> Gravitational potential energy

What's happened gravitational energy: potential energy of gravitational interaction, the formula for gravitational energy and the law gravity Newton.

Gravitational energy is the potential energy associated with the gravitational force.

Learning task

• Calculate the gravitational potential energy for two masses.

Key Points

Terms

• Potential energy is the energy of an object in its position or chemical state.
• Newton's gravitational backwater - each point universal mass attracts another with the help of a force that is directly proportional to their masses and inversely proportional to the square of their distance.
• Gravity is the net force on the ground that pulls objects toward the center. Created by rotation.

Example

What will be the gravitational potential energy of a 1 kg book at a height of 1 m? Since the position is set close to the earth's surface, the gravitational acceleration will be constant (g = 9.8 m/s 2), and the energy of the gravitational potential (mgh) reaches 1 kg ⋅ 1 m ⋅ 9.8 m/s 2 . This can also be seen in the formula:

If you add the mass and the earth's radius.

Gravitational energy reflects the potential associated with the force of gravity, because it is necessary to overcome the earth's gravity in order to do work on lifting objects. If an object falls from one point to another inside gravitational field, then gravity will do positive work, and the gravitational potential energy will decrease by the same amount.

Let's say we have a book left on the table. When we move it from the floor to the top of the table, a certain external intervention works against the gravitational force. If it falls, then this is the work of gravity. Therefore, the process of falling reflects the potential energy accelerating the mass of the book and transforming into kinetic energy. As soon as the book touches the floor, the kinetic energy becomes heat and sound.

The gravitational potential energy is affected by the height relative to a specific point, the mass and strength of the gravitational field. So the book on the table is inferior in gravitational potential energy to the heavier book below. Remember that height cannot be used in calculating gravitational potential energy unless gravity is constant.

local approximation

The strength of the gravitational field is affected by location. If the distance change is insignificant, then it can be neglected, and the force of gravity can be made constant (g = 9.8 m/s 2). Then for the calculation we use a simple formula: W = Fd. The upward force is equated to weight, so work is related to mgh, resulting in the formula: U = mgh (U is potential energy, m is the mass of the object, g is the acceleration of gravity, h is the height of the object). The value is expressed in joules. The change in potential energy is conveyed as

General formula

However, if we encounter major changes in distance, then g cannot remain constant and calculus and the mathematical definition of work must be applied. To calculate the potential energy, one can integrate the gravitational force with respect to the distance between the bodies. Then we get the formula for gravitational energy:

U = -G + K, where K is the constant of integration and is equal to zero. Here the potential energy goes to zero when r is infinite.

Introduction to Uniform Circular Motion and Gravity
Irregular circular motion
Speed, acceleration and force
Types of forces in nature
Newton's law of universal gravity

In connection with a number of features, and also in view of the special importance, the question of the potential energy of the forces of universal gravitation must be considered separately and in more detail.

We encounter the first feature when choosing the reference point for potential energies. In practice, one has to calculate the motions of a given (trial) body under the action of universal gravitational forces created by other bodies of different masses and sizes.

Let us assume that we have agreed to consider the potential energy equal to zero in a position in which the bodies are in contact. Let the test body A, when interacting separately with balls of the same mass, but different radii, first be removed from the centers of the balls at the same distance (Fig. 5.28). It is easy to see that when the body A moves before it comes into contact with the surfaces of the bodies, the gravitational forces will do different work. This means that we must consider the potential energies of the systems to be different for the same relative initial positions of the bodies.

It will be especially difficult to compare these energies with each other in cases where the interactions and movements of three or more bodies are considered. Therefore, for the forces of universal gravitation, such an initial level of counting of potential energies is sought, which could be the same, common, for all bodies in the Universe. It was agreed to consider such a common zero level of potential energy of the forces of universal gravitation the level corresponding to the location of bodies at infinitely large distances from each other. As can be seen from the law of universal gravitation, the forces of universal gravitation themselves vanish at infinity.

With such a choice of the energy reference point, an unusual situation is created with the determination of the values ​​of potential energies and the performance of all calculations.

In the cases of gravity (Fig. 5.29, a) and elasticity (Fig. 5.29, b), the internal forces of the system tend to bring the bodies to zero. As bodies approach the zero level, the potential energy of the system decreases. The zero level really corresponds to the lowest potential energy of the system.

This means that for all other positions of the bodies, the potential energy of the system is positive.

In the case of universal gravitational forces and when choosing zero energy at infinity, everything happens the other way around. The internal forces of the system tend to move the bodies away from the zero level (Fig. 5.30). They do positive work when the bodies move away from the zero level, i.e., when the bodies approach each other. At any finite distances between the bodies, the potential energy of the system is less than at In other words, the zero level (at corresponds to the highest potential energy. This means that for all other positions of the bodies, the potential energy of the system is negative.

In § 96, it was found that the work of the forces of universal gravitation when moving a body from infinity to a distance is equal to

Therefore, the potential energy of the universal gravitational forces must be considered equal to

This formula expresses another feature of the potential energy of the forces of universal gravitation - the relatively complex nature of the dependence of this energy on the distance between bodies.

On fig. 5.31 shows a graph of dependence on for the case of attraction of bodies by the Earth. This graph has the form of an isosceles hyperbola. Near the surface of the Earth, the energy changes relatively strongly, but already at a distance of several tens of Earth radii, the energy becomes close to zero and begins to change very slowly.

Any body near the Earth's surface is in a kind of "potential well". Whenever it turns out to be necessary to free the body from the action of the forces of the earth's gravity, special efforts must be made in order to "pull" the body out of this potential hole.

In the same way, all other celestial bodies create such potential holes around themselves - traps that capture and hold all not very fast moving bodies.

Knowing the nature of the dependence on makes it possible to significantly simplify the solution of a number of important practical problems. For example, you need to send a spacecraft to Mars, Venus, or any other planet in the solar system. It is necessary to determine what speed should be reported to the ship when it is launched from the surface of the Earth.

In order to send a ship to other planets, it must be removed from the sphere of influence of the forces of earth's gravity. In other words, you need to raise its potential energy to zero. This becomes possible if the ship is given such kinetic energy that it can do work against the forces of gravity, equal to where the mass of the ship,

mass and radius of the earth.

It follows from Newton's second law that (§ 92)

But since the speed of the ship before launch is zero, we can simply write:

where is the speed reported to the ship at launch. Substituting the value for A, we get

Let us use for an exception, as already done in § 96, two expressions for the force of terrestrial attraction on the surface of the Earth:

Hence - Substituting this value into the equation of Newton's second law, we obtain

The speed required to bring the body out of the sphere of influence of the forces of the earth's gravity is called the second cosmic velocity.

In the same way, one can pose and solve the problem of sending a ship to distant stars. To solve such a problem, it is already necessary to determine the conditions under which the ship will be taken out of the sphere of influence of the forces of attraction of the Sun. Repeating all the arguments that were carried out in the previous problem, we can obtain the same expression for the speed reported to the ship at launch:

Here a is the normal acceleration that the Sun informs the Earth and which can be calculated from the nature of the Earth's motion in orbit around the Sun; radius of the earth's orbit. Of course, in this case it means the speed of the ship relative to the Sun. The speed required to take a ship out of the solar system is called the third escape velocity.

The method we have considered for choosing the origin of potential energy is also used in calculations of the electrical interactions of bodies. The concept of potential wells is also widely used in modern electronics, solid state theory, atomic theory, and atomic nucleus physics.

Ticket 1

1. . The change in the kinetic energy of the system is equal to the work of all internal and external forces acting on the bodies of the system.

2. Angular moment of a material point with respect to the point O is determined by the vector product

Where is the radius vector drawn from the point O, is the momentum of the material point. J*s

3.

Ticket 2

1. Harmonic oscillator:

The kinetic energy is written as

And the potential energy is

Then the total energy has a constant value Let us find pulse harmonic oscillator. Differentiate the expression by t and, multiplying the result obtained by the mass of the oscillator, we obtain:

2. The moment of force relative to the pole is a physical quantity determined by the vector product of the radius of the vector drawn from the given pole to the point of application of force on the force vector F. newton meter

Ticket 3

1. ,

2. Oscillation phase total - the argument of a periodic function that describes an oscillatory or wave process. Hz

3.

Ticket number 4

Expressed in m/(s^2)

Ticket number 5

, F = –grad U, where .

Potential energy of elastic deformation (springs)

Find the work done when the elastic spring is deformed.
Elastic force Fupr = –kx, where k is the coefficient of elasticity. The force is not constant, so the elementary work is dA = Fdx = –kxdx.
(The minus sign indicates that work has been done on the spring). Then , i.e. A = U1 - U2. Assume: U2 = 0, U = U1, then .

On fig. 5.5 shows a diagram of the potential energy of a spring.

Rice. 5.5
Here E = K + U is the total mechanical energy of the system, K is the kinetic energy at the point x1.

Potential energy in gravitational interaction

The work of the body during the fall A = mgh, or A = U - U0.
We agreed to assume that on the Earth's surface h = 0, U0 = 0. Then A = U, i.e. A = mgh.

For the case of gravitational interaction between masses M and m, located at a distance r from each other, the potential energy can be found by the formula .

On fig. 5.4 shows a diagram of the potential energy of the gravitational attraction of the masses M and m.

Rice. 5.4
Here the total energy is E = K + E. From here it is easy to find the kinetic energy: K = E – U.

Normal acceleration is a component of the acceleration vector directed along the normal to the motion trajectory at a given point on the body motion trajectory. That is, the normal acceleration vector is perpendicular to the linear speed of movement (see Fig. 1.10). Normal acceleration characterizes the change in speed in the direction and is denoted by the letter n. The normal acceleration vector is directed along the radius of curvature of the trajectory. ( m/s 2)

Ticket number 6

Ticket 7

1)Moment of inertia of the Rod -

Hoop - L = m*R^2

Disk -

2) According to the Steiner theorem (Huygens-Steiner theorem), the moment of inertia of the body J relative to an arbitrary axis is equal to the sum of the moment of inertia of this body Jc relative to the axis passing through the center of mass of the body parallel to the considered axis, and the product of the body mass m per square distance d between axles:

where m- total body weight.

Ticket 8

1) The equation describes the change in the motion of a body of finite dimensions under the action of a force in the absence of deformation and if it moves forward. For a point, this equation is always true, so it can be considered as the basic law of motion of a material point.

Ticket 9

1) The sum of the kinetic and potential energy of the bodies that make up a closed system and interact with each other by gravitational and elastic forces remains unchanged.

2) - a curve in phase space made up of points representing a state dynamic system consecutively moments of time during the entire time of evolution.

Ticket 10

1. Moment of impulse- vector physical quantity equal to the product of the radius vector drawn from the axis of rotation to the point of application of the impulse, by the vector of this impulse

2. Angular velocity of rotation of a rigid body relative to a fixed axis- limit (at Δt → 0) of the ratio of small angular displacement Δφ to a small time interval Δt

Measured in rad/s.

Ticket 11

1. Center of mass of a mechanical system (MC)- a point whose mass is equal to the mass of the entire system, the acceleration vector of the center of mass (in the inertial frame of reference) is determined only by external forces acting on the system. Therefore, when finding the law of motion of a system of points, we can assume that the vector of the resultant external forces is applied to the center of mass of the system.
The position of the center of mass (center of inertia) of a system of material points in classical mechanics is determined as follows

MS momentum change equation:

Law of conservation of momentum MS: in a closed system, the vector sum of the impulses of all bodies included in the system remains constant for any interactions of the bodies of this system with each other.

2. Angular acceleration of rotation of a rigid body relative to a fixed axis- pseudovector physical quantity equal to the first derivative of the pseudovector of the angular velocity with respect to time.

Measured in rad / s 2.

Ticket 12

1. Potential energy of attraction of two material points


Potential energy of elastic deformations - stretching or compressing the spring leads to the storage of its potential energy of elastic deformation. The return of the spring to the equilibrium position leads to the release of the stored energy of elastic deformation.

2. Impulse of mechanical system- vector physical quantity, which is a measure of the mechanical movement of the body.

measured in

Ticket 13

1. Conservative forces. The work of gravity. Elastic force work.
In physics, conservative forces (potential forces) are forces whose work does not depend on the type of trajectory, the point of application of these forces and the law of their motion, and is determined only by the initial and final positions of this point.
The work of gravity.
Work of elastic force

2. Define the relaxation time of damped oscillations. Specify the unit for this quantity in SI.
The relaxation time is the time interval during which the amplitude of damped oscillations decreases by a factor of e (e is the base of the natural logarithm). Measured in seconds.

3. A disk with a diameter of 60 cm and a mass of 1 kg rotates around an axis passing through the center perpendicular to its plane with a frequency of 20 rpm. What work must be done to stop the disk?

Ticket 14

1. Harmonic vibrations. Vector diagram. Addition of harmonic oscillations of one direction of equal frequencies.

Harmonic oscillations are oscillations in which a physical quantity changes over time according to a harmonic (sinusoidal, cosine) law.

There is a geometric way to represent harmonic vibrations, which consists in depicting vibrations as vectors on a plane. The circuit thus obtained is called a vector diagram (Fig. 7.4).

Let's choose an axis. From the point O, taken on this axis, we set aside the length vector, which forms an angle with the axis. If we bring this vector into rotation with an angular velocity , then the projection of the end of the vector onto the axis will change with time according to the law . Therefore, the projection of the end of the vector onto the axis will make harmonic oscillations with an amplitude equal to the length of the vector; with a circular frequency equal to the angular velocity of rotation, and with an initial phase equal to the angle formed by the vector with the axis X at the initial time.

The vector diagram makes it possible to reduce the addition of oscillations to the geometric summation of vectors.

Consider the addition of two harmonic oscillations of the same direction and the same frequency, which have the following form:

Let's represent both fluctuations with the help of vectors and (fig. 7.5). Let's build the resulting vector according to the vector addition rule. It is easy to see that the projection of this vector onto the axis is equal to the sum of the projections of the terms of the vectors. Therefore, the vector represents the resulting oscillation. This vector rotates with the same angular velocity as the vectors , so that the resulting motion will be a harmonic oscillation with frequency , amplitude and initial phase . According to the law of cosines, the square of the amplitude of the resulting oscillation will be equal to

2. Define the moment of force about the axis. Specify the units of this quantity in SI.

The moment of force is a vector physical quantity equal to the vector product of the radius vector drawn from the axis of rotation to the point of application of the force by the vector of this force. It characterizes the rotational action of a force on a rigid body. The moment of force relative to an axis is a scalar value equal to the projection onto this axis of the vector moment of force relative to any point on the axis. SI: measured in kg * m 2 / s 2 = N * m.

3. A projectile weighing 100 kg flies out of a gun weighing 5 tons when fired. The kinetic energy of the projectile at the departure of 8 MJ. What is the kinetic energy of the gun due to recoil?

Ticket 15

1. The law of conservation of mechanical energy of a mechanical system.

The total mechanical energy of a closed system of bodies, between which only conservative forces act, remains constant.

In a conservative system, all forces acting on a body are potential and, therefore, can be represented as

where is the potential energy of a material point. Then Newton's second law:

where is the mass of the particle, is the vector of its velocity. Scalarly multiplying both sides of this equation by the particle velocity and taking into account that , we obtain

By elementary operations, we obtain

It follows from this that the expression under the sign of differentiation with respect to time is preserved. This expression is called the mechanical energy of a material point.

2. Define the kinetic energy of a rigid body as it rotates around a fixed axis. Specify the units of this quantity in SI.

3. A ball weighing m=20 g is introduced with an initial speed V=20 m/s into a very massive target with sand, which moves towards the ball with a speed U=10 m/s. Estimate how much heat is released during full braking of the ball.

Ticket 16

1. Moment of force about the axis- a vector physical quantity equal to the vector product of the radius vector drawn from the axis of rotation to the point of application of the force by the vector of this force. eat

Angular momentum of the MS relative to the fixed axis- a scalar value equal to the projection onto this axis of the angular momentum vector, defined relative to an arbitrary point 0 of this axis. The value of angular momentum does not depend on the position of point 0 on the z-axis.

Basic equation of dynamics rotary motion

2. Acceleration vector - a vector quantity that determines the rate of change in the speed of the body, that is, the first derivative of the speed with respect to time and shows how much the speed vector of the body changes when it moves per unit time.

Measured in m/s 2

Ticket 17

1) The moment of force is a vector physical quantity equal to the vector product of the radius vector drawn from the axis of rotation to the point of application of the force by the vector of this force. Characterizes the rotational action of force on a rigid body.

The angular momentum relative to the fixed axis z is the scalar value Lz, which is equal to the projection onto this axis of the angular momentum vector, determined relative to an arbitrary point 0 of this axis, characterizes the amount of rotational motion.

2) The displacement vector is a directed straight line segment connecting the initial position of the body with its final position. Displacement is a vector quantity. The displacement vector is directed from the starting point of the movement to the end point. The displacement vector module is the length of the segment that connects the start and end points of the movement. (m).

3)

Ticket 18

Uniform rectilinear movement called a movement in which material point for any equal intervals of time makes the same movement along a given given straight line. The speed of uniform movement is determined by the formula:

Radius of curvature RR trajectories at a point AA is the radius of the circle along the arc of which the point moves in this moment time. The center of this circle is called the center of curvature.

Physical quantity, which characterizes the change in speed in the direction, - normal acceleration.

.

The physical quantity characterizing the change in speed modulo, - tangential acceleration.

Ticket 21

3)

Ticket number 22

The coefficient of sliding friction is the ratio of the friction force to the normal component of the external forces acting on the surface of the body.

The coefficient of sliding friction is derived from the formula for the force of sliding friction

Since the support reaction force is the mass multiplied by the free fall acceleration, the coefficient formula is:

Dimensionless quantity

Ticket number 23

The space in which conservative forces act is called the potential field. Each point of the potential field corresponds to a certain value of the force F acting on the body, and a certain value of the potential energy U. This means that there must be a connection between the force F and U, on the other hand, dA = -dU, therefore Fdr = -dU, hence:

Projections of the force vector on the coordinate axes:

The force vector can be written in terms of projections: , F = –grad U, where .

A gradient is a vector showing the direction of the fastest change in a function. Therefore, the vector is directed towards the fastest decrease in U.

If only conservative forces act on the system, then we can introduce for it the concept potential energy. Any arbitrary position of the system, characterized by setting the coordinates of its material points, we will conditionally take as zero. The work done by conservative forces during the transition of the system from the considered position to zero is called potential energy of the system in first position

The work of conservative forces does not depend on the transition path, and therefore the potential energy of the system at a fixed zero position depends only on the coordinates of the material points of the system in the considered position. In other words, the potential energy of the system U is a function of only its coordinates.

The potential energy of the system is not uniquely defined, but up to an arbitrary constant. This arbitrariness cannot affect physical conclusions, since the course of physical phenomena may depend not on the absolute values ​​of the potential energy itself, but only on its difference in various states. The same differences do not depend on the choice of an arbitrary constant.

Let the system move from position 1 to position 2 along some path 12 (Fig. 3.3). work BUT 12 performed by conservative forces during such a transition can be expressed in terms of potential energies U 1 and U 2 in states 1 And 2 . For this purpose, let us imagine that the transition is made through position O, i.e., along the path 1O2. Since the forces are conservative, then BUT 12 = BUT 1O2 = BUT 1O + BUT O2 = BUT 1O - BUT 2O. By definition of potential energy U 1 = A 1 O , U 2 = A 2O. In this way,

A 12 = U 1 – U 2 , (3.10)

i.e., the work of conservative forces is equal to the decrease in the potential energy of the system.

Same job BUT 12 , as shown earlier in (3.7), can be expressed in terms of the kinetic energy increment by the formula

BUT 12 = TO 2 – TO 1 .

Equating their right-hand sides, we get TO 2 – TO 1 = U 1 – U 2 , whence

TO 1 + U 1 = TO 2 + U 2 .

The sum of the kinetic and potential energies of a system is called its total energy E. In this way, E 1 = E 2 , or

Eº K+U= const. (3.11)

In a system with only conservative forces, the total energy remains unchanged. Only transformations of potential energy into kinetic energy and vice versa can occur, but the total energy supply of the system cannot change. This position is called the law of conservation of energy in mechanics.

Let us calculate the potential energy in some simplest cases.

a) Potential energy of a body in a uniform gravitational field. If a material point located at a height h, will fall to the zero level (i.e., the level for which h= 0), then gravity will do work A=mgh. Therefore, on top h material point has potential energy U=mgh+C, where FROM is an additive constant. An arbitrary level can be taken as zero, for example, the level of the floor (if the experiment is carried out in a laboratory), sea level, etc. Constant FROM is equal to potential energy at zero level. Setting it equal to zero, we get

U=mgh. (3.12)

b) Potential energy of a stretched spring. The elastic forces that occur when a spring is stretched or compressed are central forces. Therefore, they are conservative, and it makes sense to talk about the potential energy of a deformed spring. They call her elastic energy. Denote by x spring extension,T. e. difference x = l – l 0 lengths of the spring in the deformed and undeformed states. Elastic force F depends on stretch. If stretching x not very large, then it is proportional to it: F = – kx(Hooke's law). When the spring returns from the deformed to the undeformed state, the force F does the job

If the elastic energy of the spring in the undeformed state is assumed to be equal to zero, then

c) Potential energy of gravitational attraction of two material points. According to Newton's law of universal gravitation, the gravitational force of attraction of two point bodies is proportional to the product of their masses mm and is inversely proportional to the square of the distance between them:

where G is gravitational constant.

The force of gravitational attraction, as a central force, is conservative. It makes sense for her to talk about potential energy. When calculating this energy, one of the masses, for example M, can be considered as stationary, and the other as moving in its gravitational field. When moving mass m from infinity, gravitational forces do work

where r- distance between masses M And m in final state.

This work is equal to the loss of potential energy:

Usually potential energy at infinity U¥ is taken equal to zero. With such an agreement

The quantity (3.15) is negative. This has a simple explanation. Attractive masses have maximum energy at an infinite distance between them. In this position, the potential energy is considered to be zero. In every other position it is smaller, i.e. negative.

Let us now assume that, along with conservative forces, dissipative forces also act in the system. The work of all forces BUT 12 during the transition of the system from position 1 to position 2 is still equal to the increment of its kinetic energy TO 2 – TO one . But in the case under consideration, this work can be represented as the sum of the work of conservative forces and the work of dissipative forces. The first work can be expressed in terms of the loss of potential energy of the system: Therefore

Equating this expression to the increment of kinetic energy, we obtain

where E=K+U is the total energy of the system. Thus, in the case under consideration, the mechanical energy E system does not remain constant, but decreases, since the work of dissipative forces is negative.