Two things in the world - the sky above the head and the inner
the law of man is what concerns me the most.
Immanuel Kant
The phenomena taking place in the starry sky, as a rule, have an indelible impression: the rising of the orange-red Sun, the shining of the stars and the Moon, solar and lunar eclipses, meteors, comets, rainbows and halos. Naturally, they are depicted by artists on their canvases. reproductions of these works are increasingly reproduced in popular science literature on astronomy (see, for example, books S.I. Dubkova"History of Astronomy" Magic world Stars”, “Shining Abysses of Space”, published in Moscow by the publishing house “Bely Gorod” in 2002 and 2004). In most cases, they either illustrate specific texts or carry subtext. semantic load. For example, a picture O. Bulgakova"Feast under the Moon" can be shown, accompanied by the question: "At what time of the night is the feast?" if we recall the conditions for the visibility of the moon above the horizon, then the answer is easy to find: old moon visible in the early morning hours, so that the feast was clearly delayed.
Currently works of art have become quite affordable thanks to CDs with collections different museums(Moscow State Museum of Fine Arts named after A.S. Pushkin, St. Petersburg Hermitage, Russian Museum, national museums Europe, etc.). The quality of the illustrations of some, for example, "5555 works famous artists”, is not high, but they contain significant textual material. Other discs, on the contrary, contain very high-quality illustrations, but there is little text (see, for example, the disc “ State Museum fine arts them. A.S. Pushkin. – Game World, GSC, 1997). It's good that many CDs allow you to make color copies. I offer several plots related to our natural luminaries - the Sun, the Moon, the stars.
Interesting images of the Sun in different time days, in different seasons and at different latitudes:
- at sunrise: W. Turner"Ulysses ridicules Polyphemus"; W. van Gogh « Mountain landscape after sunrise"; C.-F. Daubigny"Morning. The sun in the fog";
- during the day: Y.Podlyasky"Birth of the main"; P. Mondrian"Mill under the sun"; Claude Monet"Parliament in London" A.-P. Ryder"Siegfried and the daughters of the Rhine"; A.Russo"Snake Charmer"; P. de Chavannes"Saint Genevieve";
- on entry: Z. Sudkovsky"Sunset on the sea"; A.Bliokh"Holiday. Scarlet Sails»; V.Vasnetsov"After the battle of Igor Svyatoslavovich with the Polovtsy"; A.Mamedov"Girls on the Senezh", P. Ossovsky"Ways-roads"; K.Lorren « Sea port»;
- in spring: N.M.Romadin"Spring in the North Caucasus", 1978; V. Ulovich Zamoskvorechye. Spring, 1986;
- in summer: A. Kuprin"Beassal Valley"; M.Chagall"Lovers over St. Paul", 1971;
– in winter: M. Germashev Street in Zamoskvorechye. Winter"; V. Grinberg"Sink at the blue bridge"; M. Ivanov"Winter landscape. Ostrovsky Street (Malaya Ordynka)"; A.Z. Davydov"Winter of forty-two", 1983; Yu.Yu.Klever"Forgotten Cemetery", 1890; A. Mylnikov"Leningrad, 1941", 1974; Art van der Neer"Winter view on the river", 1763; L.Tikhomirov Frost and Sun, 1988;
- in autumn: P. Rubens « Autumn landscape with a view of the castle Sten":
– in polar latitudes ( A.A. Shumilkin"Drilling in the tundra"), in medium ( W. van Gogh"Vineyards in Arles") and equatorial ( A.Tekle"Spring in Kokadam").
The solar corona - the glow of the upper layers of the Sun's atmosphere during eclipses - was depicted in their paintings I.Glazunov("Prince Igor") and Raphael("Crucifixion"). Pictures with the image starry sky a little, but constellations can sometimes be recognized ( V.Vasnetsov"Snow Maiden"; W. van Gogh"Cafe Terrace at Night").
Vincent van Gogh. Red vineyards in Arles
Based on the pictures, you can make tasks like:
How long has it been since sunrise in van Gogh's Red Vineyards at Arles?
Solution. Knowing the latitude of Arles and, having determined the height of the Sun above the horizon by the angular dimensions, it is possible to find part of the daily path of the luminary and determine the time that has elapsed since its sunrise. The moment of sunrise is the appearance of the edge of the disk of the Sun from behind the horizon. We find from the geographical atlas that the latitude of Arles is almost 44 °. We find part of the daily path of the Sun above the horizon as the hypotenuse of a triangle, the leg of which is equal to the height of the Sun above the horizon, and the opposite angle is 46 ° as the angle between the planes of the celestial equator and the mathematical horizon. It can be seen from the drawing that AU = ρ + AD, where ρ is the radius of the Sun in the figure, and AD is the distance from the center of the Sun to the horizon, so the hypotenuse AB will be equal to (ρ + AD) : cos 44°.
By measuring directly in the figure ρ = 10 mm, AD= 20 mm, we find AB= 41.7 mm.
Length AB in arc minutes we find, if we remember that the angular radius of the Sun is 0.25 ° (or 15 ′) and make up the ratio:
The time that the Sun spends on the passage of the path AB, we find by dividing the angular path of the Sun in the sky by the speed of its movement across the sky:
360°: 24 h = 15°/1 h = 15'/1 min 62.6': 15'/1 min = 4.2 min.
D.M.Utenkov. Evening by the sea. Fragment
Is there an astronomical error in the picture D.M.Utenkova"Evening by the sea", if the depicted place lies approximately at the latitude of Chelyabinsk?
Solution. From the image of the moon, you can determine its phase, i.e. share of the illuminated part relative to the diameter: Ф = D.B.:AB. In the picture F = 0.29.
The phase of the moon is related to the angular distance φ of the moon from the sun by the formula: Ф = 0.5 (1 - cosφ), which is derived from geometric considerations.
Let AB is the diameter of the moon perpendicular to the line of sight TM. ∠MTS = ∠EMV= φ as angles with mutually perpendicular sides. From Δ EMD:
cosφ = cos∠ EMV = MD / ME \u003d (MB - DB) / ME= = 1 – 2F.
Substituting numerical data, we get:
φ = ∠ MTS= 65°.
It remains to figure out how to measure the angular distances in the picture: for this you need to know that the moon is visible from the Earth at an angle of 0.5 °. By the phase of the Moon we find φ, and by it - the angular distance of the Moon from the Sun, equal to 180 - φ. then from the center of the image of the moon on a line perpendicular to the segment connecting its horns, we set aside as many diameters as 0.5 ° times fits into 180 - φ (after all, the Sun is located just on this line). And sometimes it turns out that in the picture the Sun should be above the horizon, but in the picture it is no longer visible.
Indeed, we measure the distance of the moon from the figure H from the horizon along a straight line connecting its center with the Sun: H= 4.6 cm. Using the angular scale, i.e. number of degrees celestial sphere per unit length of the figure, we find that H= 3.3°. This means that the angular distance of the Sun from the horizon line along the Moon-Sun line is 65° - 3.3° = 62°. It is known that astronomical twilight sets in at a solar altitude of –16°; absolute darkness. At h= –6° civil twilight sets in, when you can still see something, and only the brightest stars are visible in the sky. The altitude of the Sun obtained by us for any latitude in Russia is significantly greater than –16°, so the details of the landscape should not be visible, since there should already be dark night. So the artist painted the moon too high.
(The artist most likely depicted the Russian North. I wonder if it is possible to determine the latitude of the place? - Ed.)
B.M. Kustodiev. Freezing day
In the picture B. Kustodieva"Frosty day" The Sun is depicted low above the horizon, and its height in degrees is easy to determine. If additional elements related to time are introduced into the task (for example, to assume that the picture depicts noon on the winter solstice, when the declination of the Sun is known, then you can use the formula for the height of the luminary at the climax, expressed through the latitude of the place of observation φ and the declination of the luminary + δ ( h\u003d 90 - φ + δ), determine the latitude of the depicted area.
A.I. Kuindzhi. Night. Fragment
Solution. We measure the diameter of the Sun in millimeters in the figure. let it be, for example, 5 mm. This means that 1° of the celestial sphere occupies 10 mm in the figure. We measure the distance from the center of the solar disk to the horizon line (we find one of its points as the intersection of two converging to the horizon parallel lines, and we draw the horizon line itself through the obtained point so that it is perpendicular to the line lowered onto it from the center of the Sun) and we get 52 mm, which means that the angular distance of the Sun from the horizon is 5.2 °. To find the latitude of the place, we use the above formula for the height of the luminary h at the upper culmination, assuming δ = -23.5 ° (we set the day of the winter solstice), and we get φ = 61.3 °, i.e. latitude somewhere in the region of Veliky Ustyug.
Similar tasks can be made for pictures I. Aivazovsky"Nice at night" Moonlight night" And A. Kuindzhi"Moonlit night on the Dnieper".
If the moon is depicted in a certain phase, then you can determine the time of day. For example, in the picture A.I. Kuindzhi The "night" Moon is visible in the form of a narrow crescent, which is possible either after sunset or before sunrise. Taking into account the shape of the sickle (the letter "P"), we determine that the time is evening.
Similar tasks can be formulated from pictures A.I. Kuindzhi"Moon sickle against the backdrop of sunset" and "Night", I.K. Aivazovsky " Windmill», G.Karusa"View of Dresden from the terrace in Brüll", I.I. Levitan"Twilight. Moon", A.Russo"Sleeping Gypsy" M.A. Vrubel"Pan" and "Princess Volkhova", F.Goya"Giant", A. Zhabsky"Youth", D.M.Utenkova"Old Wharf". The fact is that the age of the Moon, or the number of days after the new moon, is related to its type: the greater the age, the more complete the crescent is, which in a week becomes a crescent, and after 15 days - a full disk, and the angular removal of the Moon gradually increases. from the sun. As already shown above, Ф = 0.5 (1 + cosφ).
Questions can also be qualitative, for example: Why does the moon near the horizon acquire Orange color? (I. Levitan."Twilight. Haystacks "", Z.K.Tsereteli"Arba").
The listed examples can be used both in astronomy lessons and during circle classes, as well as tasks for astronomy olympiads.
Vladimir Fyodorovich Kartashov- Associate Professor of the Chelyabinsk State Pedagogical University, graduated from the Chelyabinsk State Pedagogical University in 1966, teaching experience 42 years. He graduated from the graduate school of the Astrophysical Institute of the Academy of Sciences of the Kazakh SSR in 1969 and worked there until 1978, then for 5 years at the Research Institute of Vocational Education of the Academy of Sciences of the USSR in Kazan, and since 1983 he has been teaching astrophysics and astronomy at the Chechen State Pedagogical University. defended his dissertation for the title of Candidate of Physical and Mathematical Sciences in 1974 at the GAO of the USSR Academy of Sciences in Pulkovo (St. Petersburg), the author of over 200 scientific works on astronomy and methods of teaching it, and 12 books on astronomy. He considers astronomy both a profession and a hobby: he collects and uses in his work everything related to the science of the stars. He was awarded a diploma of the Ministry of Education of the USSR, medals of VDNKh of the USSR.
In 1967, a series of stamps were issued in the Soviet Union with paintings by the famous science fiction artist Andrei Sokolov. Among the landscapes of distant planets and fantastic interstellar ships on one of the stamps there was a reproduction of the painting "Selenodesists", in which two astronauts salute the blue globe of the Earth hanging above the lunar horizon. The name of the picture, however, was changed and the stamp was simply written "On the Moon". But attentive philatelists immediately drew attention to a strange shadow in the lower right corner of the picture, which for some reason went beyond the image, as if closing the continuation of the title. Yes, and on the original picture of this shadow was not ...
A.Sokolov. Selenodesists
It turns out that when the circulation was printed, the signature on the stamp was really different: "On the Moon. Rising Earth". But then someone noticed a mistake - after all, everyone knows that the Moon is facing the Earth on one side, which means that the Earth does not move across the lunar firmament and cannot rise or set. An overprint in the form of a shadow was made on the entire print run ...
Only there was no mistake in this signature! Let's take a closer look at the movement of the moon.
The Moon revolves around the Earth in an elliptical orbit, making one revolution (relative to the stars) in 27.3 days, and in the same time the Moon makes one revolution around its axis. However, the movement along the orbit is uneven - at the point most distant from the Earth (apogee) the Moon moves more slowly, at the closest point (perigee) - faster. Naturally, the angular velocity of the Moon's movement along its orbit also changes, but the angular velocity of the Moon's rotation around its axis is constant, it is this difference that leads to a slight "swing" of the Moon in longitude relative to the direction to the Earth. 
- this is clearly seen in the animation of two images of the moon.
In addition, the lunar orbit is slightly inclined to the ecliptic plane, and this also leads to a slight change in the position of the center of the Moon's disk, but in the latitudinal direction. These small wiggles (up to 8° in longitude and up to 7° in latitude in each direction) are called librations. They lead to the fact that in the sky of the Moon the Earth constantly moves along a complex curve in the region of 16°x14°. And given that the size of the Earth's disk when observed from the Moon is about 2°, it is easy to see that in the regions of the Moon near the edge of the visible hemisphere, the Earth will both rise and set. True, it would be quite difficult for the astronauts depicted in the picture to notice a change in the position of the Earth - after all, sunrise (as well as sunset) lasts at least two days!
SOKOLOV
Andrei Konstantinovich
(29.09.1931- 16.03.2007)
- fantasy artist folk artist RSFSR.
He graduated from the Moscow Architectural Institute in 1955, since that time - a participant in art exhibitions.
Since childhood, he loved science fiction and he devoted his first works in the genre of science fiction painting to Bradbury's novel Fahrenheit 451.
After the launch of the first artificial Earth satellite in 1957, all the work of A. Sokolov is devoted to the theme of space exploration - he is the first artist in the world who began to draw outer space without leaving the workshop.
Tempera drawings on cardboard and canvases, painted in oil on canvas, are distinguished by a detailed description of the details of the construction of spaceships, landscapes, and cosmic phenomena. Some of his paintings are like serials: the stages of building a space station, landing on the Moon, Mars, Venus, satellites of the planets.
A. Sokolov's creativity influenced the work of other figures of science and culture. Influenced by his painting "Elevator to Space" famous science fiction Arthur Clarke wrote the book Fountains of Paradise. The story "Five Pictures" Ivan Efremov dedicated to Sokolov.
Sokolov's canvases were exhibited at the Smithsonian Institution (USA), in Dresden Gallery, in the museums of Berlin, Tokyo, Minsk, many Russian cities.
Task 1 (quiz)
Tasks a, b and c - this is the game "The fourth is extra". What is superfluous in each case from the point of view of astronomy? Why?
a) Leo, Taurus, Capricorn, Dragon.
Answer : The Dragon is a non-zodiacal constellation among the zodiac.
b) Neptune, Uranus, Pluto, Jupiter.
Answer : Pluto - dwarf planet among the giant planets.
c) Black Sea, White Sea, East Sea, North Sea.
Answer : The East Sea is a lunar sea among earthly ones.
d) By replacing one letter, turn the planet into a state.
Answer : Uranium - Iran.
e) The name of which of the months of the year is translated as "the tenth"? What is it on the account in our calendar and why?
Answer : december, twelfth month; the name came from Latin; March was the first month of the year in the Roman calendar.
Evaluation criteria : in paragraphs a, b, c to 1 point for each correct answer and 1 point for its justification; in paragraph d for the correct answer - 1 point; in paragraph e, depending on the completeness of the answer, - up to 3 points.
Max per task 10 points.
Task 2 (dictionary)
Explain the meaning of astronomical terms:
- a) gnomon;
- b) radiant;
- c) refractor;
- d) constellation;
- d) solstice.
Answers :
- a) gnomon - a vertical pole or column that allows (by shadow) to determine the height of the Sun above the horizon, the moment of true noon and the direction of the meridian;
- b) radiant - a point on the celestial sphere, which, due to perspective, seems to be a source of meteors (from where “shooting stars” seem to fly out);
- c) refractor - a type of telescope with a lens objective;
- d) constellation - a section of the celestial sphere within the established boundaries
A response like "a group of stars" or "a pattern of stars" is unfaithful.
- e) solstice (summer or winter) - the moment when the Sun reaches its northernmost or southernmost position on the ecliptic.
Evaluation criteria : By 1 point for the correct (at least in his own words) explanation of the meaning of each term.
Max per task 5 points.
Task 3 (gallery)
What space bodies are shown in the photographs?
Answers :
- dwarf planet Pluto;
- galaxy Andromeda Nebula (M31);
- planet Mercury;
- The moon is in the growing phase, ashen light is visible.
Evaluation criteria: according to 1 point for each correct answer; in paragraph d for mentioning the ashen light - additionally 1 point.
Max per task 5 points.
Task 4
Let's assume that today the height of the Sun at noon in Cape Town (33o 55' S, 18o 29' E) is the highest possible during the year. In which of the following places does the sun not rise today?
- Anadyr (64⁰ 44′ N, 177⁰ 31′ E);
- Mirny (66⁰ 33′ S, 93⁰ 00′ E);
- Murmansk (68⁰ 58′ N, 33⁰ 05′ E);
- Reykjavik (64⁰ 09′ N, 21⁰ 53′ W);
- Stockholm (59⁰ 20′ N, 18⁰ 04′ E);
- Tiksi (71⁰ 38′ N, 128⁰ 52′ E).
Answer : Cape Town is in southern hemisphere, south of the tropic. The highest midday height of the Sun there occurs on the day of the winter solstice (December 21–22). On this date, the polar night occurs in points located north of the Northern polar circle(66⁰ 34′ N). There are two such points on the list: Murmansk and Tiksi.
Evaluation criteria : for the correct answer with full justification - 4 points; if the answer is justified, but only one of the points is correctly indicated - 3 points; for determining the date on which the task action occurs, – 1 point; for mentioning the polar circle and the polar night - 1 point.
Max per task 4 points.
Task 5
From what constellation do aliens fly to us? Justify your answer.
Answer : Alien guests are flying from the constellation Taurus. To the right and above the "spaceship" is visible part of this constellation - the Pleiades star cluster.
Alien spaceship shut down main star constellation - Aldebaran.
Evaluation criteria: for the correct answer with full justification - 4 points; for mentioning Aldebaran and the Pleiades 1 point; for the correct answer with justification like “found on the map” – 2 points; for a correct answer without justification - 1 point.
IN reference materials there is a star map and a table bright stars. The task thus tests the student's ability to work with sources of information.
Max per task 4 points.
Task 6
Find astronomical errors in a painting Soviet artist Andrey Sokolov "Moon. Traces of astronauts in the lunar dust.
Answer : several astronomical errors were made in the picture.
- First, the Earth is not shown in phase. Judging by the shadows from the rocks and astronauts, the sun shines from right side and is located quite high above the horizon. That's why Earth should also be illuminated from the top right and look like a sickle or half-disk.
- Secondly, the apparent diameter of the Earth is greatly exaggerated. The Earth is about three and a half times the diameter of the Moon, so the Earth in the lunar sky should be only three and a half times the size of the Moon in the Earth's sky.
- Thirdly, the Moon has no atmosphere, so inside the shadows, where the light of the Sun does not fall, it should be completely dark, no details can be seen.
Evaluation criteria : By 1 point for finding each of the errors; 1 point for the correct substantiation of the answer (at least one of the three points).
Max per task 4 points.
Task 7
Talk future designers of spacecraft. Petya dreams: "I will build a ship that will fly to the moon in a second." Kolya: "And I will build a ship that will fly to Mars in an hour." Vasya: "And I am a ship that will fly to Alpha Centauri in a year." What do you think, which of these projects will be implemented? Justify your answer.
Solution : the highest speed possible in nature is the speed of light in vacuum, which is 300 thousand km/s. The distance to the Moon - about 400 thousand km - light overcomes in 4/3 s. The distance to Alpha Centauri is about 4 years. Therefore, the projects of Petya and Vasya are obviously not feasible.
The distance between Mars and Earth at average opposition is 0.5 AU. e. (from the reference table). To overcome such a distance in an hour, the ship must fly approximately 1.25 million km per minute, or 21 thousand km per second.
This is much less than the speed of light, so Kolya's project may be implemented.
The participant can use the distance known to him between the Earth and Mars in the great opposition - 57 million km. In this case, it turns out about 1 million km per minute, or 16 thousand km per second.
Evaluation criteria : for a correct answer without justification or with incorrect justification - 1 point; for pointing out the speed of light as the limit - 1 point; for the correct calculation of the time during which light propagates from the Earth to the Moon and to Alpha Centauri - 1 point; for correctly calculating the time it takes for light to travel from Earth to Mars, given its configuration - 2 points.
Max per task 4 points.
Total for work 36 points.
You have been given a star chart. Determine and list the constellations in which the Moon occurs in the sky. Nizhny Novgorod(latitude approximately 560). The inclination of the lunar orbit to the ecliptic 509/. The radius of the Earth is 6400 km, the average distance of the Moon from the Earth is 384000 km. Ball star cluster, galaxy, stellar association, constellation, open star cluster. Cross off the odd one out on this list and explain your answer. Are there places on Earth where the stars do not rise or set below the horizon? A person standing on the equator of the Earth moves at a certain speed relative to the center of the Earth. An astronaut standing on the equator of the Moon moves at a certain speed relative to the center of the Moon. Which of these two speeds is greater and by how many times, if it is known that the radius of the Moon is 4 times less than the radius of the Earth?
TASKS
All-Russian OLYMPIAD for schoolchildren
ASTRONOMY
2014 – 2015 academic year G.
(municipal stage)
Determine the radius of the ball with water, the mass of which is equal to the mass of the Earth, the Sun. What objects in the Universe have sizes close to those obtained? Globular star cluster, galaxy, stellar association, constellation, open star cluster. Cross off the odd one out on this list and explain your answer. How will a pendulum clock delivered from Earth to the surface of Mars go? Along which parallel on Earth can one walk on foot so that the Sun "stops"? To determine the mass of bodies, either lever or spring scales are used. Both those and others in conditions of weightlessness, for example, on a small artificial satellite of the Earth or on a spacecraft moving with the engines turned off, it would seem that they cannot work. What would you do if you were still offered to determine body weight under these conditions with the help of scales? Which scales - spring or lever - should be used and how? You have been given a star chart. Determine and list the constellations in which the Moon occurs in the sky of Nizhny Novgorod (latitude is approximately 560). The inclination of the lunar orbit to the ecliptic 509/. The radius of the Earth is 6400 km, the average distance of the Moon from the Earth is 384000 km.
TASKS
All-Russian OLYMPIAD for schoolchildren
ASTRONOMY
2014 – 2015 academic year G.
(municipal stage)
Which of the following astronomical phenomena - equinoxes, solstices, full moons, solar eclipses, lunar eclipses, planetary oppositions, meteor shower maxima, the appearance of bright comets, variable star brightness maxima, supernova explosions - occur every year at exactly approximately the same dates (from accurate to 1-2 days)? Flying to an unfamiliar planet, spaceship, turning off the engines, went into a circular orbit, and the astronauts began preliminary studies. Can they determine the average density of the planet's matter, using only clocks for this purpose? We will assume that the asteroid belt is a swarm of bodies enclosed in a torus (i.e. "donut") with a width of 1 astronomical unit(a. e.), revolving around the Sun at an average distance of 2.5 a. e. Assuming 1 million bodies in this belt, estimate the average distance between two adjacent bodies. Estimate the maximum possible and minimum possible value of the period of revolution of the comet around the Sun. A star located at a distance of 7 pc (parsec) has an apparent magnitude of 6m. What apparent magnitude will the same star have if observed from a distance of 70 pc? This original photo(by Chris Thomas) appeared on the APOD website on September 29, 2010. Estimate from the photo: a) the distance to the aircraft b) the direction of its flight c) the time of day when the photo was taken d) the place on Earth where it was taken from.
TASKS
All-Russian OLYMPIAD for schoolchildren
ASTRONOMY
2014 – 2015 academic year G.
(municipal stage)
Which of the following astronomical phenomena - equinoxes, solstices, full moons, solar eclipses, lunar eclipses, planetary oppositions, meteor shower maxima, the appearance of bright comets, variable star brightness maxima, supernova explosions - occur every year at exactly approximately the same dates (from accurate to 1-2 days)? On December 18, the asteroid Ceres collides with the Sun. What constellation will it be in? We will assume that the asteroid belt is a swarm of bodies enclosed in a torus (i.e., a “donut”) 1 astronomical unit (AU) wide, revolving around the Sun at an average distance of 2.5 AU. e. Assuming 1 million bodies in this belt, estimate the average distance between two adjacent bodies. To experimentally determine the acceleration of free fall on the newly discovered planet, the astronauts decided to use a small steel ball, a powerful lamp, an electric motor with a known speed, on the axis of which a cardboard disk with a narrow radial slit, a piece of black linen, a ruler with divisions and a camera is fixed. How should this set of instruments and objects be disposed of in order to complete the task? A certain exoplanet revolves around its star in a circular orbit with a radius of 108 km. The angular diameter of the disk of a star in the planet's sky is 10, and a year on this planet lasts 180 Earth days. Estimate the average density of this star. A star located at a distance of 7 pc (parsec) has an apparent magnitude of 6m. What apparent magnitude will the same star have if observed from a distance of 70 pc?
Astronomy grade 8-9, school (first) stage
Run time - 90 min
Task 1 (quiz)
Tasks a, b and c - this is the game "The fourth is extra". What is superfluous in each case from the point of view of astronomy? Why?
a) Ursa Minor, Ursa Major, Orion, Cassiopeia.
b) Leo, Taurus, Capricorn, Dragon.
c) Black Sea, White Sea, East Sea, North Sea.
d) In the name of which planet is the Greek letter hidden? Write this letter.
e) The Day of the Space Forces of Russia is celebrated annually on October 4th. In honor of what event was this date chosen?
Task 2 (dictionary)
What do the words mean:
a) an astronomical unit;
b) galaxy;
c) a meteor
Task 3 (gallery)
What space bodies are shown in the photographs?
Picture 1
Figure 2
Figure 3
Task 4
Find astronomical errors in the painting by Soviet artist Andrei Sokolov “The Moon. Traces of astronauts in the lunar dust.
Painting by Andrey Sokolov “Moon. Traces of astronauts in the lunar dust»
Task 5
Decorated the night blue
silver orange,
And it's only been a week...
There was a piece left of him.
Which heavenly body described in the riddle? Explain the changes that are happening to him. Make an explanatory drawing.
Task 6
Solve the crossword. What does the word received in the highlighted column mean?
1
2
3
4
5
6
7
8
9
10
ancient greek philosopher, which suggested that the Earth is located in the center, around which seven celestial spheres revolve.
An instrument used to observe celestial bodies.
An ancient Greek astronomer who developed his own system of the world, which dominated science for 13 centuries.
Mathematician, first put forward the assumption that the Earth has the shape of a ball.
That around which all the planets revolve.
Earth satellite.
The third planet in the solar system.
16th century Italian scholar named Giordano.
The great Polish astronomer who concluded that the Earth revolves around the Sun.
Outer space and everything that fills it.
Task 1
The Earth's globe has a diameter of 30 cm. At what height above the surface of the globe should the model of the ISS (International Space Station) be placed if the real ISS flies at an altitude of 400 km above the Earth's surface? What size will the station model be if the ISS is 60 m long? The radius of the Earth is 6,400 km.
Task 2
It is known that a photon (a quantum, i.e. a particle of light) originating in the center of the Sun reaches its surface 30 million years after its birth. Rate average speed displacement of a photon from the center of the Sun to the surface, if it is known that the radius of the Sun is about 200 times less than the distance from the Sun to the Earth, and the photon overcomes the distance from the Sun's surface to the Earth in 500 seconds.
