Since ancient times, people have been interested in the problem of the structure of the world. Back in the 3rd century BC, the Greek philosopher Aristarchus of Samos expressed the idea that the Earth revolves around the Sun, and tried to calculate the distances and sizes of the Sun and Earth from the position of the Moon. Since the evidentiary apparatus of Aristarchus of Samos was imperfect, the majority remained supporters of the Pythagorean geocentric system peace.
Almost two millennia passed, and the Polish astronomer Nicolaus Copernicus became interested in the idea of a heliocentric structure of the world. He died in 1543, and soon his life's work was published by his students. Copernicus' model and tables of positions of celestial bodies based on heliocentric system, reflected the state of affairs much more accurately.
Half a century later, the German mathematician Johannes Kepler, using the meticulous notes of the Danish astronomer Tycho Brahe on observations of celestial bodies, derived the laws of planetary motion that eliminated the inaccuracies of the Copernican model.
The end of the 17th century was marked by the works of the great English scientist Isaac Newton. Newton's laws of mechanics and universal gravitation expanded and gave theoretical basis formulas derived from Kepler's observations.
Finally, in 1921, Albert Einstein proposed general theory relativity, which most accurately describes the mechanics of celestial bodies at the present time. Newtonian formulas of classical mechanics and the theory of gravitation can still be used for some calculations that do not require great accuracy, and where relativistic effects can be neglected.
Thanks to Newton and his predecessors, we can calculate:
- what speed must the body have to maintain a given orbit ( first escape velocity)
- at what speed must a body move in order for it to overcome the gravity of the planet and become a satellite of the star ( second escape velocity)
- minimum required speed beyond limits planetary system (third escape velocity)
Ministry of Education and Science of the Russian Federation
State educational institution higher vocational education"St. Petersburg State University economics and finance"
Department of Technology Systems and Commodity Science
Concept course report modern natural science on the topic “Cosmic speeds”
Performed:
Checked:
Saint Petersburg
Cosmic speeds.
escape velocity(first v1, second v2, third v3 and fourth v4) - this is the minimum speed at which any body in free movement will be able:
v1 - become a companion celestial body(that is, the ability to orbit around the NT and not fall onto the surface of the NT).
v2 - overcome the gravitational attraction of a celestial body.
v3 - leave solar system, overcoming the gravity of the Sun.
v4 - leave the galaxy Milky Way.
First escape velocity or Circular velocity V1- the speed that must be given to an object without an engine, neglecting the resistance of the atmosphere and the rotation of the planet, in order to put it into a circular orbit with a radius equal to the radius of the planet. In other words, the first escape velocity is the minimum speed at which a body moving horizontally above the surface of the planet will not fall on it, but will move in a circular orbit.
To calculate the first escape velocity, it is necessary to consider the equality centrifugal force and gravitational forces acting on an object in a circular orbit.
where m is the mass of the object, M is the mass of the planet, G is the gravitational constant (6.67259·10−11 m³·kg−1·s−2), is the first escape velocity, R is the radius of the planet. Substituting numerical values(for the Earth M = 5.97 1024 kg, R = 6,378 km), we find
The first escape velocity can be determined through acceleration free fall- since g = GM/R², then
Second escape velocity (parabolic velocity, escape velocity)- the lowest speed that must be given to an object (for example, a spacecraft), the mass of which is negligible relative to the mass of a celestial body (for example, a planet), to overcome gravitational attraction this celestial body. It is assumed that after a body acquires this speed, it does not receive non-gravitational acceleration (the engine is turned off, there is no atmosphere).
The second cosmic velocity is determined by the radius and mass of the celestial body, therefore it is different for each celestial body (for each planet) and is its characteristic. For the Earth, the second escape velocity is 11.2 km/s. A body that has such a speed near the Earth leaves the vicinity of the Earth and becomes a satellite of the Sun. For the Sun, the second escape velocity is 617.7 km/s.
The second escape velocity is called parabolic because bodies with a second escape velocity move along a parabola.
Derivation of the formula:
To obtain the formula for the second cosmic velocity, it is convenient to reverse the problem - ask what speed a body will receive on the surface of the planet if it falls onto it from infinity. Obviously, this is exactly the speed that must be given to a body on the surface of the planet in order to take it beyond its boundaries. gravitational influence.
Let's write down the law of conservation of energy
where on the left are the kinetic and potential energies on the surface of the planet (potential energy is negative, since the reference point is taken at infinity), on the right is the same, but at infinity (a body at rest on the border of gravitational influence - the energy is zero). Here m is the mass of the test body, M is the mass of the planet, R is the radius of the planet, G is the gravitational constant, v2 is the second escape velocity.
Resolving with respect to v2, we get
There is a simple relationship between the first and second cosmic velocities:
Third escape velocity- the minimum required speed of a body without an engine, allowing it to overcome the gravity of the Sun and, as a result, go beyond the boundaries of the Solar system into interstellar space.
Taking off from the surface of the Earth and the best way using the planet's orbital motion spacecraft can reach a third of escape velocity already at 16.6 km/s relative to the Earth, and when starting from the Earth in the most unfavorable direction, it must be accelerated to 72.8 km/s. Here, for the calculation, it is assumed that the spacecraft acquires this speed immediately on the surface of the Earth and after that does not receive non-gravitational acceleration (the engines are turned off and there is no atmospheric resistance). With the most energetically favorable launch, the object’s speed should be co-directional with the speed of the Earth’s orbital motion around the Sun. The orbit of such a device in the Solar System is a parabola (the speed decreases to zero asymptotically).
Fourth cosmic speed- the minimum required speed of a body without an engine, allowing it to overcome the gravity of the Milky Way galaxy. The fourth escape velocity is not constant for all points of the Galaxy, but depends on the distance to the central mass (for our galaxy this is the object Sagittarius A*, the supermassive black hole). According to rough preliminary calculations, in the region of our Sun, the fourth cosmic speed is about 550 km/s. The value strongly depends not only (and not so much) on the distance to the center of the galaxy, but on the distribution of masses of matter throughout the Galaxy, about which there is no accurate data yet, due to the fact that visible matter constitutes only a small part of the total gravitating mass, and the rest is hidden mass.
To determine two characteristic “cosmic” velocities associated with the size and gravitational field of a certain planet. We will consider the planet to be one ball.
Rice. 5.8. Different trajectories of satellites around the Earth
First cosmic speed they call such a horizontally directed minimum speed at which a body could move around the Earth in a circular orbit, that is, turn into an artificial satellite of the Earth.
This, of course, is an idealization; firstly, the planet is not a ball, and secondly, if the planet has enough dense atmosphere, then such a satellite - even if it can be launched - will burn out very quickly. Another thing is that, say, an Earth satellite flying in the ionosphere on average height above a surface of 200 km has an orbital radius that differs from the average radius of the Earth by only about 3%.
A satellite moving in a circular orbit with a radius (Fig. 5.9) is acted upon by the gravitational force of the Earth, giving it normal acceleration
Rice. 5.9. Movement artificial satellite Earth in a circular orbit
According to Newton's second law we have
If the satellite moves close to the Earth's surface, then
Therefore, for on Earth we get
It can be seen that it is really determined by the parameters of the planet: its radius and mass.
The period of revolution of a satellite around the Earth is
where is the radius of the satellite’s orbit, and is its orbital speed.
Minimum value The orbital period is achieved when moving in an orbit whose radius is equal to the radius of the planet:
so the first escape velocity can be defined this way: the speed of a satellite in a circular orbit with a minimum period of revolution around the planet.
The orbital period increases with increasing orbital radius.
If the satellite's orbital period equal to the period The rotation of the Earth around its axis and their directions of rotation coincide, and the orbit is located in the equatorial plane, then such a satellite is called geostationary.
A geostationary satellite constantly hangs over the same point on the Earth's surface (Fig. 5.10).
Rice. 5.10. Movement of a geostationary satellite
In order for the body to leave the sphere gravity, that is, it could move to such a distance where gravity to the Earth ceases to play a significant role, it is necessary second escape velocity(Fig. 5.11).
Second escape velocity they call the lowest speed that must be imparted to a body so that its orbit in the Earth’s gravitational field becomes parabolic, that is, so that the body can turn into a satellite of the Sun.
Rice. 5.11. Second escape velocity
In order for a body (in the absence of environmental resistance) to overcome gravity and go into space, it is necessary that the kinetic energy of a body on the surface of the planet is equal to (or exceeds) the work done against the forces of gravity. Let's write the law of conservation of mechanical energy E such a body. On the surface of the planet, specifically the Earth
The speed will be minimal if the body is at rest at an infinite distance from the planet
Equating these two expressions, we get
whence for the second escape velocity we have
To impart the required speed (first or second cosmic speed) to the launched object, it is advantageous to use the linear speed of the Earth’s rotation, that is, launch it as close as possible to the equator, where this speed, as we have seen, is 463 m/s (more precisely 465.10 m/s ). In this case, the direction of launch must coincide with the direction of rotation of the Earth - from west to east. It is easy to calculate that in this way you can gain several percent in energy costs.
Depending on the initial speed, communicated to the body at the point of throwing A on the surface of the Earth, the following types of movement are possible (Fig. 5.8 and 5.12):
Rice. 5.12. Shapes of particle trajectory depending on throwing speed
The movement in the gravitational field of any other cosmic body, for example, the Sun, is calculated in exactly the same way. In order to overcome the gravitational force of the luminary and leave the solar system, an object at rest relative to the Sun and located at a distance from it, equal to the radius earth's orbit (see above), it is necessary to report the minimum speed determined from the equality
where, recall, is the radius of the Earth's orbit, and is the mass of the Sun.
This leads to a formula similar to the expression for the second escape velocity, where it is necessary to replace the mass of the Earth with the mass of the Sun and the radius of the Earth with the radius of the Earth’s orbit:
We emphasize that this is the minimum speed that must be given motionless body, located on earth's orbit so that it overcomes the gravity of the Sun.
Note also the connection
With orbital speed Earth. This connection, as it should be - the Earth is a satellite of the Sun, is the same as between the first and second cosmic velocities and .
In practice, we launch a rocket from the Earth, so it obviously participates in orbital movement around the Sun. As shown above, the Earth moves around the Sun at linear speed
It is advisable to launch the rocket in the direction of the Earth's movement around the Sun.
The speed that must be imparted to a body on Earth in order for it to leave the solar system forever is called third escape velocity .
Speed depends on which direction spaceship leaves the zone of gravity. At an optimal start, this speed is approximately = 6.6 km/s.
The origin of this number can also be understood from energy considerations. It would seem that it is enough to tell the rocket its speed relative to the Earth
in the direction of the Earth's movement around the Sun, and it will leave the solar system. But this would be correct if the Earth did not have its own gravitational field. The body must have such a speed having already moved away from the sphere of gravity. Therefore, calculating the third escape velocity is very similar to calculating the second escape velocity, but with additional condition- body on long distance from the Earth should still have a speed of:
In this equation, we can express the potential energy of a body on the surface of the Earth (the second term on the left side of the equation) in terms of the second escape velocity in accordance with the previously obtained formula for the second escape velocity
From here we find
http://www.plib.ru/library/book/14978.html - Sivukhin D.V. General course physics, volume 1, Mechanics Ed. Science 1979 - pp. 325–332 (§61, 62): formulas for all cosmic velocities (including the third) were derived, problems about the motion of spacecraft were solved, Kepler's laws were derived from the law of universal gravitation.
http://kvant.mirror1.mccme.ru/1986/04/polet_k_solncu.html - Magazine “Kvant” - flight of a spacecraft to the Sun (A. Byalko).
http://kvant.mirror1.mccme.ru/1981/12/zvezdnaya_dinamika.html - Kvant magazine - stellar dynamics (A. Chernin).
http://www.plib.ru/library/book/17005.html - Strelkov S.P. Mechanics Ed. Science 1971 - pp. 138–143 (§§ 40, 41): viscous friction, Newton's law.
http://kvant.mirror1.mccme.ru/pdf/1997/06/kv0697sambelashvili.pdf - “Kvant” magazine - gravitational machine (A. Sambelashvili).
http://publ.lib.ru/ARCHIVES/B/""Bibliotechka_""Kvant""/_""Bibliotechka_""Kvant"".html#029 - A.V. Bialko "Our planet - Earth". Science 1983, ch. 1, paragraph 3, pp. 23–26 - provides a diagram of the position of the solar system in our galaxy, the direction and speed of movement of the Sun and the Galaxy relative to the cosmic microwave background radiation.
The second “terrestrial” escape velocity is this is the speed that must be communicated to the body relative to the Earth, so that it overcomes the field of gravity, i.e. turned out to be capable of moving away from the Earth to an infinitely large distance.
Neglecting the effect on the body of the Sun, Moon, planets, stars, etc. and assuming that in the Earth-body system there are no non-conservative forces (and in fact there are some - these are atmospheric resistance forces), we can consider this system closed and conservative. In such a system, the total mechanical energy is a constant quantity.
If the zero level potential energy choose at infinity, then the total mechanical energy of the body at any point in the trajectory will be equal to zero (as the body moves away from the Earth, the kinetic energy imparted to it at the start will turn into potential. At infinity, where the potential energy of the body is zero,
goes to zero and kinetic energy E To =0. Hence, total energy E= E P + E To . = 0.)
By equating the total energy of the body at the start (on the surface of the Earth) and at infinity, we can calculate the second escape velocity. At the start, the body has positive kinetic energy 
And negative potential energy 
,m
-
body mass; M h -
mass of the Earth; 
II - the speed of the body at the start (the desired escape velocity); R h- radius of the Earth (we assume that the body acquires the required escape velocity in close proximity to the Earth’s surface).
Total body energy 
(12.16)
where 
(12.17)
The mass of the Earth can be expressed in terms of the acceleration of gravity g 0 (near the surface of the Earth): 
.
Substituting this expression into (12.17), we finally obtain
(12.18)
because 
there is the first escape velocity.
V. Equilibrium conditions for a mechanical system.
Let some body be acted upon only conservative force. This means that this body, together with the bodies with which it interacts, forms closed conservative system.
Let's find out under what conditions will the body in question be in a state of equilibrium (we formulate these conditions with
energy point of view). Equilibrium conditions from the point of view speakers we know: a body is in equilibrium if its speed and the geometric sum of all forces acting on it are equal
(12.19)
(12.20)
zero: Let the conservative force acting on the body be such that the potential energy of the body depends only on one coordinate, for example, x
. The graph of this dependence is shown in Figure 23. From the relationship between potential energy and force it follows that in a state of equilibrium Let the conservative force acting on the body be such that the potential energy of the body depends only on one coordinate, for example, derivative of potential energy with respect to
(12.21)
those. In a state of equilibrium, the body has an extreme reserve of potential energy. Let us make sure that the potential energy is in a state of stable equilibrium minimum, but is able unstable equilibrium – maximum.
3. Stable equilibrium of a system is characterized by the fact that when the system deviates from this state, forces arise returning system to its original state.
P 
When deviating from a state of unstable equilibrium, forces arise that tend to deviate the system further. further from the original position. Let's tilt the body out of position A
left(see Fig. 23). This will create strength 
, whose projection onto the axis Let the conservative force acting on the body be such that the potential energy of the body depends only on one coordinate, for example, is equal to:
(12.22)
Derivative 
at the point 
negative (angle 
- blunt). From (12.22) it follows, 
>0; direction of force 
matches with axis direction Let the conservative force acting on the body be such that the potential energy of the body depends only on one coordinate, for example,, i.e. directional force to the equilibrium positionA. The body will spontaneously, without additional impact, return to the equilibrium position. Therefore, the state A- state sustainable balance. But in this state, as can be seen from the graph, the potential energy minimal.
4. Let's tilt the body out of position B
also to the left. Projection of force 
per axis Let the conservative force acting on the body be such that the potential energy of the body depends only on one coordinate, for example,: 
it turns out negative
(
>0, since the angle 
spicy).
This means that the direction of the force 
opposite positive axis direction Let the conservative force acting on the body be such that the potential energy of the body depends only on one coordinate, for example,, i.e. force 
directed from the equilibrium position. State B, in which the potential energy is maximum, unstable.
Thus, able sustainable equilibrium potential energy of the system minimal, able unstable balance - balance maximum.
If it is known that the potential energy of some system minimal, this does not mean that the system is in equilibrium. It is also necessary that in this state the system does not have kinetic energy: 
(12.23)
So, the system is in a state of stable equilibrium if E To=0, a E P minimal. If E To=0, a E P is maximum, then the system is in unstable equilibrium.
EXAMPLES OF SOLVING PROBLEMS
Example 1. A man stands in the center of the Zhukovsky bench and rotates with it by inertia. Frequency 
Moment of inertia of the human body relative to the axis of rotation 
In arms extended to the sides, a man holds two weights weighing 
each. Distance between weights 
How many revolutions per second will a bench with a person make if he lowers his arms and distance 
between the weights will be equal 
Neglect the moment of inertia of the bench.
Solution. A person holding weights (see Fig. 24) together with the bench constitutes an isolated mechanical system, therefore the angular momentum 
this system must have a constant value.
Therefore, for our case
Where 
And 
- moment of inertia of a person and angular velocity of a bench and a person with outstretched arms. 
And 
- the moment of inertia of the human body and the angular velocity of the bench and the person with his arms down. From here 
, replacing angular velocity via frequency 
(
), we get 
The moment of inertia of the system considered in this problem is equal to the sum moment of inertia of the human body 
and the moment of inertia of weights in the hands of a person, which can be determined by the formula for the moment of inertia of a material point 
Hence, 
Where 
the mass of each weight, 
And 
the initial and final distance between them. Taking into account the comments made, we have
Substituting the numerical values of the quantities, we find
Example 2. Rod length 
and mass 
can rotate around a fixed axis passing through the upper end of the rod (see Fig. 25). A bullet with a mass of 
, flying in a horizontal direction at a speed 
, and gets stuck in the rod.
At what angle 
Will the rod deflect after impact?
Solution. The impact of a bullet should be considered as inelastic: after the impact, both the bullet and the corresponding point on the rod will move at the same speeds.
First, the bullet, hitting the rod, sets it in motion with a certain angular velocity in a negligible period of time 
and gives it some kinetic energy 
Where 
moment of inertia of the rod relative to the axis of rotation. Then the rod rotates through a certain angle, and its center of gravity rises to a certain height 
.
In a deflected position, the rod will have potential energy 
Potential energy is obtained due to kinetic energy and is equal to it according to the law of conservation of energy, i.e.
, where 
To determine angular velocity 
Let's use the law of conservation of angular momentum.
At the initial moment of impact, the angular velocity of the rod 
and therefore the angular momentum of the rod 
The bullet touched the rod with linear velocity 
, and began to go deeper into the rod, telling him angular acceleration and participating in the rotation of the rod about the axis.
Initial bullet impulse 
Where 
the distance of the point of impact of the bullet from the axis of rotation.
At the final moment of impact the rod had an angular velocity 
, and the bullet – linear speed 
equal to linear speed points of the rod located at a distance 
from the axis of rotation.
Because 
, then the final angular momentum of the bullet
Applying the law of conservation of angular momentum, we can write
Substituting the numerical values, we get
After this we find
SELF-TEST QUESTIONS
What system of bodies is called closed?
2. What system of interacting bodies is called conservative?
Under what conditions is the momentum of an individual body conserved?
Formulate the law of conservation of momentum for a system of bodies.
Formulate the law of conservation of angular momentum (for an individual body and a system of bodies).
Formulate the law of conservation of mechanical energy.
What systems are called dissipative?
What is a collision between bodies?
Which collision is called absolutely inelastic and which is called absolutely elastic?
10.What laws are satisfied during absolutely inelastic and absolutely elastic collisions of bodies forming a closed system?
11.What is the second escape velocity? Derive a formula for this speed.
Formulate the equilibrium conditions of a mechanical system.
Any object, being thrown up, sooner or later ends up on earth's surface, be it a stone, a sheet of paper or a simple feather. At the same time, a satellite launched into space half a century ago space station or the Moon continues to rotate in its orbits, as if they were not affected by our planet at all. Why is this happening? Why is the Moon not in danger of falling to the Earth, and why is the Earth not moving towards the Sun? Doesn't it work on them? universal gravity?
From school course physicists we know that universal gravity affects any material body. Then it would be logical to assume that there is some force that neutralizes the effect of gravity. This force is usually called centrifugal. Its effect can be easily felt by tying a small weight to one end of the thread and unwinding it in a circle. Moreover, the higher the rotation speed, the stronger the tension of the thread, and the slower we rotate the load, the more likely that he will fall down.
Thus, we are very close to the concept of “cosmic velocity”. In a nutshell, it can be described as the speed that allows any object to overcome the gravity of a celestial body. The role can be a planet, its or another system. Every object that moves in orbit has escape velocity. By the way, the size and shape of the orbit depend on the magnitude and direction of the speed that the given object received at the time the engines were turned off, and the altitude at which this event occurred.
There are four types of escape velocity. The smallest of them is the first. This is the lowest speed it must have for it to enter a circular orbit. Its value can be determined by the following formula:
V1=õ/r, where
µ - geocentric gravitational constant (µ = 398603 * 10(9) m3/s2);
r is the distance from the launch point to the center of the Earth.
Due to the fact that the shape of our planet is not a perfect sphere (at the poles it seems to be slightly flattened), the distance from the center to the surface is greatest at the equator - 6378.1. 10(3) m, and the least at the poles - 6356.8. 10(3) m. If you take average value- 6371. 10(3) m, then we get V1 equal to 7.91 km/s.
The more escape velocity will exceed this value, the more elongated the orbit will acquire, moving away from the Earth by all longer distance. At some point, this orbit will break, take the shape of a parabola, and the spacecraft will set off to plow the expanses of space. In order to leave the planet, the ship must have a second escape velocity. It can be calculated using the formula V2=√2µ/r. For our planet, this value is 11.2 km/s.
Astronomers have long determined what the escape velocity is, both the first and the second, for each planet of our home system. They can be easily calculated using the above formulas if you replace the constant µ with the product fM, in which M is the mass of the celestial body of interest, and f is constant of gravity(f= 6.673 x 10(-11) m3/(kg x s2).
The third cosmic speed will allow anyone to overcome the gravity of the Sun and leave their native solar system. If you calculate it relative to the Sun, you get a value of 42.1 km/s. And in order to enter solar orbit from Earth, you will need to accelerate to 16.6 km/s.
And finally, the fourth escape velocity. With its help, you can overcome the gravity of the galaxy itself. Its magnitude varies depending on the coordinates of the galaxy. For ours, this value is approximately 550 km/s (if calculated relative to the Sun).
