Celestial equator on a star map. Star coordinates. Celestial coordinates. Astronomy. Determination of geographic latitude
How can I find my star?
In addition to the Star Map, there are many other options for finding stars. Especially for you, OSR has developed several unique applications for convenient and fun search for stars - this is the OSR Star Finder mobile application and the One Million Stars browser application.
In this article we will describe in detail how to use several applications to find a star by name with coordinates RA 13h03m33.35 -49°31’38.1” dec 4.83 mag Cen.
All about coordinates
- Abbreviation RA means “Right Ascension”; "dec" means "declination". These values are similar to latitude and longitude, but refer to celestial coordinates.
- Mag means “stellar magnitude” (English magnitude) and characterizes the brightness of a star. Bright stars reaching magnitude 6.5 can be seen with the naked eye. With binoculars you can see stars up to 10 magnitude units. To see stars with larger magnitudes, you will need an amateur telescope.
- Cen, in this case, means “Centaurus” - this is one of the 88 constellations in the sky. Knowing which constellation your star is in will make it easier to find it.
OSR Star Finder App
The OSR Star Finder app makes it easy to find a star in the night sky. To do this, you just need to enter the OSR code and point the phone at the sky. If the star is not visible, then you are in the other hemisphere. In this case, the application will help you determine when the star will become visible, and will also show you where it is visible from at a given time.
Google Earth
To find a star using the free Google Earth app, follow these steps:
- In the top panel, point to the ‘Planet’ icon and select ‘Sky’ from the drop-down list
- On the left in the search window, enter the coordinates of the star in the following format: 13:03:33.35 -49:31:38.1. This information is extracted from the coordinates RA 13h03m33.35 -49°31’38.1” dec 4.83 mag Cen
You can also find a star via Google Sky from your personal page
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The purpose of the lesson: introduce students to stellar coordinates, instill the skills of determining these coordinates on a model of the celestial sphere.
Equipment: video projector, model of the celestial sphere
During the classes
Teacher: Since time immemorial, people have identified separate groups of bright stars in the starry sky, united them into constellations, giving them names that reflected the way of life and the peculiarities of their thinking. This is what ancient Chinese, Babylonian, and Egyptian astronomers did. Many of the constellation names we use today come from ancient Greece, where they evolved over centuries.
Table 1 Chronicle of names
At the Congress of the International Astronomical Union in 1922, the number of constellations was reduced to 88. At the same time, the current boundaries between them were established.
It deserves special mention. That the proximity of stars in constellations is apparent, is how an observer from Earth sees them. In fact, the stars lag behind each other at great distances, and for us their visibility is, as it were, projected onto celestial sphere- an imaginary transparent ball, in the center of which is the Earth (observer), onto the surface of which all the luminaries are projected as the observer sees them at a certain moment in time from a certain point in space. Presentation. Slide 1
Moreover, the stars in the constellations are different; they differ in apparent size and light. The brightest stars in the constellations are designated by letters of the Greek alphabet in descending order (a, b, g, d, e, etc.) of brightness.
This tradition was introduced by Alessandro Piccolomini (1508–1578), and consolidated by Johann Bayer (1572–1625).
Then John Flamsteed (1646–1719) within each constellation designated the stars by serial number (for example, the star 61 Cygnus). Stars with variable brightness are designated by Latin letters: R, S, Z, RR, RZ, AA.
Now we will look at how the location of the luminaries in the sky is determined.
Let's imagine the sky in the form of a giant globe of arbitrary radius, in the center of which the observer is located.
However, the fact that some luminaries are located closer to us, while others are further away, is not visible to the eye. Therefore, let us assume that all stars are at the same distance from the observer - on the surface celestial sphere. Presentation. Slide 1
Since the stars change their position during the day, we can conclude about the daily rotation of the celestial sphere (this is explained by the rotation of the Earth around its axis). The celestial sphere rotates around a certain axis PP` from east to west. The axis of apparent rotation of the sphere is the axis of the world. It coincides with the earth's axis or is parallel to it. The axis of the world intersects the celestial sphere at points P – north celestial pole and P`- south celestial pole. The North Star (a Ursa Minor) is located near the north pole of the world. Using a plumb line, we determine the vertical and depict it in the drawing. Presentation. Slide 1
This straight line ZZ` is called plumb line. Z – zenith, Z`- nadir. Through point O - the intersection of the plumb line and the axis of the world - we draw a straight line perpendicular to ZZ`. This is NS - noon line(N- north, S – south). Objects illuminated by the Sun at noon cast a shadow in the direction along this line.
Two mutually perpendicular planes intersect along the noon line. A plane perpendicular to a plumb line that intersects the celestial sphere in a great circle is true horizon. Presentation. Slide 1
The plane perpendicular to the true horizon passing through the points Z and Z` is called celestial meridian.
We have drawn all the necessary planes, now let's introduce another concept. Let us arbitrarily place a star on the surface of the celestial sphere M, draw through points Z and Z` and M big semicircle. This - height circle or vertical
The instantaneous position of the star relative to the horizon and the celestial meridian is determined by two coordinates: height(h) and azimuth(A). These coordinates are called horizontal.
The altitude of the luminary is the angular distance from the horizon, measured in degrees, minutes, arc seconds ranging from 0° to 90°. More height replaced by an equivalent coordinate – z – zenith distance.
The second coordinate in the horizontal system A is the angular distance of the vertical of the luminary from the point of south. Defined in degrees minutes and seconds from 0° to 360°.
Notice how the horizontal coordinates change. Light M during the day describes a daily parallel on the celestial sphere - this is a circle of the celestial sphere, the plane of which is perpendicular axis mundi.
<Отработка навыка определения горизонтальных координат на небесной сфере. Самостоятельная работа учащихся>
When a star moves along the daily parallel, the highest point of ascent is called upper climax. Moving under the horizon, the luminary will end up at a point, which will be a point lower climax. Presentation. Slide 1
If we consider the path of the star we have chosen, we can see that it is rising and setting, but there are non-setting and non-rising luminaries. (Here - relative to the true horizon.)
Let's consider the change in the appearance of the starry sky throughout the year. These changes are not as noticeable for most stars, but they do occur. There is a star whose position changes quite dramatically, this is the Sun.
If we draw a plane through the center of the celestial sphere and perpendicular to the axis of the world PP`, then this plane will intersect the celestial sphere in a great circle. This circle is called celestial equator. Presentation. Slide 2
This celestial equator intersects with the true horizon at two points: east (E) and west (W). All daily parallels are located parallel to the equator.
Now let's draw a circle through the poles of the world and the observed star. The result is a circle - a circle of declination. The angular distance of the luminary from the plane of the celestial equator, measured along the declination circle, is called the declination of the luminary (d). Declination is expressed in degrees, minutes and seconds. Since the celestial equator divides the celestial sphere into two hemispheres (northern and southern), the declination of stars in the northern hemisphere can vary from 0° to 90°, and in the southern hemisphere - from 0° to -90°.
The declination of the luminary is one of the so-called equatorial coordinates.
The second coordinate in this system is right ascension (a). It is similar to geographic longitude. Right ascension is counted from vernal equinox points (g). The Sun appears at the vernal equinox on March 21st. Right ascension is measured along the celestial equator in the direction opposite to the daily rotation of the celestial sphere. Presentation. Slide 2. Right ascension is expressed in hours, minutes and seconds of time (from 0 to 24 hours) or in degrees, minutes and seconds of arc (from 0° to 360°). Since the position of stars relative to the equator does not change when the celestial sphere moves, equatorial coordinates are used to create maps, atlases and catalogs.
Since ancient times it was noticed that the Sun moves among the stars and describes a full circle in one year. The ancient Greeks called this circle ecliptic, which has been preserved in astronomy to this day. Ecliptic inclined to the plane of the celestial equator at an angle of 23°27` and intersects with the celestial equator at two points: the vernal equinox (g) and the autumn equinox (W). The Sun travels the entire ecliptic in a year; it travels 1° per day.
The constellations through which the ecliptic passes are called zodiac. Every month the Sun moves from one constellation to another. It is virtually impossible to see the constellation in which the Sun is located at noon, since it obscures the light of the stars. Therefore, in practice, at midnight we observe the zodiacal constellation, which is the highest above the horizon, and from it we determine the constellation where the Sun is located at noon (Figure No. 14 of the Astronomy 11 textbook).
We should not forget that the annual movement of the Sun along the ecliptic is a reflection of the actual movement of the Earth around the Sun.
Let us consider the position of the Sun on a model of the celestial sphere and determine its coordinates relative to the celestial equator (repetition).
<Отработка навыка определения экваториальных координат на небесной сфере. Самостоятельная работа учащихся>
Homework.
- Know the contents of paragraph 116 of the Physics-11 textbook
- Know the contents of paragraphs 3, 4 of the textbook Astronomy -11
- Prepare material on the topic “Zodiac constellations”
Literature.
- E.P. Levitan Astronomy 11th grade – Enlightenment, 2004
- G.Ya. Myakishev and others. Physics 11th grade - Enlightenment, 2010
- Encyclopedia for children Astronomy - ROSMEN, 2000
On long winter nights, astronomers measure the zenith distances of the same stars in both culminations and, using formulas (4), (6), (9), independently find their declination (δ) and geographic latitude (φ) of the observatory. Knowing φ, they determine the declination of the luminaries for which only the upper culmination is observed. For high-precision measurements, refraction is taken into account, which is not considered here, except when the stars are located near the horizon.
At true noon, the zenith distance z of the Sun is regularly measured and the reading Sch of the star clock is noted, then its declination δ is calculated using formula (4), and its right ascension αsun is calculated from it, since
sin α =tg δ -ctg ε, (24)
where ε = 23°27" is the already known inclination of the ecliptic.
At the same time, the sidereal clock correction is determined
us = S-Sch = α -Sch, (25)
since at true noon the hour angle of the Sun t = 0 and therefore, according to formula (13), sidereal time S = α.
Noting the readings S "h of the same clock at the moments of the upper culmination of bright stars (they are visible in telescopes during the day), their right ascension is found
α=α + (S"h-Sch) (26)
and from it the right ascension of the remaining luminaries is determined in a similar way, which can also be found as
α=S"h +us. (27)
Using the equatorial coordinates (α and δ) of stars published in astronomical reference books, the geographic coordinates of places on the earth's surface are determined.
Example 1. At true noon on May 22, 1975, the zenith distance of the Sun in Pulkovo was 39°33" S (above the south point), and the sidereal clock showed 3h57m41s. Calculate the equatorial coordinates of the Sun and the sidereal clock correction for this moment. Geographic latitude of Pulkovo φ = +59 °46".
Data: z =39°33" S; Sch = 3h57m41s; φ= + 59°46".
Solution. According to formula (4), the declination of the Sun
δ =φ-z = 59°46"-39°33" = +20°13". According to formula (24)
sinα = tanδ -ctgε = tan 20°13" - ctg 23°27" = +0.3683-2.3053=+0.8490,
whence the direct ascension of the Sun is α = 58°06",2, or, converted into time units, α = 3h52m25s.
Since at true noon, according to formula (13), sidereal time S = α = 3h52m25s, and the sidereal clock showed Sch = 3h57m41s, then, according to formula (25), the clock correction
us=S-Sch=α -Sch = 3h52m25s-3h57m41s= -5m16s.
Example 2. At the moment of the upper culmination of the star α Draco at a zenith distance of 9°17" to the north, the sidereal clock showed 7h20m38s, and its correction to sidereal Greenwich time was +22m16s. The equatorial coordinates of α Draco: right ascension 14h03m02s and declination + 64°37". Determine the geographic coordinates of the observation site.
Data: star, α = 14h03m02s, δ=+64°37", zв = 9°17" N; sidereal hours Sch = 7h20m38s, us = 22m16s.
Solution. According to formula (6), geographic latitude
φ = δ-zв = + 64°37"-9° 17"= + 55°20".
According to formula (13), sidereal time at the observation location
S =α=14h03m02s, and sidereal time at Greenwich S0 = Sch+us=7h20m38s+22m16s = 7h42m54s.
Therefore, according to formula (14), geographic longitude
λ = S-S0 = 14h03m02s-7h42m54s = 6h20m08s,
or, converted to angular units, λ=95°02".
Problem 70. Determine the geographic latitude of the observation site and the declination of the star by measuring its zenith distance z or height h at both culminations - upper (in) and lower (n):
a) zв=15°06"W, zн=68°14"N;
b) zв=15°06" S, zн=68°14" N;
c) hв=+80°40" S, zн=72°24" c;
d) hв=+78°08"S, hн= + 17°40"S.
Problem 71. In an area with a geographical latitude φ = = +49°34" the star α Hydra passes its upper culmination at an altitude of +32°00" above the point of the south, and the star β Ursa Minor - north of the zenith at a distance of 24°48". What is equal to declination of these stars?
Problem 72. What is the declination of the stars that, at their highest culmination in Canberra (φ = -35°20"), are at a zenith distance of 63°39" north of the zenith and at an altitude of +58°42" above the point south?
Problem 73. In Dushanbe, the star Capella (α Aurigae) passes its upper culmination at an altitude of +82°35" with an azimuth of 180°, and the star Aldebaran (α Tauri), whose declination is +16°25", at a zenith distance of 22°08" south of zenith What is the declination of Capella?
Problem 74. Calculate the declination of the stars δ Ursa Major and Fomalhaut (α Southern Pisces), if the difference in the zenith distances of these stars and Altair (α Orel) at the upper culmination in Tashkent (φ=+41°18") is -48°35" and +38, respectively °38". Altair culminates in Tashkent at an altitude of +57°26" above the point of the south.
Problem 75. What is the declination of the stars that culminate on the horizon and at the zenith of Tbilisi, whose geographic latitude is + 41°42"? Refraction at the horizon is assumed to be 35".
Problem 76. Find the right ascension of the stars, at the moments of the upper culmination of which the sidereal clock showed 18h25m32s and 19h50m40s, if at their reading of 19h20m16s the star Altair (α Orla) with a right ascension of 19h48m21s crossed the celestial meridian south of the zenith.
Problem 77. At the moment of the upper culmination of the Sun, its right ascension was 23h48m09s, and the sidereal clock showed 23h50m01s. 46m48s before this, the star β Pegasus crossed the celestial meridian, and when the same clock read 0:07m40s, the upper culmination of the star α Andromeda occurred. What is the right ascension of these two stars?
Problem 78. On October 27, 1975, in Odessa, Mars culminated 15m50s by sidereal clock after the star Betelgeuse (α Orion) at an altitude exceeding the height of this star at the culmination by 16°33", Betelgeuse's right ascension is 5h52m28s and declination +7°24 ". What were the equatorial coordinates of Mars and near what point of the ecliptic was it located?
Problem 79. On August 24, 1975 in Moscow (φ = +55°45"), when the sidereal clock showed 1h52m22s, Jupiter crossed the celestial meridian at a zenith distance of 47°38". At 2h23m31s, according to the same clock, the star α Aries, whose right ascension is 2h04m21s, culminated. What were the equatorial coordinates of Jupiter?
Problem 80. At a point with a geographic latitude of +50°32" the midday altitude of the Sun on May 1 and August 11 was + 54°38", and on November 21 and January 21 +19°29". Determine the equatorial coordinates of the Sun on these days.
Problem 81. At true noon on June 4, 1975, the Sun passed in Odessa (φ = +46°29") at an altitude of +65°54", and 13m44s before that the star Aldebaran (α Taurus) crossed the celestial meridian at a zenith distance exceeding the midday zenith the distance of the Sun is 5°58". Determine the equatorial coordinates of the Sun and the star.
Problem 82. On October 28, 1975 at 13h06m41s maternity time at the point with λ = 4h37m11s (n=5) and φ=+41°18" the zenith distance of the Sun was 54°18". 45m45s (sidereal time) before this, the star Spica (α Virgo) was at the upper culmination, and 51m39s after it, the star Arcturus (α Bootes) was at an altitude of +68°01"S. Determine the equatorial coordinates of the Sun and Arcturus. Equation of time on this day it was 16m08s.
Problem 83. Find the geographic latitude of the area in which the stars β Perseus (δ = +40°46") and ε Ursa Major (δ = +56°14") at the moments of the upper culmination are at the same zenith distance, but the first is to the south, and the second - north of the zenith.
Problem 84. At the moments of the upper culmination, the star α Canes Venatici with a declination of +38°35" passes at the zenith, the star β Orionis is 46°50" to the south, and the star α Perseus is 11°06" to the north. At what geographic parallel were the measurements taken and why is the declination of these stars equal?
Problem 85. At the moment of the upper culmination of the Sun, the average chronometer showed 10h28m30s, and when it showed 14h48m52s, a 12-hour radio signal of the exact time was received from Greenwich. Find the geographic longitude of the observation site if the equation of time on that day was +6m08s.
Problem 86. At the moment of the upper culmination of the star ι Hercules at a zenith distance of 2°14" north of the zenith, sidereal Greenwich time was 23h02m39s. The equatorial coordinates of ι Hercules α = 17h38m03- and δ = +46°02", Determine the geographic coordinates of the observation site.
Problem 87. At the moment the stellar chronometer read 18:07:27 s, the expedition received a radio signal of the exact time, transmitted from Greenwich at 18:07:00 Greenwich sidereal time. At the moment of the upper culmination of the star γ Cassiopeia at a zenith distance of 9°08" south of the zenith, the reading of the same chronometer was 19h17m02s. The equatorial coordinates of γ Cassiopeia are α = 0h53m40s and δ = +60°27". Find the geographical coordinates of the expedition.
Problem 88. At true noon, the expedition's average chronometer reading was 11h41m37s, and at the time of receiving the 12-hour radio signal of the exact time from Moscow, the same chronometer showed 19h14m36s. The measured zenith distance of the star α Cygni (δ = +45°06") at the upper culmination turned out to be 3°26" north of the zenith. Determine the geographical coordinates of the expedition if on the day of the observations the equation of time was -5m 17s.
Problem 89. At true noon, the navigator of the ocean liner measured the altitude of the Sun, which turned out to be +75°41" with an azimuth of 0°. At this moment, the average chronometer with an adjustment of 16m.2 showed 14h12m.9 Greenwich time. The declination of the Sun, indicated in the naval astronomical yearbook, was +23°19", and the time equation is +2m55s. What geographic coordinates did the liner have, where and on what approximately days of the year was it located at that time?
Answers - Practical determination of geographic and celestial equatorial coordinates
Conversion of celestial coordinates and time systems. Sunrise and sunset
The connection between horizontal and equatorial celestial coordinates is carried out through the parallactic triangle PZM (Fig. 3), the vertices of which are the celestial pole P, the zenith Z and the luminary M, and the sides are the arc ΡΖ of the celestial meridian, the arc ΖΜ of the altitude circle of the luminary and the arc RM of its declination circle . It is obvious that ΡΖ = 90°-φ, ZM = z = 90°-h and PM = 90°-δ, where φ is the geographic latitude of the observation site, z is the zenith distance, h is the altitude and δ is the declination of the star.
In a parallactic triangle, the angle at the zenith is equal to 180°-A, where A is the azimuth of the luminary, and the angle at the celestial pole is the hour angle t of the same luminary. Then the horizontal coordinates are calculated using the formulas
cos z = sin φ sin δ + cos φ cos δ cos t, (28)
sin z · cos A = - sin δ · cos φ+cos δ · sin φ · cos t, (29)
sin z · sin A = cos δ · sin t, (30)
and equatorial coordinates - according to the formulas
sin δ = cos z sin φ - sin z cos φ cos A, (31)
cos δ · cos t = cos z · cos φ+sin z · sin φ · cos A, (32)
cos δ · sin t=sin z · sin A, (30)
where t = S - α, where α is the right ascension of the luminary and S is sidereal time.
Rice. 3. Parallax Triangle
When making calculations, according to Table 3, it is necessary to convert sidereal time intervals ΔS into average time intervals ΔT (or vice versa), and sidereal time s0 to Greenwich Mean Midnight of a given date should be borrowed from astronomical yearbook calendars (in the problems of this section, the values of s0 are given).
Let some phenomenon occur at some point on the earth's surface at the moment T according to the time accepted there. Depending on the adopted time counting system, using formulas (19), (20) or (21), the average Greenwich time T0 is found, which is the average time interval ΔT that has elapsed since Greenwich midnight (ΔT=T0). This interval according to Table 3 is translated into the sidereal time interval ΔS (i.e. ΔT→ΔS), and then at a given moment T corresponding to Greenwich mean time T0, sidereal time in Greenwich
and at this point
where λ is the geographical longitude of the place,
The conversion of sidereal time intervals ΔS into average time intervals ΔΤ = Τ0 (i.e. ΔS→ΔT) is carried out according to Table 3 by subtracting the correction.
The moments of time and azimuths of the points of sunrise and sunset are calculated using formulas (28), (29), (30) and (13), in which z=90°35" is assumed (taking into account refraction ρ = 35").
The found values of the hour angle and azimuth in the range from 180 to 360° correspond to the sunrise, and in the range from 0 to 180° - to its setting.
When calculating the sunrise and sunset, its angular radius r = 16 is also taken into account. The found hour angles t give moments in true solar time (see formula (17), which in formula (16) are translated into moments of average time, and then into accepted counting system.
The moments of sunrise and sunset of all luminaries are calculated with an accuracy not exceeding 1 m.
Converting Celestial Coordinates and Timing Systems - Example 1
In what direction was a telescope with a camera installed in advance to photograph the solar eclipse on April 29, 1976, if at a point with geographic coordinates λ = 2h58m.0 and φ = +40°14" the middle of the eclipse occurred at 15h29m.8 at a time different from Moscow at +1h? At this moment, the equatorial coordinates of the Sun are: right ascension α=2h27m.5 and declination δ= + 14°35". At Greenwich Mean Midnight on April 29, 1976, sidereal time s0=14h28m19c.
Data: observation point, λ = 2h58m.0, φ = +40°14", T=15h29m.8, Τ-Tm=1h; s0 = 14h28m19c = 14h28m.3; Sun, α=2h27m.5, δ = + 14°35".
Solution. In the middle of the eclipse, Moscow time Tm = T-1h = 14h29m.8, and therefore Greenwich mean time T0 = Tm-3h = 11h29m.8. Since Greenwich midnight, the time interval ΔТ = Т0 = 11h29m,8 has passed, which we translate according to Table 3 into the sidereal time interval ΔS = 11h31m,7, and then at the moment T0, according to formula (33), sidereal time in Greenwich
S0=s0+ΔS = 14h28m.3 + 11h31m.7 = 25h60m = = 2h0m.0
and at a given point, according to formula (14), sidereal time S = S0+λ=2h0m.0 + 2h58m.0 = 4h58m.0
and, according to formula (13), the hour angle of the Sun
t = S-α = 4h58m, 0-2h27m, 5 = 2h30m, 5,
or, translating from Table 1, t = 37°37",5 ~ 37°38". Using the tables of trigonometric functions we find:
sin φ = sin 40°14" = +0.6459,
cos φ = cos 40°14" = +0.7634;
sin δ = sin 14°35" = +0.2518,
cos δ = cos 14°35" = +0.9678;
sin t = sin 37°38" = +0.6106,
cos t = cos 37°38" = +0.7919.
Using formula (28) we calculate
cos z = 0.6459 · 0.2518 + 0.7634 · 0.9678 · 0.7919 = = +0.7477
and from the tables we find z = 41°36" and sin z = +0.6640. To calculate the azimuth we use formula (30):
from where we get two values: A = 62°52" and A = 180° - 62°52" = 117°08". At δ<φ значения A и t не слишком резко отличаются друг от друга и поэтому A=62°52".
Consequently, the telescope was aimed at a point in the sky with horizontal coordinates A=62°52" and z = 41°36" (or h = + 48°24").
Conversion of celestial coordinates and time systems - Example 2
Calculate the azimuths of points and the moments of sunrise and sunset, as well as the duration of day and night on June 21, 1975 in an area with geographic coordinates λ = 4h28m,4 and φ = +59°30", located in the fifth time zone, if at noon of this day, the declination of the Sun is δ = +23°27", and the equation of time is η = + 1m35s.
Data: Sun, δ = +23°27"; η = +1m35s = +1m.6; place, λ=4h28m.4, φ = 59°30", n = 5.
Solution. Taking into account the average refraction in the horizon ρ = 35" and the angular radius of the solar disk r = 16", we find that at the moment of sunrise and sunset the center of the solar disk is below the horizon, at the zenith distance
z = 90° + ρ + r = 90°51",
sin z = +0.9999, cos z = -0.0148, sin δ = + 0.3979,
cos δ = +0.9174, sin φ = +0.8616, cos φ = +0.5075.
Using formula (28) we find:
and according to tables
t = ± (180°-39°49",3) = ±140°10",7 and
sin t = ±0.6404.
From Table 2 we find that at sunrise its hour angle t1 = -140°10",7 = -9h20m,7, and at sunset t2 = +140°10",7 = +9h20m,7, i.e. true solar time, according to formula (17), the Sun rises at
T 1 = 12h + t1 = 12h-9h20m,7 = 2h39m,3
and goes into
T 2 =12h + t2 = 12h+9h20m,7 = 21h20m,7,
which, according to formula (16), corresponds to moments in average time
Tλ1 = T 1 + η = 2h39m,3 + 1m,6=2h41m and
Τλ2 = T 2 + η = 21h20m,7+1m,6 = 21h22m.
According to formulas (19), (20) and (21) the same moments in standard time: sunrise
Tn1 = Tλ1- λ+n = 2h41m - 4h28m + 5h = 3h13m
and sunset Tn2 = Tλ2 - λ+n = 21h22m - 4h28m + 5h = 21h54m,
and according to maternity time:
sunrise Td1=4h13m and sunset Td2=22h54m.
Day length τ = Td2-Td1 = 22h54m-4h13m = 18h41m.
At the moment of the lower culmination, the height of the Sun
hn = δ- (90°-φ) = +23°27" - (90°-59°30") = -7°03", i.e. the white night lasts instead of the usual one.
The azimuths of the sunrise and sunset points are calculated using formula (30):
which gives A = ±(180°-36°.0) = ±144°.0, since the azimuths and hour angles of the Sun are in the same quadrant. Consequently, the Sun rises at a point on the true horizon with azimuth A1 = -144°.0 = 216°.0 and sets at a point with azimuth A2 = +144°.0, located 36° on either side of the north point.
Problem 90. At what average time intervals do like and unlike star climaxes alternate?
Problem 91. How long after the upper culmination of Deneb will the upper culmination of the star γ Orionis occur, and then again the upper culmination of Deneb? The right ascension of Deneb is 20h39m44s, and γ of Orion is 5h22m27s. Express the required intervals in sidereal and mean time systems.
Problem 92. At 14h15m10s mean time, the star Sirius (α Canis Majoris) with a right ascension of 6h42m57s was at its lower culmination. At what immediate moments after this will the star Gemma (α Northern Corona) be at its upper culmination and when will its hour angle be equal to 3h16m0s? Gemma's right ascension is 15h32m34s.
Problem 93. At 4h25m0s, the hour angle of a star with a right ascension of 2h12m30s was equal to -34°26",0. Find the right ascension of the stars that at 21h50m0s will be at the upper culmination and at the lower culmination, as well as those stars whose hour angles will be equal - 1h13m20s and 5h42m50s.
Problem 94. What is the approximate value of sidereal time at average, standard and maternity midnight in Izhevsk (λ = 3h33m, n = 3) on February 8 and September 1?
Problem 95. Approximately on what days of the year are the stars Sirius (α = 6h43m) and Antares (α = 16h26m) at their upper and lower culminations at middle midnight?
Problem 96. Determine the sidereal time in Greenwich at 7h28m16s on January 9 (s0 = 7h11m39s)* and at 20h53m47s on July 25 (s0 = 20h08m20s).
Problem 97. Find sidereal time at average, zone and maternity noon, as well as at average, zone and maternity midnight in Moscow (λ = 2h30m17s, n=2) on January 15 (s0=7h35m18s).*
Problem 98. Solve the previous problem for Krasnoyarsk (λ = 6h11m26s, n = 6) and Okhotsk (λ = 9h33m10s, n=10) on the day of August 8 (s0 = 21h03m32s).
Problem 99. Calculate the hour angles of the star Deiebe (α Cygni) (α = 20h39m44s) at Greenwich at 19h42m10s on June 16 (S0=17h34m34s) and December 16 (S0=5h36m04s).
Problem 100. Calculate the hour angles of the stars α Andromeda (α = 0h05m48s) and β Leo (α= 11h46m31s) at 20h32m50s on August 3 (s0=20h43M40s) and December 5 (s0=4h52M42s) in Vladivostok (λ=8h47m31s, n = 9).
Problem 101. Find the hour angles of the stars Betelgeuse (α = 5h52m28s) and Spica (α =13h22m33s) at 1h52m36s on June 25 (s0=18h06m07s) and November 7 (s0=2h58m22s) in Tashkent (λ=4h37m11s, n=5).
Problem 102. At what points in time in Greenwich are the star Pollux at the upper culmination (α = 7h42m16s), and at the lower culmination the star Arcturus (α = 14h13m23s) on February 10 (s0=9h17m48s) and May 9 (s0=15h04m45s)?
Problem 103. Find the moments of the upper and lower culmination on March 22 (s0 = 11h55m31s) and June 22 (s0 = 17h58m14s) of the stars Capella (α = 5h13m00s) and Bega (α = 18h35m15s) on the geographic meridian λ = 3h10m0s (n = 3). Indicate moments according to sidereal, mean, zone and maternity time.
Problem 104. At what times on February 5 (s0 = 8h58m06s) and August 15 (s0 = 21h31m08s) are the hour angles of the stars Sirius (α = 6h42m57s) and Altair (α = 19h48m21s) in Samarkand (λ = 4h27m53s, n = 4) equal to 3h28m47s?
Problem 105. At what points in time on December 10 (s0 = 5h12m24s) are the hour angles of the stars Aldebaran (α = 4h33m03s) and β Cygni (α = 19h28m42s) in Tbilisi (λ = 2h59m11s, n = 3) and in Okhotsk (λ = 9h33m10s, n=10 ) are respectively equal to +67°48" and -24°32"?
Problem 106. On what geographic meridians are the stars α Gemini and γ Ursa Major located at the upper culmination on September 20 (s0=23h53m04s) at 8h40m26s Irkutsk time (n=7)? The right ascension of these stars is respectively 7h31m25s and 11h51m13s.
Problem 107. Determine the horizontal coordinates of the stars ε Ursa Major (a = 12h51m50s, δ = +56°14") and Antares (α = 16h26m20s, δ = -26°19") at 14h10m0s sidereal time in Evpatoria (φ = +45°12" ).
Problem 108. What are the horizontal coordinates of the stars Gemma (α = 15h32m34s, δ = +26°53") and Spica (α = 13h22m33s, δ = -10°54") on April 15 (s0 = 13h30m08s) and August 20 (s0 = 21h50m50s) in 21h30m maternity time at a point with geographical coordinates λ = 6h50m0s (n = 7) and φ = +71°58"?
Problem 109. To what points in the sky, determined by horizontal coordinates, should a telescope installed at a point with geographic coordinates λ = 2h59m.2 (n = 3) and φ = +41°42" be directed so that on May 4, 1975 (s0 = 14h45m02s) 22h40m standard time see
Uranus (α = 13h52m.1, δ = -10°55") and Neptune (α = 16h39m.3, δ = -20s32")?
Problem 110. At what points in time does the summer solstice point on March 22 (s0 = 11h55m31s) and June 22 (s0 = 17h58m14s) rise, culminate and set and how long is it above the horizon on the central meridian of the second time zone in places with geographic latitude φ = +37°45 "and φ = +68°20"? Express moments using sidereal and maternity time.
Problem 111. Calculate the azimuths and moments of rising, upper culmination, setting and lower culmination of the stars Castor (α = 7h31m25s, δ = +32°00") and Antares (α = 16h26m20s, δ = -26°19") on April 15 (s0 = 13h30m08s) and October 15 (s0=1h31m37s) in places on the earth's surface with geographic coordinates λ =3h53m33s (n = 4), φ = +37°45" and λ = 2h12m15s (n = 2), φ = +68°59".
Problem 112. Calculate the azimuths and moments of sunrise, upper culmination and sunset, its midday and midnight altitude, as well as the length of the day on the dates of the vernal equinox and both solstices at points with geographical coordinates λ = 2h36m.3 (n=2), φ = +59° 57", and λ = 5h53m.9 (n = 6), φ = +69°18". On consecutive dates the equation of time is respectively +7m23s, +1m35s and -2m08s.
Problem 113. At what points in time on July 30 (s0 = 20h28m03s) at a point with λ = 2h58m0s (n=3) and φ = +40°14" the following stars have horizontal coordinates A and z:
Problem 114. At a point with geographic coordinates λ= 4h37m11s (n = 5) and φ = + 41°18" on August 5, 1975 (s0= 20h51m42s), the horizontal coordinates of two stars were measured: at 21h10m at the first star A = -8°33" and z = 49°51", and at 22:50 m the second star has A = 46°07" and z = 38°24". Calculate the equatorial coordinates of these stars.
Answers - Converting celestial coordinates and time systems
There are orioles in the forests, and longitude in the vowels
In tonic verses the only measure
But it only spills once a year
In nature, duration
As in Homer's metric.
As if this Day gapes like a caesura:
Already in the morning there is peace
And difficult lengths,
Oxen in the pasture
And golden laziness
Extract wealth from reeds
a whole note.
O. Mandelstam
Lesson 4/4
Subject: Changes in the appearance of the starry sky throughout the year.
Target: Get acquainted with the equatorial coordinate system, the visible annual movements of the Sun and types of the starry sky (changes throughout the year), learn to work according to the PCZN.
Tasks
:
1. Educational: introduce the concepts of the annual (visible) movement of luminaries: the Sun, Moon, stars, planets and types of starry sky; ecliptic; zodiac constellations; equinox and solstice points. The reason for the “delay” of climaxes. Continue developing the ability to work with PKZN - finding the ecliptic, zodiac constellations, stars on the map by their coordinates.
2. Educating: promote the development of the skill of identifying cause-and-effect relationships; Only a thorough analysis of observed phenomena makes it possible to penetrate into the essence of seemingly obvious phenomena.
3. Developmental: using problem situations, lead students to an independent conclusion that the appearance of the starry sky does not remain the same throughout the year; by updating students’ existing knowledge of working with geographic maps, to develop skills in working with PKZN (finding coordinates).
Know:
1st level (standard)- geographical and equatorial coordinates, points in the annual movement of the Sun, inclination of the ecliptic.
2nd level- geographical and equatorial coordinates, points in the annual movement of the Sun, inclination of the ecliptic, directions and reasons for the displacement of the Sun above the horizon, zodiacal constellations.
Be able to:
1st level (standard)- set according to PKZN for various dates of the year, determine the equatorial coordinates of the Sun and stars, find zodiacal constellations.
2nd level- set according to PKZN for various dates of the year, determine the equatorial coordinates of the Sun and stars, find zodiacal constellations, use PKZN.
Equipment: PKZN, celestial sphere. Geographical and star map. Model of horizontal and equatorial coordinates, photos of views of the starry sky at different times of the year. CD- "Red Shift 5.1" (path of the Sun, Change of Seasons). Video film "Astronomy" (part 1, fr. 1 "Star landmarks").
Intersubject connection: Daily and annual movement of the Earth. The Moon is a satellite of the Earth (natural history, 3-5 grades). Natural and climatic patterns (geography, 6 classes). Circular motion: period and frequency (physics, 9 cells)
During the classes:
I. Student survey (8 min). You can test on the Celestial Sphere N.N. Gomulina, or:
1. At the board
:
1. Celestial sphere and horizontal coordinate system.
2. The movement of the luminary during the day and its culmination.
3. Converting hourly measures to degrees and vice versa.
2. 3 people on cards
:
K-1
1. In which side of the sky is the luminary located, having horizontal coordinates: h=28°, A=180°. What is its zenith distance? (north, z=90°-28°=62°)
2. Name three constellations visible during the day today.
K-2
1. In which side of the sky is the star located if its coordinates are horizontal: h=34 0, A=90 0. What is its zenith distance? (west, z=90°-34°=56°)
2. Name three bright stars visible to us during the day.
K-3
1. In which side of the sky is the star located if its coordinates are horizontal: h=53 0, A=270 o. What is its zenith distance? (east, z=90°-53°=37°)
2. Today the star is at its upper climax at 21:34. When is its next lower, upper climax? (after 12 and 24 hours, more precisely after 11 hours 58 m and 23 hours 56 m)
3. The rest(independently in pairs while they answer at the board)
A) Convert to degrees 21h 34m, 15h 21m 15s. answer=(21.15 0 +34.15 "=315 0 +510" =323 0 30", 15 hours 21 m 15 s =15.15 0 +21.15" +15.15" =225 0 + 315 " + 225"= 230 0 18"45")
b) Convert to hourly measure 05 o 15", 13 o 12"24". hole= (05 o 15"=5.4 m +15.4 c =21 m, 13 o 12"24"=13.4 m +12 .4 s +24 .1/15 s =52 m +48 s +1.6 s =52 m 49 s .6)
II. New material (20 min) Video film "Astronomy" (part 1, fr. 1 "Star landmarks").
b) The position of the luminary in the sky (celestial environment) is also uniquely determined - in equatorial coordinate system, where the celestial equator is taken as the reference point
. (equatorial coordinates were introduced for the first time by Jan Havelia (1611-1687, Poland), in a catalog of 1564 stars compiled in 1661-1687) - an atlas of 1690 with engravings and is now in use (textbook title).
Since the coordinates of stars do not change for centuries, this system is used to create maps, atlases, and catalogs [lists of stars]. The celestial equator is a plane passing through the center of the celestial sphere perpendicular to the axis of the world.
 |
Points E-east, W-west - the point of intersection of the celestial equator with the points of the horizon. (Points N and S are reminiscent). |
| Declension circle - a large circle of the celestial sphere passing through the poles of the world and the observed star (points P, M, P"). | |
|
Equatorial coordinates: |
c) Annual movement of the Sun. There are luminaries [Moon, Sun, Planets] whose equatorial coordinates change quickly. The ecliptic is the apparent annual path of the center of the solar disk along the celestial sphere.
Inclined to the plane of the celestial equator currently at an angle 23 about 26", more precisely at an angle: ε = 23°26'21",448 - 46",815 t - 0",0059 t² + 0",00181 t³, where t is the number of Julian centuries that have passed since the beginning of 2000. This formula is valid for the nearest centuries. Over longer periods of time, the inclination of the ecliptic to the equator fluctuates around the average value with a period of approximately 40,000 years. In addition, the inclination of the ecliptic to the equator is subject to short-period oscillations with a period of 18.6 years and an amplitude of 18.42, as well as smaller ones (see Nutation).
The apparent movement of the Sun along the ecliptic is a reflection of the actual movement of the Earth around the Sun (proven only in 1728 by J. Bradley with the discovery of annual aberration).
|
Cosmic phenomena |
Celestial phenomena arising as a result of these cosmic phenomena |
| Rotation of the Earth around its axis | Physical phenomena: 1) deflection of falling bodies to the east; 2) the existence of Coriolis forces. Displaying the true rotation of the Earth around its axis: 1) daily rotation of the celestial sphere around the axis of the world from east to west; 2) sunrise and sunset; 3) the culmination of the luminaries; 4) change of day and night; 5) daily aberration of luminaries; 6) daily parallax of luminaries |
| Rotation of the Earth around the Sun | Displays the true rotation of the Earth around the Sun: 1) annual change in the appearance of the starry sky (the apparent movement of celestial bodies from west to east); 2) the annual movement of the Sun along the ecliptic from west to east; 3) change in the midday height of the Sun above the horizon during the year; a) change in the duration of daylight hours throughout the year; b) polar day and polar night at high latitudes of the planet; 5) change of seasons; 6) annual aberration of luminaries; 7) annual parallax of luminaries |
.jpg) |
The constellations through which the ecliptic passes are called.
The number of zodiac constellations (12) is equal to the number of months in a year, and each month is designated by the sign of the constellation in which the Sun is located in that month. 13th constellation Ophiuchus is excluded, although the Sun passes through it. "Red Shift 5.1" (path of the Sun). |
|
|
|
| - vernal equinox point. 21 March
(day equals night). Sun coordinates: α ¤ =0 h, δ ¤ =0 o The designation has been preserved since the time of Hipparchus, when this point was in the constellation ARIES → is now in the constellation PISCES, IN 2602 it will move to the constellation AQUARIUS. |
|
| -summer solstice day. 22nd of June
(longest day and shortest night). Sun coordinates: α ¤ =6 h, ¤ =+23 about 26" The designation has been preserved since the time of Hipparchus, when this point was in the constellation Gemini, then in the constellation Cancer, and since 1988 it has moved to the constellation Taurus. |
|
 |
- autumn equinox day. 23 September
(day is equal to night). Sun coordinates: α ¤ =12 h, δ t size="2" ¤ =0 o The designation of the constellation Libra was preserved as a designation of the symbol of justice under the emperor Augustus (63 BC - 14 AD), now in the constellation Virgo, and in 2442 it will move to the constellation Leo. |
| - winter solstice. December 22
(shortest day and longest night). Sun coordinates: α ¤ =18 h, δ ¤ =-23 about 26" During the period of Hipparchus, the point was in the constellation Capricorn, now in the constellation Sagittarius, and in 2272 it will move to the constellation Ophiuchus. |
Although the position of the stars in the sky is uniquely determined by a pair of equatorial coordinates, the appearance of the starry sky at the observation location at the same hour does not remain unchanged.
Observing the culmination of the luminaries at midnight (the Sun at this time is in the lower culmination with a right ascension on a luminary different from the culmination), one can notice that on different dates at midnight, different constellations pass near the celestial meridian, replacing each other. [These observations at one time led to the conclusion that the right ascension of the Sun has changed.]
Let's choose any star and fix its position in the sky. At the same place, the star will appear in a day, more precisely in 23 hours and 56 minutes. A day measured relative to distant stars is called stellar
(to be completely precise, the sidereal day is the period of time between two successive upper culminations of the vernal equinox). Where do the other 4 minutes go? The fact is that due to the movement of the Earth around the Sun, for an observer on Earth, it shifts against the background of stars by 1° per day. To “catch up” with him, the Earth needs these 4 minutes. (picture on the left)
Each subsequent night the stars move slightly to the west, rising 4 minutes earlier. Over the course of a year it will shift by 24 hours, that is, the appearance of the starry sky will repeat itself. The entire celestial sphere will make one revolution in a year - the result of the reflection of the Earth's revolution around the Sun.
So, the Earth makes one revolution around its axis in 23 hours 56 minutes. 24 hours - the average solar day - the time the Earth rotates relative to the center of the Sun.
III. Fixing the material (10 min)
1. Work on PKZN (in the course of presenting new material)
a) finding the celestial equator, ecliptic, equatorial coordinates, equinox and solstice points.
b) determination of the coordinates of, for example, stars: Capella (α Aurigae), Deneb (α Cygnus) (Capella - α = 5 h 17 m, δ = 46 o; Deneb - α = 20 h 41 m, δ = 45 o 17")
c) finding stars by coordinates: (α=14.2 h, δ=20 o) - Arcturus
d) find where the Sun is today, in what constellations in the fall. (now the fourth week of September is in Virgo, the beginning of September is in Leo, Libra and Scorpio will pass in November)
2. Additionally:
a) The star culminates at 14:15. When is its next lower or upper culmination? (at 11:58 and 23:56, that is, at 2:13 and 14:11).
b) the satellite flew across the sky from the initial point with coordinates (α=18 h 15 m, δ=36 о) to the point with coordinates (α=22 h 45 m, δ=36 о). What constellations did the satellite fly through?
IV. Lesson summary
1. Questions:
a) Why is it necessary to introduce equatorial coordinates?
b) What is remarkable about the days of the equinox and solstice?
c) At what angle is the plane of the Earth's equator inclined to the plane of the ecliptic?
d) Is it possible to consider the annual movement of the Sun along the ecliptic as evidence of the Earth’s revolution around the Sun?
Homework:§ 4, self-control questions (p. 22), p. 30 (paragraphs 10-12).
(it is advisable to distribute this list of works with explanations to all students for the year).
You can give a task" 88 constellations
"(one constellation for each student). Answer the questions:
- What is the name of this constellation?
- At what time of year is it best to observe it at our (given) latitude?
- What type of constellation does it belong to: non-ascending, non-setting, setting?
- Is this constellation northern, southern, equatorial, zodiacal?
- Name interesting objects of this constellation and indicate them on the map.
- What is the name of the brightest star in the constellation? What are its main characteristics?
- Using a moving star chart, determine the equatorial coordinates of the brightest stars in the constellation.
Lesson completed members of the Internet Technologies circle - Prytkov Denis(10 cells) and Pozdnyak Victor(10 cells), Changed 23.09.2007
of the year
2. Grades
| "Planetarium" 410.05 mb | The resource allows you to install the full version of the innovative educational and methodological complex "Planetarium" on a teacher's or student's computer. "Planetarium" - a selection of thematic articles - are intended for use by teachers and students in physics, astronomy or natural science lessons in grades 10-11. When installing the complex, it is recommended to use only English letters in folder names. | ||
| Demo materials 13.08 MB | The resource represents demonstration materials of the innovative educational and methodological complex "Planetarium". | Equatorial coordinate system 460.7 kb
Practical work No. 1
"Determining the coordinates of stars"
(working with the coordinate grid of the map)
| Star | Constellation | Interesting facts about the star |
| Sirius | α Canis Majoris | The brightest star and closest to Earth (9 light years) |
| Epsilon | ξ Auriga | The diameter of the star is 3000 times larger than the diameter of the Sun |
| Declension δ = Right Ascension α = | ||
| Alpha | α Hercules | The volume is 10 15 times the volume of the Sun, and light takes 1200 years to reach the Earth |
| Declension δ = Right Ascension α = | ||
| α Cassiopeia | The star's substance is 2 million times denser than water | |
| Declension δ = Right Ascension α = | ||
| Tay | τ China | Most similar to the Sun |
| Declension δ = Right Ascension α = | ||
| Rigel | β Orionis | Farthest from Earth (1400 light years) |
| Declension δ = Right Ascension α = | ||
| Betelgeuse | α Orion | The density of a star is 30 times less than the density of air |
| Declension δ = Right Ascension α = |
| Grade |
Now let's get acquainted with the methods of orientation by the Sun.
1. The noon line is always directed from north to south. With its help you can always determine the sides of the horizon.
2. At the moment of true noon, the shadow of objects is always directed to the north, and the Sun is above the point of the south. Knowing the time of true noon, it is easy to determine the sides of the horizon.
Knowing the time of true noon, you can navigate using a clock. Holding the watch in a horizontal position, point the hour hand to the place on the horizon above which the Sun is located. They don't pay attention to the minute hand. The interval between the end of the hour hand and the point indicating true noon for a given observation location is divided in half. The direction from the center of the dial through the resulting middle will indicate the south point.
the surface of the platform is horizontal, the plumb line will coincide with the line drawn on the bar.
Having installed the rod (gnomon) perpendicular to the surface of the horizontal platform you have chosen, at about eleven o’clock, mark the position of the end of the gnomon’s shadow. With a radius equal to the length of this shadow, with the center at the base of the gnomon, thin out the arc.
You know that before noon the length of the shadow shortens, but after noon it begins to lengthen. Observe when the shadow from the gnomon, lengthening, reaches the arc again, and mark this point on the arc. Divide the distance between the resulting points A and B in half and connect the middle of the arc - point C to the base of the rod. This will be the noon line.
To be sure that the noon line is drawn correctly, repeat everything from the beginning, but a little earlier or later than the first time. If both lines coincide, then the noon line is determined correctly.
The next day, after checking your watch with the exact time signal, track at what time, local time, the shadow of the gnomon coincides with the noon line. This will be the time of true noon, since it is at this moment that the height of the Sun above the horizon is greatest, and the shadow from the gnomon is smallest. You will see that true noon does not coincide with 12 o'clock - - the reading of noon on the clock. This is not surprising, because the clock shows maternity or standard time, and the gnomon shows the time of noon according to the movement of the Sun.
The time determined by the Sun is called true solar time, and the time interval between two true noons is called a true solar day.
It is clear that when orienting by the Sun you should use solar time.
Observation #2
"Circumpolar constellations"
(observation of the geometric path of stars)
On a starry night, notice the location of the circumpolar constellations in the northern sky: Ursa Major, Ursa Minor and Cassiopeia. Draw their relative positions.
Observe the location of these constellations from this place every 2 weeks.
| Conclusion: | |||
| Grade | |||
Test work No. 1 (self-control)
Constellations. Star cards. Celestial coordinates
Option 1
1. Determine the equatorial coordinates of the following stars from the star map: 1) α Libra; 2) β Lyrae.
2. Why does the North Star hardly change its position relative to the horizon?
Option 2
1. Find on the star map and name the objects that have the coordinates: 1) α = 15 hours 12 minutes, δ = - 9°; 2) α - 3 hours 40 minutes,
2. At what points does the celestial equator intersect with the horizon line?
Option 3
1. Determine the equatorial coordinates of the following stars from the star map: 1) α Ursa Major; 2) γ Orion.
2. How is the axis of the world located relative to the earth's axis? relative to the plane of the celestial meridian?
Option 4
1. In what constellation is the Moon located, if its coordinates
α = 20 hours 30 minutes, δ = -20°?
2. At what points does the celestial meridian intersect with the horizon?
Option 5
1. Determine the equatorial coordinates of the following stars from the star map: 1) α Perseus; 2) β China.
2. What is the height of the zenith point above the horizon?
Option 6
1. Determine from the star map the constellation in which the M 3 1 galaxy is located if its coordinates are α = 0 h 40 min, δ = +41°.
2. How does the horizon plane lie relative to the surface of the globe?
ORIENTATION BY THE MOON
The moon, like the stars, can serve as a reliable guide to help determine the sides of the horizon. Remember two ways to navigate:
1) The Full Moon is at its greatest height above the horizon at midnight. At this time, it is above the south point and provides enough light to clearly notice the shadow of objects. At midnight, the shadow of objects is shortest and directed north. Before midnight the shadow is directed to the northwest, after midnight to the northeast.
You've probably noticed that the orientations of the Sun and Moon during a full moon are very similar.
2) The young Moon is observed in the western sky immediately after sunset. During the night, describing an arc in the southern sky, the Moon descends to the east. It is at its greatest height above the horizon at midnight. At this moment it is located above the south point.
In the middle latitudes of the northern hemisphere, the hump of the young Moon faces west in all phases.
SUN ORIENTATION
The sun is as reliable a guide as the stars. However, in order to be able to navigate by the Sun, you need to learn how to determine solar time and use it. Let's explain this.
First of all, you need to determine the direction of the noon line. To do this, you need to choose a horizontal area (in the yard, on the balcony, on the windowsill) where sunlight falls. The levelness of the site can be checked using a level or spirit level. It is easy to make a spirit level yourself. Take two straight rectangular planks and nail one to the other at right angles. Draw a line in the middle of the vertical bar and hang the weight on the thread. If
The Ursa Major, like all the stars in the sky, makes a daily revolution around the celestial pole counterclockwise with a period of 24 hours.
Imagine a huge dial in the sky with the center at the celestial pole (almost at the North Star) and the number 6 above the north point. The hand of such a clock passes from the North Star through the two outermost stars of the Big Dipper. The hand moves one division of the celestial Dial within two hours.
To determine the time, you must first calculate the date of the month from the beginning of the year with decimals. Every three days are counted as one tenth of a month. For example, October 3 corresponds to the number 10.1. This number must be added to the clock readings, and the sum multiplied by 2. The resulting product should be subtracted from the number 55.3, which depends on the specific position of the indicated constellations. The number 55.3 must be remembered. The formula for calculating the time of night is shown in the figure.!
To make the above more understandable, let’s solve the problem: let’s say on October 18 you noticed that the hand of the sidereal clock was pointing to the number 6. What time was it?
Solution. October is the tenth month of the year, therefore October 18 corresponds to the number 10.6. Adding this number to the clock reading and multiplying by two, we get: (10.6 + 6)2 = 32.2. The resulting number must be subtracted from 55.3: 55.3-33.2 = 22.1.
Answer: the observation was carried out at 10:60 pm.
Practice solving similar problems.
Observation #3
“Determination of the geographical latitude of the observation site using an eclimeter”
For observation, make a homemade device - an eclimeter, from cardboard with a radius of 10 cm. Degree divisions are applied to its semicircular part, and a thin but strong thread is attached to the center of the diameter (see figure). Attach a bead to the end of the thread. If the diameter of the eclimeter is directed towards the observed star, then the thread will pass through a division that will correspond to the height of the star above the horizon h.
