DEFINITION

Law gravity opened by I. Newton:

Two bodies are attracted to each other with , which is directly proportional to their product and inversely proportional to the square of the distance between them:

Description of the law of gravity

The coefficient is the gravitational constant. In the SI system, the gravitational constant has the value:

This constant, as can be seen, is very small, so the gravitational forces between bodies with small masses are also small and practically not felt. However, the motion of cosmic bodies is completely determined by gravity. The presence of universal gravitation or, in other words, gravitational interaction explains what the Earth and planets “hold” on, and why they move around the Sun along certain trajectories, and do not fly away from it. The law of universal gravitation allows us to determine many characteristics of celestial bodies - the masses of planets, stars, galaxies and even black holes. This law allows you to calculate the orbits of the planets with great accuracy and create mathematical model Universe.

With the help of the law of universal gravitation, it is also possible to calculate cosmic velocities. For example, the minimum speed at which a body moving horizontally above the Earth's surface will not fall on it, but will move in a circular orbit is 7.9 km/s (the first space velocity). In order to leave the Earth, i.e. to overcome its gravitational attraction, the body must have a speed of 11.2 km / s, (the second cosmic velocity).

Gravity is one of the most amazing natural phenomena. In the absence of gravitational forces, the existence of the Universe would be impossible, the Universe could not even arise. Gravity is responsible for many processes in the Universe - its birth, the existence of order instead of chaos. The nature of gravity is still not fully understood. To date, no one has been able to develop a worthy mechanism and model of gravitational interaction.

Gravity

A special case of the manifestation of gravitational forces is gravity.

Gravity is always directed vertically downward (towards the center of the Earth).

If the force of gravity acts on the body, then the body performs. The type of movement depends on the direction and module of the initial speed.

We deal with the force of gravity every day. , after a while it is on the ground. The book, released from the hands, falls down. Having jumped, a person does not fly into outer space and descends to the ground.

Considering the free fall of a body near the Earth's surface as a result of the gravitational interaction of this body with the Earth, we can write:

whence the acceleration free fall:

The free fall acceleration does not depend on the mass of the body, but depends on the height of the body above the Earth. The globe is slightly flattened at the poles, so bodies near the poles are slightly closer to the center of the earth. In this regard, the acceleration of free fall depends on the latitude of the area: at the pole it is slightly greater than at the equator and other latitudes (at the equator m / s, at the North Pole equator m / s.

The same formula allows you to find the free fall acceleration on the surface of any planet with mass and radius .

Examples of problem solving

EXAMPLE 1 (the problem of "weighing" the Earth)

Exercise The radius of the Earth is km, the acceleration of free fall on the surface of the planet is m/s. Using these data, estimate the approximate mass of the Earth.
Solution Acceleration of free fall at the surface of the Earth:

whence the mass of the Earth:

In the C system, the radius of the Earth m.

Substituting numerical values ​​into the formula physical quantities Let's estimate the mass of the Earth:

Answer Mass of the Earth kg.

EXAMPLE 2

Exercise An Earth satellite moves in a circular orbit at an altitude of 1000 km from the Earth's surface. How fast is the satellite moving? How long will it take the satellite to make one full turn around the Earth?
Solution According to , the force acting on the satellite from the side of the Earth is equal to the product of the mass of the satellite and the acceleration with which it moves:

The force acting on the satellite from the earth gravitational attraction, which according to the law of universal gravitation is equal to:

where and are the masses of the satellite and the Earth, respectively.

Since the satellite is at a certain height above the surface of the Earth, the distance from it to the center of the Earth:

where is the radius of the earth.

Gravity is the force with which the Earth attracts a body near its surface. .

The phenomena of gravity can be observed everywhere in the world around us. A ball thrown up falls down, a stone thrown in a horizontal direction will end up on the ground after a while. An artificial satellite launched from the Earth, due to the effect of gravity, does not fly in a straight line, but moves around the Earth.

Gravity always pointing vertically down toward the center of the earth. It is denoted by the Latin letter F t (t- heaviness). The force of gravity is applied to the center of gravity of the body.

To find the center of gravity of an arbitrary shape, you need to hang the body on threads at its different points. The point of intersection of all directions marked by the thread will be the center of gravity of the body. Center of gravity of bodies correct form is located in the center of symmetry of the body, and it is not necessary that it belongs to the body (for example, the center of symmetry of the ring).

For a body near the surface of the Earth, the force of gravity is:

where is the mass of the Earth, m- body mass , R is the radius of the earth.

If only this force acts on the body (and all others are balanced), then it makes a free fall. The acceleration of this free fall can be found by applying Newton's second law:

(2)

From this formula, we can conclude that the acceleration of free fall does not depend on the mass of the body m, therefore, it is the same for all bodies. According to Newton's second law, gravity can be defined as the product of a body's mass and acceleration (in this case, the acceleration due to gravity g);

Gravity, acting on the body, is equal to the product of the mass of the body and the acceleration of free fall.

Like Newton's second law, formula (2) is valid only in inertial frames of reference. On the surface of the Earth, only systems associated with the Earth's poles, which do not participate in its daily rotation, can be inertial reference systems. All other points earth's surface moving along circles with centripetal accelerations and the frames of reference associated with these points are non-inertial.

Due to the rotation of the Earth, the acceleration of free fall at different latitudes is different. However, free fall accelerations in different regions of the globe differ very little and differ very little from the value calculated by the formula

Therefore, in rough calculations, the non-inertial reference frame associated with the Earth's surface is neglected, and the free fall acceleration is assumed to be the same everywhere.

1. What letter denotes the force of gravity and in what units is it measured in Si? 2. What letter denotes body weight and in what units is it measured in C? 3. What letter denotes density and in what units is it measured in C? 4. Write down the formula for calculating gravity. 5. In what units is body mass measured in C? 6. Formula for calculating body weight? 7. What force is called gravity? 8. What is deformation? 9. In what units is the volume of a body measured in C and what letter is it denoted by? 10. What is called body weight? 11. What is the measure of the interaction of bodies? 12. What is the free fall acceleration? 13. Write down the formula for calculating the elastic force? 14. What instrument is used to measure force?


Answers: 1) Ft. (N) 2)P (N) 3)p (kg/m 3) 4)Fgr. \u003d gm 5) (kg) 6) P \u003d gm 7) The force with which the Earth attracts a body to itself. 8) Change in the shape and size of the body. 9) V (m 3) 10) The force with which the body, as a result of attraction to the Earth, acts on a support or suspension. 11) Strength 12) g \u003d 9.8N / kg \u003d 10H / kg 13) F control \u003d K (l-l 0) 14) Dynamometer For 14 (+) - 3 points For 12 (+) - 2 points For 10 (+) - 1 point Less than 10(+) - 0 points






A woman with a cart is easier for a mare; If you don’t grease, you won’t go; Things went like clockwork; You can't hold an eel in your hands; Skis glide through the weather; A rusty plow is cleared only when plowing; What is round rolls easily; The well rope frays the log house; Mow, spit, while dew, dew down - and we are home.


1) R=20H+80H=100H R=80H-20H=60H Answer: 100H; 60H. 2) Given: Solution: F 1 =1000H R=F 1 - F 2 R=1000H – 700H=300H F 2 =700H Answer: R=300H R-? 3) Given: SI: Solution: m=500 g 0.5 kg Ft.=gm Ft.=10N/kg*0.5 kg=5H g=10H/kg N/kg Ft. N Answer: Ftyazh = 5N. 4) Given: SI Solution: P=600N N m=P/g m=600H/10H/kg=60 kg g=10H/kg H/kg Answer: m=60 kg m-? kg 5) Given: SI Solution: V=20 l 0.02 m 3 P=mg m=800 kg/m 3*0.02 m 3=16 kg p=800 kg/m 3 kg/m 3 m=pV P=16kg*10N/kg=160N. g=10H/kg H/kg Answer: P=160H P-? H

Investigating the normal acceleration that occurs when the Moon moves around the Earth, I. Newton came to the conclusion that all bodies in nature are attracted to each other with a certain force, called the force of gravity. In this case, the acceleration caused by the action of a given force is inversely proportional to the square of the distance between the considered bodies acting on each other.

Let us assume that two point bodies having masses $m_1\ and\ m_2$ are at a distance $r$ from each other. These bodies interact with forces:

According to Newton's third law, the moduli of forces are:

From what has been said above about acceleration and based on (2), we obtain:

\[\frac(m_1K_1)(r^2)=\frac(m_2K_2)(r^2)\left(3\right).\]

Formula (3) will be valid if $K_1$=$\gamma m_2$ and $K_2$=$\gamma m_1$, where $\gamma $ is some constant. Then:

where $\gamma =6.67\cdot (10)^(-11)\frac(H\cdot m^2)((kg)^2)$ is the gravitational constant.

The formulation of the law of universal gravitation

Definition

The force of attraction between two material points is directly proportional to the product of the masses of these points and inversely proportional to the square of the distance between them:

Strictly speaking, formula (4) can be used to calculate the gravitational force between homogeneous balls with masses $m_1(\ and\ m)_2$, assuming that $r$ is the distance between the centers of the balls.

In order to find the gravitational forces that act on one body from the side of another body, while the bodies cannot be considered point bodies, proceed as follows. Both bodies are theoretically divided into elements that can be taken as point masses. The gravitational forces that act on one selected element of the first body from all the elements of the other body are found, and the force that acts on the considered point of the first body is obtained. Then the operation is repeated for each point of the first body. The resulting forces are added taking into account their directions. The result is the gravitational force with which the second body acts on the first. Such a task is very difficult.

Gravity

Definition

Gravity(the force of attraction to the Earth) is a special case of the appearance of the force of universal gravitation. Let's denote gravity as $F_t$. In accordance with the law of universal gravitation, this force is equal to:

where $m$ is the mass of the body attracted to the Earth; $M$ - mass of the Earth; $R$ - radius of the Earth; $h$ - the height of the body above the Earth's surface.

The force of gravity is directed towards the center of the earth. In problems, if the size of the Earth is much larger than the bodies under consideration, it is considered that the force of gravity is directed vertically downwards.

Gravity imparts an acceleration to bodies located near the surface of the Earth, which is called the acceleration of free fall, denoted as $\overline(g)$. According to Newton's second law, we have:

\[\overline(g)=\frac((\overline(F))_t)(m)\left(6\right).\]

Taking into account expression (5), we have:

\[\left|\overline(g)\right|=\gamma \frac(M)((\left(R+h\right))^2)\left(7\right).\]

Directly on the surface of the Earth (at $h=0$) the value of the free fall acceleration is equal to:

the magnitude of the gravitational acceleration calculated from (8) is approximately equal to $g\approx 9.8\ \frac(m)(c^2).$ It should be known that even at the surface of the Earth, the gravitational acceleration modulus is not the same everywhere, since The earth is not a perfect sphere, and it rotates on its axis and moves along a curved path around the sun.

Using Newton's second law and expression (8), the force of gravity is written as:

\[(\overline(F))_t=m\overline(g)\left(9\right).\]

Examples of problems with a solution

Example 1

Exercise. What is the gravitational force of two bodies whose masses are $(m=10)^4\ kg$ if the distance between their centers is $r=100$m? Consider the bodies as uniform spheres.

Solution. Since, according to the condition of the problem, the mass of bodies has spherical symmetry (homogeneous balls), then to calculate the gravitational force, you can use the formula:

Taking into account the equality of the masses of the bodies, we transform the expression (1.1) into the form:

Calculate the required force:

Answer.$F=6.67\cdot (10)^(-7)$N

Example 2

Exercise. Some body located at the Earth's pole was thrown vertically upwards with a speed $v_0$. To what height ($h$) will this body rise? Assume that the radius of the Earth ($R$) and the free fall acceleration ($g$) are known. Ignore air resistance.

Solution. We will solve the problem on the basis of the conservation law mechanical energy, since there are no resistance forces, the system is conservative. The body at the time of the throw has kinetic energy:

The potential energy of the interaction of the body and the Earth on the surface of the latter is equal to:

where $M$ is the mass of the Earth. When the body reaches the point of maximum lift, it has only potential energy:

From the law of conservation of energy we have:

Taking into account that

Answer.$h=\frac(R)(\frac(2gR)(v^2_0)-1)$

In physics, there are a huge number of laws, terms, definitions and formulas that explain everything natural phenomena on earth and in the universe. One of the main ones is the law of universal gravitation, which was discovered by the great and well-known scientist Isaac Newton. Its definition looks like this: any two bodies in the Universe are mutually attracted to each other with a certain force. The formula for universal gravitation, which calculates this force, will look like this: F = G*(m1*m2 / R*R).

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History of the discovery of the law

For a very long time people have studied the sky. They wanted to know all its features, all that reign in the inaccessible space. A calendar was compiled from the sky, important dates and dates of religious holidays were calculated. People believed that the center of the entire Universe is the Sun, around which all celestial subjects revolve.

A truly stormy scientific interest in space and astronomy in general appeared in the 16th century. Tycho Brahe, the great astronomer, during his research observed the movements of the planets, recorded and systematized observations. By the time Isaac Newton discovered the law of universal gravitation, the Copernican system had already been established in the world, according to which everything celestial bodies revolve around the star in fixed orbits. The great scientist Kepler, on the basis of Brahe's research, discovered the kinematic laws that characterize the motion of the planets.

Based on Kepler's laws, Isaac Newton opened his and found out, what:

  • The movements of the planets indicate the presence of a central force.
  • The central force causes the planets to move in their orbits.

Formula parsing

There are five variables in Newton's law formula:

How accurate are the calculations

Since Isaac Newton's law refers to mechanics, calculations do not always accurately reflect the real force with which bodies interact. Furthermore , this formula can only be used in two cases:

  • When the two bodies between which the interaction occurs are homogeneous objects.
  • When one of the bodies is material point, and the other is a uniform ball.

Gravity field

According to Newton's third law, we understand that the forces of interaction of two bodies are the same in value, but opposite in its direction. The direction of forces occurs strictly along a straight line that connects the centers of mass of two interacting bodies. The interaction of attraction between bodies occurs due to the gravitational field.

Description of interaction and gravity

Gravity has very long-range interaction fields. In other words, its influence extends over very large, cosmic scale distances. Due to gravity, people and all other objects are attracted to the earth, and the earth and all planets solar system are attracted to the sun. Gravity is the constant influence of bodies on each other, it is a phenomenon that determines the law of universal gravitation. It is very important to understand one thing - the more massive the body, the more gravity it has. The Earth has a huge mass, so we are attracted to it, and the Sun weighs several million times more than the Earth, so our planet is attracted to the star.

Albert Einstein, one of the greatest physicists, argued that gravity between two bodies is due to the curvature of space-time. The scientist was sure that space, like tissue, can be pressed through, and the more massive the object, the more it will push through this tissue. Einstein was the author of the theory of relativity, which states that everything in the universe is relative, even such a quantity as time.

Calculation example

Let's try, using the already known formula of the law of universal gravitation, solve a physics problem:

  • The radius of the Earth is approximately equal to 6350 kilometers. We take the acceleration of free fall as 10. It is necessary to find the mass of the Earth.

Solution: The free fall acceleration at the Earth will be equal to G*M / R^2. From this equation, we can express the mass of the Earth: M = g * R ^ 2 / G. It remains only to substitute the values ​​\u200b\u200bin the formula: M = 10 * 6350000 ^ 2 / 6, 7 * 10 ^-11. In order not to suffer with degrees, we bring the equation to the form:

  • M = 10* (6.4*10^6)^2 / 6.7 * 10^-11.

Having calculated, we get that the mass of the Earth is approximately equal to 6 * 10 ^ 24 kilograms.