The height of the sun above the horizon: change and measurement. Sunrise in December. Constellations. Star cards. Celestial coordinates What is the height of the midday sun on June 22
What kind of transport can you make a "round the world" trip faster?
(with return to departure point):
by plane along the equator (average speed 800 km / h),
on a ship at 60 ° S sh. (average speed 40 km / h) or
on dog sleds at 80 ° S. sh. (average speed 30 km / h).
Answer:
By plane - 50 hours, 360 * 111.3 \u003d 40068 km 40068: 800 \u003d 50 hours.
on a sea vessel - 502 hours, 360 * 55.8 \u003d 20088 km 20088: 40 \u003d 502 hours
on skis - 233 hours, 360 * 19.4 \u003d 6984 km 8984: 30 \u003d 233 hours
How long will each of these trips take (excluding stops)?
№1 On which of the parallels: 50 N; 40 N; in the southern tropic; at the equator; 10 S The sun will be lower above the horizon at noon on the summer solstice. Justify your answer.
Decision:
1) June 22, the sun is at its zenith over 23.5 N. and the sun will be lower above the parallel furthest from the northern tropic.
2) This will be the southern tropic, because the distance will be 47.
№2 On which of the parallels: 30 N; 10 N; equator; 10 S, 30 S the sun will be at noon higher over the horizon on the day of the winter solstice. Justify your answer.
Decision:
1) 30 S
2) The midday height of the sun at any parallel depends on the distance from the parallel, where the sun is at its zenith on that day, i.e. 23.5 S
A) 30 S - 23.5 S \u003d 6.5 S
B) 10 - 23.5 \u003d 13.5
№3 On which of the parallels: 68 N; 72 N; 71 S; 83 S - is the polar night shorter? Justify your answer.
Decision:
The duration of the polar night increases from 1 day (at the parallel of 66.5 N) to 182 days at the pole. The polar night is shorter at the parallel of 68 N., because. it is further from the pole.
# 4 In which city: Delhi or Rio de Janeiro, the sun is higher above the horizon at noon of the vernal equinox?
Decision:
2) Closer to the equator of Rio de Janeiro because its latitude is 23 S, and Delhi is 28.
So the sun is higher in Rio de Janeiro.
№5 Determine the geographical latitude of the point, if it is known that on the days of the equinox the midday sun stands there above the horizon at a height of 63 (the shadow of objects falls to the south.) Write down the course of the solution.
Decision:
The formula for determining the height of the sun H
90 - Y \u003d H
where Y is the difference in latitude between the parallel where the sun is at its zenith on a given day and
the desired parallel.
90– (63 - 0) \u003d 27 S lat.
# 6 Determine the height of the Sun above the horizon on the summer solstice at noon in St. Petersburg. Where else on this day will the Sun be at the same height above the horizon?
1) 90 – (60 – 23,5) = 53,5
2) The midday height of the Sun above the horizon is the same on parallels located at the same distance from the parallel where the Sun is at its zenith. St. Petersburg is 60 - 23.5 \u003d 36.5 from the northern tropic
At this distance from the northern tropic, there is a parallel of 23.5 - 36.5 \u003d -13 Or 13 S.
# 7 Determine the geographic coordinates of the point on the globe where the Sun will be at its zenith when London is celebrating the New Year. Write down your train of thought.
Solution: From December 22 to March 21, there are 3 months or 90 days. During this time, the Sun moves 23.5. The Sun moves 7.8 in a month. In one day 0.26.
23.5 - 2.6 \u003d 21 S
London is located on the prime meridian. At this moment, when New Years are celebrated in London (0 o'clock), the sun is at its zenith above the opposite meridian i.e. 180. This means that the geographical coordinates of the desired point are 28 S 180 E. d. or h. etc.
No. 8. How will the length of the day on December 22 in St. Petersburg change if the angle of inclination of the axis of rotation relative to the plane of the orbit increases to 80. Write down your thoughts.
Solution 1) Therefore, the Arctic circle will have 80, the northern circle will deviate from the existing one by 80 - 66.5 \u003d 13.5 2) The length of the day on December 22 in St. Petersburg will increase.
# 9 Determine the geographical latitude of a point in Australia, if it is known that on September 21 at noon local solar time, the height of the Sun above the horizon is 70. Write down the line of reasoning.
Solution: 90 - 70 \u003d 20 S.
№ 10 If the Earth would stop rotating around its own axis, then there would be no change of day and night on the planet. Name three more changes in the nature of the Earth in the absence of axial rotation.
Solution: a) the shape of the Earth would change, since there would be no polar compression
b) there would be no Coriolis force - the deflecting action of the Earth's rotation. The trade winds would have a meridional direction. c) there would be no ebb and flow
# 11 Determine which parallels on the day of the summer solstice the Sun is above the horizon at an altitude of 70.
Decision
1) 90 - (70 + (- 23.5) \u003d 43.5 N.
23,5+- (90 – 70)
2) 43,5 – 23,5 = 20
23.5 - 20 \u003d 3.5 sec.
Problem 3
Z - zenith point * - Polaris
the angle at which the North Star is visible to the area of \u200b\u200bthe horizon
the angle between the zenith point and the Pole Star.
On the days of the equinoxes, the height of the noon Sun above the horizon for different latitudes is determined by the formula:
In summer, when the Sun is above the tropic of each hemisphere, its height at noon increases by 23 ° 27 ", ie.
Thus, for the city of Kiev on June 21, the height of the Sun is 61 ° 27 ". In winter, when the Sun moves to the opposite hemisphere, its height decreases accordingly and reaches a minimum on the days of the solstice, when it should be reduced by 23 ° 27", i.e. ...
Assignment 32
St. Petersburg and Kiev are located almost on the same meridian. On June 22, at noon, the Sun in St. Petersburg rises above the horizon by 53 ° 30, and in Kiev at this moment - by 61.5 °. What is the distance between cities in degrees and kilometers?
Answer:
The distance between Kiev and St. Petersburg is 8 °, and in kilometers it is 890.4 km.
Assignment 33
The height of the Sun above the horizon was measured from the ship on February 20. It was 50 °. The sun was in the south. At what latitude is the ship located if on that day the Sun stood at its zenith at latitude 1105 "S?
Answer:
The ship was at 28 ° 55 "N.
Assignment 34
In the Northern Hemisphere, where there are tourists, the Sun at noon is above the horizon at an angle of 53030 ". On the same day, the midday Sun is at its zenith at 12 ° 20" N. What degree of latitude are the tourists at?
Answer:
Tourists are at 48 ° 50 "N lat.
Questions for the blitz tournament "I believe - I don't believe"
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Question |
Answer |
I do not believe |
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Capital of the Old Russian state |
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What did the Europeans call the indigenous people of America? |
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Earth's air shell |
Atmosphere |
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Giant waves |
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Scientific name for the yellow race |
Mongoloid |
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The path along which the planets move. |
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Conditional line connecting the poles? |
Meridian |
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A device for determining the sides of the horizon. |
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Where is it colder on average - at the North or South Pole? |
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The largest island on earth |
Greenland |
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Basic law of the state |
Constitution |
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The deepest river on Earth |
Amazon |
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The smallest ocean |
Arctic |
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The map legend contains |
Symbols |
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Capital of India |
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This pole on the globe is at the bottom |
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What do the Greeks call their country? |
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The holy book of Muslims |
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A set of irregularities in the earth's surface |
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Longest parallel. |
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Who made the first trip around the world? |
Magellan |
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The reduced model of the Earth is ... |
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Azimuth is measured in ... |
Degrees |
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Latitude happens ... |
North and South |
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A number that shows how many times the distance on the ground is reduced when displayed on a map or plan. |
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Earth and other celestial bodies revolve around him |
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Which round-the-world trip will be shorter: along the equator or 60 o s. sh. |
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Sacred animal in India |
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How many seas wash Russia |
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Which points of the earth have only one geographic coordinate |
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An imaginary line that divides the Earth into northern and southern hemispheres? |
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Longitude happens ... |
western and eastern |
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What is the name of the line that cannot be reached? |
Horizon |
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The side of the horizon with an azimuth of 0 o. |
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Ability to find the sides of the horizon. |
Orientation |
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Largest ocean |
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What state does Russia have the longest border with? |
Kazakhstan |
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Capital of Poland |
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Where did Arabic numerals come from? |
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An image on the plane of a small area of \u200b\u200bthe earth's surface in a reduced form using conventional symbols ... |
Site plan |
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Distance in degrees from the equator to any point on Earth? |
Geographic latitude |
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Modern name of Persia |
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The excess of one point on the earth's surface above sea level? |
Absolute height |
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Largest geographic latitude? |
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Imagine a straight line passing through the center of the earth and crossing the earth's surface at the poles? |
Earth axis |
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"Address" of any object on the surface of the Earth |
geographic coordinates |
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All parallels have the same ... |
Circle shape |
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Circumference, conventionally drawn along the surface of the Earth parallel to the equator? |
Parallel |
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Reduced image of the Earth's surface on a plane using conventional symbols? |
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Distance in degrees from the prime meridian to any point on Earth? |
Geographic longitude |
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The angle between the direction to the north and the direction to the object, measured in degrees clockwise |
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At what points do all the meridians "meet"? |
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What is called the second language of geography? |
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Can a polar bear catch a penguin? |
No, at opposite poles |
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Rocks consist of ... |
minerals |
Life on our planet depends on the amount of sunlight and heat. It is scary to imagine, even for a moment, what it would be like if there were no such star in the sky as the Sun. Every blade of grass, every leaf, every flower needs warmth and light, like people in the air.
The angle of incidence of the sun's rays is equal to the height of the sun above the horizon
The amount of sunlight and heat that enters the earth's surface is directly proportional to the angle of incidence of the rays. The sun's rays can hit the Earth at an angle of 0 to 90 degrees. The angle of the rays hitting the earth is different, because our planet has the shape of a ball. The larger it is, the lighter and warmer it is.
Thus, if the beam travels at an angle of 0 degrees, it only glides along the surface of the earth, without heating it. Such an angle of incidence occurs at the North and South Poles, beyond the Arctic Circle. At right angles, the sun's rays fall on the equator and on the surface between the South and
If the angle of the sun's rays on the ground is straight, this indicates that
Thus, the rays on the surface of the earth and the height of the sun above the horizon are equal to each other. They depend on the geographical latitude. The closer to zero latitude, the closer to 90 degrees the angle of incidence of the rays, the higher the sun is above the horizon, the warmer and brighter.
How the sun changes its height above the horizon
The sun's height above the horizon is not constant. On the contrary, it is always changing. The reason for this lies in the continuous movement of the planet Earth around the star Sun, as well as the rotation of the planet Earth around its own axis. As a result, the day is followed by the night, and the seasons of each other.
The area between the tropics receives the most heat and light, here day and night are almost equal in duration, and the sun is at its zenith 2 times a year.
The surface beyond the Arctic Circle receives the least heat and light; here there are concepts like night, which last about six months.
Days of the autumn and spring equinox
Highlighted 4 main astrological dates, which determine the height of the sun above the horizon. September 23 and March 21 are the days of the autumn and spring equinox. This means that the height of the sun above the horizon in September and March these days is 90 degrees.
The southern one is equally illuminated by the sun, and the length of the night is equal to the length of the day. When astrological autumn comes in the Northern Hemisphere, in the Southern Hemisphere, on the contrary, it is spring. The same can be said for winter and summer. If in the Southern Hemisphere it is winter, in the Northern Hemisphere it is summer.
Summer and winter solstice days
June 22 and December 22 are summer days and December 22 is the shortest day and longest night in the Northern Hemisphere, and the winter sun is at its lowest altitude over the horizon for the entire year.
Above latitude 66.5 degrees, the sun is below the horizon and does not rise. This phenomenon, when the winter sun does not rise on the horizon, is called the polar night. The shortest night happens at 67 degrees latitude and lasts only 2 days, and the longest one happens at the poles and lasts 6 months!
December is the month of the year with the longest nights in the Northern Hemisphere. People in Central Russia wake up to work in the dark and return at night too. This is a difficult month for many, as the lack of sunlight affects the physical and mental well-being of people. For this reason, depression may even develop.
In Moscow in 2016, the sunrise on December 1 will be at 08.33. The length of the day will be 7 hours 29 minutes. it will be very early for the horizon, at 16.03. The night will be 16 hours 31 minutes. Thus, it turns out that the longitude of the night is 2 times longer than the length of the day!
This year, the winter solstice is December 21. The shortest day will last exactly 7 hours. Then the same situation will last for 2 days. And already from December 24, the day will go to profit slowly but surely.
On average, one minute of daylight will be added per day. At the end of the month, the sun will rise in December at exactly 9 o'clock, which is 27 minutes later than December 1
June 22 is the day of the summer solstice. Everything happens exactly the opposite. For the whole year, it is on this date that the longest day in duration and the shortest night. This is with regards to the Northern Hemisphere.
In the South, the opposite is true. Interesting natural phenomena are associated with this day. A polar day begins beyond the Arctic Circle, the sun does not set over the horizon at the North Pole for 6 months. In St. Petersburg, in June, the mysterious white nights begin. They last from about mid-June for two to three weeks.
All these 4 astrological dates can vary by 1-2 days, since the solar year does not always coincide with the calendar year. Also, offsets occur in leap years.
Sun height above the horizon and climatic conditions
The sun is one of the most important climate-forming factors. Depending on how the height of the sun above the horizon over a specific area of \u200b\u200bthe earth's surface has changed, climatic conditions and seasons change.
For example, in the Far North, the sun's rays fall at a very small angle and only glide along the surface of the earth, not heating it at all. Under the condition of this factor, the climate here is extremely harsh, there is permafrost, cold winters with icy winds and snows.
The higher the sun is above the horizon, the warmer the climate. For example, at the equator it is unusually hot and tropical. Seasonal fluctuations are also practically not felt in the equator area, in these areas there is eternal summer.
Measuring the height of the sun above the horizon
As they say, all ingenious is simple. So it is here. A device for measuring the height of the sun above the horizon is elementary. It is a horizontal surface with a pole in the middle 1 meter long. On a sunny day at noon, the pole casts the shortest shadow. With the help of this shortest shadow, calculations and measurements are carried out. You need to measure the angle between the end of the shadow and the line that connects the end of the pole with the end of the shadow. This value of the angle will be the angle at which the sun is above the horizon. This device is called a gnomon.
Gnomon is an ancient astrological instrument. There are other devices for measuring the height of the sun above the horizon, such as sextant, quadrant, astrolabe.
1. What is the latitude of the place of observation, if on June 22 the Sun is at noon at an altitude of 58 ° 34 "?90 ° - 58 ° 34 "\u003d 31 ° 26"
2. From Moscow (n \u003d 2) the plane took off at 23h45 and arrived in Novosibirsk (n \u003d 5) at 06h 08min. How long was he in flight?
24-00 - 23-45 + 6-08 \u003d 6-23 time spent on the flight, excluding standard time
Time difference between Moscow and Novosibirsk \u003d 3 hours. 6-23 - 3 hours \u003d 3-23
3-23 hours flight time
3. What is the declination of the zenith point? What is the noon height of the Sun in Krasnozersk (φ \u003d 53 ° 58 "N) on March 21?
4. A telegram was sent from Vladivostok (n \u003d 9) at 2:20 pm to St. Petersburg (n \u003d 2), where it was delivered to the addressee at 11:25 am. How long has elapsed from the moment the telegram was sent to its delivery to the addressee?
The time difference between Vladivostok and St. Petersburg \u003d 7 hours. When it is 14-20 in Vladivostok, 7-20 in St. Petersburg. 11-25 - 7-20 \u003d 4-05.
Therefore, the delivery took 4 hours 05 minutes.
5. At 1832 hours local time, the navigator of the ship received the Moscow time signal transmitted at 11h. Determine the longitude of the ship if the longitude of Moscow is known (2h30 m).
2 hours \u003d 30 °; 60 time minutes corresponds to 15 °, therefore 30 time minutes corresponds to 7.5 °. Accordingly, the longitude of Moscow is 37.5 ° E.
The time difference between the ship and Moscow is 7 hours 32 minutes.
60 time minutes corresponds to 15 °; therefore 7 o'clock corresponds to 105 ° longitude; 30 time minutes corresponds to 7.5 °; 4 time minutes corresponds to 1 °; 2 time minutes corresponds to 0.5 °. Thus, 7h 32m corresponds to 113 °.
The ship is located 113 ° east of Moscow.
Hence, the longitude of the ship is 113 + 37.5 \u003d 150.5 ° E.
6. In what place on Earth is the Sun at its zenith twice a year? Explain the answer.
2 times a year, the Sun is at its zenith over the territory located between the tropics.
06.22 The Sun moves from the northern tropic to the south, 22.12 The Sun moves from the southern tropic.
7. On what day of the year was the observation carried out in Novosibirsk (φ \u003d 55 °), if the midday height of the Sun was 32 ° 15 "?
90 - φ - declination of the Sun \u003d 32 ° 15 "
90 - 55 - declination of the Sun \u003d 32 ° 15 "
90 - 55 - 32 ° 15 "\u003d Sun declination
2 ° 45 "\u003d declination of the sun.
The minimum value of the noon height of the Sun in Novosibirsk is 90 ° - 55 ° - 23.5 ° \u003d 11.5 °
The noon height of the Sun in Novosibirsk on the day of the equinox is 90 ° - 55 ° \u003d 35 °
Therefore, when the Sun's midday altitude is 32 ° 15 ", the declination will be negative. That is, on this day the Sun is located in the southern hemisphere
23.5 ° corresponds to 1410 arc minutes
The sun moves 1410 arc minutes in 93 days
The sun moves 15 arc minutes in 1 day. 2 ° 45 "corresponds to 165". To move 2 ° 45 "the Sun needs 11 days. Therefore, the Sun is 11 days away from the autumnal equinox. 23.09 - 11 days \u003d 12.09.
Therefore, observations in Novosibirsk were carried out on September 12
8. Determine the local time in Novosibirsk (λ \u003d 5h32 m), if the clock shows an average Moscow time (n \u003d 2) 18h38min.
Novosibirsk is located east of Moscow.
\u003d 5h32m means that Novosibirsk is at a distance from Greenwich at this time.
60 time minutes corresponds to 15 °; therefore, 5 hours corresponds to 75 ° longitude; 30 time minutes corresponds to 7.5 °; 4 time minutes corresponds to 1 °; 2 time minutes corresponds to 0.5 °. Thus, 5h 32m corresponds to 83 ° longitude.
Consequently, the longitude of Novosibirsk is 83 ° E.
The average Moscow time corresponds to 30 ° E, because Moscow belt is the 2nd, the middle meridian is a multiple of 15 °.
Thus, the difference in longitudes between Novosibirsk time and Moscow average is 53 °.
60 time minutes corresponds to 15 °; therefore, 3 o'clock corresponds to 45 ° longitude;
53 ° - 45 ° \u003d 8 °
7.5 ° corresponds to 30 time minutes; 0.5 ° corresponds to 2 time minutes
Thus, 53 ° longitude corresponds to 3h 32m
18h38m + 3h 32m \u003d 22h10m - local time in Novosibirsk.
9. In the fall, the hunter went into the forest in the direction of the North Star. How should he come back, guided by the position of the Sun?
The direction to the North Star is the direction to the north. Autumn astronomically falls on a period close to the day of the autumnal equinox. Therefore, day and night are approximately equal. Therefore, on the way to the forest (and this is morning) the Sun should be on the right in the direction of travel. On the way back, the hunter goes south in the evening, therefore, the Sun is in the west. The sun should be on the right.
10. Where the Sun is higher on the same day: in Novosibirsk (φ \u003d 55 °), or in Moscow (φ \u003d 55 ° 45 "). What is the difference in the heights of the Sun?
On the same day, the Sun has the same declination for points located in the same hemisphere between the corresponding tropic and the pole. Hence, the altitude depends on the latitude of the location. The lower the latitude, the higher, other things being equal, the Sun's midday height is higher. The difference in the heights of the Sun for 2 points when measured on one day differs by the difference in latitude
On the same day, the noon height of the Sun is higher in Novosibirsk
On the same day, the noon height of the Sun is 45 "higher in Novosibirsk than in Moscow.
11. Determine the local time at the point, the geographic longitude of which is 7h46 m, if the clock in Moscow (λ \u003d 2h30 m) shows the time 18h38min.
The point is located east of Moscow.
λ \u003d 2h30m means that Moscow is at a distance from Greenwich at this time.
60 time minutes corresponds to 15 °; therefore, 2 hours corresponds to 30 ° longitude; 30 time minutes corresponds to 7.5
λ \u003d 7h46m means that the point is away from Greenwich at this time
60 time minutes corresponds to 15 °; therefore, 7 o'clock corresponds to 105 ° longitude;
4 time minutes corresponds to 1 °, therefore 44 time minutes corresponds to 11 °.
0.5 ° corresponds to 2 time minutes
longitude of the point 105 ° + 11 ° + 0.5 ° \u003d 116.5 ° E.
Thus, the difference in longitudes between Moscow time and this point is 116.5 ° - 37.5 ° \u003d 79 °
60 time minutes corresponds to 15 °; therefore, 75 ° longitude corresponds to 5 hours;
4 time minutes corresponds to 1 °; therefore 4 ° corresponds to 16 time minutes.
Consequently, the difference between Moscow and the point is the time difference 5h16m.
18h38m + 5h 16m \u003d 23h54m - local time at this point.
12. Between what points does the Sun rise and set on the winter solstice?
22.12 The sun rises at point south-w and sets at point south-w
13. In Moscow (λ \u003d 2h30 m, n \u003d 2) the clock shows the time 18h50 min. What is the local and standard time at this moment in Omsk (λ \u003d 4h54 m, n \u003d 5)?
The difference between Moscow and Omsk in standard time is 3 hours.
Omsk east of Moscow. Therefore, 18h50min + 3h \u003d 21h50min
Standard time in Omsk 21h50min
60 time minutes corresponds to 15 °; therefore, 2 hours corresponds to 30 ° longitude; 30 time minutes corresponds to 7.5
Thus, 2h 30m corresponds to 37.5 ° E.
60 time minutes corresponds to 15 °; therefore 4 hours corresponds to 60 ° longitude;
4 time minutes corresponds to 1 °, therefore 52 minutes corresponds to 13 ° longitude
2 time minutes corresponds to 0.5 ° longitude
Thus, 4h54 m corresponds to 73.5 ° E.
The longitude difference between Moscow and Omsk is 73.5 ° E. - 37.5 ° E \u003d 36 ° longitude.
15 ° longitude corresponds to 1 hour; 1 ° longitude corresponds to 4 time minutes.
Thus, 36 ° longitude corresponds to 2 hours 24 minutes.
18h50min + 2h24min \u003d 21h14min
Local time in Omsk 21h14min
14. Between what points does the sun rise and set on the summer solstice?
06/22 The sun rises at point s-w and sets at point s-w
15. What is the longitude of the place of observation, if the observer noticed that the solar eclipse began at 13h52m, and should be at 7h15m GMT?
13h52m - 7h15m \u003d 6h37m is the distance of the observation site from Greenwich.
15 ° longitude corresponds to 1 hour; 6 hours corresponds to 90 ° longitude
1 ° longitude corresponds to 4 time minutes; 36 minutes corresponds to 9 ° longitude
60 arc minutes correspond to 4 time minutes
15 arc minutes corresponds to 1 time minute
Consequently, the longitude of the observation site is 99 ° 15 "E.
16. At what geographic latitude does the midday height of the Sun not exceed 23 ° 26 "?
Maximum noon elevation occurs in the northern hemisphere on the summer solstice and in the southern hemisphere on the winter solstice. On this day, the solar declination is + 23 ° 26 ".
h \u003d 90 ° - φ + 23 ° 26 "; therefore at h \u003d 23 ° 26" φ \u003d 90 ° - 23 ° 26 "+ 23 ° 26" \u003d 90 °
The midday height of the Sun does not exceed 23 ° 26 "at the latitude of the North Pole 22.06 and South Pole 22.12.
Purpose: to form the ability to navigate by the sun, to determine the midday line, the height of the midday sun above the horizon.
Equipment:
gnomon (an even pole 1-1.5 m long), a vertical goniometer-eclimeter or a protractor with a plumb line, a thin rod or a piece of twine 2 m long.
Guidelines
During the year, the height of the sun above the horizon changes: June 22 - on the day of the summer solstice - it occupies the highest position, December 22 - on the winter solstice - the lowest, and on the days of the equinox - March 21 and September 23 - intermediate. In the Northern and Southern Hemispheres, the change in the height of the midday sun has the opposite direction.
Working process
Exercise 1... Definition of the midday line.
Place the gnomon vertically on a flat area closer to noon. Fix the end of the shadow falling from it with the first peg and with a radius (point 1) equal to the length of the shadow and draw a circle with the other peg. Pay close attention to how the shadow will shorten. After a certain time, the shadow will begin to lengthen and touch the circle a second time, but at a different point (point 2) (see Fig. 1).
Figure: 1. Definition of the midday line
Drive the second peg into this point. Pull the twine from the first peg to the second peg. Find the middle of this line. Drive in the third peg. Connect this peg with twine to the base of the gnomon. This will be the midday line that shows the north direction and coincides with the local meridian. Check compass heading.
Assignment 2... Determining the height of the sun above the horizon.
Install the rail so that one end rests against the base of the third peg, and the other rests on the upper end of the gnomon, forming an angle with the horizontal surface. Determine its value using an eclimeter or vertical protractor. This will determine the height of the sun above the horizon at noon.
Assignment 3... Answer the questions.
1. How the height of the sun above the horizon changes during the day
and years?
2. Determine the time of solar noon by the clock. Does the time of noon (12 hours) coincide with the sun? Explain the reason.
Orientation in space
Purpose: to teach the techniques of orientation in space by local signs and a compass.
Equipment:
compass, measuring tape or 15-meter tape measure, mechanical wrist watch, school rangefinder, tablet.
Guidelines
Orientation in space is the determination on the terrain of its location or standing point relative to the sides of the horizon, surrounding terrain objects, as well as directions and distances of movement.
Orientation in space includes:
1) correlation of real terrain with a plan and a map;
2) determination of the sides of the horizon on the ground and its position in relation to terrain objects: a settlement, a river, a railway, etc .;
3) determining the distance on the ground and their graphic expression on paper.
4) selection of the required direction of movement.
Working process
Exercise 1... Determination of the direction of the sides of the horizon by compass.
The most accurate way of general orientation on the terrain is compass orientation. In order to determine the direction of the sides of the horizon using the compass, you need to do the following:
1. Remove all metal objects at a distance of 1-2 m from the compass;
2. Set the compass in a horizontal plane on the palm of your hand or tablet;
3. Rotating the compass in a horizontal plane, align the northern end of the magnetic needle of the compass with the letter C. In this position, the compass is oriented and now you can determine the sides of the horizon from it.
Assignment 2... Orientation by the sun with a watch.
With the help of a mechanical wristwatch, you can determine the direction of the north-south line at a given time. To do this, you need to do the following:
1. put the watch in a horizontal plane and point the hour hand at the sun;
2. mentally build the angle between the small hour hand
and number 11 on the clock face. The bisector of this angle will be the local meridian.
Azimuth movement
Purpose: teach the techniques of orientation in space and determining the direction of movement in azimuth.
Equipment:
compass, measuring tape or 10-15-meter tape measure, mechanical wrist watch, school rangefinder, tablet.
Guidelines
Using a compass, you can determine the sides of the horizon, the direction of movement in azimuth. Azimuth is the angle between north and direction to a given object, which is counted clockwise.
For example, knowing that the azimuth from point A to point B is 45º (A \u003d 45º), you, having orientated the compass, determine the bearing and go in the right direction.
When moving, it is either set or determined. To determine the azimuth of movement from one point (standing point) to another, a map is needed.
For orientation on the ground, it is important to be able to determine not only the direction, but also the distance. Distance is measured using various methods: counting steps and movement time, visual, instrumental. Visual (by eye) distance assessment is the observation of terrain objects and their visibility depending on the distance from the observer (see Table 1). This method allows you to determine the distance approximately, this requires constant training.
Table 1
Eye measurement of distances
| Distance | Objects Observed |
| 10 km | Pipes of large factories |
| 5 km | General outlines of houses (without doors and windows) |
| 2.5 miles | The outlines of windows and doors are barely visible |
| 2 km | Tall lonely trees; man is a subtle point |
| 1,500 m | Large cars on the road, a person is also distinguished in the form of a point |
| 1 200 m | Individual trees of medium size |
| 1,000 m | Telegraph poles; separate logs are distinguishable in buildings |
| 700 m | The figure of a man without details of clothing is already looming |
| 400 m | Human hand movements are noticeable, the color of clothing varies, binders on window frames |
| 200 m | Head outline |
| 150 m | Hands, eye line, clothing details |
| 70 m | Dotted eyes |
Working process
Exercise 1... Determination of azimuth 90º, 145º, 225º using a compass.
Walk a short distance in these directions. To
do not stray from the chosen direction of movement, write down noticeable objects of the terrain, these will be landmarks of the direction in which you should move.
Assignment 2... Determination of the distance to the selected terrain objects.
For accurate determination of distances in professional activity, tape measures, measuring tapes, theodolites, radio direction finders are used
and other tools. In ordinary life, non-instrumental methods are used.
1. Select an object in an open area and visually determine the distance to it, using Table 1.
2. To more accurately determine the distance by eye, you can use a technique that is based on a simple mathematical calculation. Take the ruler in your hand, direct it to a distant object, the height of which is known to you, let's say 10 m. By moving the ruler in your fingers, we will achieve such a position when a segment of the ruler, say 10 cm, completely covers this object. Determine the distance from the eye to the ruler. It is about 70 cm.Now you know three values, but
the distance to the object is not known. Let's draw up a formula in which the length of the ruler relates to the height of the object X in the same way as the length of the outstretched arm to the distance to the object. Let's solve the proportion:
10 m: X \u003d 10 cm: 70 cm,
10 m: X \u003d 0.1 m: 0.7 m,
X \u003d 70 m.
This method is convenient to use when determining the distance to inaccessible objects located, for example, on the other side of the river.
Assignment 3... Measuring distance in steps.
You need to know your stride length. Set aside a 50 m segment on a flat area. Walk this distance several times
and determine the arithmetic mean of the number of steps.
For example, 71 + 74 + 72 \u003d 217 steps. Divide the total number of steps by 3 (217: 3 \u003d 72). The average number of strides is 72. Divide 50 m by 72 strides and you get your average stride length of about 55 cm.
You can measure the distance to any accessible object in steps. For example, if you took 690 steps, i.e. 55 cm × 690 \u003d 37 m.
Write in your journal and compare the results for determining distances in different ways. Determine the degree of accuracy for each method.
In a given locality, each star always culminates at the same height above the horizon, because its angular distance from the pole of the world and from the celestial equator remains unchanged. The sun and moon change the altitude at which they culminate. From this we can conclude that their position relative to the stars (declination) changes. We know that the Earth moves around the Sun and the Moon around the Earth. Let us trace how the position of both luminaries in the sky changes as a result.
If we observe the time intervals between the upper culminations of the stars and the Sun by the exact clock, then we can be sure that the intervals between the culminations of the stars on four minutesshorter than the intervals between the climaxes of the Sun. This is explained by the fact that during one revolution around the axis (day) the Earth passes approximately 1/365 of its path around the Sun. It seems to us that the Sun is shifting against the background of the stars to the east - in the direction opposite to the daily rotation of the sky. This shift is about 1 °. To turn at such an angle, the celestial sphere needs another 4 minutes, by which the culmination of the Sun is "delayed". Thus, as a result of the Earth's orbital motion, the Sun in a year describes a large circle in the sky relative to the stars, called ecliptic (fig. 17).
Since the Moon makes one revolution to set the rotation of the sky in a month and therefore passes not 1 °, but about 13 ° in a day, its climax is delayed every day not by 4 minutes, but by 50 minutes.
Determining the height of the Sun at noon, they noticed that it happens twice a year at the celestial equator, at the so-called equinox points. It happens on days spring and autumn equinox(around March 21 and around September 23). The horizon plane divides the celestial equator in half (Fig. 18). Therefore, on the days of the equinox, the paths of the Sun above and below the horizon are equal, therefore, the lengths of the day and night are equal.
What is the declination of the Sun on the days of the equinoxes?
Moving along the ecliptic, the Sun on June 22 moves farthest from the celestial equator towards the north pole of the world (23 ° 27 "). At noon for the northern hemisphere of the Earth it is highest above the horizon (this value is higher than the celestial equator, see Fig. 17 and 18) .The longest day, it is called summer solstice.
The great circle of the ecliptic crosses the great circle of the celestial quator at an angle of 23 ° 27 ". At the same time, the Sun is below the quator in winter solstice, December 22 (see Fig. 17 and 18). Thus, on this day, the height of the Sun at the upper climax decreases by 46 ° 54 "compared to June 22, and the day is the shortest. (You know from the course of physical geography that the differences in the conditions of illumination and heating of the Earth by the Sun determine its climatic zones and the changing seasons.)
The deification of the Sun in antiquity gave rise to myths describing periodically recurring events of the "birth", "resurrection" of the "sun god" throughout the year: the dying of nature in winter, its rebirth in spring, etc. Christian holidays bear traces of the cult of the Sun.
The path of the sun runs through 12 constellations called zodiacal (from the Greek word zoon - animal), and their combination is called the zodiac belt. It includes the following constellations: Fish, Aries, calf, Twins, Cancer, a lion, Virgo, Libra, Scorpio, Sagittarius, Capricorn, Aquarius... Each zodiac constellation, the Sun, passes for about a month. The vernal equinox (one of the two intersections of the ecliptic with the celestial equator) is in the constellation Pisces.
It is clear that at midnight the zodiacal constellation opposite to the one in which the Sun is located passes the upper culmination. For example, in March, the Sun passes through the constellation Pisces, and at midnight the constellation Virgo culminates.
So, we have made sure that the apparent motion of the Moon, which revolves around the Earth, and the Sun, around which the Earth revolves, is detected and described in the same way. And on the basis of these observations alone, it is impossible to decide whether the Sun moves around the Earth or the Earth around it.
The planets move against the background of the starry sky in a more complex way. They move in one direction or the other, sometimes slowly writing out loops (Fig. 19). This is due to the combination of their true motion with the motion of the Earth. In the starry sky, the planets (translated from the ancient Greek as "wandering") do not occupy a permanent place, just like the Moon and the Sun. Therefore, on the map of the starry sky, the position of the Sun, Moon and planets can be indicated only for a certain moment.
An example of solving the problem
A task. Determine the midday height of the Sun in Arkhangelsk and Ashgabat during the summer and winter solstices.
Notice how the difference in the midday heights of the Sun on the days of the solstices (for each city) is related to the difference in its declination on these dates.
Compare the difference in the height of the Sun on the same day in these two cities with the difference in their geographic latitudes. Make a conclusion.
How, knowing the height of the Sun at noon in one of the cities, on the day of the summer solstice, can you calculate its height in another city?
Exercise 4
1. At what latitude does the Sun climax at its zenith on the summer solstice?
2. On what days of the year does the Sun reach its zenith for an observer at the Earth's equator?
3. Determine the geographical latitude of the point at which on the day of the winter solstice the climax of the Sun occurs in the point of the south.
Assignment 3
1. Find the 12 zodiac constellations on the star map. Using a moving map of the starry sky, determine which of them will be visible above the horizon on the evening of observation.
2. According to the "School Astronomical Calendar" find the coordinates of the planets at a given time and determine on the map in which constellation they are located. Find them in the sky in the evening.
