What speed is called the second cosmic speed. IV. Calculation of the second escape velocity (for Earth)

Second escape velocity (parabolic velocity, release velocity, escape velocity)- the lowest speed that must be given to an object (for example, a spacecraft), the mass of which is negligible compared to the mass of a celestial body (for example, a planet), in order to overcome the gravitational attraction of this celestial body and leaving a closed orbit around it. It is assumed that after a body acquires this speed, it no longer receives 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 cosmic velocity is called parabolic because bodies that have a speed at launch exactly equal to the second cosmic velocity move in a parabola relative to the celestial body. However, if a little more energy is given to the body, its trajectory ceases to be a parabola and becomes a hyperbola. If it is a little less, then it turns into an ellipse. IN general case they are all conic sections.

If a body is launched vertically upward with a second space force or more high speed, it will never stop or start falling back.

The same speed is acquired at the surface of a celestial body by any cosmic body that is infinitely long distance rested and then began to fall.

The second cosmic speed was first achieved by a USSR spacecraft on January 2, 1959 (Luna-1).

Calculation

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 the limits of its gravitational influence.

m v 2 2 2 − G m M R = 0 , (\displaystyle (\frac (mv_(2)^(2))(2))-G(\frac (mM)(R))=0,) R = h + r (\displaystyle R=h+r)

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- mass of the test body, M- mass of the planet, r- radius of the planet, h - length from the base of the body to its center of mass (height above the surface of the planet), G- gravitational constant, v 2 - second escape velocity.

Solving this equation for v 2, we get

v 2 = 2 G M R . (\displaystyle v_(2)=(\sqrt (2G(\frac (M)(R)))).)

There is a simple relationship between the first and second cosmic velocities:

v 2 = 2 v 1 . (\displaystyle v_(2)=(\sqrt (2))v_(1).)

The square of the escape velocity is equal to twice the Newtonian potential at a given point (for example, on the surface of a celestial body):

v 2 2 = − 2 Φ = 2 G M R . (\displaystyle v_(2)^(2)=-2\Phi =2(\frac (GM)(R)).)

Any object, being thrown up, sooner or later ends up on the 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 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 untwisting 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 greater the likelihood that it 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=√µ/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 the cosmic velocity exceeds this value, the more elongated the orbit will acquire, moving away from the Earth to an ever greater distance. At some point, this orbit will break, take the shape of a parabola, and spacecraft will go to explore outer 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 the gravitational constant (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).

The first escape velocity is the minimum speed at which a body moving horizontally above the surface of the planet will not fall onto it, but will move in a circular orbit.

Let's consider the motion of a body in a non-inertial frame of reference - relative to the Earth.

In this case, the object in orbit will be at rest, since two forces will act on it: centrifugal force and gravitational force.

where m is the mass of the object, M is the mass of the planet, G is the gravitational constant (6.67259 10 −11 m? kg −1 s −2),

The first escape velocity, R is the radius of the planet. Substituting numerical values (for Earth 7.9 km/s

The first escape velocity can be determined through the acceleration of gravity - since g = GM/R?, then

The second cosmic velocity is the lowest speed that must be given to an object whose mass is negligible compared to the mass of a celestial body in order to overcome the gravitational attraction of this celestial body and leave a circular orbit around it.

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. 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, v 2 is the second escape velocity.

There is a simple relationship between the first and second cosmic velocities:

The square of the escape velocity is equal to twice the Newtonian potential at a given point:

You can also find the information you are interested in in the scientific search engine Otvety.Online. Use the search form:

V. Equilibrium conditions for a mechanical system.

Let only a conservative force act on a certain body. 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 zero:

(12.19)

(12.20)

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

derivative of potential energy with respect to x equal to zero.

(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, and in a state of 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 x 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 x, 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 x:

it turns out negative (
>0, since the angle
spicy).

This means that the direction of the force
opposite positive axis direction x, 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 through frequency (
), we get

The moment of inertia of the system considered in this problem is equal to the sum of the 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, imparting angular acceleration to it and participating in the rotation of the rod around its 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.

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.

Cosmic velocity (first v1, second v2, third v3 and fourth v4) is the minimum speed at which any body in free movement will be able:

v1 - become a satellite of a celestial body (that is, the ability to orbit around the NT and not fall on the surface of the NT).

v2 - overcome the gravitational attraction of a celestial body.

v3 - leave the solar system, overcoming the gravity of the Sun.

v4 - leave the Milky Way galaxy.

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 of the centrifugal force and the gravitational force 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 the acceleration of gravity - 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 the gravitational attraction of 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 escape velocity is called parabolic because bodies with a second escape velocity move along a parabola.

Derivation of the formula:

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 making the best use of the orbital motion of the planet, a spacecraft can reach a third of escape velocity already at 16.6 km/s relative to the Earth, and when launching 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 is 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 makes up only a small part of the total gravitating mass, and the rest is hidden mass .

What speed is called the second cosmic speed. IV. Calculation of the second escape velocity (for Earth)

Calculation

More on topic 15. Derivation of formulas for the 1st and 2nd cosmic velocities:

V. Equilibrium conditions for a mechanical system.