A system that consists of two identical satellites. Instructions for performing control work. Star sizes. The density of their matter
Masses of stars. As we have seen from the example of the Sun, the mass of a star is the most important characteristic on which the physical conditions in its depths depend. Direct determination of the mass is possible only for binary stars.
Binary stars are called visual binaries if their duality can be seen through direct telescope observations.
An example of a visual double star, visible even to the naked eye, is Ursa Major, the second star from the end of the "handle" of its "ladle". With normal vision, a second faint star is visible very close to it. It was noticed by the ancient Arabs and called Alcor(Rider). They gave a name to a bright star Mizar. Mizar and Alcor are separated from each other in the sky by 11". Through binoculars, you can find a lot of such stellar pairs.
Systems with the number of stars n≥3 are called multiples. So, with binoculars, it can be seen that ε Lyra consists of two identical stars of the 4th magnitude with a distance between them of 3 ". When observed through a telescope, ε Lyra is a visual quadruple star. However, some stars turn out to be only optical-double, i.e., the proximity of such two stars is the result of their random projection onto the sky. In fact, they are far apart in space. If, when observing stars, it turns out that they form a single system and circulate under the action of forces of mutual attraction around a common center of mass, then they are called physical double.
Many double stars were discovered and studied by the famous Russian scientist V. Ya. Struve. The shortest known orbital periods for visual binary stars are several years. Pairs with circulation periods of tens of years have been studied, and pairs with periods of hundreds of years will be studied in the future. The closest star to us, a Centauri, is a double star. The circulation period of its constituents (components) is 70 years. Both stars in this pair are similar in mass and temperature to the Sun.
The main star is usually not in the focus of the visible ellipse described by the satellite, because we see its orbit in a distorted projection (Fig. 73). But knowledge of geometry makes it possible to restore the true shape of the orbit and measure its semi-major axis a in seconds of arc. If the distance D to the binary star is known in parsecs and the semi-major axis of the orbit of the satellite star in seconds of arc, equal to a", then in astronomical units it will be equal to:
since D pc \u003d 1 / p ".
Comparing the motion of the satellite of the star with the motion of the Earth around the Sun (for which the period of revolution is T = 1 year, and the semi-major axis of the orbit is a = 1 AU), we can write according to Kepler's III law:
where m 1 and m 2 are the masses of the components in a pair of stars, M and M are the masses of the Sun and the Earth, and T is the rotational period of the pair in years. Neglecting the mass of the Earth in comparison with the mass of the Sun, we get the sum of the masses of the stars that make up the pair in the masses of the Sun:
To determine the mass of each star, it is necessary to study the motion of the components relative to the surrounding stars and calculate their distances A 1 and A 2 from the common center of mass. Then we get the second equation m 1:m 2 =A 2:A 1 and from the system of two equations we find both masses separately.
Double stars in a telescope are often a beautiful sight: the main star is yellow or orange, and the satellite is white or blue.
If the components of a binary star come close to each other during mutual circulation, then even in the most powerful telescope they cannot be seen separately. In this case, duality can be determined from the spectrum. Such stars will be called spectral double. Due to the Doppler effect, the lines in the spectra of stars will shift in opposite directions (when one star moves away from us, the other approaches). The shift of the lines changes with a period equal to the period of the pair's revolution. If the brightnesses and spectra of the stars that make up the pair are similar, then in the spectrum of a binary star, a periodically repeating splitting of spectral lines is observed(Fig. 74). Let the components occupy positions A 1 and B 1 or A 3 and B 3, then one of them moves towards the observer, and the other away from him (Fig. 74, I, III). In this case, a splitting of the spectral lines is observed. In an approaching star, the spectral lines will shift to the blue end of the spectrum, and in a receding star, to the red. When the components of a binary star occupy positions A 2 and B 2 or A 4 and B 4 (Fig. 74, II, IV), then both of them move at right angles to the line of sight and there will be no bifurcation of the spectral lines.
If one of the stars glows weakly, then only the lines of the other star will be visible, shifting periodically.
The components of a spectroscopic binary star can alternately block each other during mutual circulation. Such stars are called eclipsing binaries or Algols, after the name of their typical representative β Perseus. During eclipses, the total brightness of the pair, the components of which we do not see separately, will weaken (positions B and D in Fig. 75.) At the rest of the time, in the intervals between eclipses, it is almost constant (positions A and C) and the longer, the shorter the duration of eclipses and the greater the radius of the orbit. If the satellite is large but produces little light itself, then when a bright star eclipses it, the total brightness of the system will decrease only slightly.
The ancient Arabs called β Perseus Algolem(corrupted el gul), which means "devil". It is possible that they noticed her strange behavior: for 2 days 11 hours, the brightness of Algol is constant, then in 5 hours it weakens from 2.3 to 3.5 magnitudes, and then in 5 hours its brightness returns to its previous value.
An analysis of the curve of apparent stellar magnitude as a function of time makes it possible to determine the size and brightness of the stars, the size of the orbit, its shape and inclination to the line of sight, as well as the masses of the stars. Thus, eclipsing binaries, also observed as spectroscopic binaries, are the best studied systems. Unfortunately, relatively few such systems are known so far.
The periods of known spectroscopic binary stars and Algols are mostly short, about a few days.
In general, the duality of stars is a very common phenomenon. Statistics show that up to 30% of all stars are probably binary.
The masses of stars determined by the described methods differ much less than their luminosities: approximately from 0.1 to 100 solar masses. Very large masses are extremely rare. Usually stars have a mass less than five solar masses.
It is the mass of stars that determines their existence and nature as a special type of celestial bodies, which are characterized by a high temperature of the interior (over 10 7 K). With a smaller mass, the temperature inside celestial bodies does not reach those values that are necessary for the occurrence of thermonuclear reactions.
The evolution of the chemical composition of matter in the Universe took place and is taking place at the present time mainly due to the stars. It is in their depths that an irreversible process of synthesis of heavier chemical elements from hydrogen takes place.
Problem solution example
A task. A binary star has an orbital period of 100 years. The major semiaxis of the visible orbit is a = 2.0" and the parallax is ρ = 0.05". Determine the sum of the masses and the masses of the stars separately if the stars are separated from the center of mass by distances related as 1:4.
Exercise 21
1. Determine the sum of the masses of the double star Capella if the semi-major axis of its orbit is 0.85 AU. e., and the period of circulation is 0.285 years.
2. If a star with the same mass as the Sun moved along the Earth's orbit, what would be the period of its revolution?
2. Sizes of stars. The density of their matter
Let's use a simple example to show how the sizes of stars of the same temperature can be compared, for example, the Sun and Capella (α Aurigae). These stars have the same spectra, color and temperature, but Capella's luminosity is 120 times that of the Sun. Since at the same temperature the brightness of a unit surface of the stars is also the same, it means that the surface of the Capella is 120 times larger than the surface of the Sun, and its diameter and radius are greater than the solar ones. 
once.
To determine the size of other stars allows knowledge of the laws of radiation.
So, in physics it is established that the total energy radiated per unit time from 1 m 2 of the surface of a heated body is equal to: i = σТ 4, where σ is the coefficient of proportionality, and T is the absolute temperature *. The relative linear diameter of stars with a known temperature T is found from the formula
* (The Stefan-Bolydmann law was established by the Austrian physicists J. Stefan (experimentally) and L. Boltzmann.)
where r is the radius of the star, i is the radiation of a unit surface of the star, r, i, T refer to the Sun, and L= l. From here
within the radius of the sun.
The results of such calculations of the sizes of luminaries were fully confirmed when it became possible to measure the angular diameters of stars using a special optical instrument (stellar interferometer).
Stars of very high luminosity are called supergiants. Red supergiants turn out to be similar in size (Fig. 76). Betelgeuse and Antares are hundreds of times larger than the Sun in diameter. More distant from us, VV Cephei is so large that the solar system with planetary orbits up to and including the orbit of Jupiter would fit inside it! Meanwhile, the masses of supergiants are only 30-40 times greater than the solar mass. As a result, even the average density of red supergiants is thousands of times less than the density of room air.
For the same luminosity, the sizes of stars are the smaller, the hotter these stars are. The smallest among ordinary stars are red dwarfs. Their masses and radii are tenths of the sun, and the average density is 10-100 times higher than the density of water. There are even fewer red white dwarfs - but these are already unusual stars.
Close to us and bright Sirius (having a radius of about twice that of the Sun) has a satellite that revolves around it with a period of 50 years. For this binary star, the distance, orbit, and masses are well known. Both stars are white, almost equally hot. Consequently, the surfaces of the same area radiate the same amount of energy from these stars, but in terms of luminosity, the satellite is 10,000 times weaker than Sirius. This means that its radius is less than √10000= 100 times, i.e. it is almost the same as the Earth. Meanwhile, its mass is almost like that of the Sun! Consequently, a white dwarf has an enormous density - about 10 9 kg/m 3 . The existence of a gas of such density was explained as follows: usually the density limit is set by the size of atoms, which are systems consisting of a nucleus and an electron shell. At a very high temperature in the interiors of stars and with the complete ionization of atoms, their nuclei and electrons become independent of each other. With the colossal pressure of the overlying layers, this "crumble" of particles can be compressed much more strongly than a neutral gas. Theoretically, the possibility of the existence under certain conditions of stars with a density equal to the density of atomic nuclei is admitted.
We see once again on the example of white dwarfs how astrophysical research expands our understanding of the structure of matter; it is not yet possible to create in the laboratory such conditions as are found inside stars. Therefore, astronomical observations help develop the most important physical concepts. For example, Einstein's theory of relativity is of great importance for physics. Several consequences follow from it, which can be verified from astronomical data. One of the consequences of the theory is that in a very strong gravitational field, light oscillations should slow down and the lines of the spectrum shift towards the red end, and this shift is the greater, the stronger the gravitational field of the star. Redshift has been detected in the spectrum of the satellite of Sirius. It is caused by the action of a strong gravitational field on its surface. Observations confirmed this and a number of other consequences of the theory of relativity. Similar examples of the close relationship between physics and astronomy are characteristic of modern science.
Problem solution example
A task. How many times greater is Arcturus than the Sun if the luminosity of Arcturus is 100 and the temperature is 4500 K?
Exercise 22
1. How many times does Rigel have a greater luminosity than the Sun if its parallax is 0.0069 "and the apparent magnitude is 0.34?
2. What is the average density of a red supergiant if its diameter is 300 times greater than the sun's and its mass is 30 times greater than the mass of the Sun?
5 . A piece of ice with mass m1 = 5 kg is floating in water in a vertical vessel, into which a piece of lead with mass m2 = 0.1 kg is frozen. What amount of heat must be imparted to this system so that the remainder of the ice with lead begins to sink? The temperature of the water in the vessel is 0 ˚С. The specific heat of melting of ice is 333 kJ/kg, the density of water is ρ0=1000 kg/m3, ice is ρl=900 kg/m3, and lead is ρb=11300 kg/m3.
m 1 = 5 kg m 2 = 0.1 kg t= 0 ˚С λ = 333 kJ/kg ρ0 = 1000 kg/m3 ρl = 900 kg/m3 ρsv=11300 kg/m3 |
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Answer: 1.39 MJ |
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Option 2
1 . A beam with a length of 10 m and a mass of 900 kg is lifted at a constant speed in a horizontal position on two parallel cables. Find the tension forces of the cables if one of them is fixed at the end of the beam, and the other at a distance of 1 m from the other end.
L= 10 m m= 900 kg b= 1 m g= 9.8 m/s2 |
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F 1 - ? F 2 – ? | Answer: 3.92 kN; 4.90 kN |
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2. Around a fixed charge of 10 nC, a charge of the opposite sign moves along a circle with a radius of 1 cm. The charge completes one revolution in 2p seconds. Find the ratio of charge to mass for a moving charge. Electrical constant ε0 = 8.85 10-12 F/m.
Q= 10 nC T= 2π c R= 1 cm κ = 9 109 m/F |
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Answer: 11nC/kg |
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3. The period of revolution of Jupiter around the Sun is 12 times the corresponding period of revolution of the Earth. Considering the orbits of the planets as circular, find how many times the distance from Jupiter to the Sun exceeds the distance from the Earth to the Sun.
T yu = 12 T h |
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R Yu: R h–? | Answer: ≈ 5,2 |
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4 . A lead bullet pierces a wooden wall, and its speed changes from 400 m/s at the beginning to 100 m/s at the moment of departure. What part of the bullet has melted if 60% of the lost mechanical energy is used to heat it? The temperature of the bullet before impact was 50 ˚С, the melting point of lead was 327 ˚С, the specific heat capacity of lead su = 125.7 J/kg K, the specific heat of fusion of lead l= 26.4 kJ/kg.
t= 50 ˚С t pl \u003d 327 ˚С l = 26.4 kJ/kg With= 125.7 J/kg K Q= 0.6Δ E | Q= 0.6Δ E ;
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Answer: 0,38 |
5. A stream of light with a wavelength l= 0.4 µm, the power of which P = 5 mW. Determine the strength of the saturation photocurrent in this photocell if 5% of all incident photons knock out electrons from the metal.
R= 5 mW η = 0,05 h = 6.63 10-34 J s c = 3 108 m/s e= 1.6 10-19 C | ;
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N - ? | Answer: 80 uA |
Option 3
1 . A 40 W monochromatic light source emits 1.2.1020 photons per second. Determine the wavelength of the radiation. Planck's constant h = c = 3 108 m/s.
R= 40 W n= 1.2.1020 1/s h = 6.63 10-34 J s c = 3 108 m/s |
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λ = ? | Answer: 5.9.10-7 m |
2 . steel ball radius r= 2 cm lies at the bottom of a river deep h\u003d 3 m. What is the minimum work that needs to be done to raise the ball to a height H= 2 m above the water surface? Density of water ρ o = 1000 kg/m3, steel density ρ = 7800 kg/m3.
r= 2 cm h= 3 m H= 2 m ρ = 7800 kg/m3 ρ 0 = 1000 kg/m3 g= 9.8 m/s2 |
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A- ? | Answer: 11.8 J |
; 3. According to the Rutherford-Bohr theory, an electron in a hydrogen atom moves in a circular orbit with radius R = 0.05 nm. What is its speed in this case? Mass of an electron me = 9.11 10-31 kg, elementary charge e= 1.6 10-19 C, electrical constant ε0 = 8.85 10-12 F/m.
R= 0.05 nm κ = 9 109 m/F e= 1.6 10-19 C me = 9.1 10-31 kg |
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Answer: 2250 km/s |
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4. The star system consists of two identical stars located at a distance of 500 million km from each other. The mass of each star is 1.5.1034 kg. Find the period of revolution of stars around a common center of mass.
d= 500 million km M = 1.5.1034 kg G= 6.67 10-11 m3/(kg s2) |
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Answer: 1.6 106 s |
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5. 2 liters of water are poured into an aluminum kettle at a temperature t\u003d 20 ˚С and put on an electric stove with efficiency \u003d 75%. Tile power N\u003d 2 kW, mass of the kettle M= 500 g. After what time will the mass of water in the kettle decrease by ∆ m= 100 g? The specific heat of evaporation of water is 2.25 MJ/kg, its specific heat capacity is 4190 J/kg, and the specific heat capacity of aluminum is 900 J/kg.
V= 2 l t= 20 ˚С tk= 100 ˚С η = 0,75 N= 2 kW M= 500 g ∆ m= 100 g r = 2.25 MJ/kg With= 4120 J/kg K WithA= 900 J/kg K ρ0 = 1000 kg/m3 |
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τ – ? | Answer: 10 min 21 s |
Option 4
1. At what distance from the center of the moon is the body attracted to the earth and to the moon with the same force? Assume that the mass of the Moon is 81 times less than the mass of the Earth, and the distance between their centers is 380 thousand km.
81M l = M h L = 380 thousand km | ,
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Answer: 38 thousand km |
2. A square is cut out of a homogeneous disk with a radius of 105.6 cm, as shown in the figure. Determine the position of the center of mass of a disk with such a notch.
R= 105.6 cm |
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x- ? | Answer: 10 cm to the left of the center of the circle |
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3. The gas was in a pressure vessel P = 0.2 MPa at temperature t = 127 ˚С. Then 1/6 of the gas was released from the vessel, and the temperature of the remaining part of the gas was lowered by D t = 10 ˚С. What is the pressure of the remaining gas?
P= 0.2 MPa t = 127 ˚С D t = 10 ˚С ∆m = m/6 |
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pk – ? | Answer: 0.16 MPa |
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4 . Determine the wavelength of a photon having an energy equal to the kinetic energy of an electron accelerated by a potential difference D j = 2 V. Elementary charge e h = 6.63 10-34 J s, speed of light c = 3 108 m/s.
D j = 2 V e= 1.6 10-19 C h = 6.63 10-34 J s c = 3 108 m/s |
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λ – ? | Answer: 621 nm |
5. Horizontal magnetic field with induction AT= 0.52 T is directed parallel to the inclined plane, from which it slides at a constant speed υ =
5 m/s charged body with mass m =
2 mg. Find the charge of this body if the angle of inclination of the plane to the horizon is 30˚, and the coefficient of friction of the body on the plane k = 0,5.
AT= 0.52 T υ = 5 m/s m = 2 mg g= 9.8 m/s2 |
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q - ? | Answer: 1 µC |
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Option 5
1. A mass of 17 kg is suspended from the midpoint of a horizontally stretched weightless wire 40 m long. As a result, the wire sagged by 10 cm. Determine the tension in the wire.
m= 17 kg h= 10 cm L= 40 m g= 9.8 m/s2 |
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Answer: ≈17 kN |
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m= 4 g q1 = 278 nC α = 45˚ r= 6 cm κ = 9 109 m/F g= 9.8 m/s2 |
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q2 – ? | Answer: 56.4 nC | ||
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3. Considering the orbits of the planets as circular, find the ratio of the linear velocities of the Earth and Jupiter around the Sun υЗ: υО. The period of revolution of Jupiter around the Sun is 12 times the corresponding period of revolution of the Earth.
T yu = 12 T h |
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υЗ: υЮ - ? | Answer: ≈ 2,3 |
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4. Steam hammer mass M= 10 t falls from a height h= 2.5 m per iron bar weighing m= 200 kg. How many times does it have to fall for the temperature of the blank to rise by ∆ t= 40 ˚С? 60% of the energy released during impacts is used to heat the blank. The specific heat capacity of iron is 460 J/kg.
M= 10 t h= 2.5 m m= 200 kg ∆t= 40 ˚С η = 0,6 With= 460 J/kg K g= 9.8 m/s2 | ,
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Answer: 25 |
5. Electromagnetic radiation with wavelength l = 50 nm pulls out photoelectrons from the surface of titanium in a vacuum, which fall into a uniform magnetic field with induction B = 0.1 T Find the radius of the circle along which the electrons begin to move if their velocity is perpendicular to the magnetic field induction lines, and the work function of the electrons from the titanium surface is 4 eV. elementary charge e= 1.6 10-19 C, Planck's constant h = 6.63 10-34 J s, speed of light c = 3 108 m/s.
The period of revolution of Venus around the Sun is T B = 0.615 T W = 224.635 days = 224.635 24 3600 s = 1.941 10 7 s.
In this way,
r \u003d 2/3 \u003d 1.17 10 11 m.
Answer: r=1.17 10 11 m.
Example 2: Two stars with masses m 1 and m 2 , located at a distance r, revolve around the center of mass of the stars. What is the orbital period of the stars?
Solution: 1) Let us first determine the position of the center of mass of a system of two stars relative to the first star r 1 (point C in the figure)
r 1 \u003d (m 1 0 + m 2 r) / (m 1 + m 2) \u003d m 2 r / (m 1 + m 2).
2) For the first star, the equation of motion (1) has the form:
m 1 v 1 2 / r 1 = G m 1 m 2 / r 2
Replacing, according to (2), the speed v 1 , we obtain an expression for the period of revolution:
T \u003d 2π r 1/2.
After replacing r 1 we get the answer:
T \u003d 2π r 1/2.
Example 3: What are the first and second cosmic velocities for a cosmic body with a mass of 10 30 tons and
radius 8 10 8 km?
Solution: 1) The first space velocity must be reported to the spacecraft so that it turns into an artificial satellite of the space body. According to expression (3): v 1 =(GM/R) 1/2. Substituting the numerical values we get:
v 1 \u003d 1/2 \u003d 2.9 10 5 m / s.
2) When the apparatus is informed of the second cosmic velocity, it forever leaves the zone of attraction of the planet. It can be determined using the law of conservation and transformation of energy - the kinetic energy imparted to the apparatus is spent on overcoming the gravitational attraction of the apparatus to the planet.
According to expression (4): v 2 \u003d (2GM / R) 1/2 \u003d 4.1 10 5 m / s.
Answers: v 1 \u003d 2.9 10 5 m / s.
v 2 \u003d 4.1 10 5 m / s.
Example 4: Determine the angular diameter of Jupiter α at the moment of closest approach of the Earth and Jupiter
(in radians and minutes of arc) .
Solution: In the figure: D=2R – diameter of Jupiter;
r \u003d r South-N - r W-N - the distance of closest approach of the Earth and Jupiter; α is the angular diameter of Jupiter.
It is easy to get from the figure: (2R /2)/r = tg(α/2)≈ α/2 and:
α \u003d 2R / (r S-N - r W-N)).
Radius of Jupiter R = 71398 km and distances Jupiter-Sun r S-N = 778.3 million km and Earth-Sun
r W-N = 149.6 million km we take from table 1.
α \u003d 2 71398 10 3 / [(778.3– 149.6) 10 9] \u003d 0.2275 10 -3 rad.
Considering that π=3.14 rad corresponds to 180 60 minutes of arc, it is easy to get that
α \u003d 0.2275 10 -3 rad. \u003d 0.7825΄.
Answer: α \u003d 0.2275 10 -3 rad. \u003d 0.7825΄.
Task conditions.
1. Determine the first and second cosmic velocities on the surface of the Sun.
2. Determine the first and second cosmic velocities on the surface of Mercury.
3. Determine the first and second cosmic velocities on the surface of Venus.
4. Determine the first and second cosmic velocities on the surface of Mars.
5. Determine the first and second cosmic velocities on the surface of Jupiter.
6. Determine the first and second cosmic velocities on the surface of Saturn.
7. Determine the first and second cosmic velocities on the surface of Uranus.
8. Determine the first and second cosmic velocities on the surface of Neptune.
9. Determine the first and second cosmic velocities on the surface of Pluto.
10. Determine the first and second cosmic velocities on the surface of the moon.
11. Determine the length of the year on Mars.
12. Determine the length of the year on Mercury.
13. Determine the length of the year on Venus.
14. Determine the length of the year on Jupiter.
15. Determine the length of the year on Saturn.
16. Determine the length of the year on Uranus.
17. Determine the length of the year on Neptune.
18. Determine the length of the year on Pluto.
19. The period of rotation of two stars with masses m 1 = 2 10 32 kg and m 2 = 4 10 34 kg around a common center of mass is 3.8 years. What is the distance between the stars?
20. The period of rotation of two stars with masses m 1 =2 10 30 kg and m 2 =4 10 31 kg around a common center of mass is 4.6 years. What is the distance between the stars?
21. Two stars at a distance of r = 7 10 13 m rotate around a common center of mass with a period equal to T = 7.2 years. What is the mass of one of the stars m 1 if the mass of the second star m 2 is 4 10 32 kg?
22. Two stars at a distance of r = 5 10 10 m rotate around a common center of mass with a period equal to T = 12 years. What is the mass of one of the stars m 1 if the mass of the second star m 2 is 8 10 33 kg?
23. Determine the apparent angular diameters of Neptune at the moments of greatest
and closest approaches of Earth and Neptune.
24. Determine the apparent angular diameters of Mars at the moments of greatest
and closest encounters between Earth and Mars.
25. Determine the apparent angular diameters of Venus at the moments of greatest
and closest approaches of Earth and Venus.
26. Determine the apparent angular diameters of Saturn at the moments of the greatest and least approaches of the Earth and Saturn.
27. The period of revolution of the minor planet Ceres around the Sun is 4.71 Earth years, and Mars - 1.88 Earth years. What is the average distance from the Sun to Ceres?
28. The period of revolution of the minor planet Pallas around the Sun is 4.6 Earth years, and Venus is 227.7 Earth days. What is the average distance from the Sun to Pallas?
29. A supernova exploded in a galaxy with a redshift in the spectrum corresponding to a removal velocity of 20,000 km/s. Determine the distance to this star.
30. A globular star cluster is located at a distance of 320 Mpc from us. How fast is it moving away from us?
4.2. INTERACTIONS
Basic formulas and laws.
1. The law of universal gravitation F = G m 1 m 2 / r 2 (1),
where m 1 and m 2 are the masses of interacting bodies,
r is the distance between them,
G \u003d 6.6726 10 -11 m 3 / (kg s 2) - gravitational constant.
2. When a bunch of matter of mass m rotates around a central body of mass M, the bunch decay (its fragmentation) begins when the centrifugal force acting on the bunch begins to exceed the gravitational force between the bunch and the central body, i.e., when
m ω 2 r≥ G m M / r 2 (2).
3. Coulomb's law: F = k q 1 q 2 /(ε r 2) (3) ,
where k \u003d 1 / (4πε 0) \u003d 9 10 9 N m 2 / Kl 2; ε 0 \u003d 8.85 10 -12 C 2 / (N m 2) - electrical constant; ε is the dielectric constant of the substance; q 1 and q 2 - electric charges of interacting bodies; r is the distance between them.
4. Ampere force: F A \u003d I B ℓ sinα (4),
where I is the current strength in a conductor of length ℓ, located in a magnetic field with induction B; α is the angle between the direction of the current (vector ℓ ) and vector AT .
5. Lorentz force: F L \u003d q B v sinα (5),
where q is the electric charge of a particle flying into a magnetic field with induction B at a speed v at an angle α to the induction vector AT.
6. Equation of motion of a charged particle of mass m and charge q in an electric field of intensity E:
m a= q E (6)
Examples of problem solving
Example 1: Determine how many times the force of attraction on Earth is greater than the force of attraction on Mars.
Solution: According to formula (1), the force of attraction to the Earth of a body of mass m:
F Z \u003d G m M Z / R Z 2,
where МЗ and RЗ are the mass and radius of the Earth, respectively.
Similarly, for the force of gravity on Mars:
F M \u003d G m M M / R M 2.
Dividing these two equalities one by the other, we obtain after reducing the same values:
F Z / F M \u003d M Z R M 2 / (R Z 2 M M).
Let's take the values of the masses and radii of the planets from Table 1.
M H = 5.976 10 24 kg; R W \u003d 6371 km \u003d 6.371 10 6 m;
M M \u003d 0.6335 10 24 kg; R M \u003d 3397 km \u003d 3.397 10 6 m.
Substituting, we get:
F Z / F M \u003d (5.976 10 24 / 0.6335 10 24) (3.397 10 6 / 6.371 10 6) 2 \u003d 2.7
Answer: 2.7 times.
Example 2: When flying to Venus, the spacecraft passes a point where the forces of attraction of the apparatus to the Earth and to Venus mutually compensate each other. How far from the Earth is this point? When calculating, neglect the action of all other cosmic bodies. Assume that the Earth and Venus are at the minimum distance from each other.
Solution: The sum of the gravitational forces to the Earth and to Venus must be equal to zero, otherwise, the modules of these forces must be equal: F G = F B:
G m M Z / r Z 2 \u003d G m M V / r V 2 (I),
where МЗ and МВ are the masses of the Earth and Venus, respectively, and
r W and r B are the distances of the spacecraft with mass m from the Earth and from Venus, respectively. We take into account that
r B = R SV - r S, where R SV is the distance from the Earth to Venus, which is equal to R S - R SV - the difference between the distances Earth-Sun R SV and Venus-Sun R SV. Substitute everything in expression (I):
M Z / r Z 2 \u003d M B / (R ZS - R BC - r Z) 2,
from where we can easily get the answer:
r W \u003d (R ZS - R VS) / (1 + 
) .
Distances and masses are taken from Table 1.
M Z \u003d 5.976 10 24 kg; M B \u003d 4.8107 10 24 kg; R ZS = 149.6 million km; R BC \u003d 108.2 million km.
r W \u003d (R ZS - R VS) / (1 + 
)=
(149,6-108,2)/(1+)=
41.4 / 1.8972 = 21.823 million km
Answer: r W = 21.823 million km.
Example 3: A proton flies at a speed v=5 10 4 m/s into a magnetic field with induction B=0.1mT perpendicular to the lines of force. Define:
A) the radius of the circle described by the proton;
C) proton revolution period;
Solution: A charged particle flying into a magnetic field perpendicular to the lines of force moves in a circle.
Its motion is described by the equation of motion:
m v 2 /r = q v B.
From this relation it is easy to obtain an expression for the radius r= m v/(q B) (I).
If we take into account that the rotation speed v is related to the period Т by the relation: v=2π r/T, then from (I) we obtain r=2π r m/(T q B), whence the rotation period is equal to:
T \u003d m 2π / (q B) (II).
Taking the magnitude of the charge q=1.6 10 -19 C and the mass
m=1.67 10 -27 kg of proton in the table of reference data and substituting them into (I-II), we find:
r \u003d 1.67 10 -27 5 10 4 / (1.6 10 -19 0.1 10 -3) \u003d 5.22m.
T \u003d 1.67 10 -27 6.28 / (1.6 10 -19 0.1 10 -3) \u003d 6.55 s.
r = 5.22m. T \u003d 6.55 s.
Task Conditions
31. How many times do the forces of attraction of the Earth to Jupiter and to the Sun differ at the time when the Earth is on a straight line connecting the centers of Jupiter and the Sun?
32. How many times do the forces of attraction of the Earth to Saturn and the Sun differ at the time when the Earth is on a straight line connecting the centers of Saturn and the Sun?
33. Determine at what point (counting from the Earth) on the straight line connecting the centers of the Earth and the Sun the rocket must be located so that the resulting forces of attraction of the Earth and the Sun are equal to zero.
34. With what acceleration does the Earth “fall” on the Sun as it moves around the Sun?
35. Determine at what point (counting from the Earth) on a straight line connecting the centers of the Earth and the Moon should be a rocket. that the resultant forces of attraction of the Earth and the Moon be equal to zero.
36. How many times do the forces of attraction of the Moon to the Earth and to the Sun differ at the time when the Moon is on a straight line connecting the centers of the Earth and the Sun?
37. How many times is the force of electrostatic repulsion of two protons located at a certain distance greater than their gravitational attraction?
38. How many times is the force of electrostatic repulsion of two α-particles located at a certain distance greater than their gravitational attraction?
39. Around a massive star with a mass of M=4 10 23 kg, a bunch of matter rotates at a distance of 10 6 km. At what angular velocity does fragmentation (breaking into parts) of the bunch begin?
40. Around a massive star with a mass of M=4 10 25 kg, a bunch of matter rotates at a distance of 10 7 km. At what angular velocity does fragmentation (breaking into parts) of the bunch begin?
41. Around a massive star with a mass of M=4 10 24 kg, a bunch of matter rotates at a speed of 100 m/s. Determine the distance between the star and the bunch at which fragmentation (breaking into parts) of the bunch occurs.
42. Two bodies with equal negative electric charges repel each other in the air with a force of 5 microns. Determine the number of excess electrons in each body if the distance between charges is 5 cm.
43. A charge equal to q 1 \u003d 2 μC is placed in a medium with a permittivity ε \u003d 2 at a distance of 8 cm from another charge q 2. Determine the sign and magnitude of the charge q 2 if the charges are attracted with a force F=0.5mH.
44. Two point electric charges interact in air at a distance of r 1 = 3.9 cm with the same force as in a non-conductive liquid at a distance of r 2 = 3 cm. What is the dielectric constant of the liquid ε.
45. A proton is accelerated by an electric field with a strength of E \u003d 2000 V / m.
With what acceleration is the particle moving?
46. A charged body with mass m=10mg and charge q=2μC moves in an electric field with an acceleration a=20m/s 2 . What is the electric field strength?
47. At what angle α to the lines of induction of a uniform magnetic field should a conductor with an active length be located 
\u003d 0.2 m, through which a current flows with a force of I \u003d 10A, so that a field with an induction of B \u003d 10 μT acts on a conductor with a force of F \u003d 10 μN?
48. Determine the length of the active part of a rectilinear conductor placed in a uniform magnetic field with an induction V = 1mTl at an angle α = 60 0 to the induction lines, if at a current strength I = 8A the conductor is acting
the force is F=2mN.
49. Determine the force acting from a uniform magnetic field with induction B = 0.1mTl, on a conductor with a length 
\u003d 0.4 m, through which current flows with a force of I \u003d 100 A and which is located at an angle α \u003d 45 0 to
lines of induction.
50. An electron flies into a uniform magnetic field with induction B=0.1mT at a speed v=5 10 6 m/s perpendicular to its lines of induction. Define
the radius of the circle along which the particle is moving.
51. α-particle flies into a uniform magnetic field with an induction B=100mkT at a speed v=3 10 5 m/s perpendicular to the lines of force. Determine the maximum force acting on the particle from the side of the field.
52. A proton and an α-particle fly into a uniform magnetic field with induction B=2mT perpendicular to its lines of induction. Determine the periods of revolution of these particles in a magnetic field
53. According to Bohr's theory, the hydrogen atom consists of a proton and an electron revolving around the proton in a circular orbit. The radius of the Bohr orbit in a hydrogen atom is 0.53·10 -10 m. What is the speed of an electron in an atom?
54. A proton flies into an electric field with a strength of 200 V / m in the direction of the lines of force with an initial speed v 0 = 3 10 5 m / s. Determine the momentum of the proton after 5 sec.
55. A particle with an electric charge q = 0.1 μC flies into a uniform magnetic field with an induction B = 0.1 mT perpendicular to its lines of force with a speed v = 3 10 3 m / s. With what force does the magnetic field act on the particle?
56. How many times does the force of attraction on Jupiter differ from the force of attraction on the Sun?
57. What is the mass of a star if its radius is 100 times greater than the earth's, and the force of attraction on its surface is 80 times greater than the same force on Earth?
58. What is the mass of a star if its radius is 1000 times greater than the Martian one, and the force of attraction on its surface exceeds the same force on Mars by 5 times?
59. How many times does the force of attraction on Jupiter differ from the force of attraction on Saturn?
60. What is the mass of a star if its radius is 500 times greater than the radius of Venus, and the force of attraction on its surface exceeds the same force on Venus by 7 times?
4.3. LAWS OF CONSERVATION OF MOMENTUM,
TORQUE AND MECHANICAL ENERGY
Basic formulas and laws
1. p \u003d m v - the momentum of the body - the characteristic of the act
body movement..
2. The law of conservation of momentum: the total momentum of a closed system of bodies is preserved: Σ i p i =const.
3. L=I ω=r p sinα - angular momentum - characteristic of rotational motion.
I is the moment of inertia of the body, ω is its angular velocity.
4. The law of conservation of angular momentum: the total angular momentum of a closed system of bodies is preserved:
Σ i L i = const.
5. E K \u003d m v 2 / 2 - the kinetic energy of the body - the energy of translational motion.
E K = I ω 2 /2 is the kinetic energy of a body rotating about a fixed axis.
E K = m v 2 /2 + I ω 2 /2 is the kinetic energy of the rolling body.
6. Е Р =f(r) – potential energy of the body; depends on the position of the body in relation to other bodies.
E P =G m 1 m 2 /r is the energy of the gravitational interaction of two bodies;
E P =m g h-potential energy of the body in the Earth's gravity field;
Е Р = к Δх 2 /2 potential energy of an elastically deformed body
(k is the coefficient of elasticity (stiffness));
E R \u003d k q 1 q 2 / (ε r) is the energy of the electrostatic interaction of charged bodies, where
k \u003d 1 / (4πε 0) \u003d 9 10 9 N m 2 / Kl 2; ε 0 \u003d 8.85 10 -12 C 2 / (N m 2) - electrical constant;
7. The law of conservation of mechanical energy: the total mechanical energy E of a closed system of bodies is preserved: E=Σ i (E K + E R) i = const.
If the system is not closed, then work is done against external forces, or work on the system is done by external forces. Both of these cases lead to a change in the total energy of the system: A=ΔE.
8. А=F s cosα – work of force F .
А=q Δφ=ΔU is the work of moving an electric charge q by an electric field (U = E P is the potential energy of a charge in an electric field; φ is the potential of a given point of the field; Δφ and ΔU are the potential differences and potential energies of two points of the field).
Examples of problem solving
Example 1: What is the mass of a particle carrying an electric charge q = 1 μC, if in an electric field with a potential difference Δφ = 100V its speed changed from v 1 = 100 m / s to v 2 = 300 m / s?
Solution: The work of the electric field forces leads to a change in the kinetic energy of the particle: A \u003d ΔE K or
q Δφ \u003d m v 2 2 /2 - m v 1 2 /2.
From this expression we get:
m \u003d 2 q Δφ / (v 2 2 -v 1 2) \u003d 2 10 -6 100 / (300 2 -100 2) \u003d 2.5 10 -9 kg.
Answer: m=2.5 10 -9 kg.
Example 2: What speed will two identical particles acquire, located at a distance r 1 \u003d 1 cm and having a mass m \u003d 1 mg and an electric charge q \u003d 2 μC each, when they expand to a distance r 2 \u003d 5 cm?
Solution: At the initial moment of time, the total energy E 1 of a system of two particles is the potential energy of their electrostatic repulsion:
E 1 \u003d to q 1 q 2 / r \u003d to q 2 / r 1.
At a distance r 2, the total energy E 2 is the sum of the potential energy of the electrostatic interaction and the kinetic energies of the particles:
E 2 \u003d k q 2 / r 2 + 2 m v 2 / 2.
In accordance with the law of conservation of energy: E 1 \u003d E 2, that is
to q 2 /r 1 = to q 2 /r 2 + 2 m v 2 /2.
From this expression it is easy to get:
v= 
Let's substitute the values: r 1 \u003d 1 cm \u003d 0.01 m; r 2 \u003d 5 cm \u003d 0.05 m; m=1mg=10 -6 kg; k \u003d 9 10 9 N m 2 / Cl 2; q \u003d 2 μC \u003d 2 10 -6 C and we get v \u003d 1.7 10 3 m / s.
Answer: v=1.7 10 3 m/s.
Example 3: A platform with sand with a total mass of M = 1000 kg stands on rails on a horizontal section of the track. A projectile hits the sand and gets stuck in it. At the moment of hitting the platform, the speed of the projectile was v 1 =200m/s and was directed from top to bottom at an angle α =60 0 to the horizon. Determine the mass of the projectile m if, as a result of the hit, the platform began to move at a speed v 2 = 0.5 m/s.
Solution: For horizontal x-components of impulses, the momentum conservation law can be applied.
Before impact, the projectile momentum p 1x =m v 1 cosα; platform momentum p 2x =0; and the resulting x-component of the projectile-platform momentum is:
p 1x + p 2x \u003d mv 1 cosα.
After impact, the momentum of the platform and projectile P x =(m+M) v 2 . According to the law of conservation of momentum:
p 1x + p 2x \u003d P x or m v 1 cosα \u003d (m + M) v 2.
From this expression we finally get:
m \u003d M v 2 / (v 1 cosα -v 2) \u003d 1000 0.5 / (200 0.5 - 0.5) \u003d 5.02 kg
Answer: m = 5.02 kg.
Example 4: A homogeneous thin rod of mass M=200 g and length ℓ=50 cm can freely rotate in a horizontal plane about a vertical axis passing through the center of the rod. A plasticine ball with a mass m=10 g, flying horizontally and perpendicular to the rod, falls into one of the ends of the rod and sticks to it, as a result of which the rod begins to rotate with an angular velocity ω=3 rad/s. Determine the speed of the plasticine ball at the moment of impact.
Solution: According to the law of conservation of angular momentum, the sum of the momentum of the rod and the ball before the impact must be equal to their sum after the impact.
Before the impact: the angular momentum of the ball relative to the axis of rotation of the rod at the moment of impact L 1 = m v (ℓ/2); angular momentum of the rod L 2 =0.
After the impact: the momentum of the rod and the ball is equal to
L \u003d (I 1 +I 2) ω,
where I 1 \u003d m (ℓ / 2) 2 is the moment of inertia of a ball with mass m and I 2 \u003d M ℓ 2 /12 is the moment of inertia of a rod with mass M relative to the axis of rotation, respectively.
Thus, L 1 + L 2 =L or
m v (ℓ/2) =(I 1 +I 2) ω= ω.
It follows from this expression that: v=ℓ ω /2.
Substituting ℓ=0.5m; ω=3 rad/s; m=0.01kg; M=0.2kg, we get v=5.75m/s.
Answer: v=5.75m/s.
Example 5: When a star of radius R 1 =10 6 km, slowly rotating at the speed of points on the surface v 1 =10m/s, turns into a neutron star (pulsar), its radius decreases N=10 5 times. What will be the period T of pulses of electromagnetic radiation of the pulsar?
Solution: The period of the pulsar radiation pulses will be equal to its period of revolution around its own axis, which can be determined using the law of conservation of angular momentum: I 1 ω 1 = I 2 ω 2, where I 1 =2 M R 1 2 /5 is the moment of inertia of the stellar ball of radius R 1 and mass M; ω 1 \u003d v 1 / R 1 - angular velocity of rotation of the star; I 2 \u003d 2 M R 2 2 / 5 is the moment of inertia of a neutron star of radius R 2 and mass M; ω 2 = 2π/T is the angular velocity of rotation of the neutron star; Thus, we can write:
2 M R 1 2 v 1 / (5 R 1) \u003d 2 M R 2 2 2π / (5 T)
and after reductions and taking into account that: N= R 1 /R 2 , we get:
T \u003d 2π R 1 / (v 1 N 2) \u003d 0.0628 s.
Answer: T \u003d 0.0628s.
Example 6: A wagon with a mass m=12t stopped, having run into a spring buffer and compressing the buffer spring by Δх=4cm. Determine the speed of the car if the stiffness of the spring k = 4 10 8 N/m.
Solution: We apply the law of conservation and transformation of energy: the kinetic energy of the car is converted into the potential energy of a compressed spring:
m v 2 /2 = to Δх 2 /2,
from where we get:
v=Δх 
=4 10 -2 
\u003d 7.3 m / s.
Answer: v \u003d 7.3 m / s.
Example 7: What is the kinetic energy of a ball of mass m = 8.55 kg, which rolls without slipping at a speed v = 5 m / s?
Solution: In the absence of slippage v=ω r or
ω = v/r; moment of inertia of the ball I=2 m R 2 /5. Substituting these expressions, and then numerical data, into the formula for the kinetic energy of a rolling ball:
E K \u003d m v 2 / 2 + I ω 2 / 2 \u003d m v 2 / 2 + m v 2 / 5 \u003d 0.7 m v 2,
we get E K \u003d 150 J.
Answer: E K \u003d 150 J.
Task Conditions
61. A particle with an electric charge q=2 μC and mass m=3 10 -6 kg flies into a uniform electric field along the line of tension with a speed v 1 =5 10 4 m/s. What potential difference must a particle pass in order for its speed to increase to v 2 = 10 5 m/s?
62. What speed can a particle with mass m=2 10 -8 kg and electric charge q=2 10 -12 C, at rest, be informed by an accelerating potential difference of U=100 V?
63. What work is required to be done in order to bring two electric charges q 1 \u003d 2 μC and q 2 \u003d 4 μC, located at a distance r 1 \u003d 1.2 m, closer to
distance r 2 \u003d 0.4 m?
64. Two point electric charges q 1 \u003d 3 μC and q 2 \u003d 5 μC are at a distance r 1 \u003d 0.25 m. How much will the interaction energy of these charges change if they are brought closer to a distance r 2 \u003d 0.1 m?
65. A platform with sand with a total weight of M = 1000 kg stands on rails on a horizontal section of the track. A projectile of mass m=10 kg hits the sand and gets stuck in it. Neglecting friction, determine how fast
the platform will move if at the moment of impact the projectile speed v = 200 m/s, and its direction is from top to bottom at an angle α 0 = 30 to the horizon.
66. A projectile weighing m = 20 kg at the top of the trajectory had a speed v = 250 m / s. At this point, it broke into two pieces. A smaller part with a mass m 1 =5kg received a speed u 1 =300m/s in the same direction. Determine the speed of the second, larger part of the projectile after the break.
67. A projectile with a mass of m = 20 kg at the top of the trajectory had a speed of v = 300 m / s. At this point, it broke into two pieces. Most of the projectile weighing m 1 =15kg received a speed u 1 =100m/s in the same direction. Determine the speed of the second, smaller part of the projectile after the break.
68. A bullet weighing m=10g, flying horizontally at a speed v=250m/s, hit a wooden ball hanging on a thread weighing M=1kg and got stuck in it. To what height did the ball rise after the impact?
69. A bullet weighing m=10g, flying horizontally at a speed v=250m/s, hit a wooden ball hanging on a thread weighing M=1.5kg and got stuck in it. At what angle did the ball deviate as a result of this height?
70. A bullet of mass m = 15g, flying horizontally, hit a wooden ball hanging on a thread with mass M = 2.5kg and got stuck in it. As a result, the ball deviated by an angle equal to 30 0 . Determine the speed of the bullet.
71. A bullet of mass m=10g, flying horizontally at a speed of v=200m/s, hit a wooden ball hanging on a thread and got stuck in it. What is the mass of the ball if the ball, having been pumped out after the impact, has risen to a height h = 20 cm?
Mass - one of the most important physical characteristics of stars - can be determined by its effect on the motion of other bodies. Such other bodies are the satellites of some stars (also stars), circulating with them around a common center of mass.
If you look at Ursa Major, the second star from the end of the "handle" of her "ladle", then with normal vision you will see a second faint star very close to it. She was noticed by the ancient Arabs and called Alcor (Horseman). They named the bright star Mizar. They can be called a double star. Mizar and Alcor are separated from each other by . With binoculars, you can find a lot of such stellar pairs. So, Lyra consists of two identical stars of the 4th magnitude with a distance of 5 between them.
Rice. 80. The orbit of the satellite of a double star (v Virgo) relative to the main star, whose distance from us is 10 pc. (The points mark the measured positions of the satellite in the indicated years. Their deviations from the ellipse are due to observational errors.)
Binary stars are called visual binaries if their duality can be seen through direct telescope observations.
In the Lyra telescope - a visual quadruple star. Systems with a number of stars are called multiple.
Many of the visual binaries turn out to be optical binaries, that is, the proximity of such two stars is the result of their random projection onto the sky. In fact, they are far apart in space. And during long-term observations, one can be convinced that one of them passes by the other without changing direction at a constant speed. But sometimes, when observing stars, it turns out that a weaker companion star revolves around a brighter star. The distances between them and the direction of the line connecting them systematically change. Such stars are called physical binary, they form a single system and circulate under the action of forces of mutual attraction around a common center of mass.
Many double stars were discovered and studied by the famous Russian scientist V. Ya. Struve. The shortest known orbital period for visual binary stars is 5 years. Pairs with circulation periods of tens of years have been studied, and pairs with periods of hundreds of years will be studied in the future. The closest star to us, a Centauri, is a double star. The circulation period of its constituents (components) is 70 years. Both stars in this pair are similar in mass and temperature to the Sun.
The main star is usually not in the focus of the visible ellipse described by the satellite, because we see its orbit in a distorted projection (Fig. 80). But knowledge of geometry makes it possible to restore the true shape of the orbit and measure its semi-major axis a in seconds of arc. If the distance to the binary star is known in parsecs and the semi-major axis of the orbit of the satellite star in seconds of arc, equal to then in astronomical units (since it will be equal to:
The most important characteristic of a star, along with its luminosity, is its mass. Direct determination of the mass is possible only for binary stars. By analogy with § 9.4, comparing the motion of the satellite
stars with the motion of the Earth around the Sun (for which the period of revolution is 1 year, and the semi-major axis of the orbit is 1 AU), we can write according to Kepler's third law:
where are the masses of the components in a pair of stars, are the masses of the Sun and the Earth, and the orbital period of the pair in years. Neglecting the mass of the Earth in comparison with the mass of the Sun, we get the sum of the masses of the stars that make up the pair in the masses of the Sun:
To determine the mass of each star separately, it is necessary to study the motion of each of them relative to the surrounding stars and calculate their distances from the common center of mass. Then we have the second equation:
To and from the system of two equations we find both masses separately.
Double stars in a telescope are often a beautiful sight: the main star is yellow or orange, and the satellite is white or blue. Imagine a wealth of colors on a planet revolving around one of a pair of stars, where the red Sun shines in the sky, then the blue one, then both together.
The masses of stars determined by the described methods differ much less than their luminosities, approximately from 0.1 to 100 solar masses. Large masses are extremely rare. Usually stars have a mass less than five solar masses. We see that from the point of view of luminosity and temperature, our Sun is an ordinary, average star, nothing special stands out.
(see scan)
2. Spectral binary stars.
If the stars in mutual circulation come close to each other, then even in the most powerful telescope they cannot be seen separately, in this case the duality can be determined from the spectrum. If the plane of the orbit of such a pair almost coincides with the line of sight, and the rotation velocity is high, then the velocity of each star in the projection onto the line of sight will change rapidly. In this case, the spectra of binary stars are superimposed on each other, and since the difference in the velocities of these
Rice. 81. Explanation of the bifurcation, or fluctuation, of lines in the spectra of spectral binary stars.
stars is large, then the lines in the spectrum of each of them will shift in opposite directions. The shift value changes with a period equal to the period of rotation of the pair. If the brightness and spectra of the stars that make up the pair are similar, then a periodically repeating splitting of spectral lines is observed in the spectrum of the binary star (Fig. 81). Let the components occupy positions, or then one of them moves towards the observer, and the other - away from him (Fig. 81, I, III). In this case, a splitting of the spectral lines is observed. In an approaching star, the spectral lines will shift to the blue end of the spectrum, and in a receding star, to the red. When the components of a binary star occupy positions or (Fig. 81, II, IV), then both of them move at right angles to the line of sight and there will be no bifurcation of the spectral lines.
If one of the stars glows weakly, then only the lines of the other star will be visible, shifting periodically.
One of Mizar's components is itself a spectroscopic binary.
3. Eclipsing binary stars - Algols.
If the line of sight lies almost in the plane of the orbit of a spectral binary, then the stars of such a pair will alternately block each other. During eclipses, the overall brightness of a pair whose components we cannot see individually will weaken (positions B and D in Fig. 82). In the rest of the time, in the intervals between eclipses, it is almost constant (positions A and C) and the longer, the shorter the duration of the eclipses and the greater the radius of the orbit. If the satellite is large, but itself gives little light, then when the bright
the star eclipses it, the total brightness of the system will decrease only slightly.
The brightness minima of eclipsing binary stars occur when their components move across the line of sight. An analysis of the apparent magnitude curve as a function of time makes it possible to determine the size and brightness of the stars, the size of the orbit, its shape and inclination to the line of sight, as well as the masses of the stars. Thus, eclipsing binaries, also observed as spectroscopic binaries, are the most well studied systems. Unfortunately, relatively few such systems are known so far.
Eclipsing binary stars are also called Algols, after the name of their typical representative Perseus. The ancient Arabs called Perseus Algol (spoiled el gul), which means "devil". It is possible that they noticed her strange behavior: for 2 days 11 hours, the brightness of Algol is constant, then in 5 hours it weakens from 2.3 to 3.5 magnitudes, and then in 5 hours its brightness returns to its previous value.
The periods of known spectroscopic binary stars and Algols are mostly short, about a few days. Altogether, stellar binaries are very common Statistics show that up to 30% of all stars are probably binaries Obtaining a variety of data on individual stars and their systems from the analysis of spectroscopic binaries and eclipsing binaries are examples of the unlimited possibility of human cognition
Rice. 82. Changes in the apparent brightness of Lyra and the motion pattern of its satellite (The shape of stars that are close to each other, due to their tidal effect, can differ greatly from spherical)
Conditions of the 1st round and the 2nd round
5-7 grades, 8-9 grades
1. Which of the listed astronomical phenomena - equinoxes, solstices, full moons, eclipses of the Sun, eclipses of the Moon, opposition of planets, maxima of meteor showers, the appearance of bright comets, maxima of the brightness of variable stars, supernovae - occur every year exactly at approximately the same dates (accurate to 1-2 days)?
In crystal dew
even the shadows are rounded,
In Silver River
half moon at the bottom.
Who will bring the news
embroidered brocade with letters?
Furrowed brows,
finally extinguish the candle...
10th grade, 11th grade
1. In 2010, the opposition of Saturn will occur on March 22.
2. In the 20th century, there were 14 transits of Mercury across the solar disk:
II round
5-7 grades, 8-9 grades
10th grade, 11th grade
m, and during the greatest elongation
–4.4m
SOLUTIONS
I round
5-7 grades, 8-9 grades
1. Which of the listed astronomical phenomena - equinoxes, solstices, full moons, eclipses of the Sun, eclipses of the Moon, opposition of planets, maxima of meteor showers, the appearance of bright comets, maxima of the brightness of variable stars, supernovae - occur every year exactly at approximately the same dates (accurate to 1-2 days)?
Solution. Every year, those astronomical phenomena are repeated that are associated only with the movement of the Earth in orbit around the Sun, that is, the equinoxes, solstices and maxima of meteor showers. These phenomena repeat on approximately the same dates, for example, the spring equinox falls on March 20 or 21, since there are leap years in our calendar. In meteor showers, the inaccurate repetition of the dates of maxima is also associated with the drift of their radiants. The rest of the mentioned phenomena either have a periodicity different from the Earth year (full moons, solar eclipses, lunar eclipses, planetary oppositions, variable star brightness maxima), or are generally non-periodic (appearance of bright comets, supernova explosions).
2. In the textbook of astronomy by Belarusian authors A.P. Klishchenko and V.I. Shuplyak, such a scheme of a lunar eclipse is placed. What is wrong with this diagram?
Solution. The moon should be almost three times smaller than the diameter of the earth's shadow at the distance of the lunar orbit. The night side of our satellite, of course, should be dark.
3. Yesterday there was an occultation of the Pleiades star cluster by the Moon. Could there be a solar eclipse tomorrow? Moon eclipse?
Solution. Eclipses occur when the full moon or new moon is near the ecliptic. The Pleiades are located about 5 degrees north of the ecliptic, and the Moon can cover them only when it is at the greatest distance from the nodes of its orbit. It will be near the ecliptic only in a week. Therefore, neither a solar nor a lunar eclipse can happen tomorrow.
4. Here are the lines from the poem of the classical Chinese poet Du Fu "River Moon" (translated by E.V. Balashov):
In crystal dew
even the shadows are rounded,
In Silver River
half moon at the bottom.
Who will bring the news
embroidered brocade with letters?
Furrowed brows,
finally extinguish the candle...
It is not difficult to guess that the Chinese call the Milky Way the Silver River. What month of the year was this observation made?
Solution. So, the "half of the moon" is visible against the background of the Milky Way. Moving near the ecliptic, the Moon crosses the Milky Way twice a month: on the border of Taurus and Gemini and on the border of Scorpio and Sagittarius, that is, near the solstices. "Half of the Moon" can be both growing and aging and located both 90 o to the west of the Sun, and 90 o to the east. In both cases, it turns out that the Sun is located on the ecliptic near the equinoxes. So, the observation was made in March or September.
10th grade, 11th grade
Where on Earth can Saturn be seen at its zenith this year?
What will be the height of Saturn above the horizon at local midnight on March 22 when observed from Moscow (latitude 55 about 45 ')?
Solution. Since the opposition of Saturn almost coincides in time with the spring equinox, the planet itself is in 2010 near the autumnal equinox, that is, on the celestial equator (d=0 o). Therefore, it passes through the zenith for an observable located at the Earth's equator.
On March 22, Saturn will be located on the celestial sphere opposite the Sun, so at local midnight it will be at the top culmination. We apply the formula for calculating the height of the star at the culmination: h \u003d (90 o - f) + d, h \u003d 34 o 15 '.
2. * In the 20th century, there were 14 transits of Mercury across the solar disk:
Why are passages observed only in May and November? Why are November passages much more frequent than May ones?
Solution. An inner planet can be projected for an earthly observer onto the disk of the Sun only when, at the moment of inferior conjunction, it is near the plane of the ecliptic, that is, near the nodes of its orbit. The nodes of Mercury's orbit are oriented in space so that the Earth is on the same line with them in May and November.
Mercury's orbit is essentially elliptical. In November, near the perihelion of its orbit, the planet is closer to the Sun (and farther from the Earth), and therefore it is projected onto the disk of the Sun more often than in May, near aphelion.
3. What is the percentage difference between the amount of sunlight falling on the Moon in the phase of the first quarter and in the phase of the full moon?
Solution. The illumination of the lunar surface is inversely proportional to the square of the distance from the Sun to the Moon. In the phase of the first quarter, the Moon is at a distance of about 1 AU. from the Sun, in the phase of the full moon - an average of 384,400 km further.
4. During the great (perihelion) opposition, the apparent angular diameter of Mars reaches 25”, during the aphelion it is only 13”. Determine the eccentricity of Mars' orbit from these data. The semi-major axis of the orbit of Mars is 1.5 AU, the orbit of the Earth is considered a circle.
Solution. The apparent angular diameter of Mars is inversely proportional to the distance between the Earth and the planet. At aphelion, Mars is located at a distance a m (1+e) from the Sun, at perihelion - at a distance a m (1-e). The distance between Earth and Mars at aphelion and perihelion opposition is related as
(a m (1+e)-1)/(a m (1-e)-1).
On the other hand, this ratio is 25/13. Let's write the equation and solve it for e:
(a m (1+e)-1)/(a m (1-e)-1)=25/13, e=0.1.
II round
5-7 grades, 8-9 grades
1. Can Venus be observed in the constellation Gemini? In the constellation Canis Major? In the constellation of Orion?
Solution. Venus can be observed in the zodiac constellation Gemini. It can also be observed in the northern part of the constellation Orion, since it is only a few degrees south of the ecliptic, and the deviation of Venus from the ecliptic can be up to 8 °. Venus was visible in the constellation Orion in August 1996. In the constellation Canis Major, far from the ecliptic, Venus cannot be.
2. The star rose at 00:01 local time. How many times will it cross the horizon at this point on that day?
Solution. A sidereal day, equal to the period of rotation of the Earth relative to the fixed stars, is slightly shorter than the solar day and is approximately 23 hours 56 minutes. Therefore, this star will have time to go beyond the horizon during this day and rise again at 23 hours 57 minutes local time, that is, it will cross the horizon twice more (unless, of course, the star does not go back beyond the horizon in the remaining three minutes).
3. Explain why, whatever the magnification of the telescope, we cannot see the disks of distant stars through its eyepiece.
Solution. The minimum angular size of an object visible through a telescope (its "resolving power") is determined by the size of the lens and the properties of the earth's atmosphere through which the star's light passes. The wave nature of light leads to the fact that even a completely point source will be visible through a telescope as a disk surrounded by a system of rings. The size of this disk is the smaller, the larger the diameter of the telescope objective, but even for large telescopes it is about 0.1 arc second. In addition, the image is blurred by the Earth's atmosphere, and the sizes of the "shudder disks" of stars are rarely less than one arc second. The true angular diameters of distant stars are much smaller, and we cannot see them with a telescope, no matter what magnification we use.
4. Describe the view of the starry sky from one of the Galilean satellites of Jupiter. Will it be possible to see the Earth and the Moon separately with the naked eye?
Solution. The main luminaries in the sky of the Galilean satellites of Jupiter will be the Sun and Jupiter itself. The sun will be the brightest luminary in the sky, although it will be much weaker and smaller than on Earth, since Jupiter and its satellites are 5 times farther from the Sun than our planet. Jupiter, on the contrary, will have huge angular dimensions, but it will still shine weaker than the Sun. In this case, Jupiter will be visible only from half of the satellite's surface, remaining motionless in the sky, since all Galilean satellites, like the Moon to the Earth, are turned to Jupiter on one side. In its movement across the sky, the Sun will set behind Jupiter on each revolution, and solar eclipses will occur, and only when observed from the most distant satellite, Callisto, an eclipse may not occur.
In addition to the Sun and Jupiter, other satellites of this planet will be clearly visible in the sky, during oppositions with the Sun it is very bright (up to -2 m) will be Saturn, and other, more distant planets of the solar system will become a little brighter: Uranus, Neptune and Pluto. But the planets of the terrestrial group will be seen worse, and the point is not so much in their brilliance, but in a small angular distance from the Sun. So, our Earth will be an inner planet, which, even during the greatest elongation, will move away from the Sun by only 11 ° . However, this angular distance may be sufficient for observations from the surface of Jupiter's satellite, which is devoid of a dense atmosphere that scatters the light of the Sun. During the greatest elongation, the distance from the Jupiter system to the Earth will be
Here a and a 0 - radii of the orbits of Jupiter and the Earth. Knowing the distance from the Earth to the Moon (384400 km), we get the maximum angular distance between the Earth and the Moon, equal to 1 ¢ 43.8² , which in principle is sufficient for their resolution with the naked eye. However, the brightness of the Moon at this moment will be +7.5 m, and it will not be visible to the naked eye (the brightness of the Earth will be about +3.0 m). The Earth and Moon will be much brighter near upper conjunction with the Sun (–0.5 m and +4.0 m respectively), but at this time it will be difficult to see them in the rays of the daylight.
10th grade, 11th grade
1. How will the pendulum clock delivered from Earth to the surface of Mars go?
Solution. Acceleration of free fall on the surface of the planet g equals
where M and R - the mass and radius of the planet. The mass of Mars is 0.107 of the mass of the Earth, and its radius is 0.533 of the radius of the Earth. As a result, the free fall acceleration g on Mars is 0.377 of the same value on Earth. Clock oscillation period T with pendulum length l equals
and the pendulum clock on Mars will run 1.629 times slower than on our planet.
2. Suppose that today the Moon in the phase of the first quarter covers the star Aldebaran (a Taurus). What season is it now?
2 Decision. The star Aldebaran is located near the ecliptic in the constellation Taurus. The sun passes through this region of the sky in late May - early June. The Moon in its first quarter phase is 90 degrees from the Sun.° to the east and is located in that place in the sky where the Sun will come in three months. Therefore, now the end of February - the beginning of March.
3. The brightness of Venus at the time of superior conjunction is -3.9 m, and during the greatest elongation –4.4 m. What is the brightness of Venus in these configurations as seen from Mars? The distance from Venus to the Sun is 0.723 AU, and from Mars to the Sun 1.524 AU.
3 Solution The phase of Venus is 1.0 at superior conjunction and 0.5 at greatest elongation whether we are observing from Earth or Mars. Thus, we only need to calculate how much the distance to Venus will change in one configuration or another if the observation point moves from Earth to Mars. Denote by a 0 is the radius of the orbit of Venus, and through a is the radius of the orbit of the planet from which observations are made. Then the distance to Venus at the time of its superior conjunction will be equal to a+a 0 , which is 1.723 a.u. for the Earth and 2.247 a.u. for Mars. Then the magnitude of Venus at the time of superior conjunction on Mars will be
m 1 =–3.9 + 5 lg (2.247/1.723) = –3.3.
The distance to Venus at the moment of greatest elongation is
and is 0.691 a.u. for the Earth and 1.342 a.u. for Mars. The magnitude of Venus at its greatest elongation is
m 2 = –4.4 + 5 lg (1.342/0.691) = –3.0.
Interestingly, Venus shines on Mars (like Mercury on Earth) at its greatest elongation is weaker than at superior conjunction.
4. A binary system consists of two identical stars with a mass of 5 solar masses, circulating in circular orbits around a common center of mass with a period of 316 years. Will it be possible to resolve this pair visually with a TAL-M telescope with an objective diameter of 8 cm and an eyepiece magnification of 105 X if the distance to it is 100 pc?
4 Decision. Let's determine the distance between the stars according to the III generalized Kepler's law:
Here a- semi-major axis of the orbit (equal to the distance between the stars in the case of a circular orbit), T- circulation period, and M- the total mass of two bodies. Let's compare this system with the Sun-Earth system. The total mass of the two stars is 10 times the mass of the Sun (the mass of the Earth makes a negligible contribution), and the period exceeds the period of the Earth's revolution by 316 times. As a result, the distance between the stars is 100 AU. From a distance of 100 pc these two stars will be visible no more than 1² from each other. It will not be possible to resolve such a close pair with the TAL-M telescope, no matter what magnification we use. It is easy to verify this by calculating the size of the diffraction disks of these stars using the well-known formula for green-yellow rays:
where D is the lens diameter in centimeters. Here we have not taken into account the influence of the earth's atmosphere, which will further aggravate the picture. So, this pair will be visible in the TAL-M telescope only as a single star.