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 13:06m41s decree 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

Star charts, celestial coordinates and time (§)

I. Determine the equatorial coordinates of the following stars from the star map:

• 1. b Ursa Major,
• 2. g Orion,
• 3. in China.

Answer. 1) b =11 hours, d =+620;

• 2)b =5 h 20 m, d =+60;
• 3) b =0 h 40 m, d = - 190 301

II. Find on the star map and name objects that have coordinates:

• 1) b =15 h 12 m, d = -9 0;
• 2)b =3 h 40 m, d =+48 0;

Answer. 1) in Libra and 2) d Perseus.

III. Find on the star map the three brightest stars located no further than 10 0 from the ecliptic and having a right ascension from 10 a.m. to 5 p.m. Determine their equatorial coordinates.

Answer. b Leo (b =10h 5m, d =+120); b Virgo (b =13h 20m, d =-110); b Scorpio (b =16h 25m, d =-260).

IV. Using PKZN, determine the declination and altitude at the upper culmination of the star Arcturus. Calculate the height of this star using the formula

(taking d from the table in an astronomy textbook), compare the results obtained and indicate with what accuracy the required quantities are determined from the star chart.

Answer. With c =570 301 we find from the map d =+190, h =500. Using the formula we get: h =510,571 (with d =190,271).

Composition of the solar system (§)

I. Having learned from the school astronomical calendar the coordinates of the planets observed today (at a given moment in time), plot their positions on the star map, indicate in which constellations these planets are visible.

• · Using a moving map, indicate in which constellations these planets are visible.
• · Using a moving star chart, determine which of these planets are observed today at 10 p.m. and in which part of the sky.
• · Determine the rising and setting times of these planets today, and calculate the duration of their visibility.
• · Having learned from the school astronomical calendar the coordinates of the planets observed in the middle of two adjacent months, plot their positions on the star map and, having determined the direction of movement among the stars using an overhead circle, indicate whether each of these planets is moving forward or backward.

(Note: Regardless of the date, the overlay circle must be positioned so that the planet's path is above the horizon. If the planet is moving from west to east, its motion is direct.)

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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.

1. Know the contents of paragraph 116 of the Physics-11 textbook
2. Know the contents of paragraphs 3, 4 of the textbook Astronomy -11
3. Prepare material on the topic “Zodiac constellations”

Literature.

1. E.P. Levitan Astronomy 11th grade – Enlightenment, 2004
2. G.Ya. Myakishev and others. Physics 11th grade - Enlightenment, 2010
3. Encyclopedia for children Astronomy - ROSMEN, 2000

Key questions: 1. The concept of constellation. 2. Difference between stars in brightness (luminosity), color. 3. Magnitude. 4. Apparent daily motion of stars. 5. celestial sphere, its main points, lines, planes. 6. Star map. 7. Equatorial SC.

Demonstrations and TSO: 1. Demonstration moving sky map. 2. Model of the celestial sphere. 3. Star atlas. 4. Transparencies, photographs of constellations. 5. Model of the celestial sphere, geographical and star globes.

For the first time, stars were designated by letters of the Greek alphabet. In the constellation atlas of Baiger in the 18th century, the drawings of the constellations disappeared. The magnitudes are indicated on the map.

Ursa Major - (Dubhe), (Merak), (Fekda), (Megrets), (Aliot), (Mizar), (Benetash).

Lyra - Vega, Lebedeva - Deneb, Bootes - Arcturus, Auriga - Capella, B. Canis - Sirius.

The Sun, Moon and planets are not indicated on the maps. The path of the Sun is shown on the ecliptic in Roman numerals. Star maps display a grid of celestial coordinates. The observed daily rotation is an apparent phenomenon - caused by the actual rotation of the Earth from west to east.

Proof of Earth's rotation:

1) 1851 physicist Foucault - Foucault pendulum - length 67 m.

2) space satellites, photographs.

Celestial sphere- an imaginary sphere of arbitrary radius used in astronomy to describe the relative positions of luminaries in the sky. The radius is taken as 1 Pc.

88 constellations, 12 zodiac. It can be roughly divided into:

1) summer - Lyra, Swan, Eagle 2) autumn - Pegasus with Andromeda, Cassiopeia 3) winter - Orion, B. Canis, M. Canis 4) spring - Virgo, Bootes, Leo.

Plumb line intersects the surface of the celestial sphere at two points: at the top Z - zenith- and at the bottom Z" - nadir.

Mathematical horizon- a large circle on the celestial sphere, the plane of which is perpendicular to the plumb line.

Dot N mathematical horizon is called north point, dot S - point south. Line N.S.- called noon line.

Celestial equator called a great circle perpendicular to the axis of the world. The celestial equator intersects the mathematical horizon at points of the east E And west W.

Heavenly meridian called the great circle of the celestial sphere passing through the zenith Z, celestial pole R, south celestial pole R", nadir Z".

Homework: § 2.

Constellations. Star cards. Celestial coordinates.

1. Describe what daily circles the stars would describe if astronomical observations were carried out: at the North Pole; at the equator.

The apparent motion of all stars occurs in a circle parallel to the horizon. The North Pole of the world when observed from the North Pole of the Earth is at the zenith.

All stars rise at right angles to the horizon in the eastern part of the sky and also set below the horizon in the western part. The celestial sphere rotates around an axis passing through the poles of the world, located exactly on the horizon at the equator.

2. Express 10 hours 25 minutes 16 seconds in degrees.

The Earth makes one revolution in 24 hours - 360 degrees. Therefore, 360 o corresponds to 24 hours, then 15 o - 1 hour, 1 o - 4 minutes, 15 / - 1 minute, 15 // - 1 s. Thus,

1015 o + 2515 / + 1615 // = 150 o + 375 / +240 / = 150 o + 6 o +15 / +4 / = 156 o 19 / .

3. Determine the equatorial coordinates of Vega from the star map.

Let's replace the name of the star with a letter designation (Lyra) and find its position on the star map. Through an imaginary point we draw a circle of declination until it intersects with the celestial equator. The arc of the celestial equator, which lies between the point of the vernal equinox and the point of intersection of the circle of declination of a star with the celestial equator, is the right ascension of this star, measured along the celestial equator towards the apparent daily rotation of the celestial sphere. The angular distance measured along the declination circle from the celestial equator to the star corresponds to the declination. Thus, = 18 h 35 m, = 38 o.

We rotate the overlay circle of the star map so that the stars cross the eastern part of the horizon. On the limb, opposite the mark of December 22, we find the local time of its sunrise. By placing the star in the western part of the horizon, we determine the local time of sunset of the star. We get

5. Determine the date of the upper culmination of the star Regulus at 21:00 local time.

We install the overhead circle so that the star Regulus (Leo) is on the line of the celestial meridian (0 h - 12 h scale of the overhead circle) south of the north pole. On the dial of the applied circle we find the mark 21 and opposite it on the edge of the applied circle we determine the date - April 10.

6. Calculate how many times brighter Sirius is than the North Star.

It is generally accepted that with a difference of one magnitude, the apparent brightness of stars differs by approximately 2.512 times. Then a difference of 5 magnitudes will amount to a difference in brightness of exactly 100 times. So 1st magnitude stars are 100 times brighter than 6th magnitude stars. Consequently, the difference in the apparent magnitudes of two sources is equal to unity when one of them is brighter than the other (this value is approximately equal to 2.512). In general, the ratio of the apparent brightness of two stars is related to the difference in their apparent magnitudes by a simple relationship:

Luminaries whose brightness exceeds the brightness of stars 1 m, have zero and negative magnitudes.

Magnitudes of Sirius m 1 = -1.6 and Polaris m 2 = 2.1, we find in the table.

Let us take logarithms of both sides of the above relationship:

Thus, . From here. That is, Sirius is 30 times brighter than the North Star.

Note: using the power function, we will also get the answer to the question of the problem.

7. Do you think it is possible to fly on a rocket to any constellation?

A constellation is a conventionally defined area of ​​the sky within which there are luminaries located at different distances from us. Therefore, the expression “fly to a constellation” is meaningless.

The units of the hour measure of angles should not be confused with units of the measure of time that are identical in name and designation, since angles and time intervals are dissimilar quantities. The hour measure of angles has simple relationships with the degree measure:

	corresponds to 15°;		1° corresponds to 4Ш;
\ T
						1/15s.
For translate		quantities				hourly measures in

degree and	back there are tables (Table V in
AE or adj.	1 of this book).
Geographical			coordinates		sometimes called
ronomic					definitions.
§ 2. Equatorial coordinates of luminaries
Position		celestial bodies		convenient to define

vatorial coordinate system. Let's imagine that

the sky is			huge			sphere, in the center of which is
for the sphere, we can we-
too hard to build
coordinate
			parallels
globe. If pro-
passing through the Northern
before crossing with the imagination
		heavenly
then you will get diametrically
	opposite
ki of Northern R and South
called
is		geometric axis						equatorial
	coordinates Continuing the plane of the earth

ra, until it crosses the celestial sphere, we get the line of the celestial equator on the sphere.

The earth rotates around its axis from west to east

drain, and its full turnover takes one day. To an observer on Earth it seems that the celestial sphere is

rotates with all visible luminaries
in the opposite	direction, i.e. from the east
west. It seems to us that the Sun is every day
around the Earth: in the morning it		rises		eastern
part of the horizon, and			Over the horizon

west. In the future, instead of the actual rotation of the Earth around its axis, we will consider the daily rotation of the celestial sphere. It occurs clockwise when viewed from the North Pole.

It is easier to visually imagine the celestial sphere if you look at it from the outside, as shown in Fig. 2. In addition, it shows the trace of the intersection of the plane of the earth's orbit, or the plane of the ecliptic, with the celestial sphere. The Earth completes its orbit around the Sun in one year. A reflection of this annual revolution is the visible annual movement of the Sun along the celestial sphere in the same plane, that is, along the ecliptic J F JL - F J T . Every day, the Sun moves among the stars along the ecliptic to the east by about one degree of arc, completing a full revolution in a year. The ecliptic intersects with the celestial equator at two diametrically opposite points, called equinox points: T - the vernal equinox and - the autumn equinox. When the Sun is at these points, then everywhere on Earth it rises exactly in the east, sets exactly in the west, and day and night are equal to 12 hours. Such days are called equinoxes, and they fall on March 21 and September 23 with no deviation from these dates less than one day.

The planes of geographic meridians, extended until they intersect with the celestial sphere, form celestial meridians at the intersection with it. There are countless celestial meridians. Among them, it is necessary to select the initial one in the same way as on Earth the meridian passing through the Greenwich Observatory is accepted as the zero one. In astronomy, such a reference line is taken to be the celestial meridian passing through the point of the vernal equinox and called the circle of declination of the point of the vernal equinox. Celestial meridians passing through the positions of the luminaries are called the circles of declination of these luminaries,

In the equatorial coordinate system, the main circles are the celestial equator and the declination circle of the Y point. The position of any luminary in this coordinate system is determined by right ascension and declination.

Rectal descent is the spherical angle at the celestial pole between the circle of declination of the vernal equinox and the circle of declination of the luminary, calculated in the direction opposite to the daily rotation of the celestial sphere.

Right ascension is measured by the arc of the celestial

niya of the celestial sphere, therefore a does not depend on the daily rotation of the celestial sphere.

and the direction towards the luminary. The declination is measured by the corresponding arc of the declination circle from the celestial equator to the place of the luminary. If the luminary is in the northern hemisphere (north of the celestial equator), its declination is assigned the name N, and if it is in the southern hemisphere, the name 5. When solving astronomical problems, the plus sign is assigned to the declination value, which is the same as the latitude of the observation site. In the Northern Hemisphere of the Earth, the northern declination is considered positive, and the southern declination is considered negative. The declination of the luminary can vary from 0 to ±90°. The declination of each point on the celestial equator is 0°. The declination of the North Pole is 90°.

Any luminary makes a complete revolution around the celestial pole during the day along its daily parallel together with the celestial sphere, therefore b, like a, does not depend on its rotation. But if the luminary has additional movement (for example, the Sun or a planet) and moves across the celestial sphere, then its equatorial coordinates change.

The values ​​of a and b are related to the observer, as if located at the center of the Earth. This allows you to use the equatorial coordinates of luminaries anywhere on Earth.

§ 3. Horizontal coordinate system

The center of the celestial sphere can be moved to any

point in space.

in particular,

fit with the point of intersection of the main axes

ta. In this case, vertical

tool (Fig.

geometric

horizontal

coordinates

At the intersection with the sky

sheer

forms

observer.

passing

heavenly

perpendicular-

direction

called

plane

true

horizon and at the intersection

surface

heavenly

true

horizon

designations

countries of the world adopted traditional

transcription: N (north), S (south), W (west)

Through a plumb line you can draw

countless

new set

vertical

planes. At the intersection

with surface

celestial sphere

form

circles called verticals. Any vertical

that passes through the location of the luminary is called the vertical of the luminary.

		RRH					characterize
	as a line parallel to the axis of rotation
Then the plane of the celestial equator QQ\ will be parallel
	plane		earth's equator. vertical,
					PZP\ZX ,	is
temporarily heavenly				meridian			observations,
or meridian			observer. Meridian				observer

The observer's meridian with the plane of the true horizon is called the noon line. The closest point of intersection of the midday to the North Pole

through the points of east and west is called the first vertical. Its plane is perpendicular to the plane of the observer's meridian. The celestial sphere is usually

				meridian plane				observer
coincides with the drawing plane.
The main coordinate circles in the horizontal
the system is served by the true horizon and							meridian
giver. According to the first of these circles							the system received
its name.		Coordinates
	are				and anti-aircraft		distance.
A z i m u t		s v e t i l a			A - spherical
zenith point between the observer's meridian
				astronomy				count down
					meridian		observer, but

Since, ultimately, astronomical azimuths of directions are determined for geodetic purposes, it is more convenient to immediately adopt a geodetic account of azimuths in this book. They are measured by arcs of the true horizon from the point of north to the vertical of the luminary along the course of the

the center of the sphere between the direction to the zenith and the direction to the luminary. The zenith distance is measured by the vertical arc of the luminary from the zenith point to the place of the luminary. The zenith distance is always positive and varies in value from 0 to 180°.

The rotation of the Earth around its axis from west to east causes the visible daily rotation of the luminaries around the celestial pole along with the entire celestial sphere. This