In the basic curriculum There is no astronomy, but it is recommended to hold an Olympiad in this subject. In our city of Prokopyevsk, the text of the Olympiad problems for grades 10-11 was compiled by Evgeniy Mikhailovich Ravodin, Honored Teacher of the Russian Federation.
To increase interest in the subject of astronomy, tasks are offered at the first and second levels of difficulty.
We provide the text and solutions to some tasks.
Problem 1. At what speed and direction should an airplane fly from Novokuznetsk airport in order to, moving along the parallel of 54°N, arrive at its destination at the same hour local time as when departing from Novokuznetsk?
Problem 2. The disk of the Moon is visible at the horizon in the form of a semicircle, convex to the right. In which direction are we looking, approximately at what time, if the observation occurs on September 21? Justify the answer.
Task 3. What is an “astronomical staff”, what is it intended for and how is it designed?
Problem 5. Is it possible to observe a 2 m spacecraft descending on the Moon using a school telescope with a lens diameter of 10 cm?
Problem 1. Vega's magnitude is 0.14. How many times brighter is this star than the Sun if the distance to it is 8.1 parsecs?
Task 2. In ancient times, when solar eclipses were “explained” by the capture of our star by a monster, eyewitnesses found confirmation of this in the fact that during a partial eclipse they observed light reflections “resembling the shape of claws” under the trees and in the forest. How can such a phenomenon be scientifically explained?
Problem 3. How many times is the diameter of the star Arcturus (Bootes) greater than the Sun if the luminosity of Arcturus is 100 and the temperature is 4500 K?
Problem 4. Is it possible to observe the Moon a day before a solar eclipse? And the day before the lunar day? Justify the answer.
Problem 5. A spaceship of the future, having a speed of 20 km/s, flies at a distance of 1 pc from a spectral binary star, whose spectral oscillation period is equal to a day, and the semi-major axis of the orbit is 2 astronomical units. Will a spaceship be able to escape the gravitational field of a star? Take the mass of the Sun as 2*10 30 kg.
Solving problems at the municipal stage of the Astronomy Olympiad for schoolchildren
The earth rotates from west to east. Time is determined by the position of the Sun; therefore, in order for the plane to be in the same position relative to the Sun, it must fly against the rotation of the Earth at a speed equal to the linear speed of points on the Earth at the latitude of the route. This speed is determined by the formula:
; r = R 3 cos?
Answer: v= 272 m/s = 980 km/h, fly west.
If the Moon is visible from the horizon, then in principle it can be seen either in the west or in the east. The convexity to the right corresponds to the phase of the first quarter, when the Moon lags behind the Sun in its daily motion by 90 0. If the moon is at the horizon in the west, then this corresponds to midnight, the sun is at its lower culmination, and exactly in the west this will happen on the days of the equinoxes, therefore, the answer is: we look to the west, approximately at midnight.
An ancient device for determining angular distances on celestial sphere between the luminaries. It is a ruler on which a traverse is movably fixed, perpendicular to this ruler, and marks are fixed at the ends of the traverse. At the beginning of the line there is a sight through which the observer looks. By moving the traverse and looking through the sight, he aligns the marks with the luminaries, between which the angular distances are determined. On the ruler there is a scale on which you can determine the angle between the luminaries in degrees.
Eclipses occur when the Sun, Earth and Moon are on the same line. Before a solar eclipse, the Moon will not have time to reach the Earth-Sun line. But at the same time, within a day he will be close to her. This phase corresponds to the new moon, when the Moon faces the Earth with its dark side, and is also lost in the rays of the Sun - therefore not visible.
A telescope with a diameter D = 0.1 m has angular resolution according to the Rayleigh formula;
500 nm (green) - wavelength of light (the wavelength to which the human eye is most sensitive is taken)
Angular size of the spacecraft;
l- device size, l= 2 m;
R - distance from the Earth to the Moon, R = 384 thousand km
, which is less than the resolution of the telescope.
Answer: no
To solve, we apply a formula that relates the apparent magnitude m with absolute magnitude M
M = m + 5 - 5 l g D,
where D is the distance from the star to the Earth in parsecs, D = 8.1 pc;
m - magnitude, m = 0.14
M is the magnitude that would be observed from a given star at a standard distance of 10 parsecs.
M = 0.14 + 5 - 5 l g 8.1 = 0.14 + 5 - 5*0.9 = 0.6
The absolute magnitude is related to the luminosity L by the formula
l g L = 0.4 (5 - M);
l g L = 0.4 (5 - 0.6) = 1.76;
Answer: 58 times brighter than the Sun
During a partial eclipse, the Sun appears as a bright crescent. The spaces between the leaves are small holes. They, working like holes in a camera obscura, give multiple images of sickles on Earth, which can easily be mistaken for claws.
Let's use the formula, where
D A - diameter of Arcturus in relation to the Sun;
L = 100 - Arthur's luminosity;
T A = 4500 K - Arcturus temperature;
T C = 6000 K - temperature of the Sun
Answer: D A 5.6 solar diameters
Eclipses occur when the Sun, Earth and Moon are on the same line. Before a solar eclipse, the Moon will not have time to reach the Earth-Sun line. But at the same time, within a day he will be close to her. This phase corresponds to the new moon, when the moon faces the earth with its dark side, and is also lost in the rays of the Sun - therefore not visible.
A day before a lunar eclipse, the Moon does not have time to reach the Sun-Earth line. At this time it is in the full moon phase and therefore visible.
v 1 = 20 km/s = 2*10 4 m/s
r = 1 pc = 3*10 16 m
m o = 2*10 30 kg
T = 1 day = year
G = 6.67 * 10 -11 N * m 2 / kg 2
Let's find the sum of the masses of spectroscopic binary stars using the formula m 1 + m 2 = * m o = 1.46 * 10 33 kg
Let's calculate the escape velocity using the formula for the second cosmic velocity (since the distance between the components of a spectral binary star - 2 AU is much less than 1 pc)
2547.966 m/s = 2.5 km/h
Answer: 2.5 km/h, the speed of the starship is higher, so it will fly away.
Problem 1
The focal length of the telescope lens is 900 mm, and the focal length of the eyepiece used is 25 mm. Determine the magnification of the telescope.
Solution:
The magnification of the telescope is determined from the relation: , where F– focal length of the lens, f– focal length of the eyepiece. Thus, the magnification of the telescope will be 
once.
Answer: 36 times.
Problem 2
Convert the longitude of Krasnoyarsk to hourly units (l=92°52¢ E).
Solution:
Based on the relationship between the hourly unit of angle and the degree unit:
24 hours =360°, 1 hour =15°, 1 minute =15¢, 1 s = 15², and 1°=4 minutes, and taking into account that 92°52¢ = 92.87°, we get:
1 hour · 92.87°/15°= 6.19 hours = 6 hours 11 minutes. e.d.
Answer: 6 hours 11 minutes e.d.
Problem 3
What is the declination of a star if it culminates at an altitude of 63° in Krasnoyarsk, whose latitude is 56° N?
Solution:
Using the relationship connecting the height of the luminary at the upper culmination, culminating south of the zenith, h, declination of the luminary δ and latitude of the observation site φ , h = δ + (90° – φ ), we get:
δ = h + φ – 90° = 63° + 56° – 90° = 29°.
Answer: 29°.
Problem 4
When it is 10 hours 17 minutes 14 seconds in Greenwich, at some point the local time is 12 hours 43 minutes 21 seconds. What is the longitude of this point?
Solution:
Local time is mean solar time, and local Greenwich time is universal time. Using the relationship relating the mean solar time T m, universal time T0 and longitude l, expressed in hourly units: T m = T0 +l, we get:
l = T m – T 0 = 12 hours 43 minutes 21 seconds. – 10 hours 17 minutes 14 seconds = 2 hours 26 minutes 07 seconds.
Answer: 2h 26 min 07 s.
Problem 5
After what period of time do the moments of maximum distance of Venus from the Earth repeat if its sidereal period is 224.70 days?
Solution:
Venus is the lower (inner) planet. The planetary configuration at which the inner planet is at its maximum distance from the Earth is called superior conjunction. And the period of time between successive configurations of the same name on the planet is called the synodic period S. Therefore, it is necessary to find the synodic period of the revolution of Venus. Using the equation of synodic motion for the lower (inner) planets, where T– sidereal, or sidereal period of revolution of the planet, TÅ – sidereal period of rotation of the Earth (sidereal year), equal to 365.26 average solar days, we find:
=583.91 days.
Answer: 583.91 days.
Problem 6
The sidereal period of Jupiter's revolution around the Sun is about 12 years. What is the average distance of Jupiter from the Sun?
Solution:
The average distance of a planet from the Sun is equal to the semi-major axis of the elliptical orbit a. From Kepler's third law, comparing the motion of a planet with the Earth, for which taking the sidereal period of revolution T 2 = 1 year, and the semimajor axis of the orbit a 2 = 1 AU, we obtain a simple expression for determining the average distance of the planet from the Sun in astronomical units based on the known sidereal period of revolution, expressed in years. Substituting the numerical values we finally find:
Answer: about 5 AU
Problem 7
Determine the distance from Earth to Mars at the moment of its opposition, when its horizontal parallax is 18².
Solution:
From the formula for determining geocentric distances 
, Where ρ
– horizontal parallax of the luminary, RÅ = 6378 km – the average radius of the Earth, let’s determine the distance to Mars at the moment of opposition:
» 73×10 6 km. Dividing this value by the value of the astronomical unit, we get 73 × 10 6 km / 149.6 × 10 6 km » 0.5 AU.
Answer: 73×10 6 km » 0.5 AU
Problem 8
The horizontal parallax of the Sun is 8.8². At what distance from Earth (in AU) was Jupiter when its horizontal parallax was 1.5²?
Solution:
From the formula it is clear that the geocentric distance of one star D 1 is inversely proportional to its horizontal parallax ρ
1, i.e. . A similar proportionality can be written for another luminary for which the distance D 2 and horizontal parallax are known ρ
2: . Dividing one ratio by the other, we get . Thus, knowing from the conditions of the problem that the horizontal parallax of the Sun is 8.8², while it is located at 1 AU. from Earth, you can easily find the distance to Jupiter from the known horizontal parallax of the planet at this moment:
=5.9 a.u.
Answer: 5.9 a.u.
Problem 9
Determine the linear radius of Mars if it is known that during great opposition its angular radius is 12.5² and its horizontal parallax is 23.4².
Solution:
Linear radius of luminaries R can be determined from the relation, r is the angular radius of the star, r 0 is its horizontal parallax, R Å is the radius of the Earth, equal to 6378 km. Substituting the values from the problem conditions, we get: 
= 3407 km.
Answer: 3407 km.
Problem 10
How many times is the mass of Pluto less than the mass of the Earth, if it is known that the distance to its satellite Charon is 19.64 × 10 3 km, and the satellite’s orbital period is 6.4 days. The distance of the Moon from the Earth is 3.84 × 10 5 km, and its orbital period is 27.3 days.
Solution:
To determine the masses of celestial bodies, you need to use Kepler's third generalized law: 
. Since the masses of the planets M 1 and M 2 significantly less than the masses of their satellites m 1 and m 2, then the masses of the satellites can be neglected. Then this Kepler law can be rewritten as follows: 
, Where A 1 – semimajor axis of the orbit of the satellite of the first planet with mass M 1, T 1 – period of revolution of the satellite of the first planet, A 2 – semimajor axis of the orbit of the satellite of the second planet with mass M 2, T 2 – period of revolution of the satellite of the second planet.
Substituting the corresponding values from the problem conditions, we get:
= 0,0024.
Answer: 0.0024 times.
Problem 11
The Huygens space probe landed on Saturn's moon Titan on January 14, 2005. During the descent, he transmitted to Earth a photograph of the surface of this celestial body, on which formations similar to rivers and seas are visible. Estimate the average temperature on the surface of Titan. What kind of liquid do you think the rivers and seas on Titan might consist of?
Note: The distance from the Sun to Saturn is 9.54 AU. The reflectivity of the Earth and Titan is assumed to be the same, and the average temperature on the Earth's surface is 16°C.
Solution:
The energies received by Earth and Titan are inversely proportional to the square of their distances from the Sun r. Some of the energy is reflected, some is absorbed and goes to heat the surface. Assuming that the reflectivity of these celestial bodies is the same, then the percentage of energy spent on heating these bodies will be the same. Let us estimate the surface temperature of Titan in the black body approximation, i.e. when the amount of energy absorbed is equal to the amount of energy emitted by a heated body. According to the Stefan-Boltzmann law, the energy emitted by a unit surface per unit time is proportional to the fourth power of the absolute temperature of the body. Thus, for the energy absorbed by the Earth we can write 
, Where r h – distance from the Sun to the Earth, T h is the average temperature on the Earth’s surface, and Titan – 
, Where r c – distance from the Sun to Saturn with its satellite Titan, T T is the average temperature on the surface of Titan. Taking the relation, we get: 
, from here 
94°K = (94°K – 273°K) = –179°C. At such low temperatures, the seas on Titan may consist of liquid gas, such as methane or ethane.
Answer: From liquid gas, for example, methane or ethane, since the temperature on Titan is –179°C.
Problem 12
What is the apparent magnitude of the Sun as seen from the nearest star? The distance to it is about 270,000 AU.
Solution:
Let's use Pogson's formula: 
, Where I 1 and I 2 – brightness of sources, m 1 and m 2 – their magnitudes, respectively. Since brightness is inversely proportional to the square of the distance to the source, we can write 
. Taking logarithm of this expression, we get 
. It is known that the apparent magnitude of the Sun from Earth (from a distance r 1 = 1 a.u.) m 1 = –26.8. You need to find the apparent magnitude of the Sun m 2 from a distance r 2 = 270,000 a.u. Substituting these values into the expression, we get:
, hence ≈ 0.4 m.
Answer: 0.4 m.
Problem 13
Annual parallax of Sirius (a Canis Major) is 0.377². What is the distance to this star in parsecs and light years?
Solution:
Distances to stars in parsecs are determined from the relation , where π is the annual parallax of the star. Therefore = 2.65 pcs. So 1 pc = 3.26 sv. g., then the distance to Sirius in light years will be 2.65 pc · 3.26 sv. g. = 8.64 sv. G.
Answer: 2.63 pcs or 8.64 sv. G.
Problem 14
The apparent magnitude of the star Sirius is –1.46 m, and the distance is 2.65 pc. Determine the absolute magnitude of this star.
Solution:
Absolute magnitude M related to apparent magnitude m and distance to the star r in parsecs with the following ratio: 
. This formula can be derived from Pogson's formula 
, knowing that absolute magnitude is the magnitude that a star would have if it were at a standard distance r 0 = 10 pcs. To do this, we rewrite Pogson’s formula in the form 
, Where I– the brightness of a star on Earth from a distance r, A I 0 – brightness from a distance r 0 = 10 pcs. Since the apparent brightness of a star will change in inverse proportion to the square of the distance to it, i.e. , That 
. Taking logarithms, we get: either or .
Substituting the values from the problem conditions into this relation, we obtain:
Answer: M= 1.42 m.
Problem 15
How many times is the star Arcturus (a Boötes) larger than the Sun if the luminosity of Arcturus is 100 times greater than the solar one and the temperature is 4500° K?
Solution:
Star luminosity L– the total energy emitted by a star per unit time can be defined as , where S is the surface area of the star, ε is the energy emitted by the star per unit surface area, which is determined by the Stefan-Boltzmann law, where σ is the Stefan-Boltzmann constant, T– absolute temperature of the star’s surface. Thus, we can write: , where R– radius of the star. For the Sun we can write a similar expression: 
, Where L c – luminosity of the Sun, R c – radius of the Sun, T c is the temperature of the solar surface. Dividing one expression by the other, we get:
Or you can write this relationship this way: 
. Taking for the Sun R c =1 and L with =1, we get 
. Substituting the values from the problem conditions, we find the radius of the star in radii of the Sun (or how many times the star is larger or smaller than the Sun):
≈ 18 times.
Answer: 18 times.
Problem 16
In the spiral galaxy in the constellation Triangulum, Cepheids are observed with a period of 13 days, and their apparent magnitude is 19.6 m. Determine the distance to the galaxy in light years.
Note: The absolute magnitude of a Cepheid with the indicated period is equal to M= – 4.6 m.
Solution:
From the relation , relating the absolute magnitude M with apparent magnitude m and distance to the star r, expressed in parsecs, we get: = 
. Hence r ≈ 690,000 pc = 690,000 pc · 3.26 light. city ≈2,250,000 St. l.
Answer: approximately 2,250,000 St. l.
Problem 17
The quasar has a redshift z= 0.1. Determine the distance to the quasar.
Solution:
Let's write down Hubble's law: , where v– radial velocity of removal of the galaxy (quasar), r- distance to it, H– Hubble constant. On the other hand, according to the Doppler effect, the radial velocity of a moving object is equal to 
, с is the speed of light, λ 0 is the wavelength of the line in the spectrum for a stationary source, λ is the wavelength of the line in the spectrum for a moving source, is the red shift. And since the red shift in the spectra of galaxies is interpreted as a Doppler shift associated with their removal, Hubble's law is often written in the form: . Expressing the distance to the quasar r and substituting the values from the problem conditions, we get:
≈ 430 Mpc = 430 Mpc · 3.26 light. g. ≈ 1.4 billion St.L.
Answer: 1.4 billion St.L.
Examples of solving problems in astronomy
§ 1. The star Vega is located at a distance of 26.4 sv. years from Earth. How many years would it take for a rocket to fly towards it at a constant speed of 30 km/s?
The speed of the rocket is 10 0 0 0 times less than the speed of light, so the astronauts will fly to Begi 10,000 times longer.
Solutions:
§ 2. At noon your shadow is half the size of your height. Determine the height of the Sun above the horizon.
Solutions:
Sun height h measured by the angle between the horizon plane and the direction towards the luminary. From a right triangle, where the legs are L (shadow length) and H (your height), we find
§ 3. How different is the local time in Simferopol from Kyiv time?
Solutions:
in winter
That is, in winter, local time in Simferopol is ahead of Kiev time. In the spring, the hands of all clocks in Europe are moved forward 1 hour, so Kiev time is 44 minutes ahead of local time in Simferopol.
§ 4. The Amur asteroid moves along an ellipse with an eccentricity of 0.43. Could this asteroid hit the Earth if its rotation period around the Sun is 2.66 years?
Solutions:
An asteroid could hit Earth if it crosses orbitEarth, that is, if the distance is at perihelion rmin =< 1 а. o .
Using Kepler's third law, we determine the semimajor axis of the asteroid's orbit:
where a 2- 1 a. o .- semimajor axis of the Earth's orbit; T 2 = 1 year period
rotation of the Earth:
Rice. P. 1.
Answer.
Asteroid Amur will not cross the Earth's orbit, so it cannot collide with the Earth.
§ 5. At what height above the Earth’s surface should a geostationary satellite hovering over one point rotate? Earth?
Rose LS (X - N ІЛ
1. Using Kepler's third law we determine the semimajor axis of the satellite’s orbit:
where a2 = 3 80000 km is the semimajor axis of the Moon’s orbit; 7i, = 1 day - the period of rotation of the satellite around the Earth; T”2 = 27.3 days - the period of revolution of the Moon around the Earth.
a1 = 41900 km.
Answer. Geostationary satellites rotate from west to east in the equatorial plane at an altitude of 35,500 km.
§ 6. Can astronauts from the surface of the Moon see the Black Sea with the naked eye?
Rozv "yazannya:
We determine the angle at which the Black Sea is visible from the Moon. From a right triangle, in which the legs are the distance to the Moon and the diameter of the Black Sea, we determine the angle:
Answer.
If it is daytime in Ukraine, then the Black Sea can be seen from the Moon, because its angular diameter is greater than the resolution of the eye.
§ 8. On the surface of which terrestrial planet will the weight of astronauts be the least?
Solutions:
P = mg ; g =GM /R 2,
where G - gravitational constant; M is the mass of the planet, R - radius of the planet. The least weight will be on the surface of the planet where the free acceleration is lessfalls. From the formula g = GM/R we determine that on Mercury # = 3.78 m/s2, on Venus # = 8.6 m/s2, on Mars # = 3.72 m/s2, on Earth # = 9.78 m/s2.
Answer.
The weight will be the smallest on Mars, 2.6 times less than on Earth.
§ 12. When, in winter or summer, more midday hits the window of your apartment solar energy? Consider the cases: A. The window faces south; B. The window faces east.
Solutions:
A. The amount of solar energy that a unit surface area receives per unit time can be calculated using the following formula:
E =qcosi
where q - solar constant; i is the angle of incidence of sunlight.
The wall is located perpendicular to the horizon, so in winter the angle of incidence of the sun's rays will be less. So, strange as it may seem, in winter more energy comes into the window of your apartment from the Sun than in summer.
Would. If the window faces east, then the sun's rays at noon never illuminate your room.
§ 13. Determine the radius of the star Vega, which emits 55 times more energy than the Sun. The surface temperature is 1,1000 K. What appearance would this star have in our sky if it were shining in the place of the Sun?
Solutions:
The radius of the star is determined using formula (13.11):
where Dr, = 6 9 5 202 km - radius of the Sun;
Temperature of the surface of the Sun.
Answer.
The star Vega has a radius twice that of the Sun, so in our sky it would appear as a blue disk with an angular diameter of 1°. If Vega shone instead of the Sun, then the Earth would receive 55 times more energy than it does now, and the temperature on its surface would be above 1000°C. Thus, the conditions on our planet would become unsuitable for any form of life.
I will again use the brochure “Didactic Material on Astronomy” written by G.I. Malakhova and E.K. Strout and published by the Prosveshcheniye publishing house in 1984. This time the first tasks of the final test work on page 75.
To visualize formulas, I will use the LaTeX2gif service, since the jsMath library is not able to render formulas in RSS.
Task 1 (Option 1)
Condition: The planetary nebula in the constellation Lyra has an angular diameter of 83″ and is located at a distance of 660 pc. What are the linear dimensions of the nebula in astronomical units?
Solution: The parameters specified in the condition are related to each other by a simple relationship:
1 pc = 206265 AU, respectively:
Task 2 (Option 2)
Condition: The parallax of the star Procyon is 0.28″. The distance to the star Betelgeuse is 652 light years. of the year. Which of these stars and how many times is farther from us?
Solution: Parallax and distance are related by a simple relationship:
Next, we find the ratio of D 2 to D 1 and find that Betelgeuse is approximately 56 times further than Procyon.
Task 3 (Option 3)
Condition: How many times has the angular diameter of Venus, observed from Earth, changed as a result of the planet moving from minimum distance to the maximum? The orbit of Venus is considered to be a circle with a radius of 0.7 AU.
Solution: We find the angular diameter of Venus for the minimum and maximum distances in astronomical units and then their simple ratio:
We get the answer: it decreased by 5.6 times.
Task 4 (Option 4)
Condition: What angular size will our Galaxy (whose diameter is 3 × 10 4 pc) be seen by an observer located in the M 31 galaxy (Andromeda nebula) at a distance of 6 × 10 5 pc?
Solution: An expression connecting the linear dimensions of an object, its parallax and angular dimensions is already in the solution to the first problem. Let’s use it and, slightly modifying it, substitute the required values from the condition:
Problem 5 (Option 5)
Condition: Resolution of the naked eye is 2′. What size objects can an astronaut discern on the surface of the Moon when flying above it at an altitude of 75 km?
Solution: The problem is solved similarly to the first and fourth:
Accordingly, the astronaut will be able to distinguish surface details measuring 45 meters.
Problem 6 (Option 6)
Condition: How many times is the Sun larger than the Moon if their angular diameters are the same and their horizontal parallaxes are respectively 8.8″ and 57′?
Solution: This is a classic problem of determining the size of luminaries by their parallax. The formula for the relationship between the parallax of a luminary and its linear and angular dimensions repeatedly came across higher. As a result of reducing the repeating part, we get: 
The answer is that the Sun is almost 400 times larger than the Moon.
Keys to Olympiad assignments in astronomy 7-8 GRADES
Task 1. An astronomer on Earth observes a total lunar eclipse. What can an astronaut observe on the Moon at this time?
Solution: If there is a total lunar eclipse on Earth, an observer on the Moon will be able to see a total solar eclipse - the Earth will cover the solar disk.
Task 2. What evidence of the sphericity of the Earth could have been known to ancient scientists?
Solution: Evidence of the sphericity of the Earth, known to ancient scientists:
the rounded shape of the edge of the earth's shadow on the disk of the Moon during lunar eclipses;
the gradual appearance and disappearance of ships as they approach and move away from the shore;
change in the altitude of the North Star when changing the latitude of the observation site;
Removing the horizon as you ascend upward, for example, to the top of a lighthouse or tower.
Task 3.
On an autumn night, a hunter walks into the forest towards North Star. Immediately after sunrise he returns back. How should a hunter navigate by the position of the sun?
Solution: The hunter walked into the forest to the north. Returning, he should move south. Since the Sun is near the equinox in autumn, it rises close to the east point. Therefore, you need to walk so that the Sun is on the left.
Task 4.
What luminaries are visible during the day and under what conditions?
Solution: The Sun, Moon and Venus are visible to the naked eye, and stars up to 4 m - using a telescope.
Task 5. Determine which celestial objects do not change their right ascension, declination, azimuth and altitude due to the daily rotation of the Earth? Do such objects exist? Give an example:
Solution: If the star is located in the North or South Pole of the world, all four coordinates for an observer anywhere on Earth will be unchanged due to the rotation of the planet around its axis. Near the North Pole of the world there is such a star - Polaris.
Keys to Olympiad assignments in astronomy GRADE 9
Task 1. The steamer, having left Vladivostok on Saturday, November 6, arrived in San Francisco on Wednesday, November 23. How many days was he on the road?
Solution: On its way to San Francisco, the steamer crossed the international date line from west to east, subtracting one day. The number of days on the way is 23 – (6 – 1) = 18 days.
Task 2. The altitude of a star located on the celestial equator at the time of its upper culmination is 30. What is the height of the Celestial Pole at the observation location? (You can make a drawing for clarity).
Solution: If the star is at its highest culmination on the celestial equator,h = 90 0 - . Therefore, the latitude of the place = 90 0 – h = 60 0 . The height of the Celestial Pole is equal to the latitudeh p = = 60 0
Problem 3 . On March 4, 2007 there was a complete Moon eclipse. What and where was the Moon in the sky two weeks immediately after sunset?
Solution . A lunar eclipse occurs during the full moon phase. Since a little less than two weeks pass between the phases of the full moon and the new moon, then two weeks immediately after sunset, the Moon will be visible in the form of a narrow crescent above the horizon on its western side.
Problem 4 . q = 10 7 J/kg, solar mass 2 * 10 30 kg, and the luminosity is 4 * 10 26
Solution . Q = qM = 2*10 37 t = Q: L = 2 *10 37 /(4* 10 26 )= 5 * 10 10
Task 5. How to prove that the Moon is not made of cast iron, if it is known that its mass is 81 times less than the mass of the Earth, and its radius is approximately four times less than that of the Earth? Consider the density of cast iron to be approximately 7 times the density of water.
Solution . The simplest thing is to determine the average density of the Moon and compare it with the table density value for different materials: p =m/V. Then, substituting the mass and volume of the Moon into this expression in fractions of Earth sizes, we get: 1/81:1/4 3 =0.8.The average density of the Moon is only 0.8 of the Earth’s density (or 4.4 g/cm 3 -true value of the average density of the Moon 3.3 g/cm 3 ). But this value is also less than the density of cast iron, which is approximately 7g/cm 3 .
Keys to Olympiad tasks in astronomy 10-11 GRADES
Task 1. The sun at the north pole rose on the meridian of Yekaterinburg (λ= 6030` east). Where (approximately) will it rise next?
Solution: With sunrise, polar day began at the North Pole. The next time the Sun will rise at the beginning of the next polar day, i.e. exactly one year later.
If in a year the Earth made an integer number of revolutions around its axis, then the next sunrise would also be on our meridian. But the Earth makes about a quarter more revolutions (hence the leap year).
This quarter turn corresponds to the rotation of the Earth by 90 0 and since its rotation occurs from west to east, the sun will rise on the meridian with longitude 60.5 0 e.d. – 90 0 = - 29.5 0 , i.e. 29.5 0 w.d. At this longitude is the eastern part of Greenland.
Task 2. Travelers noticed that according to local time the lunar eclipse began at 5 hours 13 minutes, while according to the astronomical calendar this eclipse should begin at 3 hours 51 minutes Greenwich time. What is the geographic longitude of the place where the travelers are observed?
Solution: The difference in geographical longitude of two points is equal to the difference in the local times of these points. In our problem, we know the local time at the point where the lunar eclipse was observed at 5 hours 13 minutes and the local Greenwich (Worldwide) time of the beginning of the same eclipse at 3 hours 51 minutes, i.e. local prime meridian time.
The difference between these times is 1 hour 22 minutes, which means that the longitude of the place where the lunar eclipse was observed is 1 hour 22 minutes east longitude, because The time at this longitude is greater than Greenwich.
Task 3. At what speed and in what direction should a plane fly at the latitude of Yekaterinburg so that local solar time stops for the plane's passengers?
Solution: The plane must fly west at the speed of the Earth's rotationV= 2πR/T
At the latitude of YekaterinburgR = R eq cos , E 57 0
V= 2π 6371 cos 57 0 /24 3600 = 0.25 km/s
Task 4. At the end of the 19th century. Some scientists believed that the source of the sun's energy was chemical combustion reactions, in particular the combustion of coal. Assuming that the specific heat of combustion of coalq = 10 7 J/kg, solar mass 2 * 10 30 kg, and the luminosity is 4 * 10 26 W, provide strong evidence that this hypothesis is incorrect.
Solution: Heat reserves excluding oxygen areQ = qM = 2 *10 37 J. This supply will last for a whilet = Q: L = 2* 10 37 / 4* 10 26 = 5* 10 10 c = 1700 years. Julius Caesar lived more than 2000 years ago, dinosaurs froze out about 60 million years ago, so that due to chemical reactions the Sun cannot shine. (If someone talks about a nuclear power source, that will be great.)
Task 5. Try to find a complete answer to the question: under what conditions does the change of day and night occur anywhere on the planet?
Solution: To ensure that there is no change of day and night anywhere on the planet, three conditions must be met simultaneously:
a) the angular velocities of orbital and axial rotation must coincide (the length of the year and sidereal day are the same),
b) the axis of rotation of the planet must be perpendicular to the orbital plane,
c) the angular velocity of orbital motion must be constant, the planet must have a circular orbit.