Observatory talks (faq)

Observatory talks (faq)

This homepage, as well as our observatory, shall serve the care and promotion of popular astronomy, with the aim to make the spectacular events and developments in this field more understandable to the interested and open-minded person.

Observatory Talks (FAQ)

Here we give answers to questions frequently asked in the guided tours at the observatory, and explanations of astronomical terms.

The position of the moon

Why is the position of the moon at the equator or on the southern hemisphere of the earth different from ours?

The horizon is the reference line for the observer. An observer on the north pole stands in such a way that his head points in the direction of the north, his feet to the south. Its horizon is parallel to the equator. An observer at the equator is exactly perpendicular to an observer at the north pole. Therefore its horizon is also perpendicular to the horizon of the observer at the north pole. Thus it comes that in equator proximity the points of the moon crescent lie with the rising or setting of the moon frequently horizontally to the horizon. At the south pole an observer stands in such a way that his head points to the south and the feet to the north. He therefore sees the moon from exactly the opposite direction as an observer at the North Pole. So, while on the northern hemisphere the illuminated side of the waxing crescent is to the right of the observer, an observer on the southern hemisphere sees it to his left.

The different lengths of the year

The earth orbits the sun in a slightly elliptical orbit in the period of one year. Thereby the earth axis is not perpendicular on the orbit, but inclined by 23,5. Gravitational forces, mainly from sun and moon, try to straighten the axis. This leads to a circular movement of the axis corresponding to the wobbling movement of a spinning top. One revolution lasts about 25800 years, this period is also called an platonic year. This means that the north pole shifts during this time period. The today’s pole in the small wagon wanders z.B. in 14000 years near the star Vega in the Lyra. The vernal equinox changes of course. It was z.B. 2000 years ago on the border of the constellation Pisces and Aries, today it is on the western edge of Pisces, and in 600 years it will merge into the constellation Aquarius. The winter constellation Orion will be in the summer sky for the northern hemisphere in 13000 years.

Depending on the choice of the reference point slightly different lengths of the year result. So the so called tropical year, also called astronomical year, the time between 2 passages of the middle sun through the middle vernal point, thus the sun course from vernal point to vernal point, it lasts 365,242190 middle sun days. The sidereal year (lat. sidus = star) refers to a fixed star in the earth’s orbital plane (ecliptic) and is 365.256363 mean. solar days long. The wobbling motion of the earth’s axis described above, which is called precession by the way, is the reason why the tropical year is approx. 20 minutes shorter than the sidereal one. The anomalistic year is the time difference of the passage of the earth through the perihelion of its orbit, and is 365.259636 mean. solar days.

The Julian year was used in the historically developed calendar, and was fixed with 365,25 days, which however over a larger period of time led to large deviations to the real solar orbit. The reform of Pope Gregory in 1572-85 gave rise to our present calendar year, the so-called Gregorian year with 365,2425 or 365 + 1/4 – 3/400 mean. Sunny days. The difference between the tropical and the Gregorian year is only 26 seconds and will add up to one day only in 3320 years.

The different lengths of the month

The earth rotation determines the day, the moon the month, and the sun the year. The sidereal month denotes the moon orbit from fixed star to fixed star and is 27.32 days = 27 days 7 hours 43 min. and 11.5 seconds long. The tropical month is almost 7 s shorter than a sidereal one and marks the passages through the vernal equinox. The synodic month has the length of 29.53 days, it describes the orbital period with respect to the Sun, i.e. from lunar phase to lunar phase (z.B. full moon to full moon). It is longer than the sidereal one, since the sun moves daily by apparently approx. 1° moves. The relation of both periods is : 1 / sid. month – 1/ sid. year = 1 / syn. Month. The time between two successive passes of the Moon through the same node is called the draconic month. It is important for the calculation of eclipses and has a time span of well 27 d and 5 h. The anomalistic month is the time span between two passes through its perigee. It is 5 1/2 hours longer than the tropical month.

Determination of Mars distance during opposition

If ve and vm are the orbital velocities of Earth and Mars around the Sun, then the following applies ve> vm. For the angular velocities we and wm of Earth and Mars is valid: w = v / R. Rm and Re are the orbital radii of Mars and Earth. These data lead to the following relation:

vm – ve = wm × Rm – we × Re = – dw × Rme = – dw × (Rm – Re ) ,

where is – dw the retrograde movement of the relative angular velocity.

wm × Rm – we × Re = – dw × Rm + dw × Re wm × Rm + dw × Rm = we × Re + dw × Re Rm × (wm + dw) = Re × (we + dw) Rm / Re = (we + dw) / (wm + dw) .

The value dw can be determined by observation at the Mars position.

we = 360° / 365.25 d = 0.9856° / dwm = 360° / 687 d = 0.5240° / dRm / Re = (0,9856 + dw) / (0,5240 + dw) .

Note: This calculation is based on the orbit on circular paths. However, since both orbits are elliptical, the calculation is not exact, but more a thought experiment with a good approximation to reality. For the amateur, it may be an interesting exercise to measure the daily right ascension change of the planet at a Mars position. The following example shows the accuracy of the result. Mars opposition was on 24.12.07 at 21 o’clock. The right ascension at this time was 92.9224°. 24 hours later the right ascension was 92.4839°. The difference is therefore 0.4385. Substituting this into the above calculation gives : Rm / Re = 0.9856+0.4385 / 0.5240+0.4385 = 1.4796 AE. The exact value is 1.5237 AU . So the calculated value is 97,1% of the exact value.

The determination of the orbital velocity of the earth around the sun

This extremely important and interesting value can be determined from a simple consideration. One must apply for this only the experience of that, which hurried already times with the umbrella somewhat faster through the rain. In order not to get wet legs here, one must tilt the umbrella a little forward in direction of movement. The optimal tilt angle, determined by trial and error in this case, is given by the ratio of the walking speed of the umbrella carrier to the falling speed of the rain. Conversely, by applying simple geometric laws, if we know the angle of inclination, we can determine the corresponding velocity.

Applied to the astronomical problem we have the following situation. One knows the speed of the light which reaches us from the stars: c = 299792 km/sec. If you now observe a star at two suitable times 6 months apart, you have to move the telescope by 20.5 arcseconds in the direction of the Earth’s motion each time. So you know a velocity vector and the acute angle of a right-angled triangle, from this you can get the relation for the velocity component of the earth v = c * tg 20.5 and the result is 29.80 km/sec.

This effect is known as Aberration of the light known.

The coorbital moons of Saturn 1980 S1 and 1980 S3

For satellites in an orbit an amazing rule is valid: If one accelerates the satellite, its orbit radius becomes larger, it becomes slower thereby. Similarly, when a satellite is braked, the orbit radius becomes smaller and the satellite becomes faster.

The two moons 1980 S1 ("Janus") and 1980 S3 ("Epimetheus") orbit Saturn on a common orbit. Due to small differences in the orbital radius the orbital velocities of the moons are different. The inner moon travels slightly faster ahead, the outer slower behind. The faster moon moves away up to half of the orbit and then approaches the slower one from behind. During the approach then the gravitation becomes effective, which accelerates on the rear moon and brakes on the front moon. According to the mentioned rule their orbits change: The rear moon increases its orbit radius and becomes slower, the front moon changes to the faster inner orbit. Now both move away from each other again, the cycle starts anew with reversed roles.

Calculation of orbit data:

The dimensions of the moons (in km) are S1: 220 x 200 x 160, S3: 140 x 120 x 100.

Orbital radii: R1 = 151 400 km ; R3 = 151 450 km .

The orbital period can be calculated according to the 3. Kepler’s law (T2 / R3 = constant = k) can be calculated, if the constant and R are known. The constant k = 4×π2 / G×M is calculated from the gravitational constant G and the mass M of the central body, in this case Saturn’s. But it can also be determined directly by the known data of the moon Mimas. For Mimas applies: orbital radius R = 185 600 km; orbital period T = 0.9424 days.

The Saturn constant results from this to k = 1.037 × 10-15 .

Now one can determine the orbital period of the moons.

From T2 = (k×R3) we get T1 = 0.69432 days; T3 = 0.69466 days (about 16 h 40 min).

The orbital velocities result then to

v1 = 2 × π × R1 / T1 = 15.8575 km/s

v3 = 2 × π × R3 / T3 = 15.8549 km/s.

The difference are 2,6 m/s

10 km/h, d.h. about the speed of jogging.

When both moons meet? 1/Tsyn = 1/T1 – 1/T3 , Tsyn = 1402,2 days = 3,8 years.

During this time of nearly 4 years the moons have made a little more than 2000 orbits around Saturn, the inner 1 orbit more than the outer one.

Why do we always see the same side of the moon?

As a result of the tidal effect that the Earth’s gravity exerts on the Moon, the Moon’s rotation has been progressively decelerated, so that in one orbit around the Earth it rotates only once on its own axis. This is called a bound rotation. Therefore from the earth always the same side is visible. Because orbital period and moon rotation are not exactly equal, the face of the moon wavers a little bit; this is called libration. Because of this libration to the east and west and to the north and south almost 59 % of the lunar surface is visible. The remaining 40 % of the lunar surface could be observed for the first time in 1959 by the Russian lunar probe Lunik 3.

Because of the bound rotation an observer on the moon sees the earth always at the same place of the sky, apart from the slight variations by the libration. So the earth never rises or sets on the moon and an observer on the far side of the moon can never see the earth.

Energy and mass

Nowadays almost every child knows the famous formula E = mc^2 . For some this is a mnemonic used in conversation to remind people of the equivalence, for others the dimensions and values of this relationship are also of interest. If you put into the formula the mass "m" in gram(g); the speed of light "c" in (cm/s) , you get as result the energy in (erg). This is a very small amount of energy , measured by our energy data for daily use , since 10^7 erg = 1 joule = 1 Ws ( watt-second) correspond to. The Energy of 1 g mass corresponds to : E = 8.998 * 10^20 erg = 8.998 * 10^13 Ws = 8.998 * 10^10 KWs (kilowatt seconds) or equal to 24.994.444 or. just 25 million Kilowatt hours. – For500 g mass then the proud value of 12 , 5 billion KWh results. – In nuclear power plants, however, only a fraction of this energy can be converted. In the hydrogen bomb, however, the conversion is largely realized.

A comparison of the energy of a rest mass according to E=mc^2 with the chemical energy shows the following example: How big is the energy content of 1 g matter?? and how does it compare to the burning of 1 g of coal? ( 1 g coal produces 7000 cal when burning, 1 cal = 4,18 J ) E = (10^-3 kg ) ( 3*10^8 m/s )^2 = 9*10^13 (J). The released energy when burning 1g coal is 7000 * 4,18 = 2,9*10^4 (J). So the energy of the rest mass is 3.1*10^9 times larger than the chemical energy. This example shows that the release of only one thousandth of the rest energy is still million times more than the energy yield of conventional energy sources.

Sidereal time

What is actually understood by the term sidereal time ? Sidereal time measurements are made, as the name already says, at stars.A sidereal day is the time interval between 2 passages of a star by the local meridian (more exactly the passage of the vernal equinox) A sidereal day indicates thus the rotation of the earth in relation to the starry sky.

Observatory talks (faq)

Due to the progress of the sun on its orbit (the ecliptic) from west to east, the sidereal day is somewhat shorter than the solar day. Its length is : 23h 56m 04,1 s measured in mean solar time.The sidereal time is u.a. important to determine the position of a star in the sky. The hour angle of the star is the angular distance of the star to the local meridian.The local sidereal time is equal to the hour angle of the vernal equinox. Between the important quantities right ascension, sidereal time and hour angle exists the relation:

sidereal time = hour angle + right ascension

Movement in the starry sky

If one looks into a moonless, dark, clear starry sky, one sees with good eyes on a moonless night, under good viewing conditions, approx. 3000 stars above the horizon. If one does not leave it with the one view and looks after approx. 1 hour again, so the attentive observer can determine by the position of constellations or – if he does not know them – by particularly bright objects that the image has shifted. The stars move in the direction of west. The shift is most clearly perceived near the horizon, namely on the western and eastern horizon. New stars appear in the east, while in the west the stars visible 1 hour ago have disappeared below the horizon.

But not enough with it, if one over several weeks always at the same time, z.B. always at 22 o’clock, looks for certain stars or constellations in the night sky, one also finds out that they shift to the west, and not insignificantly by 30° per month.

In the daytime sky we all know the course of the sun, which rises in the east and sets in the west. Exactly this is correct however only on 2 days in the year, at the autumn and at the spring beginning. In the remaining time the point of setting shifts in the direction of southwest up to the beginning of winter, in order to then move back again, beyond the west in the direction of northwest up to the beginning of summer and/or the beginning of spring. the summer solstice. For the rising point of the sun the corresponding pendulum process in direction south-east over east to north-east is valid. In connection with this, the altitude of the sun at noon also changes considerably, causing the seasons to appear.

The people before ca. 2000 years have interpreted these observations to mean that all celestial objects revolve around the earth. In the 2. Jh. of our era, Ptolemy summarized the astronomical knowledge of his time in the so-called "Ptolemaic World System", in which the Earth was the center of the universe. This view of the world lasted for almost 1500 years until Copernicus put an end to this idea and placed the sun at the center of the planetary system. Thus began the breathtaking realization of the structure and the mode of action of our universe.

But let’s return to our look at the night sky, and look ca. two weeks after our observation in the moonless night again into the starry sky, so now the moon stands in the sky. Now if you are lucky enough to see the Moon If you follow the sun on several successive evenings, you will notice two serious changes. The first one is the change of the luminous shape of the moon, which is familiar to everyone and which is unmistakable in the sky during the phases of the waxing and waning moon. The second, less noticeable change is the position of the moon among the stars. While the moon, like all other celestial bodies, rises in the east and sets in the west, it moves during one day around ca. 13° from west to east through the stars. Thus it sets every day later.

If you follow the arrangement of the stars among themselves over weeks and months, you can see that some of the brighter points of light slowly change their position among the other stars. These points of light are called "convertible stars" or "planets". Altogether 5 planets can be seen with the naked eye. Two of them (Venus and Mercury) appear only at dusk or dawn. They wander between the stars from west to east and thus move away from the sun, but only up to a certain distance, then they reverse their direction of travel, and disappear behind the sun, only to reappear at dawn and repeat the same game in the morning sky. The other 3 planets ( Mars, Jupiter and Saturn ) are also visible at night. They travel through the constellations mostly from west to east, but sometimes they simply turn around in the opposite direction to continue their earlier course a few weeks later after passing through a loop.

One can well imagine that the disentanglement of all these complicated movements was not easy and that on the way to the solution some errors had to be overcome. The solution is common knowledge today: Our earth rotates around its own axis, the ca.is inclined 23.5° against the earth’s orbit; seen from the north pole its direction of rotation is counterclockwise. Furthermore, the earth is on a slightly elliptical orbit in the same direction of motion around the sun, together with 7 other planets which also orbit the sun at different distances with different speeds in more or less elliptical orbits. The sun as gravitational center unites thus all bodies in its sphere of influence, which reaches far beyond the most distant planet Neptune, to a system, which one calls the solar system. From these orbits of the bodies of the solar system explain themselves quite naturally the movement sequences, which we observe in the star sky.

If we use a telescope, we can observe other planets (Uranus, Neptune), dwarf planets (Pluto, Ceres, Eris), moons and small bodies of the solar system and follow their orbits. With a lot of patience, an amateur astronomer can, within a few decades of his life, even detect motion outside our solar system, such as double stars orbiting each other.

The constellations and time

The idea of summarizing the irregularly distributed constellations in the sky in pictures dates back to the early advanced civilizations. The position of the stars was among other things a reference point for the agriculture, in order to z.B. to begin with the Aussat.

Sun, planets, moon and stars rise in the east, reach their highest point in the south and set in the west. This is due to the rotation of the earth around its axis and, if you look at the north pole from above, counterclockwise. One revolution takes approx. 23 hours and 56 minutes.

At the same time the earth moves in an average distance of ca. 150 million kilometers in an elliptical but nearly circular orbit around the sun. For this it needs approx. 365 ¼ days, advancing on its orbit by a little less than 1° per day.

Thus a terrestrial observer sees the following:

a.) The celestial objects move parallel to the celestial equator from east to west and advance per hour ca. 15° in the night sky to the west ahead.

b.) After 24 hours, they are up from the previous day by approx. 1° to the west and set a little bit earlier every day.

c.) On its annual orbit the night side of the earth, i.e. the side facing away from the sun, points in a different direction each time, so that other objects become visible.

In this way the constellations can be used for approximate date determination: Orion at midnight in the south at the beginning of December, the Great Leo at the beginning of March, the Bootes at the beginning of May, the Summer Triangle (Lyra, Swan, and Eagle) at the beginning of July, Pegasus at the beginning of September.

The lunar course

The position of the moon always raises questions for the observer on earth, since during a month and over the year its position in the sky is subject to strong fluctuations. Since the moon’s orbit is approx. 5° from the Earth’s orbit, the Moon never moves more than 5° from the Sun’s apparent orbit, the ecliptic. So, in a rough consideration, the moon’s orbit can be equated to the sun’s orbit. As known, the sun in its apparent motion needs 365.25 days for one revolution around the earth, but the moon needs only 27.32 days (sidereal revolution time), this results in a ratio of 13.3. The sun moves on its annual orbit every day for the earth observer by 1° thus to the east (in relation to the stars), but the moon is 13.3 times as fast. If you take the new moon as reference point – it is at the same position as the sun – it describes approximately the same diurnal arc as the sun. Due to the described motion conditions the moon stands after 1 day where the sun will stand only in 13.3 days, after 2 days where the sun will stand in 26 days etc. We can, in a first approximation, calculate the above 13.3 days round up to 14 days, so 2 weeks. So the crescent moon, 7 days after new moon, stands where the sun will be in 14 weeks, or about a quarter of a year. Thus in spring the moon describes the large diurnal arc, which the sun describes in summer, and in autumn the flat diurnal arc of the winter sun. The crescent moon, 7 days after new moon, describes the position of the sun fourteen weeks, so ca. a quarter of a year later. The full moon, 14 days after new moon, shows the position of the sun 28 weeks, so ca. half a year later.

Observation objects for summer

Until today one has approx. 1000 visible PN cataloged. The best summer objects are presented in the following matrix.

13 secm=1110.9 NGC 6210Her16 h

2014"129.3 NGC 6543Dra17 h

1920"10,98,8Also known as Cats EyeM 57Lyr18 h

3076"14,79,7Ring nebula in the LeyerNGC 6804Aq19 h

-35"14,412,2 NGC 6818 19 h

3017"159.9 NGC 6826Cyg19 h

1325"10,29,8Also called flasherM 27Vul19 h

30402"13,87,6Dumbbell NebulaNGC 6891Del20 h

-15"11,811.7 NGC 6894Cyg20 h

-40"17,514,4 NGC 6905Del20 h

2740"13,911,9 NGC 6781 19 h

-90"16,211,8 NGC 7293 22 h

131612.7-Sunflower NebulaNGC 7662And23 h

Double and multiple stars

The observation of multiple stars is always a good idea when the brightness conditions at the observing site are not optimal.

This includes observations in moonlight and twilight.

A small selection of clear and interesting objects are:

Epsilon Lyrae in the constellation Lyra ( 4-fold constellation )

Mu Bootis Bootes

31 Cygni Swan

Beta Cygni Swan

17 Draconis Dragon

100 Herculis Hercules

Up to which apparent brightness can you see with a 12 cm amateur telescope?

How many magnitudes more in brightness can I see with my 12 cm telescope than with the naked eye??

Let us assume that with the max. pupil diameter of 8 mm in absolute darkness, i.e. without disturbing influences z.B. of the moon or other light sources , up to m = 6 mag can be seen, then a ratio of the incident light

from ( 120 / 8 )^2 = 225.

The difference of the observable size classes thus results in m1 – m2 = 2.5 * log 225 = 5,88 mag.

One would have to be able to recognize thus by the telescope still 225 times lichtschwachere objects, than with the naked eye, under the indicated conditions, thus approx.12.mag , (6mag + 5,88 mag ).

Dangers from the cosmos

At intervals of several thousand or millions of years, we see ourselves threatened by asteroids or even by nearby supernovae.

It is not long ago that we have come to the realization due to refined measuring technique that the orbit of the earth, lies in the orbit of a whole quantity of asteroids ( sog. earth orbital cross).The probability of a collosion becomes uncomfortably large with it.

Such near collisions are more frequent than most people realize.One occurred z. B. at 3. January 1993 and was captured by NASA astronomers by radar. On the photos of the asteroid Toutatis can be seen that it consists of rock cores, which have a diameter of more than 3 km each. It approached the earth up to 3.5 million km.

On 23. March 1989 an asteroid flew with approx. 0,8 km diameter even closer to the earth.( approx. 3 times the distance to the moon).

At the end of 1992 was to be heard,,that a gigantic comet will hit the earth in the year 14. August 2126, and possibly wipe out all life on our planet.The probability was estimated with 1 : 10 000.This is the Swift- Tuttle- comet. Parts of the comet have already hit the earth. Every 130 years, it completes a full orbit around the sun, leaving a stream of meteors and particles in its path into space. If the earth crosses the stream, we see in the sky the shooting star stream of the Perseids, which already arranged the purest fireworks in the sky.

In Jan.In 1991, a NASA committee estimated that there are about 1,000 to 4,000 Earth-orbiting asteroids that are greater than 0.8 km in diameter, and therefore pose a threat to human civilization. However, convincing radar records exist of only 150 of these large asteroids. The sum of all earth orbital crosses over 90 m is said to be 300 000. Rarer but more spectacular are supernova explosions in the neighborhood of the earth. A SN releases enormous amounts of energy, more than 100 billion stars. All alone with the bursts of X-rays it produces, it causes serious disturbances in any nearby star system. And in the worst case can destroy all life in this system. Unfortunately, a SN explosion takes place without warning and spreads at the speed of light.All that is left for a civilization is to carefully monitor the stars in its neighborhood, which are on the cusp of SN.

65 million years ago, in the Cretaceous period, the Earth’s surface was hit by a comet or asteroid, causing profound changes in the Earth’s atmosphere that led to the extinction of the dinosaurs on Earth.This is quite obvious, considering that the object is approx. 8 km in size, and hit the earth’s surface at 30 km/s ( ten times faster than a rifle bullet), creating a crater of ca.180 km diameter produced.One found in whole Mexico, Haiti, up to Florida debris particles of this impact, which took place in the Mexican state Yucatan near the village Chicxulup. Among the amazing features of life on earth is that this event that wiped out the dinosaurs is only one of several accurately documented mass extinctions. For example, a mass extinction that took place 250 million years ago wiped out 96% of all plant and animal species within a radius of.

In total there have been 5 such mass extinctions among animals and plants.

Every 26 million years there is such a mass destruction of species, as the paleontologists have shown.This can be shown for 10 cycles ( with the exception of 2 ) over a period of 260 million years.In one of these cycles died, at the end of the Cretaceous 65 million years ago. years ago, the dinos died out, in another, at the end of the Eocene 35 million years ago, many species of land mammals became extinct.

Since the time cycle of 26 million years can be explained neither by geological, biological, nor also astronomical data, there is a theory and/or. Hypothesis that our sun is in a binary star system, in which the sister star ( called Nemesis, or the dead star) is responsible for the periodic mass extinction on Earth. This is based on the assumption that this mass-rich invisible partner orbits our sun every 26 million years, and during the passage of Oort’s cloud, directs objects into the solar system by its gravity, which can cause this damage to the earth. Currently we are exactly in the middle between 2 death cycles, which means Nemesis, if it exists, is at the farthest point of its orbit (probably several light years away). This left us with 10 million years until its next critical position.At the moment we know 160 secured impact craters.The youngest major crater with a diameter of 23 km is the nordlinger Ries in southern Germany. It was formed 14.7 million years ago, when a rock about 500m in size hit the region between the Franconian and Swabian Alb Mountains. Statistics from Nasa suggest such an impact every 500,000 years.The famous Barringer crater in Arizona with a diameter of 1.2 km, left about 45 000 years ago the impact of a ca. 50 m large lump from iron and nickel, from whose remains still today small fragments are found.Such an event extinguishes all life in ca.50 km radius. From a Ries – event a whole continent would be affected.

It looks differently, if a ca. 10 km large chunk crashes to earth.The heat flash extinguishes the life in the environment of the then 150 km large crater in the periphery of thousands km.The shock wave would circle the earth several times.Large amounts of dust would be hurled up into the stratosphere, enveloping the Earth in a dark mantle for many years, photosynthesis could no longer take place as a result food chains would collapse, there would be a sog. Faunal section, as it has occurred several times in the paleontological time scale. If we think about cosmically caused catastrophes on the earth, we must consider also the end of the sun.Our sun is a middle aged star. It is ca. 5 billion years old, and is thus in the middle of its life, which is estimated to be 10 billion years old. When it exhausts its hydrogen supply, it begins to burn helium, and at the same time it expands to the enormous dimensions of a red giant, reaching as far as the orbit of Mars. Thus the earth disappears completely in the solar atmosphere, and is burned and extinguished by the enormous temperature. This scenario is described by astronomers as follows. When the last day of the earth dawns, the ice caps of the Arctic and Antarctic will melt, the coasts will be flooded, while the temperature of the water rises, it evaporates and leads to a thick cloud formation. Finally the oceans begin to boil, and the atmosphere escapes into space A catastrophe of unimaginable extent takes its course.

It remains to mention that our galaxy the Milky Way also does not remain eternally in such a way as it is, thus also an end has. When we look up at the night sky today, we glimpse in the ca. 3000 visible stars only a tiny fraction of our galaxy, on a fraction of the so-called Orion arm. The remaining approx. 200 billion stars are so far away that they appear to us only as a blurred bright band stretching across the night sky. At a distance of about 2 million light-years from the Milky Way, is our nearest galactic neighbor, the great Andromeda Galaxy, which is about. 2-3 times bigger than the Milky Way.

Both galaxies are heading towards each other at 125 km/s, and collide in ca. 5 to 10 million years together. Thus our galaxy is dissolved and destroyed. In periods of billions of years, our universe will then also move towards its end and be doomed to death.

The light pollution

Observatory talks (faq)

Due to the flood of light in large cities and, unfortunately, even in smaller settlements and streets, observing the starry sky is becoming increasingly difficult, i.e. the brightness of objects in the sky is becoming more and more difficult for the observer

depending on position and viewing direction, and above all the distance to the light source, partly considerably reduced.A quantitative description of the relations is given in "Walker’s Law" (Sterne und Weltraum 01 / 2002 , To think about).

After that, the following relationship exists:

lg L = – 4,7 – 2,5 lg d / km + lg F / Lum

there is "F" the total luminous flux in "lumen" and "d" the distance in "km" ; "L" is the light pollution in relation to the natural night sky brightness. The value refers to the view in the direction of the city at a zenith distance of 45°.

The limit for the application of the equation is : d> as the radius of the.

In the industrialized countries, for the total flow of light "F" per inhabitant("E")

500 lumens assumed.

The equation can then be rewritten to population "E" and distance "d" from the city center.

lg L = – 4.7 – 2.5 lg d + lg 500*E

For the Hofingen Observatory on the outskirts of Stuttgart , ca. 20 km as the crow flies from the city center, this results in a population of S. of 582 000 (E):

lg L = – 4,7 – 2,5 lg 20 + lg 291 000 000 = – 4,7 – 2,5*1,30 + 8,46 = 0,5 ,

and with it L= 3,16

The effective sky brightness "H" is calculated according to

H = 1 + L

The effective diameter of a telescope aperture , related to 45° zenith distance in the direction of Stuttgart is then calculated according to the factor:

1 / H^1/2

For the telescopes of the observatory this results in a value of 49%.

The decrease in visual limiting brightness results from :

m1 – m2 = – 2.5 mag * lg H

related to the observatory this results in a reduction of visibility

of m = 1,55 mag

In the above quoted source also a table is given , which gives an assignment of limiting magnitudes to the number of visible stars per year for a horizon of

Since our horizon on the observatory is at d = – 41.2 according to our latitude of 48.8°, the values can be used without making a serious mistake.


Number of visible

Stars brighter than G


It is quite remarkable how 1,5 likes Size class reduction affects the number of visible stars.

Finally a concrete example to determine the visibility limit in the field.

As an object here sensibly the small cars chosen, because its immediate proximity to the poles guarantees permanent visibility in the night sky of the northern hemisphere.

ISS Space Station

This image shows the European module on the International Space Station ISS

Observatory talks (faq)

The legendary figures around the constellations " Pegasus , Andromeda , Cassiopeia , Perseus , and Cepheus

The constellations originate from the Greek mythical circle around Perseus, which roughly follows the following pattern.A king had a daughter.The oracle prophesies that she will give birth to a son who will kill him.The king thereupon locks up his daughter in an underground chamber. Zeus then turned into golden rain and penetrated through the ceiling of the vault and took her as his wife. When she gave birth to a son, the king locked her and her child in a chest and threw them into the sea. Stranded on an island, the king’s brother found the chest, opened it and took it back to his castle. The daughter, named Danae, recovered quickly, and because of her beauty and noble way of thinking, quickly found recognition in the country. Her son, who was named PERSEUS In the meantime, he grew up to be a strong and courageous young man. The brother of the king wanted to marry Danae, and therefore wanted her son Perseus remove from her life. To achieve this he demanded him to procure the head of Medusa. Medusa had such a horrible face that everyone turned to stone at the sight of her. With the help of the gods managed Perseus however, to cut off Medusa’s head. From the blood shed thereby then PEGASUS have originated.

On the flight to his homeland, with his winged shoes he came to Ethiopia.

There reigned a queen named CASSIOPEIA. She wanted to be more beautiful than the mermaids and was therefore under the curse that every day a sea monster came ashore and took a human being. The curse could be lifted only if Cassiopeia her beautiful daughter ANDROMEDA would sacrifice. Then the citizens forced their king KEPHEUS to sacrifice Andromeda. They tied her to a rock face on the coast to leave her for dead. When now Perseus When he saw the girl on the rock during his flight, he inquired about her fate, killed the sea monster and got married Andromeda. Perseus After that he had to fight many battles to defend his wife and his mother. With the head of Medusa, however, he turned all his opponents into stone After returning home from Perseus, the king remembered the oracle’s words, and quietly, unrecognized, set out to escape his fate foretold by the oracle, and to escape from the oracle Perseus to be safe. At a great and important competition, where the unrecognized king was among the spectators, and Perseus was taking part in a discus competition, the discus slipped from his hand and killed a man in the crowd. It was the escaped king.

Visibility in the night sky

Observatory talks (faq)

The so called seeing of the night sky is easily determined from the constellation of the little chariot.In the picture the magnitudes of the individual stars are indicated. With a good eye, on a moonless night, you can see stars up to the 6. See greatness.

Day length

Question Have all places on earth the same day length over the time span of a year?

Answer : NO. At the equator z.B. the length of the day is shorter than at most other latitudes.

Every place on earth would have exactly 50% of the yearly hours of daylight, if the earth would revolve in a circular orbit around the sun, the earth would be a point object, and there would be no atmosphere. Due to the atmospheric refraction and the disk shape of the sun, whose center is below the horizon, if you can still see a part of the disk, the annual day length will be above 50%. This effect has the least influence at the equator.where the sun would rise.- and setting at the steepest point of the horizon. But he appears most strongly near the polar circles where the sun grazes the horizon for the longest time. The distance of the earth from the sun is greatest at the beginning of July, and its orbital velocity is lowest. This is the reason why in the northern hemisphere the spring and summer ca. lasts 8 days longer than in the southern hemisphere.

The Zodiacal Light

The term Zodiacal Light ( ZL ) refers to the illumination of the sky above the rise.- or setting position of the sun. In the tropics around the equator, the almost triangular washed-out illuminated area is observable almost daily. In our latitudes the brightening can be observed only in spring in the evening sky and in autumn in the morning sky.The symmetry plane of the ZK lies in the ecliptic ( apparent sun orbit ) and thus in the so-called zodiac ( Greek Zodiacus ) . From this the name is derived.The reason for the limited observation possibilities at our latitudes is the position of the ecliptic to our horizon. Only in the mentioned seasons the ecliptic rises so steeply above the horizon that the CC does not set in the twilight, and can be observed through the haze layer near the ground.The sun forms the largest possible angle with the horizon, if the northernmost point of the ecliptic stands in the meridian, thus on the midday line.

( This is the case at the times : beginning of summer 12 o’clock local time ; beginning of winter 24 o’clock beginning of autumn ; 6 o’clock ; beginning of spring 18 o’clock.)

Another condition for the observation is that the sun is at the beginning (or end) of astronomical twilight . ( sun 18° below the horizon ) Therefore the times of summer and winter beginning are out of question. The apex of the ZK – triangle is about 90 – 100° away from the sun. A narrow light bridge then continues along the night sky arc of the ecliptic to a position opposite the sun. There the ZL reaches another brightness maximum which is also called G e g e n s c h e i n . The cause of this phenomenon is attributed to a flat cloud of dust and gas distributed around the Sun in the ecliptic .The interplanetary gas cloud consists mainly of hydrogen atoms. In autumn the CC is to be observed before the beginning of dawn. The best date is not the beginning of autumn, but the 14 October. At this time the conditions for visibility are best fulfilled.

Sunset on Mars

Observatory talks (faq)

The picture shows a sunset on Mars, taken by the Mars robot Spirit.The distance of Mars from the Sun is 1.52 times greater than that of the Earth.On Earth the Sun has a mean size of approx. 30 arcminutes. Converted to the distance to Mars this results in a solar diameter on Mars of approx. 20 arcminutes. So the sun is 1/3 smaller for the observer on Mars than on Earth, which is clearly visible on this image.

observation of satellites

If one sees in the evening or in the early morning, before dawn, a star-like object (not an airplane) moving through the star constellations, mostly from W to E, with clearly perceptible speed, we see a satellite. Also the international space station can be observed, at times which are exactly known and available on the Internet on this page. It is an interesting game to determine the altitude of the object from the observation. For this one must know only the following.1.) The formula we want to derive now and 2.) The rotation speed of the earth per time unit, to the observation value at the west east movementof z. B. 125 minutes in the meridian passage to the necessary value of the hour angle in relation to the fixed star sky (i.e. the sidereal orbit). At the observation time of z.B. 125 min, one must approximate. subtract 10 min from the observed orbital period from meridian to meridian. The formula consists of the following quantities : go = acceleration due to gravity ; g’ = gravitational acceleration. in orbit ; Ro = earth radius ; R’ = orbit radius of the Sat. ; g’/go = (Ro / R’ )^2 , g’ = v^2 / R’ , v is the orbital velocity. of Sat. v =2 * R’ * pi / T , T is its orbital period. Subtract : 4 pi^2 R’^2 / T^2 * R’ = go ( Ro /R’ )^2 , or R’^3 = go * Ro^2 * T^2 / 4 * pi^2 . This determines the path radius. Calculation example : For Ro = 6400 km = 6,4 * 10^6 (m) ; go = 9,81 m/s^2 ; T = 115 min = 6900 (s) ; we get R’ = 7855 km , the Sat. Height is thereby approx. 1455 km.

What happens at 13.April 2029?

In June 2004, the minor planet 2004MN4 was discovered, which has the respectable size of ca.300 m has. First orbit determinations showed that it is on collision course with the earth at the above date.An impact on the earth would mean the devastation of an area of the size of Germany.However, due to further orbital investigations it was found that the Earth is not hit by the object, it is observed in approx. 30 000 km, i.e. in 1/12 of the distance to the moon, passing by the earth.According to Kosmos celestial year the object is defined as point-like object ca. 3. size as a point of light rushing at 42° per hour through the constellations of Sextant and Cancer.From our point of view the object is well visible.at the next flybys in 2013 and 2021 the distance will be larger.According to the above source there should be no danger for the earth until the year 2070.

How to add star brightnesses?

The faintest stars visible to the naked eye have the apparent magnitude of m = 6 .How many stars would have to stand next to each other to be perceived by the eye as one star of m = 0 ? The luminosities are logarithmically graded. The correlation is given by… m1 – m2 = – 2,5 * lg ( L1 / L2 ) described. In this case m1 = 0 , m2 = 6 , L1 = n * L2 inserted in the formula gives: – 6 = – 2,5 * lg (n), resolved by n gives : lg (n) = 6 / 2,5 = 2,4 ; lg 2,4 = 251. That means 251 stars of magnitude 6 correspond to the brightness of a star of magnitude m = 0.

With which speed we move in the universe?

The earth moves around the sun with approx. 30 km/sec, the sun moves around the center of the milky way with ca. 250 km/sec, that means one orbit in 200 million years. The Milky Way moves within the local group of our neighboring galaxies and finally the local group moves with a velocity of ca. 600 km/sec in direction of the Virgo galaxy cluster. The different movements are however not of the same sense, but affect in each case into different directions.

Objects and forces of our universe

Our universe is formed by 4 forces.

>The weak nuclear force, which is responsible among other things for the radioactive decay.

>The strong nuclear force, which holds protons and neutrons together in the atomic nuclei, and plays an important role in nuclear fusion.

>The electromagn. Force acting in charged particles and in magnetic fields.

>The gravitation, which provides for hirarchies in the universe, and among other things also the structure of the chem. Elements shaped.

stars :

Our sun is a star of medium size. The stars visible in the night sky are almost all more massive, and therefore brighter than our sun. They are all located in our galaxy of the so called. Milky Way. A star with 10 – times the mass of the sun, like z.B. the Vega in the constellation Lyra ( Lyra ) shines 10 000 times brighter than our sun. The closest star to us in the northern hemisphere, in the summer constellation "Ophiuchus", also called arrow star, because of its small distance of 5.94 light years, changes its position significantly in a few years, (in 200 years just under a full moon width or 10.4" per year) has only 20% of the mass of the sun, and therefore shines 100 times fainter than the sun, it has ca. Jupiter size, and luminosity, and is only observable in telescopes ( apparent brightness : 9.5m ) . One calls this category of stars also red dwarfs.

Red Dwarfs

…are very low mass stars. They have< 50% of the solar mass, are very numerous in the universe , and therefore embody the largest part of the total mass of the universe.Because of their low mass they are able to store their energy for a very long time, therefore they reach a lifetime of> 10 trillion years. ( The sun becomes ca. 10 billion years old )

White dwarfs

White dwarfs are unbelievably dense, they have a density which corresponds to the millionfold density of water. The sun will shrink to a white dwarf at the end of its life, and then will have a remainder of 0.6 of its original mass.

Brown dwarfs

The smallest stars have a mass of ca. 8% of the solar mass. At these masses, gravity is not strong enough to reach the temperature for nuclear reaction. They keep with it their element frequency with which they originated. Therefore, they consist mainly of hydrogen, the most abundant element in the universe. Collisions of 2 brown dwarfs can also form planetary systems.

Neutron stars

Neutron stars are even denser than white dwarfs, and considerably so. They form when massive stars collapse at the end of their lives. They correspond to the density of atomic nuclei, or 10^15 times the density of water.A typical neutron star has about 1.5 times the mass of the sun, in maximum 2 – 3 times. Its radius is about 10 km.

Black holes

….exists in the center of galaxies , with 10^6…10^9 solar masses. A black hole of 10^6 solar masses has a radius of ca. 4 solar masses.Furthermore schw. Holes when very massive stars of> 8 solar masses collapse at the end of their lives. Their number is estimated to be less than 1 in 3000 stars. Our galaxy hosts in its center a black hole of ca. 3 million solar masses.


Our neighbor galaxy M31 is moving with high speed towards the Milky Way, and will be visible in about. Collode with it 6 billion years, about the time our sun inflates into a red giant, beyond Mercury and Venus. The end of galaxies is usually not caused by the disintegration of stars, but by collisions with other galaxies of their local group. Stars are ejected by this process with speeds up to 300 km/ s as computer simulations show.

Dark matter

Galactic halos consist mainly of dark matter. One supposes under this term mass-rich particles with 10 to 100 times proton masses without electric charge, and without strong nuclear force. They are therefore subject only to weak nuclear power and especially to gravitation. Dark matter can be destroyed in collisions with other particles in the process.

The edge darkening of the sun

When observing the sun, it is noticeable that the brightness of the sun decreases in the direction of the sun’s edge.

This effect is called center-rim darkening of the sun. The reason for this phenomenon is that the view into the center of the sun is in deeper and therefore hotter resp. penetrates brighter zones of the surface structure of the sun.In the edge zone, you can’t see as far into the depth of the gas mixture of the surface, the depth of view decreases until it touches tangents. The radiation of the cooler gas masses in the edge zone is less intense, and thus darker.

History of the brightness of celestial bodies

In ancient times, the brightnesses of stars as perceived by the observer were divided into magnitude classes. The brightest stars were stars of the 1. Size , the weakest of the 6. Size. The designations of antiquity were based on the sensory impression of the observer. The connection is described by the so-called Weber-Fechner law, according to which the sensation is proportional to the logarithm of the stimulus. So……apparent brightness m is proportional log s . where s is the radiant flux reaching the observer. In order to put the empirical determination of the antiquity into a mathematical context, it was necessary to define a proportionality factor in order to obtain a good approximation to the historical scale. The algorithm valid today is:

m1 – m2 = – 2.5 log s1 / s2 ; or : s1 / s2 = 10^ – 0.4 ( m1- m2 )

for delta m =1 ………there is s1 / s2 = 10^ -0,4 = 1 : 2,512

for delta m = 5…………then s1 / s2 = 10^ – 2 = 1 : 100

The geometrical gradation has a great importance in human life . It is not that they were invented simply to create advantageous conditions. The fact that geometrically stepped physical sizes are felt as normal,has its reason in the fact that the sense organs of humans register logarithmically, which changes in the environment after natural number sizes. One feels the doubling of the initial value only as the 1.3-fold, a tenfold increase of the initial size evokes a twice as strong sensation.Nature has built into the sensory organs of man conversion organs , which allow man to compress very large stimulus areas into small sensory areas. Man has purposefully applied this fact in his own creations. geometrically staged series exist in machines, type series, and individual parts in technology.A geometrical series is perceived as harmonic.The geometrical gradation of natural numbers always means a linear gradation of logarithms. In the industrial standards there are the so-called standard numbers, which are constructed according to the same principle of a geometrical series. And are frequently used in engineering.

Radioactivity and the age of the solar system

Heavy atomic nuclei are sources of radioactive radiation. Thus, z.B.the atomic nucleus of uranium, with the charge number 92,and mass number 238 (238/92U) during radioactive decay by the spontaneous emission of an alpha particle (helium nucleus 4/2He) into the atomic nucleus of thorium (234/90 Th ) at..Beside the alpha decay there are 2 other decay types, the beta decay resp. the emission of electrons , and the gamma decay that is the emission of very short wavelength electromagnetic radiation. Radiation.The atomic nucleus of the lightest element, hydrogen (H), is called a proton. Heavy elements have an atomic nucleus consisting of protons and neutrons, which have no electric charge.The nucleus of the helium atom z.B. consists of 2 protons and 2 neutrons (4/2 He ) , and the nucleus of uranium (238/92 U ) of 92 protons and 146 neutrons. The mass number of the nuclei corresponds to the sum of the protons and neutrons bound in the nucleus. The decay of the nuclei follows the laws of probability and therefore occurs at irregular intervals, following the laws of chance. However, measurements show that in equal periods of time, the intensity of the radiation decreases to exactly half, regardless of its initial intensity. If we take for the initial intensity 1 , then after a time characteristic for the radioactive substance t = ½ , the intensity has decreased to the value ½. This characteristic time is called half-life, which is different for each radioactive substance as said before. It says when a given amount of radioactive atomic nuclei of a certain element has decayed by half. The half-lives of different radioactive atomic nuclei differ considerably, so there are z.B. Half-lives of 2.2*10^-7 sec. and 4,5*10^9 years at the ( 238/92 U ). Decay produces either a stable or unstable further decaying nucleus. The vast majority of the 45 naturally occurring radioactive elements can be assigned to 4 decay series. Thus, the uranium – radium series begins with the alpha decay of uranium (238/92 U ) After 4 steps, it reaches the radium isotope (226/88 Ra ), which is transformed by alpha decay into radon (222/86 Ra ). The series ends with the stable lead isotope (206/82 Pb ). If a mineral, or more generally, a substance, contains a sufficient amount of uranium atoms, an equilibrium is formed within the series in different concentrations depending on the random series. From this one can determine the age of the substance then very exactly. The oldest rocks found on Earth so far were measured to be 3.96 billion years old.Measurements at stone meteorides resulted in values up to 4.6 billion years. This is how astronomers gain insight into the early history of the solar system.

The 10 brightest stars in the night sky are ..

[AlphaCanis Majoris(Sirius)-1,46 m AlphaCarinae(Canopus)-0,72 m( not observable in our latitudes )AlphaCentauri(Rigil Kentaurus)-0,27 m( combined double star brightness )AlphaBootis(Arcturus)-0, 04 m AlphaLyrae(Vega)+0.03 m AlphaAurigae(Capella)+0.08 m BetaOrionis(Rigel)+0.12 m AlphaCanis Minoris(Procyon)+0.38 m AlphaEridani(Achernar)+0.46 m AlphaOrionis(Betelgeuse)+0.50 m( average magnitude). Brightness of the variable )

Resonances in the planetary system

That Saturn offers us such a splendid and varied picture, we have to owe among other things also to the structure of its ring.However, only if the ring plane is inclined to the line of sight of the observer on Earth, which is not always, but mostly the case.

Through a larger telescope, one can discover a clear gap in the inclined ring of Saturn, which is called the Cassinian division after its discoverer.

The cause of this empty swept zone can be explained by the 3.Kepler’s law, Newton’s law of gravitation,and explain the resulting resonance.

If a body is within this zone, the Cassini division, it has a certain orbital period. This is calculated according to the formula:

R^3 / T^2= G*M / 4 pi^2

(R=orbit radius,T=orbit time,G=gravitational constant,M=mass of the central body ) For the inner fringes of the cas. division has been measured for R = 1.17 * 10^8 m. This results in an orbital period of T = 4.077 * 10^4 sec. Or 0.472 days.If one compares this time with the orbital period of some moons of Saturn, the following interesting connection results:

SATURNMONDUMLAUFZEITVERH.d.UML.-TIMESMimas0,942 d2,0Enceladus1,370 d2,9Thetis1,888 d4,0Dione2,737 d5,8

Bodies in the gap are regularly disturbed by Mimas after 2 orbits, by Enceladus after 3 orbits, by Thetis after 4 orbits and so on.Due to the disturbance every body is driven out of this zone, because the mass attraction force of the moon on the body is always greatest at the same place.

Mercury resonance

This resonance is another remarkable example. The sidereal rotation period of Mercury is 58.65 days The sidereal orbital period is…………………88 days This is a ratio of 2 : 3 .

With respect to the orbital period of Mercury around the Sun, i.e. to its synodic orbital period, this means that 1 Mercury day corresponds to 2 Mercury years.In other words the synodic rotation period of Mercury corresponds to 2 orbits in its orbit. After 2 solar orbits the Mercury globe has completed 3 revolutions around its axis, and the sun is again at the same position.This constellation is also called a commensurability.It is the result of a dynamic coupling due to the action of gravitational forces on a nonspherical body

The moons of Mars

Mars has 2 moons named PHOBOS and DEIMOS (fear and terror) It is generally believed that both bodies were captured by Mars during its existence. Due to the small distance of the moon Phobos from the surface of Mars, and the resulting high orbital velocity of the moon, interesting phenomena arise. One of these phenomena is the ratio of the orbital periods of both moons. They behave like T (Phobos)/ T (Deimos) = 1 / 3,96 .So they are very close to a 1:4 resonance.

Neptune and Pluto

Also these two outermost planets of the solar system, which move in very different orbits around the sun, are synchronized. The reason is the partly small distance of their orbital path. The orbital data of Neptune and Pluto.

PlanetOrbitEccentricityOrbit in daysOrbit InclinationAverage. Distance.from the sun (AU)PLUTO0,25090465,017,2°-NEPTUN0,008660189,01,77°30

The mean period of the orbital periods of both planets around the sun behave like

T (Pluto) = 1,5 * T (Neptune)

Both planets are at their starting point every 495 years. Neptune has 3 orbits when Pluto has completed 2.

How far are the stars from us?

If the spectrum of a star and its spectral class are known, the observed brightness of the star can be used to determine its distance.From the spectral class one can determine the absolute brightness of the star with the so called Hertzsprung- Russel diagram. The result with the observed brightness ( also called apparent brightness ) into the formula a = 10 (pc) * 10^(m-M) / 5gives the distance "a".

How to determine the radius or. the size of a star?

An alternative way is the luminosity (L) of a star : L = 4*pi*S*R^2*T^4 ; T the temp. in K one gets from the spectral analysis , S is a constant with the value 5,6710^-8 (Watt/m^2*T^4) S*T^4 is according to Stefan-Boltzmann the total radiant power coming through the unit area of the surface from the inside of the star. With the distance of the object and the observed brightness one can determine L. Thus the radius can be calculated according to the above formula.Other methods are based on stellar occultations and optical methods that determine the angular diameter of a light source from the wave properties of the light.

The Earth – Moon System

This article is out of date and will be revised soon

The physical law of conservation of angular momentum can also be applied to the Earth – Moon system, and leads to revealing results. The total angular momentum of this system consists of 4 components.the intrinsic angular momentum of Earth and Moon, and the orbital angular momentum of both bodies. By observations of star occultations and lunar eclipses one determined that the orbital period of the moon becomes larger around 2.1 milliseconds in 100 years. This corresponds with the assumption of circular orbits an increase of the distance of 2.7 m in one century.The cause is to be looked for in the tidal friction. The intrinsic angular momentum of the earth decreases constantly for this reason, until finally earth and moon always turn to the same side by rotating around a common center of mass.Thereby the intrinsic angular momentum of the earth is transferred to the orbital angular momentum of the moon. The calculation shows that in the final stage the distance of the moon increases to the 1.44-fold of the today’s distance. The friction of water on the ocean floor, and the tidal wave during the tidal cycle, contribute to the deceleration of the Earth’s intrinsic angular momentum.The deceleration of the earth rotation is 0,00000002 seconds daily. 100 years ago the period of the earth rotation was 0.00073 seconds shorter. Extrapolated into the past z.B.4.5 billion years ago, results in a day length of only 5 hours,this corresponds to a distance of the moon in the order of magnitude of the orbits of today’s artificial satellites.

The Kuiper belt and the Oort cloud

The sun is surrounded by a spherical reservoir of objects, from which also the long-period comets originate. The zone is called Oort’s cloud. It is supposed to contain 1 quadrillion ( 10^12) comet nuclei, and to range from 50 000 to 100 000 AU.A comet z. B. with a perihelion distance of 5 AU and an aphelion distance of 100 000 AU has an orbital period of 11 million years. So an object from this zone can be on its way to us for 5 million years, and will reach the inner solar system only in 500 000 years in half of its orbit.

The Kuiper Belt is a disk-shaped region behind the orbit of Neptune, in ca. 30 – 55 AU distance from the sun,it consists of smaller objects and mind. a dwarf planet namely Pluto. It is thought to host more than 70,000 objects larger than 1 km in diameter, and short-period comets with orbital periods smaller than 200 years

The Asteorid Belt

The asteroid belt marks the boundary between the inner and the outer planetary system, it is located between Mars and Jupiter, and contains countless bodies in different sizes. The total mass of the bodies in this zone is estimated to be ca. 1/ 1000 of the earth mass estimated. The largest object is the small planet Ceres with a diameter of about 1000 km.

The Milky Way

When we perceive a nebulous band of light in a particularly dark and clear night sky, which stretches from Gemini to Fuhrmann across the constellations Cassiopeia, Swan, Eagle, and Sagittarius, we are looking in the direction of our disk-shaped spiral galaxy, in which the sun has its place and which we call the Milky Way. Through binoculars or a small telescope, the bright nebula resolves into a large set of faint stars all belonging to this galaxy.The constellations are of course also part of it, but they have much smaller distances to the sun and are therefore brighter and better to see than the other stars. Our galaxy has a diameter of more than 100 000 light years, and has 2 small companions, the Small and the Large Magellanic Clouds, which appear in the southern night sky as detached parts of the Milky Way.However, they are so far south that they can only be observed from near the equator in the night sky without optical aids. There are countless other galaxies in the universe. Another neighboring galaxy is the Andromeda Nebula, or M31 as it is called by astronomers, 2.5 million light years away. If you know its position in the constellation Andromeda exactly, you can see it even with the naked eye as a faint nebula, but only under ideal viewing conditions on a moonless night.

Adaptation of the eyes

When stepping from a brightly lit room into the dark at night, it takes some time for the eye to adjust to the darkness. One can see stars up to 4 after a few minutes. Perceiving size. This adaptation is based on a dilation of the pupils, but also on a real adaptation of the cones in the retina, which are centrally located in the yellow spot at the point of sharpest vision. More important for night vision, however, are the rods in the peripheral area of the retina; their adaptation, which is based on the production of rhodopsin, takes about 45 minutes and is reversed by white light within a very short time. Red light almost does not affect the dark adaptation of the rods; therefore the observatory has a red night light (just like in a photographic darkroom) to allow sufficient orientation. By the way, the rods are not sensitive to colors (at night all cats are gray). With complete dark adaptation, stars can be seen up to 6. Size class to be seen. Because the site of sharpest vision is much less sensitive to light than the rods (which, in turn, have much less visual acuity), indirect vision – that is, deliberately looking next to an object while seeing with the rods – allows us to make out low-light point-like or planar objects that disappear when we fixate on the site of sharpest vision.

The resolving power of the human eye

The resolving power of the naked eye under ideal conditions is about 0.5′ to 1′ (corresponding to 1 mm at 3-6 meters). It depends, just as with optical instruments, on the size of the pupil, corresponding to the aperture of a telescope. The spacing of the photoreceptor cells (cones) in the retinal pit, the site of sharpest vision, is matched to the resolving power of the eye. It is approx. 0,3′. In average conditions, two points can be perceived separately if their angular separation is 2′ apart. However, for faint objects and toward the edge of the visual field, where the rods are located, visual acuity decreases noticeably.

A normal-sighted eye can easily separate the nearly 12′ separated stars MIZAR + ALKOR in the Big Dipper (eye examiner).

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