its value
being 23° 27’. The obliquity of ecliptic changes with a mean annual diminution of 0’.47.
The points of the intersection of the ecliptic with the equator are called the equinoctial
points, the declination of the sun being zero at these points. The Vernal Equinox or the First
point of Aries (Y) is the point in which the sun’s declination changes from south to north.
Autumnal Equinox or the First point of Libra ( Ω ) is the point in which the sun’s declination
changes from north to south. The points at which sun’s declinations are a maximum are called
solstices. The point at which the north declination of sun is maximum is called the summer
solstice, while the point at which the south declination of the sun is maximum is known as the
winter solstice.
The earth moves eastward around the sun once in a year in a pat that is very nearly a
huge circle with a radius of about 93 millions of miles. More accurately, the path is described
as an ellipse, one focus of the ellipse being occupied by the sun.
This supports the preposition proved with reference to Fig. 1.30 (a). The relationship
is true for all positions of the star.
When the star is on the meridian, its hour angle is zero. Hence equation 1 reduces to
Star’s right ascension = local sidereal time at its transit.
A sidereal clock, therefore, records the right ascension of stars as they make their
upper transits.
The hour angle and the right ascension are generally measured in time in preference to
angular units. Since one complete rotation of celestial sphere through 360° occupies 24 hours,
we have
24 hours = 360° ; 1 hour = 15°
The difference between the local sidereal times of two places is evidently equal to the
difference in their longitudes.
Solar Apparent Time:
Since a man regulates his time with the recurrence of light and darkness due to rising
and setting of the sun, the sidereal division of time is not suited to the needs of every day life,
for the purposes of which the sun is the most convenient time measurer. A solar day is the
interval of time that elapes between two successive lower transits of the sun’s centers over the
meridian of the place. The lower transit is chosen in order that the date may change at mid-
night. The solar time at any instant is the hour angle of the sun’s centre reckoned westward
from 0h to 24h. This is called the apparent solar time, and is the time indicated by a sun-dial.
Unfortunately, the apparent solar day is not of constant length throughout the year but
changes. Hence our modern clocks and chronometers cannot be used to give us the apparent
solar time. The non-uniform length of the day is due to two reasons :
(1) The orbit of the earth round the sun is not circular but elliptical with sun at one of
its foci. The distance of the earth from the sun is thus variable. In accordance with the law of
gravitation, the apparent angular motion of the sun is not uniform – it moves faster when is
nearer to the earth and slower when away. Due to this, the sun reaches the meridian
sometimes earlier and sometimes later with the result that the days are of different lengths at
different seasons.
(2) The apparent diurnal path of the sun lies in the ecliptic. Due to this, even though
the eastward progress of the sun in the ecliptic were uniform, the time elapsing between the
departure of a meridian from the sun and its return thereto would vary because of the
obliquity of the ecliptic.
The sun changes its right ascension from 0h to 24h in one year, advancing eastward
among the stars at the rate of about 1° a day. Due to this, the earth will have to turn nearly
361° about its axis to complete one solar day, which will consequently be about minutes
longer than a sidereal day. Both the obliquity of the ecliptic and the sun’s unequal motion
cause a variable rate of increase of the sun’s right ascension. If the rate of change of the sun’s
right ascension were uniform, the solar day would be of constant length throughout the year.
Again, since the our angle of the sun (true or mean) is zero at its upper transit while
the solar time (apparent or mean) is zero as its lower transit, we have
The apparent solar time = the hour angle of the sun + 12h … (5)
The mean solar time = the hour angle of mean sun + 12h … (6)
Thus, if the hour angle of the mean sun is 15° (1 hour) the mean time is 12 + 1 = 13
hours, which is the same thing as 1 o’clock mean time in the afternoon; if the hour angle of
the mean sun is 195° (13 hours), the mean time is 12 + 13 =25 hours, which is the same as 1
o’clock mean time after the midnight (i.e., next. Day).
The equation of time is positive when the apparent solar time is more than the mean
solar time ; to get the apparent solar time, the equation of time should then be added to mean
solar time. For example, at 0h G.M.T. on 15 October 1949, the equation of the time is + 13m
12s. This means that the apparent time at 0h mean time is 0h 13m 12s. In other words, the true
sun is 13m 12s ahead of the mean sun. Similarly, the equation of time is negative when the
apparent time is less than the mean time. For example, at 0h G.M.T. on 18 Jan., 1949, the
equation of time is – 10m 47s. This means that the apparent time at 0h mean time will be 23h
49m 13s on January 17. In other words, the true sun is behind the mean sun at that time.
The value of the equation of time varies in magnitude throughout the year and its
value is given in the Nautical Almanac at the instant of apparent midnight for the places on
the meridian of Greenwich for each day of the year. For any other time it must be found by
adding or subtracting the amount by which the equation has increased or diminished since
midnight.
It is obvious that the equation of time is the value expressed in time, of the difference
at any instant between the respective hour angles or right ascensions of the true and mean
suns.
The amount of equation of the time and its variations are due to two reasons :
(1) obliquity of the ecliptic, and (2) elasticity of the orbit. We shall discuss both the effects
separately and then combine them to get the equation of time.
11. Explain the conversion of local time to standard time and vice versa.
The difference between the standard time and the local mean time at a place is equal to
the difference of longitudes between the place and the standard meridian.
If the meridian of the place is situated east of the standard meridian, the sun, while
moving apparently from east to west, will transit the meridian of the place earlier than the
standard meridian. Hence the local time will be greater than the standard time. Similarly, if
the meridian of the place is to the west of the standard meridian, the sun will transit the
standard meridian earlier than the meridian of the place and hence the local time will be lesser
than the standard time. Thus, we have
Use (+) sign if the meridian of place is to the east of the standard meridian, and (-)
Sign if it to the west of the standard meridian.
If the local time is to be found from the given Greenwich time, we have
Solution:
(a) The longitude of the place = 20° E
Longtitude of the standard meridian = 82° 30’E
∴ Difference in the longitudes = 82° 30’ – 20° = 62° 30’, E. the place being to the west of
the standard meridian.
62
Now 62° of longitude = hms
15 h = 4 8 0
30 h m s
Now 30’ of longitude = m=020
15
-----------------------------------
h m s
Total = 4 10 0