Problems in astronomy. Assignments for independent work in astronomy Assignments in astronomy
Tasks for independent work on astronomy.
Topic 1. Studying the starry sky using a moving map:
1. Set the mobile map for the day and hour of observations.
date of observation __________________
observation time ___________________
2. List the constellations that are located in the northern part of the sky from the horizon to the celestial pole.
_______________________________________________________________
5) Determine whether the constellations Ursa Minor, Bootes, Orion will set.
Ursa Minor___
Bootes___
______________________________________________
7) Find the equatorial coordinates of the star Vega.
Vega (α Lyrae)
Right ascension a = _________
Declension δ = _________
8) Specify the constellation in which the object is located with coordinates:
a=0 hours 41 minutes, δ = +410
9. Find the position of the Sun on the ecliptic today, determine the length of the day. Sunrise and sunset times
Sunrise____________
Sunset _____________
10. The residence time of the Sun at the moment of the upper climax.
________________
11. In what zodiac constellation is the Sun located during the upper climax?
12. Determine your zodiac sign
Date of Birth___________________________
constellation __________________
Topic 2. The structure of the solar system.
What are the similarities and differences between the terrestrial planets and the giant planets. Fill in the form of a table:
2. Select a planet by option in the list:
| Mercury | ||||
Make a report about the planet of the solar system according to the option, focusing on the questions:
How is the planet different from others?
What is the mass of this planet?
What is the position of the planet in the solar system?
How long is a planetary year and how long is a sidereal day?
How many sidereal days fit into one planetary year?
The average life expectancy of a person on Earth is 70 Earth years, how many planetary years can a person live on this planet?
What details can be seen on the surface of the planet?
What are the conditions on the planet, is it possible to visit it?
How many satellites does the planet have and which ones?
3. Select the appropriate planet for the corresponding description:
| Mercury | The most massive |
|
| The orbit is strongly inclined to the plane of the ecliptic |
||
| The smallest of the giant planets |
||
| A year is approximately equal to two Earth years |
||
| closest to the sun |
||
| Close to Earth in size |
||
| Has the highest average density |
||
| Spins while lying on its side |
||
| Has a system of picturesque rings |
Topic 3. Characteristics of stars.
Choose a star according to the option.
Indicate the position of the star on the spectrum-luminosity diagram.
| temperature | Parallax | density | Luminosity, | Life time t, years | distance |
|||
Required formulas:
Average density:
Luminosity: 
Lifetime:
Star distance: 
Topic 4. Theories of origin and evolution of the Universe.
Name the galaxy we live in:
Classify our galaxy according to the Hubble system:
Draw schematically the structure of our galaxy, sign the main elements. Determine the position of the Sun.
What are the names of the satellites of our galaxy?
How long does it take for light to pass through our galaxy along its diameter?
What objects are the constituent parts of galaxies?
Classify the objects of our galaxy by photographs:
What objects are the constituent parts of the universe?
Universe
Which galaxies make up the population of the Local Group?
What is the activity of galaxies?
What are quasars and how far from Earth are they?
Describe what is seen in the photographs: 
Does the cosmological expansion of the Metagalaxy affect the distance from the Earth...
to the moon; □
To the center of the Galaxy; □
To the galaxy M31 in the constellation Andromeda; □
To the center of the local cluster of galaxies □
Name three possible variants of the development of the Universe according to Friedman's theory.
Bibliography
Main:
Klimishin I.A., "Astronomy-11". - Kyiv, 2003
Gomulina N. "Open Astronomy 2.6" CD - Physicon 2005
Workbook on astronomy / N.O. Gladushina, V.V. Kosenko. - Lugansk: Educational book, 2004. - 82 p.
Additional:
Vorontsov-Velyaminov B. A.
"Astronomy" Textbook for the 10th grade of high school. (Ed. 15th). - Moscow "Enlightenment", 1983.
Perelman Ya. I. "Entertaining astronomy" 7th ed. - M, 1954.
Dagaev M. M. "Collection of problems in astronomy." - Moscow, 1980.
There is no astronomy in the basic curriculum, but it is recommended to hold an Olympiad in this subject. In our city of Prokopievsk, the text of the Olympiad problems for grades 10-11 was compiled by Evgeny Mikhailovich Ravodin, Honored Teacher of the Russian Federation.
To increase interest in the subject of astronomy, tasks of the first and second levels of complexity are offered.
Here is the text and the solution of some tasks.
Task 1. With what magnitude and direction should an airplane fly from Novokuznetsk airport in order to arrive at its destination at the same hour local time as when flying from Novokuznetsk, moving along the parallel 54 ° N?
Task 2. The disk of the Moon is visible at the horizon in the form of a semicircle, bulging to the right. In which direction are we looking, approximately at what time, if the observation takes place on September 21st? Justify the answer.
Task 3. What is an "astronomical staff", what is it intended for and how is it arranged?
Problem 5. Is it possible to observe a 2 m spacecraft descending to the Moon with a school telescope with a lens diameter of 10 cm?
Task 1. The magnitude of Vega 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 luminary by a monster, eyewitnesses found confirmation of this in the fact that during a partial eclipse they observed light glare under the trees, in the forest, "resembling the shape of claws." How can this phenomenon be scientifically explained?
Task 3. How many times the diameter of the star Arcturus (Boötes) is greater than the Sun if the luminosity of Arcturus is 100 and the temperature is 4500 K?
Task 4. Is it possible to observe the Moon a day before a solar eclipse? And a day before the moon? Justify the answer.
Problem 5. The spaceship of the future, having a speed of 20 km/s, flies at a distance of 1 pc from a spectral binary star, in which the spectrum oscillation period is equal to days, and the semi-major axis of the orbit is 2 astronomical units. Will the starship be able to escape from the star's gravitational field? Take the mass of the Sun as 2 * 10 30 kg.
Solving problems of the municipal stage of the Olympiad for schoolchildren in astronomy
The earth rotates from west to east. Time is determined by the position of the Sun; therefore, in order for the aircraft 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 velocity of the Earth's points 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 bulge to the right corresponds to the phase of the first quarter, when the Moon lags behind the Sun in daily motion by 90 0 . If the moon is near the horizon in the west, then this corresponds to midnight, the sun in the lower climax, and exactly in the west this will happen on the equinoxes, therefore, the answer is: we look to the west, approximately at midnight.
An ancient device for determining the angular distances on the celestial sphere between the stars. It is a ruler on which a traverse is movably fixed, perpendicular to this ruler, marks are fixed at the ends of the traverse. At the beginning of the ruler there is a sight through which the observer looks. Moving the traverse and looking through the sight, he aligns the marks with the luminaries, between which the angular distances are determined. The ruler has a scale on which you can determine the angle between the luminaries in degrees.
Eclipses occur when the Sun, Earth and Moon are in the same straight line. Before a solar eclipse, the Moon will not have time to reach the Earth-Sun line. But at the same time, it will be close to her in a day. This phase corresponds to the new moon, when the Moon is facing the Earth with its dark side, and besides, it is lost in the rays of the Sun - therefore it is not visible.
A telescope with a diameter D = 0.1 m has an 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)
The 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 the formula that relates the apparent stellar magnitude m with absolute magnitude M
M = m + 5 - 5 l gD,
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 the distance of a given star from a standard distance of 10 parsecs.
M = 0.14 + 5 - 5 l g 8.1 \u003d 0.14 + 5 - 5 * 0.9 \u003d 0.6
The absolute magnitude is related to the luminosity L by the formula
l g L = 0.4 (5 - M);
l g L \u003d 0.4 (5 - 0.6) \u003d 1.76;
Answer: 58 times brighter than the Sun
During a partial eclipse, the Sun appears as a bright crescent. The gaps between the leaves are small holes. They, working like holes in a camera obscura, give multiple images of sickles on Earth, which are easily mistaken for claws.
Let's use the formula where
D A is the diameter of Arcturus with respect to the Sun;
L = 100 - Arthur's luminosity;
T A \u003d 4500 K - Arcturus temperature;
T C \u003d 6000 K - the temperature of the Sun
Answer: D A 5.6 diameters of the Sun
Eclipses occur when the Sun, Earth and Moon are in the same straight line. Before a solar eclipse, the Moon will not have time to reach the Earth-Sun line. But at the same time, it will be close to her in a day. This phase corresponds to the new moon, when the moon is facing the earth with its dark side, and besides, it is lost in the rays of the Sun - therefore it is 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 phase of the full moon, and therefore is visible.
v 1 \u003d 20 km / s \u003d 2 * 10 4 m / s
r \u003d 1 pc \u003d 3 * 10 16 m
m o \u003d 2 * 10 30 kg
T = 1 day = years
G \u003d 6.67 * 10 -11 N * m 2 / kg 2
Let's find the sum of the masses of spectral binary stars using the formula m 1 + m 2 = * m o = 1.46 * 10 33 kg
We calculate the escape velocity using the second cosmic velocity formula (since the distance between the components of a spectral binary is 2 AU, much less than 1 pc)
2547.966 m/s = 2.5 km/h
Answer: 2.5 km / h, the speed of the starship is greater, so it will fly away.
I will again use the brochure "Didactic Material on Astronomy" written by G.I. Malakhova and E.K. Strout and published by the Prosveshchenie publishing house in 1984. This time, the first tasks of the final test on page 75 are handed out.
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?
Decision: The parameters specified in the condition are interconnected by a simple relation:
1 pc = 206265 AU, respectively:
Task 2 (Option 2)
Condition: The parallax of the star Procyon is 0.28″. Distance to the star Betelgeuse 652 St. of the year. Which of these stars is farthest from us and how many times?
Decision: Parallax and distance are related by a simple relation:
Next, we find the ratio of D 2 to D 1 and we get that Betelgeuse is about 56 times further than Procyon.
Task 3 (Option 3)
Condition: How many times has the angular diameter of Venus, observed from the Earth, changed as a result of the fact that the planet has moved from a minimum distance to a maximum? Consider the orbit of Venus as a circle with a radius of 0.7 AU.
Decision: 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: decreased by 5.6 times.
Task 4 (Option 4)
Condition: What angular size will our Galaxy (whose diameter is 3 × 10 4 pc) see from an observer located in the galaxy M 31 (Andromeda Nebula) at a distance of 6 × 10 5 pc?
Decision: The expression connecting the linear dimensions of the object, its parallax and angular dimensions is already in the solution of the first problem. Let's use it and, having slightly modified it, substitute the necessary values from the condition:
Task 5 (Option 5)
Condition: Resolution of the naked eye 2'. What size objects can an astronaut distinguish on the surface of the Moon, flying over it at an altitude of 75 km?
Decision: The problem is solved similarly to the first and fourth:
Accordingly, the astronaut will be able to distinguish the details of the surface with a size of 45 meters.
Task 6 (Option 6)
Condition: How many times larger is the Sun than the Moon if their angular diameters are the same and the horizontal parallaxes are respectively 8.8″ and 57′?
Decision: This is a classic problem of determining the size of the stars by their parallax. The formula for the connection between the parallax of the luminary and its linear and angular dimensions has repeatedly come across above. As a result of the reduction of the repeating part, we get: 
In response, we get that the Sun is almost 400 times larger than the Moon.
". On our site you will find a theoretical part, examples, exercises and answers to them, divided into 4 main categories, for the convenience of using the site. These sections cover: the basics of spherical and practical astronomy, the basics of theoretical astronomy and celestial mechanics, the basics of astrophysics and the characteristics of telescopes.
By clicking the mouse cursor on the right side of our site on any of the subsections in 4 categories, you will find in each of them a theoretical part, which we advise you to study before the crime to the direct solution of problems, then you will find the "Examples" item, which we added for a better understanding the theoretical part, the exercises themselves to consolidate and expand your knowledge in these areas, and also the “Answers” item to test your knowledge and correct errors.
Perhaps, at first glance, some tasks will seem outdated, since the geographical names of the countries, regions and cities mentioned on the site have changed over time, while the laws of astronomy have not undergone any changes. Therefore, in our opinion, the collection contains a lot of useful information in the theoretical parts, which contain timeless information available in the form of tables, graphs, diagrams and text. Our site gives you the opportunity to start learning astronomy from the basics and continue learning through problem solving. The collection will help you lay the foundations for your passion for astronomy and maybe one day you will discover a new star or fly to the nearest planet.
FOUNDATIONS OF SPHERICAL AND PRACTICAL ASTRONOMY
The culmination of the luminaries. View of the starry sky at various geographical parallels
In each place on the earth's surface, the height hp of the pole of the world is always equal to the geographical latitude φ of this place, i.e. hp=φ (1)
and the plane of the celestial equator and the plane of celestial parallels are inclined to the plane of the true horizon at an angle
Azimuth" href="/text/category/azimut/" rel="bookmark">azimuth AB=0° and hour angle tB = 0°=0h.
Rice. 1. Upper culmination of the luminaries
When δ>φ, the luminary (M4) at the upper climax crosses the celestial meridian north of the zenith (above the north point Ν), between the zenith Z and the north celestial pole P, and then the zenithal distance of the luminary
height hв=(90°-δ)+φ (7)
azimuth AB=180° and hour angle tB = 0° = 0h.
At the moment of the lower climax (Fig. 2), the luminary crosses the celestial meridian under the north celestial pole: the non-setting luminary (M1) is above the north point N, the setting luminary (M2 and M3) and the non-rising luminary (M4) is under the north point. At the lower climax, the height of the luminary
hn=δ-(90°-φ) (8)
its zenith distance zн=180°-δ-φ (9)
), at the geographical latitude φ=+45°58" and at the Arctic Circle (φ=+66°33"). Declination of the Chapel δ=+45°58".
Data: Chapel (α Aurigae), δ=+45°58";
northern tropic, φ=+23°27"; place with φ = +45°58";
Arctic Circle, φ=+66°33".
Decision: The declination of Capella δ = +45°58">φ of the northern tropic, and therefore formulas (6) and (3) should be used:
zv= δ-φ = +45°58 "-23°27" = 22°31"N, hv=90°-zv=90°-22°31"=+67°29"N;
therefore, the azimuth Av=180°, and the hour angle tv=0° = 0h.
At the geographical latitude φ=+45°58"=δ, the zenith distance of the Chapel is zв=δ-φ=0°, i.e. at the upper climax it is at the zenith, and its height hв=+90°, hour angle tv=0 °=0h and the azimuth AB is undefined.
The same values for the Arctic Circle are calculated using formulas (4) and (3), since the declination of the star δ<φ=+66°33":
zв = φ-δ = +66°33 "-45°58" = 20°35"S, hв=90°-zв= +90°-20°35"= +69°25"S, and therefore Aв= 0° and tv = 0°=0h,
Calculations of the height hн and zenith distance zн of the Chapel in the lower climax are carried out according to formulas (8) and (3): on the northern tropic (φ=+23°27")
hn \u003d δ- (90 ° -φ) \u003d + 45 ° 58 "- (90 ° -23 ° 27") \u003d -20 ° 35 "N,
i.e., at the lower climax, the Chapel goes beyond the horizon, and its zenith distance
zн=90°-hn=90°-(-20°35") = 110°35" N, azimuth An=180° and hour angle tн=180°=12h,
At the geographical latitude φ \u003d + 45 ° 58 "for the star hн \u003d δ-(90 ° - φ) \u003d + 45 ° 58 "-(90 ° -45 ° 58") \u003d + 1 ° 56 "N,
i.e. it is already non-setting, and its zн=90°-hn=90°-1°56"=88°04" N, An=180° and tн=180°=12h
At the Arctic Circle (φ = +66°33")
hн = δ-(90°-φ) = +45°58 "- (90°-66°33") = +22°31" N, and zн = 90°-hn = 90°-22°31" = 67°29" N,
i.e. the star also does not go beyond the horizon.
Example 2 At what geographical parallels does the star Capella (δ = + 45 ° 58 ") go beyond the horizon, is never visible, and passes in the nadir at the lower culmination?
Data: Chapel, δ=+45°58".
Decision. By condition (10)
φ≥ + (90°-δ) = + (90°-45°58"), from where φ≥+44°02", i.e. on the geographical parallel, from φ=+44°02" and to the north of it, up to the north pole of the Earth (φ=+90°), Capella is a non-setting star.
From the symmetry condition of the celestial sphere, we find that in the southern hemisphere of the Earth Capella does not rise in areas with a geographical latitude from φ=-44°02" to the geographic south pole (φ=-90°).
According to formula (9), the lower climax of the Capella in nadir, i.e. at zΗ=180°=180°-φ-δ, occurs in the southern hemisphere of the Earth, on the geographical parallel with the latitude φ=-δ=-45°58" .
Task 1. Determine the height of the celestial pole and the inclination of the celestial equator to the true horizon on the earth's equator, on the northern tropic (φ = + 23 ° 27 "), on the Arctic circle (φ = + 66 ° 33") and on the northern geographic pole.
Task 2. The declination of the star Mizara (ζ Ursa Major) is +55°11". At what zenith distance and at what height does it appear in the upper climax in Pulkovo (φ=+59°46") and Dushanbe (φ=+38°33") ?
Task 3. At what is the smallest zenith distance and the highest altitude in Evpatoria (φ = + 45 ° 12 ") and Murmansk (φ = + 68 ° 59") the stars Aliot (ε Ursa Major) and Antares (a Scorpio), whose declination is respectively equal to + 56°14" and -26°19"? Indicate the azimuth and hour angle of each star at these moments.
Task 4. At some point of observation, a star with a declination of +32°19" rises above the south point to a height of 63°42". Find the zenithal distance and altitude of this star at the same location with an azimuth of 180°.
Task 5. Solve the problem for the same star, provided that its smallest zenith distance is 63°42" north of the zenith.
Task 6. What declination must the stars have in order to pass at the zenith at the upper climax, and at the nadir, the north point and the south point of the observation point at the lower culmination? What is the geographic latitude of these places?
Task 1
The focal length of the telescope objective is 900 mm, and the focal length of the eyepiece used is 25 mm. Determine the magnification of the telescope.
Decision:
The magnification of the telescope is determined from the ratio: , where F is the focal length of the lens, f is the focal length of the eyepiece. Thus, the magnification of the telescope will be 
once.
Answer: 36 times.
Task 2
Convert the longitude of Krasnoyarsk to hours (l=92°52¢ E).
Decision:
Based on the ratio of the hourly measure of the angle and degree:
24 h = 360°, 1 h = 15°, 1 min = 15¢, 1 s = 15², and 1° = 4 min, and given that 92°52¢ = 92.87°, we get:
1 h 92.87°/15°= 6.19 h = 6 h 11 min. o.d.
Answer: 6 h 11 min. o.d.
Task 3
What is the declination of a star if it culminates at an altitude of 63° in Krasnoyarsk, whose geographic latitude is 56° N?
Decision:
Using the ratio relating 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°.
Task 4
When it is 10:17:14 in Greenwich, at some point the local time is 12:43:21. What is the longitude of this point?
Decision:
Local time is mean solar time and Greenwich local time is universal time. Using the relation relating the mean solar time T m , universal time T0 and longitude l, expressed in hours: T m = T0 +l, we get:
l = T m- T0 = 12 h 43 min 21 s. – 10 h 17 min 14 s = 2 h 26 min 07 s.
Answer: 2h 26 min 07 s.
Task 5
After what period of time do the moments of the maximum distance of Venus from the Earth repeat if its sidereal period is 224.70 days?
Decision:
Venus is the lower (inner) planet. The configuration of the planet, at which the maximum distance of the inner planet from the Earth occurs, is called the upper connection. And the time interval between successive planetary configurations of the same name 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 stellar period of the planet's revolution, TÅ is the sidereal period of the Earth's revolution (stellar year), equal to 365.26 mean solar days, we find:
=583.91 days
Answer: 583.91 days
Task 6
The sidereal period of Jupiter around the Sun is about 12 years. What is the average distance of Jupiter from the Sun?
Decision:
The average distance of the 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 the planet with the Earth, for which, assuming a sidereal period of revolution T 2 = 1 year, and the semi-major axis of the orbit a 2 \u003d 1 AU, we get a simple expression for determining the average distance of the planet from the Sun in astronomical units according to the known stellar (sidereal) period of revolution, expressed in years. Substituting the numerical values, we finally find:
Answer: about 5 AU
Task 7
Determine the distance from the Earth to Mars at the time of its opposition, when its horizontal parallax is 18².
Decision:
From the formula for determining geocentric distances 
, where ρ
- horizontal parallax of the star, RÅ = 6378 km - the average radius of the Earth, we determine the distance to Mars at the time 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
Task 8
The horizontal parallax of the Sun is 8.8². How far from Earth (in AU) was Jupiter when its horizontal parallax was 1.5²?
Decision:
From the formula it can be seen that the geocentric distance of one luminary 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 another, we get . Thus, knowing from the condition of the problem that the horizontal parallax of the Sun is 8.8², while it is located at 1 AU. from the Earth, you can easily find the distance to Jupiter from the known horizontal parallax of the planet at that moment:
= 5.9 a.u.
Answer: 5.9 a.u.
Task 9
Determine the linear radius of Mars if it is known that during the great opposition its angular radius is 12.5² and the horizontal parallax is 23.4².
Decision:
The linear radius of the luminaries R can be determined from the relationship , 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 condition of the problem, we get: 
= 3407 km.
Answer: 3407 km.
Task 10
How many times the mass of Pluto is 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 period of revolution of the satellite is 6.4 days. The distance of the Moon from the Earth is 3.84 × 10 5 km, and the period of revolution is 27.3 days.
Decision:
To determine the masses of celestial bodies, you need to use the third generalized Kepler's law: 
. Since the masses of the planets M 1 and M 2 much smaller than the masses of their satellites m 1 and m 2, then the masses of the satellites can be neglected. Then this Kepler's law can be rewritten in the following form: 
, where a 1 - semi-major axis of the orbit of the satellite of the first planet with a mass M1, T 1 - the period of revolution of the satellite of the first planet, a 2 - semi-major axis of the orbit of the satellite of the second planet with a mass M2, T 2 - the period of revolution of the satellite of the second planet.
Substituting the appropriate values from the problem statement, we get:
= 0,0024.
Answer: 0.0024 times.
Task 11
On January 14, 2005, the Huygens space probe landed on Saturn's moon Titan. During the descent, he transmitted to Earth a photograph of the surface of this celestial body, which shows formations similar to rivers and seas. 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.
Decision:
The energies received by the Earth and Titan are inversely proportional to the squares of their distances from the Sun. r. Part of the energy is reflected, part is absorbed and goes to heat the surface. Assuming that the reflectivity of these celestial bodies is the same, then the percentage of energy used to heat these bodies will be the same. Let us estimate the temperature of the surface of Titan in the blackbody approximation, i.e. when the amount of energy absorbed is equal to the amount of energy emitted by the heated body. According to the Stefan-Boltzmann law, the energy radiated by a unit surface per unit time is proportional to the fourth power of the absolute body temperature. Thus, for the energy absorbed by the Earth, we can write 
, where r h is the distance from the Sun to the Earth, T h - the average temperature on the surface of the Earth, and Titan - 
, where r c is the distance from the Sun to Saturn with its satellite Titan, T T is the average temperature on the surface of Titan. Taking the ratio, we get: 
, hence 
94°K = (94°K - 273°K) = -179°C. At such low temperatures, the seas on Titan may be composed 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.
Task 12
What is the apparent magnitude of the Sun as seen from the nearest star? The distance to it is about 270,000 AU.
Decision:
Let's use Pogson's formula: 
, where I 1 and I 2 – brightness of sources, m 1 and m 2 are their magnitudes, respectively. Since the brightness is inversely proportional to the square of the distance to the source, we can write 
. Taking the logarithm of this expression, we get 
. It is known that the apparent stellar magnitude of the Sun from the Earth (from a distance r 1 = 1 AU) m 1 = -26.8. It is required to find the apparent magnitude of the Sun m 2 from a distance r 2 = 270,000 AU Substituting these values into the expression, we get:
, hence ≈ 0.4 m .
Answer: 0.4m.
Task 13
The annual parallax of Sirius (a Canis Major) is 0.377². What is the distance to this star in parsecs and light years?
Decision:
Distances to stars in parsecs are determined from the relation , where π is the annual parallax of the star. Therefore = 2.65 pc. So 1 pc \u003d 3.26 sv. g., then the distance to Sirius in light years will be 2.65 pc · 3.26 sv. g. \u003d 8.64 St. G.
Answer: 2.63 pc or 8.64 St. G.
Task 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.
Decision:
Absolute magnitude M related to apparent magnitude m and the distance to the star r in parsecs the following ratio: 
. This formula can be derived from Pogson's formula 
, knowing that the absolute magnitude is the magnitude that the star would have if it were at a standard distance r 0 = 10 pc. To do this, we rewrite the Pogson formula in the form 
, where I is the brightness of a star on Earth from a distance r, a I 0 - brightness from a distance r 0 = 10 pc. Since the apparent brightness of the star will change inversely with the square of the distance to it, i.e. , then 
. Taking a logarithm, we get: or or .
Substituting in this relation the values from the condition of the problem, we get:
Answer: M= 1.42m.
Task 15
How many times the star Arcturus (a Boötes) is larger than the Sun, if the luminosity of Arcturus is 100 times greater than the sun, and the temperature is 4500 ° K?
Decision:
star luminosity L– the total energy emitted by the 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 is the absolute temperature of the star's surface. Thus, we can write: , where R is the radius of the star. For the Sun, we can write a similar expression: 
, where L c is the luminosity of the Sun, R c is the radius of the Sun, T c is the temperature of the solar surface. Dividing one expression by another, we get:
Or you can write this ratio like this: 
. Taking for the sun R c =1 and L c = 1, we get 
. Substituting the values from the condition of the problem, we find the radius of the star in the radii of the Sun (or how many times the star is larger or smaller than the Sun):
≈ 18 times.
Answer: 18 times.
Task 16
In a spiral galaxy in the constellation Triangulum, Cepheids with a period of 13 days are observed, 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 specified period is M\u003d - 4.6 m.
Decision:
From the relation , relating the absolute magnitude M with apparent magnitude m and the distance to the star r, expressed in parsecs, we get: = 
. Hence r ≈ 690,000 pc = 690,000 pc 3.26 St. g. ≈2,250,000 St. l.
Answer: approximately 2,250,000 St. l.
Problem 17
The quasar is redshifted z= 0.1. Determine the distance to the quasar.
Decision:
Let's write the Hubble law: , where v is the radial velocity of the galaxy (quasar) receding, r- distance to it, H is the Hubble constant. On the other hand, according to the Doppler effect, the radial velocity of a moving object is 
, c 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 redshift. And since the redshift in the spectra of galaxies is interpreted as a Doppler shift associated with their removal, Hubble's law is often written as: . Expressing the distance to the quasar r and substituting the values from the condition of the problem, we get:
≈ 430 Mpc = 430 Mpc 3.26 St. g. ≈ 1.4 billion sv.l.
Answer: 1.4 billion sv.l.