Instructions for completing the test. Instructions for completing the test The period of revolution of stars around a common center of mass
Masses of stars. As we saw from the example of the Sun, the mass of a star is the most important characteristic on which physical conditions in its depths. Direct definition mass is possible only for double stars.
Binary stars are called visual binaries if their duality can be seen by direct observation through a telescope.
An example of a visual double star, visible even to the naked eye, is c, Ursa Major, the second star from the end of the “handle” of her “bucket”. 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 the bright star a name Mizar. Mizar and Alcor are 11" apart in the sky." You can find many such star pairs through binoculars.
Systems with the number of stars n≥3 are called multiples. So, through binoculars you can see that ε Lyrae consists of two identical stars of the 4th magnitude with a distance between them of 3". When observed through a telescope, ε Lyrae is a visually quadruple star. However, some stars turn out to be only optical-dual, that is, the closeness of such two stars is the result of their random projection onto the sky. In fact, in space they are far from each other. If, when observing stars, it turns out that they form unified system and revolve under the influence 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 of visual binary stars are several years. Pairs with 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, Centauri, is a double star. The period of circulation 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 distorted in the projection (Fig. 73). But knowledge of geometry makes it possible to restore the true shape of the orbit and measure its semimajor axis a in arcseconds. If the distance D to the binary star is known in parsecs and the semimajor axis of the satellite star’s orbit in arcseconds is equal to a", then in astronomical units it will be equal to:
since D pc = 1 / p" .
Comparing the motion of the star’s satellite with the motion of the Earth around the Sun (for which the period of revolution T = 1 year, and the semi-major axis of the orbit 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 Earth, and T is the pair’s orbital period in years. Neglecting the mass of the Earth in comparison with the mass of the Sun, we obtain the sum of the masses of the stars that make up the pair, in solar masses:
To determine the mass of each star, it is necessary to study the movement 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 obtain 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 often present a beautiful sight: main star yellow or orange, and the satellite is white or blue.
If the components of a double star come close to each other during mutual rotation, then even with the most powerful telescope they cannot be seen separately. In this case, duality can be defined by a spectrum. Such stars will be called spectroscopic doubles. Due to the Doppler effect, the lines in the spectra of stars will shift in opposite directions (when one star moves away from us, another approaches). The displacement of the lines changes with a period equal to the period of revolution of the pair. If the brightnesses and spectra of the stars that make up the pair are similar, then in the spectrum of a binary star there is a periodically repeating bifurcation of spectral lines(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 bifurcation of the spectral lines is observed. An approaching star will shift its spectral lines toward the blue end of the spectrum, while a receding star will shift toward the red end. When the components of a double 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 bifurcation of the spectral lines will not work.
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 rotation. Such stars are called eclipsing binaries or algols, after their typical representative, β Persei. During eclipses, the overall brightness of the pair, the components of which we do not see individually, will weaken (positions B and D in Fig. 75.) 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 larger the orbital radius. If the satellite is large, but itself gives little light, then when bright Star eclipses it, the total brightness of the system will decrease only slightly.
The ancient Arabs named β Perseus Algolem(corrupted el gul), meaning "devil". It is possible that they noticed its 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 magnitude, and then in 5 hours its brightness returns to its previous value.
Analysis of the curve of changes in apparent stellar magnitude as a function of time makes it possible to determine the size and brightness of 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 binary stars, also observed as spectroscopic binaries, are the most well-studied systems. Unfortunately, relatively few such systems are known so far.
The periods of known spectroscopic binary stars and algols are generally short - about a few days.
In general, stellar duality is a very common phenomenon. Statistics show that up to 30% of all stars are likely binaries.
The masses of stars determined by the methods described differ much less than their luminosities: from approximately 0.1 to 100 solar masses. Very large masses are extremely rare. Stars typically have a mass less than five solar masses.
It is the mass of stars that determines their existence and nature as a special type celestial bodies, which are characterized by high subsoil temperatures (over 10 7 K) - Occurring at such temperatures nuclear reactions The transformation of hydrogen into helium is the source of energy emitted by most stars. With a smaller mass, the temperature inside celestial bodies does not reach the values necessary for thermonuclear reactions to occur.
Evolution chemical composition The transformation of matter in the Universe occurred and is currently occurring mainly due to stars. It is in their depths that the irreversible process of synthesis of heavier chemical elements from hydrogen.
Example of problem solution
Task. A binary star has an orbital period of 100 years. The semimajor axis 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 at distances in a ratio of 1:4.
Exercise 21
1. Determine the sum of the masses of the Capella binary star if the semimajor axis of its orbit is 0.85 AU. e., and the circulation period is 0.285 years.
2. If a star with the same mass as the Sun moved in Earth’s orbit, what would be its orbital period?
2. Sizes of stars. Density of their substance
We'll show you on simple example, how you can compare the sizes of stars of the same temperature, 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 per unit surface of stars is also the same, it means that the surface of Capella is 120 times larger than the surface of the Sun, and its diameter and radius are greater than the solar 
once.
Knowledge of the laws of radiation allows us to determine the sizes of other stars.
Thus, in physics it has been established that the total energy emitted per unit time from 1 m 2 of the surface of a heated body is equal to: i = σT 4, where σ is the proportionality coefficient, and T is the absolute temperature *. The relative linear diameter of stars having a known temperature T is found from the formula
* (The Stefan-Boltzmann 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 per unit surface of the star, r, i, T refer to the Sun, and L= l. From here
within the radii 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 with very high luminosity are called supergiants. Red supergiants turn out to be similar in size (Fig. 76). Betelgeuse and Antares hundreds of times bigger than the sun by diameter. The more distant VV Cepheus is so large that the Solar System with the orbits of the planets up to and including the orbit of Jupiter could fit inside it! Meanwhile, the masses of supergiants are only 30-40 times greater than the Sun. As a result, even the average density of red supergiants is thousands of times less than the density of room air.
With the same luminosity, the sizes of stars are smaller, the hotter these stars are. The smallest ordinary stars are red dwarfs. Their masses and radii are tenths of solar masses, and their average densities are 10-100 times higher than the density of water. Even fewer red dwarfs are white dwarfs - but these are already unusual stars.
Close and bright Sirius (with a radius about twice that of the Sun) has a satellite orbiting it every 50 years. For this binary star, the distance, orbit and masses are well known. Both stars are white and almost equally hot. Consequently, surfaces of the same area emit the same amount of energy from these stars, but the luminosity of the satellite is 10,000 times fainter than Sirius. This means that its radius is √10000= 100 times smaller, i.e. it is almost the same as the Earth. Meanwhile, its mass is almost like that of the Sun! Hence, 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 limit to density is set by the size of atoms, which are systems consisting of a nucleus and an electron shell. At very high temperature in the interior of stars and with the complete ionization of atoms, their nuclei and electrons become independent of each other. Under the colossal pressure of the overlying layers, this “crumb” of particles can be compressed much more strongly than neutral gas. Theoretically, the possibility of the existence, under certain conditions, of stars with a density equal to the density of atomic nuclei is allowed.
We see once again in 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 the conditions that exist inside stars. That's why astronomical observations help the development of the most important physical concepts. For example, Einstein's theory of relativity is of enormous importance for physics. Several consequences follow from it, which can be verified using astronomical data. One of the consequences of the theory is that in a very strong gravitational field, light vibrations should slow down and the lines of the spectrum shift towards the red end, and this shift is greater, the stronger the gravitational field of the star. A red shift was discovered in the spectrum of the moon Sirius. It is caused by action strong field gravity 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.
Example of problem solution
Task. How many times is Arcturus larger than the Sun if the luminosity of Arcturus is 100 and the temperature is 4500 K?
Exercise 22
1. How many times greater luminosity does Rigel have than the Sun if its parallax is 0.0069", and its apparent magnitude is 0.34?
2. What is the average density of a red supergiant if its diameter is 300 times that of the Sun and its mass is 30 times that of the Sun?
5 . A piece of ice with a mass m1 = 5 kg floats in a vertical vessel in water, into which a piece of lead with a mass m2 = 0.1 kg is frozen. What amount of heat must be imparted to this system so that the remaining ice with lead begins to sink? The temperature of the water in the vessel is 0 °C. Specific heat melting of ice is 333 kJ/kg, density of water ρ0=1000 kg/m3, ice ρl=900 kg/m3, lead ρbl=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 10 m long and weighing 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 is 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. A charge of the opposite sign moves around a stationary charge of 10 nC in a circle with a radius of 1 cm. The charge completes one revolution in 2p seconds. Find the charge to mass ratio for a moving charge. Electric 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 Jupiter's revolution around the Sun is 12 times longer than the corresponding period of revolution of the Earth. Assuming the orbits of the planets are 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 melted if 60% of the lost mechanical energy was used to heat it? The temperature of the bullet before impact was equal to 50 ˚С, the melting point of lead was 327 ˚С, the specific heat capacity of lead court = 125.7 J/kg K, the specific heat of fusion of lead l= 26.4 kJ/kg.
t= 50 ˚С t pl = 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 of 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 electrons out of 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 µA |
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 with radius r= 2 cm lies on the river bottom deep h= 3 m. What is the minimum work required to raise the ball to a height N= 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 a radius R = 0.05 nm. What is its speed in this case? Electron mass 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. 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 the stars around the 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 were poured into an aluminum kettle at a temperature t= 20 ˚С and placed on an electric stove with efficiency = 75%. Tile power N= 2 kW, weight 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 a body attracted to the Earth and the Moon with equal force? Accept 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 from a uniform disk with a radius of 105.6 cm, as shown in the figure. Determine the position of the center of mass of the disk with such a cutout.
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 vessel under pressure 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 was 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 IN= 0.52 T directed parallel inclined plane, from which it slides at a constant speed υ =
5 m/s charged body 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 is k = 0,5.
IN= 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 load weighing 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 force of 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. Assuming the orbits of the planets are circular, find the ratio of the linear velocities of motion of the Earth and Jupiter around the Sun υZ: υY. The period of Jupiter's revolution around the Sun is 12 times longer than the corresponding period of revolution of the Earth.
T yu = 12 T h |
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υЗ: υУ – ? | Answer: ≈ 2,3 |
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4. Steam hammer weighing M= 10 t falls from a height h= 2.5 m per iron bar weighing m= 200 kg. How many times must it 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. Specific heat 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 in a vacuum from the surface of titanium, which fall into a uniform magnetic field with induction B = 0.1 T. Find the radius of the circle along which the electrons will begin to move if their speed is perpendicular to the induction lines magnetic field, and the work function of 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 orbital period of Venus around the Sun is T V = 0.615 T W = 224.635 days = 224.635 24 3600 s = 1.941 10 7 s.
Thus,
r = 2/3 =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 stars?
Solution: 1) Let us first determine the position of the center of mass of the system of two stars relative to the first star r 1 (t.C in the figure)
r 1 = (m 1 0 + m 2 r)/(m 1 + m 2) = 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 the expression for the circulation period:
T= 2π r 1/2.
After replacing r 1 we get the answer:
T= 2π r 1/2.
Example 3: What are the first and second escape velocities for a cosmic body weighing 10 30 tons and
with a radius of 8 10 8 km?
Solution: 1) The first escape velocity must be reported spacecraft, so that it turns into an artificial satellite of a cosmic body. According to expression (3): v 1 = (GM/R) 1/2. Substituting the numerical values we get:
v 1 = 1/2 =2.9 10 5 m/s.
2) When the device reaches the second escape velocity, it forever leaves the planet’s gravitational zone. It can be determined using the law of conservation and transformation of energy - kinetic energy, reported to the apparatus, is spent on overcoming the gravitational attraction of the apparatus to the planet.
According to expression (4): v 2 = (2GM/R) 1/2 = 4.1 10 5 m/s.
Answers: v 1 =2.9 10 5 m/s.
v 2 =4.1 10 5 m/s.
Example 4: Determine the angular diameter of Jupiter α at the moment of closest approach between the Earth and Jupiter
(in radians and arcminutes).
Solution: In the figure: D=2R – diameter of Jupiter;
r =r Yu-N – r Z-N - the distance of closest approach of the Earth and Jupiter;
α is the angular diameter of Jupiter.
From the figure it is easy to obtain: (2R /2)/r = tan(α/2)≈ α/2 and:
α = 2R/(r Yu-N – r Z-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 is taken from Table 1.
α = 2 71398 10 3 /[(778.3–149.6) 10 9 ] = 0.2275 10 -3 rad.
Considering that π=3.14 rad corresponds to 180 60 arc minutes, it is easy to obtain that
α = 0.2275 10 -3 rad. = 0.7825΄.
Answer: α = 0.2275 10 -3 rad = 0.7825΄.
1. Determine the first and second escape velocities on the surface of the Sun.
2. Determine the first and second escape velocities on the surface of Mercury.
3. Determine the first and second escape velocities on the surface of Venus.
4. Determine the first and second escape velocities on the surface of Mars.
5. Determine the first and second escape velocities on the surface of Jupiter.
6. Determine the first and second escape velocities on the surface of Saturn.
7. Determine the first and second escape velocities on the surface of Uranus.
8. Determine the first and second escape velocities on the surface of Neptune.
9. Determine the first and second escape velocities on the surface of Pluto.
10. Determine the first and second escape 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 located 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 located 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 the closest approach of the Earth and Neptune.
24. Determine the apparent angular diameters of Mars at the moments of greatest
and the closest approach between Earth and Mars.
25. Determine the apparent angular diameters of Venus at the moments of greatest
and the smallest approaches of Earth and Venus.
26. Determine the apparent angular diameters of Saturn at the moments of greatest and least approach of the Earth and Saturn.
27. Circulation period small planet Ceres's orbit around the Sun is 4.71 Earth years, and that of Mars is 1.88 Earth years. At what average distance from the Sun is Ceres?
28. The period of revolution of the small planet Pallas around the Sun is 4.6 Earth years, and that of Venus is 227.7 Earth days. At what average distance from the Sun is Pallas?
29. In a Galaxy with a red shift in the spectrum corresponding to a removal velocity of 20,000 km/s, a supernova exploded. Determine the distance to this star.
30. A globular star cluster is located at a distance of 320 Mpc from us. At what speed is it moving away from us?
4.2.
INTERACTIONS
Basic formulas and laws. 1. Law universal gravity
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=6.6726 10 -11 m 3 /(kg s 2) – gravitational constant.
2. When a clump of substance with mass m rotates around a central body with mass M, the disintegration of the clump (its fragmentation) begins when the centrifugal force acting on the clump begins to exceed the gravitational force between the clump 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=1/(4πε 0)=9 10 9 N m 2 /Cl 2;
ε 0 =8.85 10 -12 C 2 / (N m 2) – electrical constant; ε – dielectric constant of the substance; q 1 and q 2 – electric charges of interacting bodies; r is the distance between them. ℓ 4. Ampere force: F A =I B ℓ sinα (4), IN .
where I is the current strength in a conductor of length ℓ located in a magnetic field with induction B; α- angle between the direction of current (vector
) and vector 5. Lorentz force: F L =q B v sinα (5), where q is the electric charge of a particle flying into a magnetic field with induction B at speed IN.
v at an angle α to the induction vector:
6. Equation of motion of a charged particle of mass m and charge q in an electric field of strength E m E (6)
a
= q Examples of problem solving Example 1: Determine how many times the force of gravity on Earth
more power
gravity on Mars.
Solution: According to formula (1), the force of attraction to the Earth of a body of mass m:
F Z = G m M Z / R Z 2,
where MZ and RZ are the mass and radius of the Earth, respectively.
Similarly, for the force of gravity on Mars:
F M = G m M M / R M 2.
Dividing these two equalities by one another, we obtain after reducing the same quantities:
F Z / F M = 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 Z =5.976 10 24 kg;
R W =6371km=6.371 10 6 m;
M M =0.6335 10 24 kg;
Example 2: When flying to Venus, the spacecraft passes a point where the forces of attraction of the spacecraft to the Earth and to Venus cancel each other out. At what distance from the Earth is this point? When calculating, neglect the action of all other cosmic bodies. Assume that the Earth and Venus are at a minimum distance from each other.
Solution: The sum of the gravitational forces towards the Earth and towards Venus must be equal to zero, or otherwise, the modules of these forces must be equal: F З = F B:
G m M Z / r Z 2 = G m M B / r B 2 (I),
where MZ and MV are the masses of the Earth and Venus, respectively, and
r W and r B are the distances of a spacecraft of mass m from the Earth and from Venus, respectively. Let's take into account that
r B = R ZV - r Z, where R ZV is the distance from the Earth to Venus, which is equal to R ZS - R VS - the difference between the Earth-Sun distances R ZS and Venus-Sun R VS. Let's substitute everything into expression (I):
M Z / r Z 2 = M V / (R ZS - R VS - r Z) 2,
from where we can easily get the answer:
r З = (R ЗС - R ВС)/(1 + 
) .
We take distances and masses from Table 1.
M Z = 5.976 10 24 kg; M B =4.8107 10 24 kg; R ZS = 149.6 million km; R BC =108.2 million km.
r З = (R ЗС - R ВС)/(1 + 
)=
(149,6-108,2)/(1+)=
41.4/1.8972 = 21.823 million km
Answer: r Z = 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=0.1 mT perpendicular to the lines of force. Define:
A) the radius of the circle described by the proton;
B) proton orbital 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 relationship it is easy to obtain an expression for the radius r= m v/(q B) (I).
If we take into account that the speed of revolution v is related to the period T by the relation: v=2π r/T, then from (I) we obtain r=2π r m/(T q B), from which the period of revolution is equal to:
Т= m 2π /(q B) (II).
Taking the charge values q=1.6 10 -19 C and mass
m=1.67 10 -27 kg proton in the table of reference data and substituting them in (I-II), we find:
r=1.67 10 -27 5 10 4 /(1.6 10 -19 0.1 10 -3)=5.22 m.
T=1.67 10 -27 6.28/(1.6 10 -19 0.1 10 -3)=6.55s.
r =5.22m. T =6.55s.
Problem conditions
31. How many times do the forces of attraction of the Earth to Jupiter and to the Sun differ at the moment of time when the Earth is on the 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 to the Sun differ at the moment of time when the Earth is on the 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 should be located so that the resulting gravitational forces 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 the straight line connecting the centers of the Earth and the Moon the rocket should be located. so that the resulting gravitational forces of the Earth and the Moon are 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 moment of time when the Moon is on the 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. A clump of matter rotates around a massive star with a mass of M = 4 10 23 kg at a distance of 10 6 km. At what angular velocity does fragmentation (breakup into parts) of the bunch begin?
40. A clump of matter rotates around a massive star with a mass of M = 4 10 25 kg at a distance of 10 7 km. At what angular velocity does fragmentation (breakup into parts) of the bunch begin?
41. A clump of matter rotates around a massive star with a mass of M = 4 10 24 kg at a speed of 100 m/s. Determine the distance between the star and the clump at which fragmentation (breakup into parts) of the clump occurs.
42. Two bodies having equal negative electric charges repel in air with a force of 5 μN. Determine the number of excess electrons in each body if the distance between the charges is 5 cm.
43. A charge equal to q 1 =2 µC is placed in a medium with dielectric constant ε =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 attract with a force F = 0.5 mN.
44. Two point electric charges interact in air at a distance r 1 = 3.9 cm with the same force as in a non-conducting liquid at a distance 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 = 2000 V/m.
With what acceleration does the particle move?
46. A charged body with mass m=10mg and charge q=2μC moves in an electric field with acceleration a=20m/s 2 . What is the electric field strength?
47. At what angle α to the induction lines of a uniform magnetic field should a conductor with an active length be located 
= 0.2 m, through which a current of force I = 10 A flows, so that a field with induction B = 10 μT acts on the conductor with a force F = 10 μN?
48. Determine the length of the active part of a straight conductor placed in a uniform magnetic field with induction B = 1 mT at an angle α = 60 0 to the induction lines, if at current strength I = 8A the conductor is acted upon
the force is F=2mN.
49. Determine the force acting from a uniform magnetic field with induction B = 0.1 mT on a conductor of length 
= 0.4 m, through which a current of force I = 100 A flows and which is located at an angle α = 45 0 to
induction lines.
50. An electron flies into a uniform magnetic field with induction B = 0.1 mT with a speed v = 5 10 6 m/s perpendicular to its induction lines. Define
the radius of the circle along which the particle moves.
51. An α particle flies into a uniform magnetic field with induction B = 100 μT 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 field.
52. A proton and an alpha particle fly into a uniform magnetic field with induction B = 2 mT perpendicular to its induction lines. 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 rotating 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 the electron in the atom?
54. A proton flies into an electric field of 200 V/m in the direction of the field lines with an initial speed v 0 =3 10 5 m/s. Determine the momentum of the proton after 5 seconds.
55. A particle with an electric charge q = 0.1 μC flies into a uniform magnetic field with induction B = 0.1 mT perpendicular to its field lines with a speed v = 3 10 3 m/s. What force does the magnetic field exert on the particle?
56. How many times does the force of gravity on Jupiter differ from the force of gravity on the Sun?
57. What is the mass of a star if its radius is 100 times greater than that of Earth, and the force of gravity on its surface exceeds the similar force on Earth by 80 times?
58.What is the mass of a star if its radius is 1000 times greater than that of Mars, and the force of gravity on its surface is 5 times greater than the similar force on Mars?
59. How many times does the force of gravity on Jupiter differ from the force of gravity 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 gravity on its surface exceeds the similar force on Venus by 7 times?
4.3. LAWS OF CONSERVATION OF MOMENTUM,
MOMENTUM OF IMPULSE AND MECHANICAL ENERGY
Basic formulas and laws
1. р=m v – body impulse - characteristic of action
body movement..
2. Law of conservation of momentum: total momentum closed system body 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. Law of conservation of angular momentum: the total angular momentum of a closed system of bodies is conserved:
Σ i L i =const.
5. E K = m v 2 /2 – kinetic energy of the body – energy of translational motion.
E K = I ω 2 /2 – kinetic energy of a body rotating about a fixed axis.
E K = m v 2 /2 + I ω 2 /2 – kinetic energy of a 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 – energy of gravitational interaction of two bodies;
E P =m g h-potential energy of the body in the Earth’s gravitational field;
Е Р = к Δх 2 /2 potential energy of an elastically deformed body
(k- coefficient of elasticity (stiffness));
Е Р =к q 1 q 2 /(ε r) - energy of electrostatic interaction of charged bodies, where
k=1/(4πε 0)=9 10 9 N m 2 /Cl 2;
ε 0 =8.85 10 -12 C 2 /(N m 2) - electrical constant; 7. Law of conservation of mechanical energy: complete mechanical energy
E of a closed system of bodies is preserved: E = Σ i (E K + E P) i = const.
If the system is not closed, then work is done against external forces, or work on the system is performed by external forces. Both of these cases lead to a change in the total energy of the system: A=ΔE.
8. A=F s cosα – work done by force F.
a
A = q Δφ = ΔU – work on moving an electric charge q by an electric field (U = E P - potential energy of a charge in an electric field; φ is the potential of a given field point; Δφ and ΔU are the potential differences and potential energies of two field points).
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 = 100 V its speed changes from v 1 = 100 m/s to v 2 = 300 m/s? Solution: Work of forces electric field
leads to a change in the kinetic energy of the particle: A = ΔE K or
q Δφ= m v 2 2 /2 - m v 1 2 /2.
From this expression we get:
m=2 q Δφ/(v 2 2 -v 1 2)=2 10 -6 100/(300 2 -100 2)=2.5 10 -9 kg.
Answer: m=2.5 10 -9 kg.
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 = k q 1 q 2 /r = k q 2 /r 1.
At a distance r 2, the total energy E 2 consists of the potential energy of electrostatic interaction and the kinetic energies of particles:
E 2 = k q 2 /r 2 + 2 m v 2 /2.
In accordance with the law of conservation of energy: E 1 = 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 obtain:
v = 
Let's substitute the values: r 1 =1cm=0.01m; r 2 =5cm=0.05m; m=1mg=10 -6 kg; k=9 10 9 N m 2 /Cl 2; q=2μC=2 10 -6 C and we get v=1.7 10 3 m/s.
Answer: v=1.7 10 3 m/s.
Example 3: A platform with sand with a total weight of M = 1000 kg stands on rails on a horizontal section of the track. A shell hits the sand and gets stuck in it. At the moment the projectile hit the platform, the velocity of the projectile was v 1 =200 m/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 law of conservation of momentum can be applied.
Before the impact, the projectile momentum p 1x =m v 1 cosα; platform impulse p 2x =0; and the resulting x-component of the momentum of the projectile-platform system is equal to:
р 1х +р 2х =mv 1 cosα.
After the impact, the momentum of the platform and the projectile P x = (m+M) v 2. According to the law of conservation of momentum:
р 1х + р 2х = Р x or m v 1 cosα=(m+M) v 2 .
From this expression we finally get:
m =M v 2 /(v 1 cosα -v 2)= 1000 0.5/(200 0.5 – 0.5) = 5.02 kg
Answer: m=5.02kg.
Example 4: A homogeneous thin rod with a mass M = 200 g and a length ℓ = 50 cm can rotate freely in a horizontal plane relative to a vertical axis passing through the center of the rod. A plasticine ball with a mass of m = 10 g, flying horizontally and perpendicular to the rod, hits 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 of 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 angular momentum of the rod and the ball before the impact must be equal to their sum after the impact.
Before impact: moment of 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 angular momentum of the rod and the ball is equal
L=(I 1 +I 2) ω,
where I 1 =m (ℓ/2) 2 is the moment of inertia of a ball with mass m and I 2 =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.
From this expression it follows 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 =10 m/s, turns into neutron star(pulsar) its radius decreases by N=10 5 times. What will be the period T of the pulses? electromagnetic radiation 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 М R 1 2 /5 is the moment of inertia of the stellar a ball of radius R 1 and mass M; ω 1 = v 1 / R 1 - angular velocity of rotation of the star; I 2 =2 M R 2 2 /5 – moment of inertia of a neutron star of radius R 2 and mass M; ω 2 = 2π/T-angular velocity of rotation of the neutron star; Thus, we can write:
2 M R 1 2 v 1 /(5 R 1)=2 M R 2 2 2π /(5 T)
and after reductions and taking into account that: N= R 1 /R 2, we get:
T=2π R 1 /(v 1 N 2)=0.0628s.
Answer: T=0.0628s.
Example 6: A car weighing m=12t stopped by running into a spring buffer and compressing the buffer spring by Δx=4cm. Determine the speed of the car if the spring stiffness k = 4 10 8 N/m.
Solution: Let us 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 Δx 2 /2,
from where we get:
v=Δх 
=4 10 -2 
=7.3m/s.
Answer: v=7.3m/s.
Example 7: What is the kinetic energy of a ball with 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 the numerical data, into the formula for the kinetic energy of a rolling ball:
E K = m v 2 /2 + I ω 2 /2 = m v 2 /2 + m v 2 /5 = 0.7 m v 2,
we get E K = 150 J.
Answer: E K =150 J.
Problem conditions
61. A particle with an electric charge q=2 μC and a mass m=3 10 -6 kg flies into a uniform electric field along a line of tension with a speed v 1 =5 10 4 m/s. What potential difference must the particle pass through for its speed to increase to v 2 = 10 5 m/s?
62. What speed can be imparted to a particle with mass m=2 10 -8 kg and electric charge q=2 10 -12 C, which is at rest, by an accelerating potential difference of U=100 V?
63. What work is required to bring two electric charges q 1 = 2 μC and q 2 = 4 μC, located at a distance r 1 = 1.2 m, closer to
distance r 2 =0.4 m?
64. Two point electric charges q 1 = 3 µC and q 2 = 5 µC are located at a distance r 1 = 0.25 m. How much will the interaction energy of these charges change if they are brought closer to a distance r 2 =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 at what speed
the platform will move if at the moment of impact the projectile speed is v = 200 m/s, and its direction is from top to bottom at an angle α 0 = 30 to the horizon.
66. A projectile with a mass of m=20kg at the top point of the trajectory had a speed of v=250m/s. At this point it broke into two parts. The smaller part with a mass m 1 = 5 kg received a speed u 1 = 300 m/s in the same direction. Determine the speed of the second, larger part of the projectile after the explosion.
67. A projectile with a mass of m=20kg at the top point of the trajectory had a speed of v=300m/s. At this point it broke into two parts. Most of the projectile with mass m 1 =15 kg received speed u 1 =100 m/s in the same direction. Determine the speed of the second, smaller part of the projectile after the explosion.
68. A bullet with a mass m = 10 g, flying horizontally at a speed v = 250 m/s, hit a wooden ball with a mass M = 1 kg hanging on a thread and got stuck in it. To what height did the ball rise after the impact?
69. A bullet with a mass m = 10 g, flying horizontally at a speed v = 250 m/s, hit a wooden ball with a mass M = 1.5 kg hanging on a thread and got stuck in it. By what angle did the ball deviate as a result?
70. A bullet with a mass m = 15 g, flying horizontally, hit a wooden ball with a mass M = 2.5 kg hanging on a thread and got stuck in it. As a result of this, the ball deflected by an angle equal to 30 0. Determine the speed of the bullet.
71. A bullet with a mass m=10g, flying horizontally at a speed 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 pumped out after the impact, rose to a height of h = 20 cm?
Conditions of the 1st round and 2nd round
5-7 grades, 8-9 grades
1. Which of the following astronomical phenomena - equinoxes, solstices, full moons, solar eclipses, lunar eclipses, planetary oppositions, maxima meteor showers, the appearance of bright comets, maximum brightness of variable stars, supernova explosions - occur every year on approximately the same dates (with an accuracy of 1-2 days)?
In crystal dew
even the shadows are rounded,
In Serebryannaya Rechka
there is half a moon at the bottom.
Who will bring the news?
embroidering brocade with letters?
Frowning my eyebrows,
I'm finally extinguishing the candle...
10th grade, 11th grade
1. In 2010, Saturn's opposition will occur on March 22.
2. In the 20th century, there were 14 transits of Mercury across the disk of the Sun:
2nd 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 following astronomical phenomena - equinoxes, solstices, full moons, solar eclipses, lunar eclipses, planetary oppositions, maximum meteor showers, the appearance of bright comets, maximum brightness of variable stars, supernova explosions - occur every year on approximately the same dates (within 1-2 days)?
Solution. Those astronomical phenomena that are associated only with the movement of the Earth in its orbit around the Sun, that is, equinoxes, solstices and maxima of meteor showers, are repeated annually. These phenomena repeat on approximately the same dates, for example, the vernal equinox falls on March 20 or 21, since our calendar has leap years. For meteor showers, the inaccurate repetition of maximum dates is also due to the drift of their radiants. The rest of the mentioned phenomena either have a periodicity different from the Earth’s year (full moons, solar eclipses, lunar eclipses, planetary oppositions, maximum brightness of variable stars), or are completely non-periodic (the appearance of bright comets, supernova explosions).
2. The astronomy textbook of the Belarusian authors A.P. Klishchenko and V.I. Shuplyak contains such a diagram of a lunar eclipse. What's 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 Moon's orbit. The night side of our satellite, of course, should be dark.
3. Yesterday the Moon was observed covering the Pleiades star cluster. Could there be a solar eclipse tomorrow? Moon eclipse?
Solution. Eclipses occur when the Moon is close to the ecliptic during a full or new moon. The Pleiades are located approximately 5 degrees north of the ecliptic, and the Moon can only cover them when it is at its 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 occur tomorrow.
4. Here are the lines from the poem by the classical Chinese poet Du Fu “River Moon” (translation by E.V. Balashov):
In crystal dew
even the shadows are rounded,
In Serebryannaya Rechka
there is half a moon at the bottom.
Who will bring the news?
embroidering brocade with letters?
Frowning my eyebrows,
I'm finally extinguishing the candle...
It is not difficult to guess that the Chinese call the Silver River Milky Way. In 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. The “Half Moon” can be either growing or aging and be located either 90° west of the Sun or 90° east. In both cases, it turns out that the Sun is located on the ecliptic near the equinox points. 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 altitude of Saturn above the horizon at local midnight on March 22 when observed from Moscow (latitude 55 o 45’)?
Solution. Since Saturn's opposition almost coincides in time with spring equinox, the planet itself is located in 2010 near the point of the autumn equinox, that is, on the celestial equator (d=0 o). Therefore, it passes through the zenith for an observer located at the Earth's equator.
On March 22, Saturn will be located on celestial sphere opposite the Sun, so at local midnight it will be at its highest culmination. Let’s apply the formula to calculate the height of the luminary at its culmination: h = (90 o – f) + d, h = 34 o 15’.
2. * In the twentieth century, there were 14 transits of Mercury across the disk of the Sun:
Why are passages observed only in May and November? Why are November passages observed much more often than May ones?
Solution. The inner planet can be projected onto the disk of the Sun for an earthly observer only when, at the moment of inferior conjunction, it is near the ecliptic plane, that is, near the nodes of its orbit. The nodes of Mercury's orbit are oriented in space so that the Earth is in 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 further from the Earth), and therefore is projected onto the solar disk more often than in May, near aphelion.
3. By what percentage does the quantity differ? sunlight falling on the Moon in the first quarter phase and in the full moon phase?
Solution. The illumination of the lunar surface is inversely proportional to the square of the distance from the Sun to the Moon. In the first quarter phase, the Moon is at a distance of approximately 1 AU. from the Sun, in the full moon phase - on average 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". Using these data, determine the eccentricity of the orbit of Mars. The semimajor axis of Mars's orbit is 1.5 AU; the Earth's orbit 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 of m (1+e) from the Sun, at perihelion - at a distance of a m (1st). The distance between Earth and Mars at aphelion and perihelion opposition is related as
(a m (1+e)-1)/(a m (1st)-1).
On the other hand, this ratio is 25/13. Let's write down the equation and solve it for e:
(a m (1+e)-1)/(a m (1st)-1)=25/13, e=0.1.
2nd round
5-7 grades, 8-9 grades
1. Can Venus be observed in the constellation Gemini? In the constellation Canis Major? In the constellation 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 reach 8°. Venus was visible in the constellation Orion in August 1996. Venus cannot be located in the constellation Canis Major, far from the ecliptic.
2. The star rose at 00:01 local time. How many more times will it cross the horizon at this point in this 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. That's why this star during this day it will have time to go beyond the horizon and rise again at 23:57 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, no matter what 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 means that even a completely point source will be visible through a telescope as a disk surrounded by a system of rings. The larger the diameter of the telescope lens, the smaller the size of this disk, but even for large telescopes it is about 0.1 arcsecond. In addition, the image is blurred by the earth's atmosphere, and the size of the “jitter disks” of stars is rarely less than one arcsecond. The true angular diameters of distant stars are much smaller, and we cannot see them in 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 enormous angular dimensions, but it will still shine weaker than the Sun. In this case, Jupiter will be visible only from half the surface of the satellite, remaining motionless in the sky, since all Galilean satellites, like the Moon to the Earth, are turned to Jupiter with one side. In its movement across the sky, the Sun will set behind Jupiter at each revolution, and there will be solar eclipses, and only when observed from the most distant satellite, Callisto, an eclipse may not occur.
In addition to the Sun and Jupiter, the rest of the satellites of this planet will be clearly visible in the sky; during oppositions with the Sun they will be very bright (up to –2 m) Saturn will be, other, more distant planets will become a little brighter solar system: Uranus, Neptune and Pluto. And here are the planets terrestrial group will be less visible, and the point is not so much in their brilliance as in their small angular distance from the Sun. Thus, 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 E And E 0 - radii of the orbits of Jupiter and Earth. Knowing the distance from the Earth to the Moon (384400 km), we obtain the maximum angular distance between the Earth and the Moon equal to 1 ¢ 43.8² , which in principle is sufficient to resolve them with the naked eye. However, the Moon's brightness at this moment will be +7.5 m, and it will not be visible to the naked eye (the Earth's brightness will be about +3.0 m). The Earth and Moon will be much brighter near superior conjunction with the Sun (–0.5 m and +4.0 m respectively), but at this time they will be difficult to see in the rays of daylight.
10th grade, 11th grade
1. How will the pendulum clock run when delivered from Earth to the surface of Mars?
Solution. Acceleration of free fall on the surface of the planet g equals
Where M And R - 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 acceleration of free fall g on Mars is equal to 0.377 of the same value on Earth. Clock oscillation period T with length pendulum 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 first quarter phase covers the star Aldebaran (a Taurus). What season is it now?
2 Solution. The star Aldebaran is located near the ecliptic in the constellation Taurus. The sun passes through this area of the sky in late May - early June. The Moon in its first quarter phase is 90 degrees away from the Sun.° to the east and is located in that place in the sky where the Sun will come in three months. Therefore, it is now the end of February - beginning of March.
3. The magnitude of Venus during superior conjunction is –3.9 m, and during the greatest elongation –4.4 m. What is the brightness of Venus in these configurations when observed 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, regardless of whether we are observing from Earth or from 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. Let us denote by E 0 is the radius of the orbit of Venus, and after E - radius of the orbit of the planet from which observations are made. Then the distance to Venus at the moment of its superior conjunction will be equal to a+a 0, which is 1.723 au. for Earth and 2.247 AU. for Mars. Then the magnitude of Venus at the time of superior conjunction on Mars will be equal to
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 Earth and 1.342 AU. for Mars. The magnitude of Venus at the moment of greatest elongation is
m 2 = –4.4 + 5 lg (1.342/0.691) = –3.0.
Interestingly, Venus shines weaker on Mars (like Mercury on Earth) at greatest elongation than at superior conjunction.
4. A binary system consists of two identical stars with a mass of 5 solar masses, revolving in circular orbits around a common center of mass with a period of 316 years. Will it be possible to resolve this pair visually in 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 Solution. Let us determine the distance between stars according to Kepler’s III generalized law:
Here E- semimajor 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 us 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 Earth's orbital period 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 at no more than 1² from each other. It will not be possible to resolve such a close pair in the TAL-M telescope, no matter what magnification we use. This is easy to verify by calculating the size of the diffraction disks of these stars using the well-known formula for green-yellow rays:
Where D- lens diameter in centimeters. Here we did not take into account the influence of the earth’s atmosphere, which will further worsen the picture. So, this pair will be visible in the TAL-M telescope only as a single star.
Mass is one of the most important physical characteristics stars - can be determined by its effect on the movement of other bodies. Such other bodies are the satellites of some stars (also stars), orbiting them around a common center of mass.
If you look at Ursa Major, the second star from the end of the “handle” of its “bucket”, then with normal vision you will see a second faint star very close to it. The ancient Arabs noticed her and called her Alkor (Horseman). They gave the name Mizar to the bright star. They can be called a double star. Mizar and Alcor are separated by . You can find many such star pairs through binoculars. So, Lyrae consists of two identical stars of the 4th magnitude with a distance of 5 between them.
Rice. 80. The orbit of a satellite of a binary star (v Virgo) relative to the main star, the distance of which from us is 10 pc. (The dots mark the measured positions of the satellite in the indicated years. Their deviations from the ellipse are caused by observational errors.)
Binary stars are called visual binaries if their duality can be seen by direct observation through a telescope.
In the Lyra telescope - a visually quadruple star. Systems with a number of stars are called multiples.
Many of the visual double stars turn out to be optical double stars, that is, the proximity of such two stars is the result of their random projection onto the sky. In fact, in space they are far from each other. And over many years of 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 fainter companion star orbits a brighter star. The distances between them and the direction of the line connecting them systematically change. Such stars are called physical binaries; they form a single system and rotate under the influence of mutual attractive forces 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 of visual binary stars is 5 years. Pairs with 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, Centauri, is a double star. The period of circulation 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 distorted in the projection (Fig. 80). But knowledge of geometry makes it possible to restore the true shape of the orbit and measure its semimajor axis a in arcseconds. If the distance to the binary star is known in parsecs and the semimajor axis of the satellite star’s orbit in arcseconds is equal to then in astronomical units (since it will be equal to:
The most important characteristic of a star, along with luminosity, is its mass. Direct determination of mass is possible only for double stars. By analogy with § 9.4, comparing the motion of the satellite
stars with the movement 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 obtain the sum of the masses of the stars that make up the pair, in solar masses:
To determine the mass of each star separately, it is necessary to study the motion of each of them relative to 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 are often a beautiful sight in a telescope: the main star is yellow or orange, and the companion is white or blue. Imagine the richness of colors on a planet orbiting one of a pair of stars, where the sky shines either red or blue, or both.
The masses of stars determined by the methods described differ much less than their luminosities, from approximately 0.1 to 100 solar masses. Large masses are extremely rare. Stars typically 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, not standing out in anything special.
(see scan)
2. Spectral double stars.
If the stars come close to each other during mutual rotation, then even with the most powerful telescope they cannot be seen separately, in this case the duality can be determined by the spectrum. If the orbital plane of such a pair almost coincides with the line of sight, and the speed of revolution is high, then the speed of each star in the projection onto the line of sight will change quickly. The spectra of double stars overlap each other, and since the difference in the velocities of these
Rice. 81. Explanation of bifurcation, or fluctuations, of lines in the spectra of spectroscopic double stars.
stars is large, then the lines in the spectrum of each of them will shift in opposite directions. The magnitude of the shift changes with a period equal to the period of revolution of the pair. If the brightnesses and spectra of the stars that make up the pair are similar, then a periodically repeating bifurcation of spectral lines is observed in the spectrum of a double star (Fig. 81). Let the components take positions, or then one of them moves towards the observer, and the other away from him (Fig. 81, I, III). In this case, a bifurcation of the spectral lines is observed. An approaching star will shift its spectral lines toward the blue end of the spectrum, while a receding star will shift toward the red end. When the components of a double star occupy positions or (Fig. 81, II, IV), then both of them move at right angles to the line of sight and bifurcation of the spectral lines will not work.
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 star.
3. Eclipsing double stars - algoli.
If the line of sight lies almost in the plane of the orbit of a spectroscopic binary star, then the stars of such a pair will alternately block each other. During eclipses, the overall brightness of a pair, the components of which we do not see individually, will weaken (positions B and D in Fig. 82). 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 larger the radius of the orbit. If the satellite is large, but itself gives little light, then when bright
the star eclipses it, the total brightness of the system will decrease only slightly.
The brightness minima of eclipsing binary stars occurs when their components move across the line of sight. Analysis of the curve of changes in apparent stellar magnitude as a function of time makes it possible to determine the size and brightness of stars, the dimensions of the orbit, its shape and inclination to the line of sight, as well as the masses of the stars. Thus, eclipsing binary stars, also observed as spectroscopic binaries, are the most well studied systems. Unfortunately, relatively few such systems are known so far.
Eclipsing double stars are also called algols, after the name of their typical representative Perseus. The ancient Arabs called Perseus Algol (corrupted el gul), which means “the devil.” It is possible that they noticed its 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 magnitude, and then in 5 hours its brightness returns to its previous value.
The periods of known spectroscopic binary stars and algols are generally short - about a few days. In general, stellar binaries are a very common phenomenon Statistics show that up to 30% of all stars are likely to be binaries Obtaining a variety of data about individual stars and their systems from the analysis of spectroscopic binaries and eclipsing binaries - examples of the unlimited possibility of human knowledge
Rice. 82. Changes in the apparent brightness of Lyra and the pattern of motion of its satellite (The shape of stars located close to each other, due to their tidal influence, can differ greatly from spherical)