What determines the first space velocity. Life of wonderful names

Any object, being tossed up, sooner or later ends up on the earth's surface, whether it be a stone, a sheet of paper or a simple feather. At the same time, a satellite launched into space half a century ago, a space station or the Moon continue to rotate in their orbits, as if they were not affected by our planet at all. Why is this happening? Why does the Moon not threaten to fall to the Earth, and the Earth does not move towards the Sun? Isn't gravity acting on them?

From the school physics course, we know that universal gravitation affects any material body. Then it would be logical to assume that there is a certain force that neutralizes the effect of gravity. This force is called centrifugal. Its action is easy to feel by tying a small load to one end of the thread and spinning it around the circumference. At the same time, the greater the rotation speed, the stronger the tension of the thread, and the slower we rotate the load, the more likely it is to fall down.

Thus, we came close to the concept of "space velocity". In a nutshell, it can be described as the speed that allows any object to overcome the gravity of a celestial body. The planet, its or another system can act as a quality. Every object that moves in orbit has space velocity. By the way, the size and shape of the orbit depend on the magnitude and direction of the speed that this object received at the time the engines were turned off, and the altitude at which this event occurred.

Space velocity is of four types. The smallest of them is the first one. This is the minimum speed that it must have in order 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 ball (at the poles it is, as it were, slightly flattened), the distance from the center to the surface is greatest at the equator - 6378.1. 10 (3) m, and least of all at the poles - 6356.8. 10 (3) m. If we take the 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 at an ever greater distance. At some point, this orbit will break, take the form of a parabola, and the spacecraft will go to surf space. In order to leave the planet, the ship must have the second space 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 cosmic velocity, both the first and the second, is equal to for each planet of our native system. They are easy to calculate using the above formulas, if we 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 the near-solar orbit from the Earth, it will be necessary to accelerate to 16.6 km / s.

And, finally, the fourth cosmic speed. With its help, you can overcome the attraction of the galaxy itself. Its value varies depending on the coordinates of the galaxy. For ours, this value is approximately 550 km / s (if calculated relative to the Sun).

We - earthlings - are used to standing firmly on the ground and not flying anywhere, and if we throw some object into the air, it will surely fall to the surface. The gravitational field created by our planet is to blame for everything, which bends space-time and makes an apple thrown to the side, for example, fly along a curved path and intersect with the Earth.

The gravitational field creates around itself any object, and the Earth, which has an impressive mass, this field is quite strong. That is why powerful multi-stage space rockets are being built, capable of accelerating spacecraft to high speeds, which are needed to overcome the gravity of the planet. The value of these speeds are called the first and second cosmic speeds.

The concept of the first cosmic velocity is very simple - this is the velocity that must be given to a physical object so that it, moving parallel to the cosmic body, cannot fall on it, but at the same time remains in a constant orbit.

The formula for finding the first space velocity is not difficult: whereV G Mis the mass of the object;Ris the radius of the object;

Try to substitute the necessary values ​​\u200b\u200bin the formula (G - the gravitational constant is always equal to 6.67; the mass of the Earth is 5.97 10 24 kg, and its radius is 6371 km) and find the first space velocity of our planet.

As a result, we will get a speed equal to 7.9 km / s. But why, moving exactly at such a speed, the spacecraft will not fall to the Earth or fly away into outer space? It will not fly into space due to the fact that this speed is still too low to overcome the gravitational field, but it will just fall to Earth. But only because of the high speed, it will always "avoid" a collision with the Earth, while at the same time continuing its "fall" in a circular orbit caused by the curvature of space.

It is interesting: The International Space Station “works” on the same principle. The astronauts who are on it spend all the time in a constant and incessant fall, which does not end tragically due to the high speed of the station itself, which is why it consistently “misses” past the Earth. The speed value is calculated from .

But what if we want the spacecraft to leave our planet and not be dependent on its gravitational field? Accelerate it to the second space speed! So, the second cosmic speed is the minimum speed that must be given to a physical object so that it overcomes the gravitational attraction of a celestial body and leaves its closed orbit.

The value of the second space velocity also depends on the mass and radius of the celestial body, so it will be different for each object. For example, in order to overcome the gravitational attraction of the Earth, the spacecraft needs to gain a minimum speed of 11.2 km/s, Jupiter - 61 km/s, the Sun - 617.7 km/s.

The second escape velocity (V2) can be calculated using the following formula:

where Vis the first cosmic velocity;Gis the gravitational constant;Mis the mass of the object;Ris the radius of the object;

But if the first cosmic velocity of the object under study (V1) is known, then the task is greatly facilitated, and the second cosmic velocity (V2) is quickly found by the formula:

It is interesting: second black hole cosmic formula more299,792 km/c, which is more than the speed of light. That is why nothing, not even light, can break out of it.

In addition to the first and second comic speeds, there are third and fourth, which must be reached in order to go beyond our solar system and galaxy, respectively.

Illustration: bigstockphoto | 3DSculptor

If you find an error, please highlight a piece of text and click Ctrl+Enter.

If a certain body is given a speed equal to the first cosmic velocity, then it will not fall to the Earth, but will become an artificial satellite moving in a near-Earth circular orbit. Recall that this speed should be perpendicular to the direction to the center of the Earth and equal in magnitude
v I = √(gR) = 7.9 km/s,
where g \u003d 9.8 m / s 2− free fall acceleration of bodies near the Earth's surface, R = 6.4 × 10 6 m− radius of the Earth.

Can a body completely break the chains of gravity that “bind” it to the Earth? It turns out that it can, but for this it needs to be “thrown” with even greater speed. The minimum initial speed that must be reported to the body at the surface of the Earth in order for it to overcome the earth's gravity is called the second cosmic velocity. Let's find its meaning vII.
When the body moves away from the Earth, the force of attraction does negative work, as a result of which the kinetic energy of the body decreases. At the same time, the force of attraction also decreases. If the kinetic energy falls to zero before the force of attraction becomes zero, the body will return back to Earth. To prevent this from happening, it is necessary that the kinetic energy be kept non-zero until the force of attraction vanishes. And this can happen only at an infinitely large distance from the Earth.
According to the kinetic energy theorem, the change in the kinetic energy of a body is equal to the work done by the force acting on the body. For our case, we can write:
0 − mv II 2 /2 = A,
or
mv II 2 /2 = −A,
where m is the mass of the body thrown from the Earth, A− work of the force of attraction.
Thus, to calculate the second cosmic velocity, it is necessary to find the work of the force of attraction of the body to the Earth when the body moves away from the surface of the Earth to an infinitely large distance. Surprising as it may seem, this work is not at all infinitely large, despite the fact that the movement of the body seems to be infinitely large. The reason for this is the decrease in the force of attraction as the body moves away from the Earth. What is the work done by the force of attraction?
Let's take advantage of the feature that the work of the gravitational force does not depend on the shape of the body's trajectory, and consider the simplest case - the body moves away from the Earth along a line passing through the center of the Earth. The figure shown here shows the globe and a body of mass m, which moves along the direction indicated by the arrow.

Find a job first A 1, which makes the force of attraction in a very small area from an arbitrary point N to the point N 1. The distances of these points to the center of the Earth will be denoted by r and r1, respectively, so work A 1 will be equal to
A 1 = -F(r 1 - r) = F(r - r 1).
But what is the meaning of strength F should be substituted into this formula? Because it changes from point to point: N it is equal to GmM/r 2 (M is the mass of the Earth), at the point N 1 − GmM/r 1 2.
Obviously, you need to take the average value of this force. Since the distances r and r1, differ little from each other, then as the average we can take the value of the force at some midpoint, for example, such that
r cp 2 = rr 1.
Then we get
A 1 = GmM(r − r 1)/(rr 1) = GmM(1/r 1 − 1/r).
Arguing in the same way, we find that on the segment N 1 N 2 work is done
A 2 = GmM(1/r 2 − 1/r 1),
Location on N 2 N 3 work is
A 3 = GmM(1/r 3 − 1/r 2),
and on the site NN 3 work is
A 1 + A 2 + A 2 = GmM(1/r 3 − 1/r).
The pattern is clear: the work of the force of attraction when moving a body from one point to another is determined by the difference in the reciprocal distances from these points to the center of the Earth. Now it's easy to find and all the work BUT when moving a body from the surface of the Earth ( r = R) over an infinite distance ( r → ∞, 1/r = 0):
A = GmM(0 − 1/R) = −GmM/R.
As can be seen, this work is indeed not infinitely large.
Substituting the resulting expression for BUT into the formula
mv II 2 /2 = −GmM/R,
find the value of the second cosmic velocity:
v II = √(−2A/m) = √(2GM/R) = √(2gR) = 11.2 km/s.
This shows that the second cosmic velocity in √{2} times greater than the first cosmic velocity:
vII = √(2)vI.
In our calculations, we did not take into account the fact that our body interacts not only with the Earth, but also with other space objects. And first of all - with the Sun. Having received the initial speed equal to vII, the body will be able to overcome gravity towards the Earth, but will not become truly free, but will turn into a satellite of the Sun. However, if the body near the surface of the Earth is informed of the so-called third cosmic velocity vIII = 16.6 km/s, then it will be able to overcome the force of attraction to the Sun.
See example

"Uniform and uneven movement" - t 2. Uneven movement. Yablonevka. L 1. Uniform and. L2. t 1. L3. Chistoozernoe. t 3. Uniform movement. =.

"Curvilinear motion" - Centripetal acceleration. UNIFORM MOVEMENT OF A BODY IN A CIRCLE Distinguish: - curvilinear movement with a constant modulo speed; - movement with acceleration, tk. speed changes direction. Direction of centripetal acceleration and velocity. The movement of a point in a circle. The movement of a body in a circle with a constant modulo speed.

"Movement of bodies in a plane" - Estimate the obtained values ​​of unknown quantities. Substitute numerical data in a general solution, perform calculations. Make a drawing, depicting interacting bodies on it. Perform an analysis of the interaction of bodies. Ftr. Motion of a body on an inclined plane without friction force. Study of the motion of a body along an inclined plane.

"Support and movement" - An ambulance brought a patient to us. Slender, round-shouldered, strong, strong, fat, clumsy, agile, pale. Game situation “Council of Doctors”. Sleep on a hard bed with a low pillow. Body support and movement. Rules for maintaining correct posture. Correct posture when standing. The bones of children are soft and elastic.

"Space Speed" - V1. USSR. That's why. April 12, 1961 Message to extraterrestrial civilizations. Third cosmic speed. On board Voyager 2 is a disk with scientific information. Calculation of the first cosmic velocity at the Earth's surface. The first manned flight into space. The trajectory of Voyager 1. The trajectory of movement of bodies moving at low speed.

"Body dynamics" - What is the basis of the dynamics? Dynamics is a branch of mechanics that considers the causes of the movement of bodies (material points). Newton's laws are applicable only for inertial frames of reference. Frames of reference in which Newton's first law is satisfied are called inertial. Dynamics. What are the frames of reference for Newton's laws?

In total there are 20 presentations in the topic

02.12.2014

Lesson 22 (Grade 10)

Topic. Artificial satellites of the Earth. development of astronautics.

About the movement of thrown bodies

In 1638, Galileo's book "Conversations and Mathematical Proofs Concerning Two New Branches of Science" was published in Leiden. The fourth chapter of this book was called "On the motion of thrown bodies." Not without difficulty, he managed to convince people that in an airless space "a grain of lead should fall with the same speed as a cannonball." But when Galileo told the world that a cannonball that had flown out of a cannon in a horizontal direction had been in flight for the same amount of time as a cannonball that had simply fallen out of its muzzle to the ground, they did not believe him. Meanwhile, this is true: a body thrown from a certain height in a horizontal direction moves to the ground in the same time as if it had simply fallen vertically downward from the same height.
To verify this, we will use the device, the principle of operation of which is illustrated in Figure 104, a. After being hit with a hammer M on elastic plate P the balls begin to fall and, despite the difference in trajectories, simultaneously reach the ground. Figure 104b shows a stroboscopic photograph of falling balls. To obtain this photograph, the experiment was carried out in the dark, and the balls were illuminated at regular intervals with a bright flash of light. At the same time, the camera shutter was open until the balls fell to the ground. We see that at the same moments of time when the flashes of light occurred, both balls were at the same height and just as simultaneously they reached the ground.

Free fall time h(near the surface of the Earth) can be found by the formula known from mechanics s=at2/2. Replacing here s on the h and a on the g, we rewrite this formula in the form

whence, after simple transformations, we obtain

The same time will be in flight and the body thrown from the same height in the horizontal direction. In this case, according to Galileo, "to the uniform unhindered movement, another is added, caused by gravity, due to which a complex movement arises, consisting of uniform horizontal and naturally accelerated movements."
During the time determined by expression (44.1), moving in a horizontal direction with a speed v0(i.e., with the speed with which it was thrown), the body will move horizontally a distance

From this formula it follows that the flight range of a body thrown in a horizontal direction is proportional to the initial speed of the body and increases with the height of the throw.
To find out which trajectory the body moves in this case, let us turn to experiment. We attach a rubber tube equipped with a tip to the water tap and direct the stream of water in a horizontal direction. In this case, the water particles will move in exactly the same way as a body thrown in the same direction. Turning off or, conversely, turning the tap, you can change the initial speed of the jet and thus the range of flight of water particles (Fig. 105), however, in all cases, the water jet will have the form parabolas. To verify this, a screen with parabolas pre-drawn on it should be placed behind the jet. The water jet will exactly match the lines shown on the screen.

So, a freely falling body with a horizontal initial velocity moves along a parabolic trajectory.
By parabola the body will also move when it is thrown at some acute angle to the horizon. The flight range in this case will depend not only on the initial speed, but also on the angle at which it was directed. Conducting experiments with a jet of water, it can be established that the greatest flight range is achieved when the initial speed makes an angle of 45 ° with the horizon (Fig. 106).

At high speeds of movement of bodies, air resistance should be taken into account. Therefore, the flight range of bullets and projectiles in real conditions is not the same as it follows from the formulas that are valid for movement in an airless space. So, for example, with an initial bullet velocity of 870 m/s and an angle of 45°, in the absence of air resistance, the flight range would be approximately 77 km, while in reality it does not exceed 3.5 km.

first cosmic speed

Let us calculate the speed that must be reported to an artificial satellite of the Earth so that it moves in a circular orbit at a height h over the ground.
At high altitudes, the air is very rarefied and offers little resistance to bodies moving in it. Therefore, we can assume that the satellite is affected only by the gravitational force directed to the center of the Earth ( fig.4.4).

According to Newton's second law.
The centripetal acceleration of the satellite is given by , where h is the height of the satellite above the Earth's surface. The force acting on the satellite, according to the law of universal gravitation, is determined by the formula, where M is the mass of the earth.
Substituting the values F and a into the equation for Newton's second law, we get

From the obtained formula it follows that the speed of the satellite depends on its distance from the Earth's surface: the greater this distance, the lower the speed it will move in a circular orbit. It is noteworthy that this speed does not depend on the mass of the satellite. This means that any body can become a satellite of the Earth if it is given a certain speed. In particular, when h=2000 km=2 10 6 m speed v≈ 6900 m/s.
The minimum speed that must be imparted to a body on the surface of the Earth in order for it to become a satellite of the Earth moving in a circular orbit is called first cosmic speed.
The first cosmic velocity can be found from formula (4.7) if we take h=0:

Substituting into formula (4.8) the value G and values M and R for the Earth, you can calculate the first cosmic velocity for the Earth satellite:

If such a speed is given to a body in a horizontal direction near the surface of the Earth, then in the absence of an atmosphere it will become an artificial satellite of the Earth, circling around it in a circular orbit.
Only sufficiently powerful space rockets are able to communicate such a speed to satellites. Currently, thousands of artificial satellites are orbiting the Earth.
Any body can become an artificial satellite of another body (planet) if you tell it the necessary speed.

Movement of artificial satellites

In the works of Newton, one can find a wonderful drawing showing how it is possible to make the transition from a simple fall of a body along a parabola to an orbital motion of a body around the Earth (Fig. 107). “A stone thrown to the ground,” wrote Newton, “will deviate under the action of gravity from a straight path and, having described a curved trajectory, will finally fall to the Earth. If you throw it at a faster speed, it will fall further.” Continuing these considerations, it is easy to come to the conclusion that if a stone is thrown from a high mountain with a sufficiently high speed, then its trajectory could become such that it would never fall to the Earth at all, turning into its artificial satellite.

The minimum speed that must be given to a body near the surface of the Earth in order to turn it into an artificial satellite is called first cosmic speed.
To launch artificial satellites, rockets are used that raise the satellite to a given height and tell it the required speed in the horizontal direction. After that, the satellite is separated from the carrier rocket and continues further movement only under the influence of the Earth's gravitational field. (We neglect the influence of the Moon, the Sun and other planets here.) The acceleration imparted by this field to the satellite is the free fall acceleration g. On the other hand, since the satellite is moving in a circular orbit, this acceleration is centripetal and therefore equal to the ratio of the square of the satellite's speed to the radius of its orbit. In this way,

Where

Substituting expression (43.1) here, we obtain

We got the formula circular speed satellite , i.e., such a speed that the satellite has, moving along a circular orbit with a radius r on high h from the surface of the earth.
To find the first space velocity v1, it should be taken into account that it is defined as the speed of the satellite near the surface of the Earth, i.e. when h< and r≈R3. Taking this into account in formula (45.1), we obtain

Substituting numerical data into this formula leads to the following result:

For the first time, it was possible to tell the body such a tremendous speed only in 1957, when the first in the world was launched in the USSR under the leadership of S.P. Korolev artificial earth satellite(abbreviated AES). The launch of this satellite (Fig. 108) is the result of outstanding achievements in the field of rocket technology, electronics, automatic control, computer technology and celestial mechanics.

In 1958, the first American satellite "Explorer-1" was launched into orbit, and somewhat later, in the 60s, other countries also launched satellites: France, Australia, Japan, China, Great Britain, etc., and many The satellites were launched using American launch vehicles.
At present, the launch of artificial satellites is commonplace, and international cooperation has long been widespread in the practice of space research.
Satellites launched in different countries can be divided according to their purpose into two classes:
1. Research satellites. They are designed to study the Earth as a planet, its upper atmosphere, near-Earth space, the Sun, stars and the interstellar medium.
2. Applied satellites. They serve to satisfy the earthly needs of the national economy. This includes communication satellites, satellites for studying the natural resources of the Earth, meteorological satellites, navigation, military, etc.
AES intended for human flight include manned satellite ships and orbital stations.
In addition to working satellites in near-Earth orbits, the so-called auxiliary objects also circulate around the Earth: the last stages of launch vehicles, head fairings and some other parts that are separated from satellites when they are put into orbit.
Note that due to the enormous air resistance near the Earth's surface, the satellite cannot be launched too low. For example, at an altitude of 160 km, it is able to make only one revolution, after which it decreases and burns out in the dense layers of the atmosphere. For this reason, the first artificial Earth satellite, launched into orbit at an altitude of 228 km, lasted only three months.
As altitude increases, atmospheric drag decreases and h>300 km becomes negligible.
The question arises: what will happen if a satellite is launched at a speed greater than the first space one? Calculations show that if the excess is insignificant, then the body remains an artificial satellite of the Earth, but it no longer moves in a circle, but along elliptical orbit. With increasing speed, the satellite's orbit becomes more and more elongated, until it finally "breaks", turning into an open (parabolic) trajectory (Fig. 109).

The minimum speed that must be given to a body near the surface of the Earth in order for it to leave it, moving along an open trajectory, is called second cosmic speed.
The second cosmic velocity is √2 times greater than the first cosmic one:

At this speed, the body leaves the area of ​​gravity and becomes a satellite of the Sun.
To overcome the attraction of the Sun and leave the solar system, you need to develop even greater speed - third space. The third escape velocity is 16.7 km/s. Having approximately this speed, the automatic interplanetary station "Pioneer-10" (USA) in 1983 for the first time in the history of mankind went beyond the solar system and is now flying towards Barnard's star.

Examples of problem solving

Task 1. A body is thrown vertically upward at a speed of 25 m/s. Determine the climb height and flight time.

Given: Solution:

; 0=0+25 . t-5 . t2

; 0=25-10 . t 1 ; t 1 \u003d 2.5 s; H=0+25. 2.5-5. 2.5 2 =31.25 (m)

t-? 5t=25; t=5c

H-? Answer: t=5c; H=31.25 (m)

Rice. 1. Choice of reference system

First we must choose a frame of reference. Reference system select the one connected to the ground, the starting point of the movement is indicated by 0. The Oy axis is directed vertically upwards. The velocity is directed upwards and coincides in direction with the Oy axis. Free fall acceleration is directed downward along the same axis.

Let's write down the law of motion of the body. We must not forget that speed and acceleration are vector quantities.

The next step. Note that the final coordinate, at the end, when the body has risen to some height and then fell back to the ground, will be 0. The initial coordinate is also 0: 0=0+25 . t-5 . t2.

If we solve this equation, we get the time: 5t=25; t=5 s.

Let us now determine the maximum lifting height. First, we determine the time of lifting the body to the top point. To do this, we use the velocity equation: .

We have written the equation in general form: 0=25-10 . t1,t 1 \u003d 2.5 s.

When we substitute the values ​​​​known to us, we get that the time of lifting the body, time t 1 is 2.5 s.

Here I would like to note that the entire flight time is 5 s, and the rise time to the maximum point is 2.5 s. This means that the body rises exactly as much time as it will then fall back to the ground. Now let's use the equation we've already used, the law of motion. In this case, we put H instead of the final coordinate, i.e. maximum lifting height: H=0+25. 2.5-5. 2.5 2 =31.25 (m).

After making simple calculations, we get that the maximum height of the body will be 31.25 m. Answer: t=5c; H=31.25 (m).

In this case, we used almost all the equations that we studied in the study of free fall.

Task 2. Determine the height above ground level at which acceleration of gravity is reduced by half.

Given: Solution:

R W \u003d 6400 km; ;

H-? Answer: H ≈ 2650 km.

To solve this problem, we need, perhaps, one single data. This is the radius of the earth. It is equal to 6400 km.

Acceleration of gravity is determined on the surface of the Earth by the following expression: . It's on the surface of the earth. But as soon as we move away from the Earth at a great distance, the acceleration will be determined as follows: .

If we now divide these quantities by each other, we get the following: .

Constant values ​​are reduced, i.e. the gravitational constant and the mass of the Earth, but the radius of the Earth and the height remain, and this ratio is 2.

Transforming the equations now obtained, we find the height: .

If we substitute the values ​​in the resulting formula, we get the answer: H ≈ 2650 km.

Task 3.A body moves along an arc with a radius of 20 cm at a speed of 10 m/s. Determine the centripetal acceleration.

Given: SI Solution:

R=20 cm 0.2 m

V=10 m/s

and C - ? Answer: a C = .

Formula to calculate centripetal acceleration known. Substituting the values ​​here, we get: . In this case, the centripetal acceleration is huge, look at its value. Answer: a C =.