The most important phenomenon constantly studied by physicists is movement. Electromagnetic phenomena, laws of mechanics, thermodynamic and quantum processes - all this is a wide range of fragments of the universe studied by physics. And all these processes come down, one way or another, to one thing - to.
Everything in the Universe moves. Gravity is a common phenomenon for all people since childhood, we were born in the gravitational field of our planet, this physical phenomenon is perceived by us at the deepest intuitive level and, it would seem, does not even require study.
But, alas, the question is why and how do all bodies attract each other, remains to this day not fully disclosed, although it has been studied far and wide.
In this article we will look at what universal attraction is according to Newton - the classical theory of gravity. However, before moving on to formulas and examples, we will talk about the essence of the problem of attraction and give it a definition.
Perhaps the study of gravity became the beginning of natural philosophy (the science of understanding the essence of things), perhaps natural philosophy gave rise to the question of the essence of gravity, but, one way or another, the question of the gravitation of bodies became interested in ancient Greece.
Movement was understood as the essence of the sensory characteristic of the body, or rather, the body moved while the observer saw it. If we cannot measure, weigh, or feel a phenomenon, does this mean that this phenomenon does not exist? Naturally, it doesn't mean that. And since Aristotle understood this, reflections began on the essence of gravity.
As it turns out today, after many tens of centuries, gravity is the basis not only of gravity and the attraction of our planet to, but also the basis for the origin of the Universe and almost all existing elementary particles.
Movement task
Let's carry out thought experiment. Let's take in left hand small ball. Let's take the same one on the right. Let's release the right ball and it will begin to fall down. The left one remains in the hand, it is still motionless.
Let's mentally stop the passage of time. The falling right ball “hangs” in the air, the left one still remains in the hand. The right ball is endowed with the “energy” of movement, the left one is not. But what is the deep, meaningful difference between them?
Where, in what part of the falling ball is it written that it should move? It has the same mass, the same volume. It has the same atoms, and they are no different from the atoms of a ball at rest. Ball has? Yes, this is the correct answer, but how does the ball know what has potential energy, where is it recorded in it?
This is precisely the task that Aristotle, Newton and Albert Einstein set themselves. And all three brilliant thinkers partly solved this problem for themselves, but today there are a number of issues that require resolution.
Newton's gravity
In 1666, the greatest English physicist and mechanic I. Newton discovered a law that can quantitatively calculate the force due to which all matter in the Universe tends to each other. This phenomenon is called universal gravity. When you are asked: “Formulate the law of universal gravitation,” your answer should sound like this:
The force of gravitational interaction contributing to the attraction of two bodies is located in direct proportion to the masses of these bodies and in inverse proportion to the distance between them.
Important! Newton's law of attraction uses the term "distance". This term should be understood not as the distance between the surfaces of bodies, but as the distance between their centers of gravity. For example, if two balls of radii r1 and r2 lie on top of each other, then the distance between their surfaces is zero, but there is an attractive force. The thing is that the distance between their centers r1+r2 is different from zero. On a cosmic scale, this clarification is not important, but for a satellite in orbit, this distance is equal to the height above the surface plus the radius of our planet. The distance between the Earth and the Moon is also measured as the distance between their centers, not their surfaces.
For the law of gravity the formula is as follows:
,
- F – force of attraction,
- – masses,
- r – distance,
- G – gravitational constant equal to 6.67·10−11 m³/(kg·s²).
What is weight, if we just looked at the force of gravity?
Force is a vector quantity, but in the law of universal gravitation it is traditionally written as a scalar. In a vector picture, the law will look like this:
.
But this does not mean that the force is inversely proportional to the cube of the distance between the centers. The relation should be perceived as a unit vector directed from one center to another:
.
Law of Gravitational Interaction
Weight and gravity
Having considered the law of gravity, one can understand that it is not surprising that we personally we feel the Sun's gravity much weaker than the Earth's. Although the massive Sun has a large mass, it is very far from us. is also far from the Sun, but it is attracted to it, since it has a large mass. How to find the gravitational force of two bodies, namely, how to calculate the gravitational force of the Sun, Earth and you and me - we will deal with this issue a little later.
As far as we know, the force of gravity is:
where m is our mass, and g is the acceleration of free fall of the Earth (9.81 m/s 2).
Important! There are not two, three, ten types of attractive forces. Gravity is the only force that gives a quantitative characteristic of attraction. Weight (P = mg) and gravitational force are the same thing.
If m is our mass, M is the mass of the globe, R is its radius, then the gravitational force acting on us is equal to:
Thus, since F = mg:
.
The masses m are reduced, and the expression for the acceleration of free fall remains:
As we can see, the acceleration of gravity is truly a constant value, since its formula includes constant quantities - the radius, the mass of the Earth and the gravitational constant. Substituting the values of these constants, we will make sure that the acceleration of gravity is equal to 9.81 m/s 2.
At different latitudes, the radius of the planet is slightly different, since the Earth is still not a perfect sphere. Because of this, the acceleration of free fall at individual points on the globe is different.
Let's return to the attraction of the Earth and the Sun. Let's try to prove with an example that the globe attracts you and me more strongly than the Sun.
For convenience, let’s take the mass of a person: m = 100 kg. Then:
- The distance between a person and the globe is equal to the radius of the planet: R = 6.4∙10 6 m.
- The mass of the Earth is: M ≈ 6∙10 24 kg.
- The mass of the Sun is: Mc ≈ 2∙10 30 kg.
- Distance between our planet and the Sun (between the Sun and man): r=15∙10 10 m.
Gravitational attraction between man and Earth:
This result is quite obvious from the simpler expression for weight (P = mg).
The force of gravitational attraction between man and the Sun:
As we can see, our planet attracts us almost 2000 times stronger.
How to find the force of attraction between the Earth and the Sun? As follows:
Now we see that the Sun attracts our planet more than a billion billion times stronger than the planet attracts you and me.
First escape velocity
After Isaac Newton discovered the law of universal gravitation, he became interested in how fast a body must be thrown so that it, having overcome the gravitational field, leaves the globe forever.
True, he imagined it a little differently, in his understanding it was not a vertically standing rocket aimed at the sky, but a body that horizontally made a jump from the top of a mountain. This was a logical illustration because At the top of the mountain the force of gravity is slightly less.
So, at the top of Everest, the acceleration of gravity will not be the usual 9.8 m/s 2 , but almost m/s 2 . It is for this reason that the air there is so thin, the air particles are no longer as tied to gravity as those that “fell” to the surface.
Let's try to find out what it is escape velocity.
The first escape velocity v1 is the speed at which the body leaves the surface of the Earth (or another planet) and enters a circular orbit.
Let's try to find out the numerical value of this value for our planet.
Let's write down Newton's second law for a body that rotates around a planet in a circular orbit:
,
where h is the height of the body above the surface, R is the radius of the Earth.
In orbit, a body is subject to centrifugal acceleration, thus:
.
The masses are reduced, we get:
,
This speed is called the first escape velocity:
As you can see, escape velocity is absolutely independent of body mass. Thus, any object accelerated to a speed of 7.9 km/s will leave our planet and enter its orbit.
First escape velocity
Second escape velocity
However, even having accelerated the body to the first escape velocity, we will not be able to completely break its gravitational connection with the Earth. This is why we need a second escape velocity. When this speed is reached the body leaves the planet's gravitational field and all possible closed orbits.
Important! It is often mistakenly believed that in order to get to the Moon, astronauts had to reach the second escape velocity, because they first had to “disconnect” from the gravitational field of the planet. This is not so: the Earth-Moon pair are in the Earth’s gravitational field. Their common center of gravity is inside the globe.
In order to find this speed, let's pose the problem a little differently. Let's say a body flies from infinity to a planet. Question: what speed will be reached on the surface upon landing (without taking into account the atmosphere, of course)? This is exactly the speed the body will need to leave the planet.
Second escape velocity
Let's write down the law of conservation of energy:
,
where on the right side of the equality is the work of gravity: A = Fs.
From this we obtain that the second escape velocity is equal to:
Thus, the second escape velocity is times greater than the first:
The law of universal gravitation. Physics 9th grade
Law of Universal Gravitation.
Conclusion
We learned that although gravity is the main force in the Universe, many of the reasons for this phenomenon still remain a mystery. We learned what Newton's force of universal gravitation is, learned to calculate it for various bodies, and also studied some useful consequences that follow from such a phenomenon as the universal law of gravity.
Law of Gravity
Gravity (universal gravitation, gravitation)(from Latin gravitas - “gravity”) - a long-range fundamental interaction in nature, to which all material bodies are subject. According to modern data, it is a universal interaction in the sense that, unlike any other forces, it imparts the same acceleration to all bodies without exception, regardless of their mass. Mainly gravity plays a decisive role on a cosmic scale. Term gravity also used as the name of the branch of physics that studies gravitational interaction. The most successful modern physical theory in classical physics that describes gravity is the general theory of relativity; the quantum theory of gravitational interaction has not yet been constructed.
Gravitational interaction
Gravitational interaction is one of the four fundamental interactions in our world. Within the framework of classical mechanics, gravitational interaction is described law of universal gravitation Newton, who states that the force of gravitational attraction between two material points of mass m 1 and m 2 separated by distance R, is proportional to both masses and inversely proportional to the square of the distance - that is
.Here G- gravitational constant, equal to approximately 
m³/(kg s²). The minus sign means that the force acting on the body is always equal in direction to the radius vector directed to the body, that is, gravitational interaction always leads to the attraction of any bodies.
The law of universal gravitation is one of the applications of the inverse square law, which also occurs in the study of radiation (see, for example, Light Pressure), and is a direct consequence of the quadratic increase in the area of the sphere with increasing radius, which leads to a quadratic decrease in the contribution of any unit area to area of the entire sphere.
The simplest problem of celestial mechanics is the gravitational interaction of two bodies in empty space. This problem is solved analytically to the end; the result of its solution is often formulated in the form of three Kepler's laws.
As the number of interacting bodies increases, the task becomes dramatically more complicated. Thus, the already famous three-body problem (that is, the motion of three bodies with non-zero masses) cannot be solved analytically in general view. With a numerical solution, instability of the solutions relative to the initial conditions occurs quite quickly. When applied to the Solar System, this instability makes it impossible to predict the motion of planets on scales larger than a hundred million years.
In some special cases, it is possible to find an approximate solution. The most important case is when the mass of one body is significantly greater than the mass of other bodies (examples: the solar system and the dynamics of the rings of Saturn). In this case, as a first approximation, we can assume that light bodies do not interact with each other and move along Keplerian trajectories around the massive body. The interactions between them can be taken into account within the framework of perturbation theory, and averaged over time. In this case, non-trivial phenomena may arise, such as resonances, attractors, chaos, etc. A clear example of such phenomena is the non-trivial structure of the rings of Saturn.
Despite attempts to describe the behavior of the system from large number attracting bodies of approximately the same mass, this cannot be done due to the phenomenon of dynamic chaos.
Strong gravitational fields
In strong gravitational fields, when moving at relativistic speeds, the effects of general relativity begin to appear:
- deviation of the law of gravity from Newton's;
- delay of potentials associated with the finite speed of propagation of gravitational disturbances; the appearance of gravitational waves;
- nonlinearity effects: gravitational waves tend to interact with each other, so the principle of superposition of waves in strong fields no longer holds true;
- changing the geometry of space-time;
- the emergence of black holes;
Gravitational radiation
One of the important predictions of general relativity is gravitational radiation, the presence of which has not yet been confirmed by direct observations. However, there is indirect observational evidence in favor of its existence, namely: energy losses in the binary system with the pulsar PSR B1913+16 - the Hulse-Taylor pulsar - are in good agreement with a model in which this energy is carried away by gravitational radiation.
Gravitational radiation can only be generated by systems with variable quadrupole or higher multipole moments, this fact suggests that gravitational radiation of most natural sources directional, which significantly complicates its detection. Gravity power l-field source is proportional (v / c) 2l + 2 , if the multipole is of electric type, and (v / c) 2l + 4 - if the multipole is of magnetic type, where v is the characteristic speed of movement of sources in the radiating system, and c- speed of light. So the dominant moment will be the quadrupole moment electric type, and the power of the corresponding radiation is equal to:
Where Q ij- quadrupole moment tensor of the mass distribution of the radiating system. Constant 
(1/W) allows us to estimate the order of magnitude of the radiation power.
From 1969 (Weber's experiments) to the present (February 2007), attempts have been made to directly detect gravitational radiation. In the USA, Europe and Japan in present moment there are several operating ground-based detectors (GEO 600), as well as a project for a space gravity detector of the Republic of Tatarstan.
Subtle effects of gravity
In addition to the classical effects of gravitational attraction and time dilation, the general theory of relativity predicts the existence of other manifestations of gravity, which under terrestrial conditions are very weak and their detection and experimental verification are therefore very difficult. Until recently, overcoming these difficulties seemed beyond the capabilities of experimenters.
Among them, in particular, we can name the entrainment of inertial frames of reference (or the Lense-Thirring effect) and the gravitomagnetic field. In 2005, NASA's unmanned Gravity Probe B conducted an unprecedented precision experiment to measure these effects near Earth, but its full results have not yet been published.
Quantum theory of gravity
Despite more than half a century of attempts, gravity is the only fundamental interaction for which a consistent renormalizable quantum theory has not yet been constructed. However, at low energies, in the spirit of quantum field theory, gravitational interaction can be represented as an exchange of gravitons - gauge bosons with spin 2.
Standard theories of gravity
Due to the fact that quantum effects of gravity are extremely small even under the most extreme experimental and observational conditions, there are still no reliable observations of them. Theoretical estimates show that in the overwhelming majority of cases it is possible to limit classical description gravitational interaction.
There is a modern canonical classical theory of gravity - general theory of relativity, and many hypotheses and theories of varying degrees of development that clarify it, competing with each other (see the article Alternative theories of gravity). All of these theories make very similar predictions within the approximation in which experimental tests are currently carried out. The following are several basic, most well-developed or known theories of gravity.
- Gravity is not a geometric field, but a real physical force field described by a tensor.
- Gravitational phenomena should be considered within the framework of flat Minkowski space, in which the laws of conservation of energy-momentum and angular momentum are unambiguously satisfied. Then the motion of bodies in Minkowski space is equivalent to the motion of these bodies in effective Riemannian space.
- In tensor equations to determine the metric, the graviton mass should be taken into account, and gauge conditions associated with the Minkowski space metric should be used. This does not allow the gravitational field to be destroyed even locally by choosing some suitable reference frame.
As in general relativity, in RTG matter refers to all forms of matter (including the electromagnetic field), with the exception of the gravitational field itself. The consequences of the RTG theory are as follows: black holes as physical objects predicted in General Relativity do not exist; The universe is flat, homogeneous, isotropic, stationary and Euclidean.
On the other hand, there are no less convincing arguments by opponents of RTG, which boil down to the following points:
A similar thing occurs in RTG, where the second tensor equation is introduced to take into account the connection between non-Euclidean space and Minkowski space. Due to the presence of a dimensionless fitting parameter in the Jordan-Brans-Dicke theory, it becomes possible to choose it so that the results of the theory coincide with the results of gravitational experiments.
| Theories of gravity | ||||||||
|
Sources and notes
Literature
- Vizgin V. P. Relativistic theory of gravity (origins and formation, 1900-1915). M.: Nauka, 1981. - 352c.
- Vizgin V. P. Unified theories in the 1st third of the twentieth century. M.: Nauka, 1985. - 304c.
- Ivanenko D. D., Sardanashvili G. A. Gravity, 3rd ed. M.: URSS, 2008. - 200 p.
See also
- Gravimeter
Links
- The law of universal gravitation or “Why doesn’t the Moon fall to Earth?” - Just about the complex
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Sir Isaac Newton, having been hit on the head with an apple, deduced the law of universal gravitation, which states:
Any two bodies are attracted to each other with a force directly proportional to the product of the masses of the body and inversely proportional to the square of the distance between them:
F = (Gm 1 m 2)/R 2, where
m1, m2- body masses
R- distance between the centers of bodies
G = 6.67 10 -11 Nm 2 /kg- constant
Let us determine the acceleration of free fall on the Earth's surface:
F g = m body g = (Gm body m Earth)/R 2
R (radius of the Earth) = 6.38 10 6 m
m Earth = 5.97 10 24 kg
m body g = (Gm body m Earth)/R 2 or g = (Gm Earth)/R 2
Please note that the acceleration due to gravity does not depend on the mass of the body!
g = 6.67 10 -11 5.97 10 24 /(6.38 10 6) = 398.2/40.7 = 9.8 m/s 2
We said earlier that the force of gravity (gravitational attraction) is called weight.
On the surface of the Earth, the weight and mass of a body have the same meaning. But as you move away from the Earth, the weight of the body will decrease (since the distance between the center of the Earth and the body will increase), and the mass will remain constant (since mass is an expression of the inertia of the body). Mass is measured in kilograms, weight - in newtons.
Thanks to the force of gravity, celestial bodies rotate relative to each other: the Moon around the Earth; Earth around the Sun; The Sun around the center of our Galaxy, etc. In this case, the bodies are held by centrifugal force, which is provided by the force of gravity.
The same applies to artificial bodies (satellites) revolving around the Earth. The circle around which the satellite rotates is called the orbit.
In this case, a centrifugal force acts on the satellite:
F c = (m satellite V 2)/R
Gravity force:
F g = (Gm satellite m Earth)/R 2
F c = F g = (m satellite V 2)/R = (Gm satellite m Earth)/R 2
V2 = (Gm Earth)/R; V = √(Gm Earth)/R
Using this formula, you can calculate the speed of any body rotating in an orbit with a radius R around the Earth.
The Earth's natural satellite is the Moon. Let us determine its linear speed in orbit:
Earth mass = 5.97 10 24 kg
R is the distance between the center of the Earth and the center of the Moon. To determine this distance, we need to add three quantities: the radius of the Earth; radius of the Moon; distance from the Earth to the Moon.
R moon = 1738 km = 1.74 10 6 m
R earth = 6371 km = 6.37 10 6 m
R zł = 384400 km = 384.4 10 6 m
Total distance between the centers of the planets: R = 392.5·10 6 m
Linear speed of the Moon:
V = √(Gm Earth)/R = √6.67 10 -11 5.98 10 24 /392.5 10 6 = 1000 m/s = 3600 km/h
The Moon moves in a circular orbit around the Earth with a linear speed of 3600 km/h!
Let us now determine the period of revolution of the Moon around the Earth. During its orbital period, the Moon covers a distance equal to the length of its orbit - 2πR. Moon orbital speed: V = 2πR/T; on the other side: V = √(Gm Earth)/R:
2πR/T = √(Gm Earth)/R hence T = 2π√R 3 /Gm Earth
T = 6.28 √(60.7 10 24)/6.67 10 -11 5.98 10 24 = 3.9 10 5 s
The Moon's orbital period around the Earth is 2,449,200 seconds, or 40,820 minutes, or 680 hours, or 28.3 days.
1. Vertical rotation
Previously, a very popular trick in circuses was in which a cyclist (motorcyclist) made a full turn inside a vertical circle.
What minimum speed should a stuntman have to avoid falling down at the top?
To pass the top point without falling, the body must have a speed that creates such a centrifugal force that would compensate for the force of gravity.
Centrifugal force: F c = mV 2 / R
Gravity: F g = mg
F c = F g ; mV 2 /R = mg; V = √Rg
Again, note that body weight is not included in the calculations! Please note that this is the speed that the body should have at the top!
Let's assume that there is a circle with a radius of 10 meters in the circus arena. Let's calculate the safe speed for the trick:
V = √Rg = √10·9.8 = 10 m/s = 36 km/h
Newton's classical theory of gravity (Newton's Law of Universal Gravitation)- a law describing gravitational interaction within the framework of classical mechanics. This law was discovered by Newton around 1666. It says that strength F (\displaystyle F) gravitational attraction between two material points of mass m 1 (\displaystyle m_(1)) And m 2 (\displaystyle m_(2)), separated by distance R (\displaystyle R), is proportional to both masses and inversely proportional to the square of the distance between them - that is:
F = G ⋅ m 1 ⋅ m 2 R 2 (\displaystyle F=G\cdot (m_(1)\cdot m_(2) \over R^(2)))
Here G (\displaystyle G)- gravitational constant equal to 6.67408(31)·10 −11 m³/(kg s²) :.
Encyclopedic YouTube
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✪ Introduction to Newton's law of universal gravitation
✪ Law of Gravity
✪ physics LAW OF UNIVERSAL GRAVITY 9th grade
✪ About Isaac Newton ( Brief history)
✪ Lesson 60. The law of universal gravitation. Gravitational constant
Subtitles
Now let's learn a little about gravity, or gravitation. As you know, gravity, especially in a beginner or even in a fairly advanced course of physics, is a concept that can be calculated and the basic parameters that determine it, but in fact, gravity is not entirely understandable. Even if you are familiar with the general theory of relativity, if you are asked what gravity is, you can answer: it is the curvature of space-time and the like. However, it is still difficult to get an intuition as to why two objects, simply because they have so-called mass, are attracted to each other. At least for me it's mystical. Having noted this, let us begin to consider the concept of gravity. We will do this by studying Newton's law of universal gravitation, which is valid for most situations. This law states: the force of mutual gravitational attraction F between two material points with masses m₁ and m₂ is equal to the product of the gravitational constant G by the mass of the first object m₁ and the second object m₂, divided by the square of the distance d between them. This is a fairly simple formula. Let's try to transform it and see if we can get some results that are familiar to us. We use this formula to calculate the acceleration of gravity near the Earth's surface. Let's draw the Earth first. Just to understand what we are talking about. This is our Earth. Let's say we need to calculate the gravitational acceleration acting on Sal, that is, on me. Here I am. Let's try to apply this equation to calculate the magnitude of the acceleration of my fall to the center of the Earth, or to the center of mass of the Earth. The value indicated capital letter G is the universal gravitational constant. Once again: G is the universal gravitational constant. Although, as far as I know, although I am not an expert on this matter, it seems to me that its value can change, that is, it is not a real constant, and I assume that its value differs in different measurements. But for our needs, and also in most physics courses, it is a constant, a constant equal to 6.67 * 10^(−11) cubic meters, divided by kilogram per second squared. Yes, its dimension looks strange, but it is enough for you to understand that these are conventional units necessary to, as a result of multiplying by the masses of objects and dividing by the square of the distance, obtain the dimension of force - newton, or kilogram per meter divided by second squared. So there's no need to worry about these units: just know that we'll have to work with meters, seconds, and kilograms. Let's substitute this number into the formula for force: 6.67 * 10^(−11). Since we need to know the acceleration acting on Sal, m₁ is equal to the mass of Sal, that is, me. I wouldn’t like to expose how much I weigh in this story, so let’s leave this mass as a variable, denoting it ms. The second mass in the equation is the mass of the Earth. Let's write down its meaning by looking at Wikipedia. So, the mass of the Earth is 5.97 * 10^24 kilograms. Yes, the Earth is more massive than Sal. By the way, weight and mass are different concepts. So, the force F is equal to the product of the gravitational constant G by the mass ms, then by the mass of the Earth, and divide all this by the square of the distance. You may object: what is the distance between the Earth and what stands on it? After all, if objects touch, the distance is zero. It is important to understand here: the distance between two objects in this formula is the distance between their centers of mass. In most cases, a person's center of mass is located approximately three feet above surface of the Earth unless the person is too tall. Anyway, my center of mass may be three feet above the ground. Where is the center of mass of the Earth? Obviously in the center of the Earth. What is the radius of the Earth? 6371 kilometers, or approximately 6 million meters. Since the height of my center of mass is about one millionth the distance to the Earth's center of mass, it can be neglected in this case. Then the distance will be equal to 6 and so on, like all other quantities, you need to write it in standard form- 6.371 * 10^6, since 6000 km is 6 million meters, and a million is 10^6. We write, rounding all fractions to the second decimal place, the distance is 6.37 * 10^6 meters. The formula contains the square of the distance, so let's square everything. Let's try to simplify now. First, let's multiply the values in the numerator and move forward the variable ms. Then the force F is equal to Sal’s mass on the entire upper part, let’s calculate it separately. So 6.67 times 5.97 equals 39.82. 39.82. This is a product of significant parts, which should now be multiplied by 10 to the required degree. 10^(−11) and 10^24 have the same base, so to multiply them it is enough to add the exponents. Adding 24 and −11, we get 13, resulting in 10^13. Let's find the denominator. It is equal to 6.37 squared times 10^6 also squared. As you remember, if the number written in as a degree, is raised to another power, then the exponents are multiplied, which means that 10^6 squared is equal to 10 to the power of 6 multiplied by 2, or 10^12. Next, we calculate the square of 6.37 using a calculator and get... Square 6.37. And it's 40.58. 40.58. All that remains is to divide 39.82 by 40.58. Divide 39.82 by 40.58, which equals 0.981. Then we divide 10^13 by 10^12, which is equal to 10^1, or just 10. And 0.981 times 10 is 9.81. After simplification and simple calculations, we found that the gravitational force near the Earth’s surface acting on Sal is equal to Sel’s mass multiplied by 9.81. What does this give us? Is it now possible to calculate gravitational acceleration? It is known that force is equal to the product of mass and acceleration, therefore the gravitational force is simply equal to the product of Sal’s mass and gravitational acceleration, which is usually denoted lowercase letter g. So, on the one hand, the force of gravity is equal to 9.81 times Sal's mass. On the other hand, it is equal to Sal’s mass per gravitational acceleration. Dividing both sides of the equation by Sal’s mass, we find that the coefficient 9.81 is the gravitational acceleration. And if we included in the calculations the full record of units of dimension, then, having reduced the kilograms, we would see that gravitational acceleration is measured in meters divided by a second squared, like any acceleration. You can also notice that the resulting value is very close to the one we used when solving problems about the motion of a thrown body: 9.8 meters per second squared. This is impressive. Let's do another quick gravity problem because we have a couple of minutes left. Let's say we have another planet called Baby Earth. Let the radius of the Baby rS be half the radius of the Earth rE, and its mass mS is also equal to half the mass of the Earth mE. What will be the force of gravity acting here on any object, and how much less is it than the force of gravity? Although, let's leave the problem for next time, then I'll solve it. See you. Subtitles by the Amara.org community
Properties of Newtonian gravity
In Newtonian theory, each massive body generates a force field of attraction towards this body, which is called a gravitational field. This field is potential, and the function of gravitational potential for a material point with mass M (\displaystyle M) is determined by the formula:
φ (r) = − G M r . (\displaystyle \varphi (r)=-G(\frac (M)(r)).)IN general case, when the density of the substance ρ (\displaystyle \rho ) distributed randomly, satisfies the Poisson equation:
Δ φ = − 4 π G ρ (r) . (\displaystyle \Delta \varphi =-4\pi G\rho (r).)The solution to this equation is written as:
φ = − G ∫ ρ (r) d V r + C , (\displaystyle \varphi =-G\int (\frac (\rho (r)dV)(r))+C,)Where r (\displaystyle r) - distance between volume element d V (\displaystyle dV) and the point at which the potential is determined φ (\displaystyle \varphi ), C (\displaystyle C) - arbitrary constant.
The force of attraction acting in a gravitational field on a material point with mass m (\displaystyle m), is related to the potential by the formula:
F (r) = − m ∇ φ (r) . (\displaystyle F(r)=-m\nabla \varphi (r).)A spherically symmetrical body creates the same field outside its boundaries as a material point of the same mass located in the center of the body.
The trajectory of a material point in a gravitational field created by a much larger material point obeys Kepler's laws. In particular, planets and comets in the Solar System move in ellipses or hyperbolas. The influence of other planets, which distorts this picture, can be taken into account using perturbation theory.
Accuracy of Newton's law of universal gravitation
An experimental assessment of the degree of accuracy of Newton's law of gravitation is one of the confirmations of the general theory of relativity. Experiments on measuring the quadrupole interaction of a rotating body and a stationary antenna showed that the increment δ (\displaystyle \delta ) in the expression for the dependence of the Newtonian potential r − (1 + δ) (\displaystyle r^(-(1+\delta))) at distances of several meters is within (2 , 1 ± 6 , 2) ∗ 10 − 3 (\displaystyle (2.1\pm 6.2)*10^(-3)). Other experiments also confirmed the absence of modifications in the law of universal gravitation.
Newton's law of universal gravitation in 2007 was also tested at distances less than one centimeter (from 55 microns to 9.53 mm). Taking into account the experimental errors, no deviations from Newton's law were found in the studied range of distances.
Precision laser ranging observations of the Moon's orbit confirm the law of universal gravitation at the distance from the Earth to the Moon with precision 3 ⋅ 10 − 11 (\displaystyle 3\cdot 10^(-11)).
Connection with the geometry of Euclidean space
Fact of equality with very high accuracy 10 − 9 (\displaystyle 10^(-9)) exponent of the distance in the denominator of the expression for the force of gravity to the number 2 (\displaystyle 2) reflects the Euclidean nature of the three-dimensional physical space of Newtonian mechanics. In three-dimensional Euclidean space, the surface area of a sphere is exactly proportional to the square of its radius
Historical sketch
The very idea of the universal force of gravity was repeatedly expressed before Newton. Previously, Epicurus, Gassendi, Kepler, Borelli, Descartes, Roberval, Huygens and others thought about it. Kepler believed that gravity is inversely proportional to the distance to the Sun and extends only in the ecliptic plane; Descartes considered it to be the result of vortices in the ether. There were, however, guesses with a correct dependence on distance; Newton, in a letter to Halley, mentions Bulliald, Wren and Hooke as his predecessors. But before Newton, no one was able to clearly and mathematically conclusively connect the law of gravity (a force inversely proportional to the square of the distance) and the laws of planetary motion (Kepler’s laws).
- law of gravitation;
- law of motion (Newton's second law);
- system of methods for mathematical research(mathematical analysis).
Taken together, this triad is sufficient for a complete study of the most complex movements. celestial bodies, thereby creating the foundations of celestial mechanics. Before Einstein, no fundamental amendments to this model were needed, although the mathematical apparatus turned out to be necessary to significantly develop.
Note that Newton's theory of gravity was no longer, strictly speaking, heliocentric. Already in the two-body problem, the planet rotates not around the Sun, but around a common center of gravity, since not only the Sun attracts the planet, but the planet also attracts the Sun. Finally, it became clear that it was necessary to take into account the influence of the planets on each other.
During the 18th century, the law of universal gravitation was the subject of active debate (it was opposed by supporters of the Descartes school) and careful testing. By the end of the century, it became generally accepted that the law of universal gravitation makes it possible to explain and predict the movements of celestial bodies with great accuracy. Henry Cavendish in 1798 carried out a direct test of the validity of the law of gravity in terrestrial conditions, using extremely sensitive torsion balances. An important stage was the introduction by Poisson in 1813 of the concept of gravitational potential and the Poisson equation for this potential; this model made it possible to study the gravitational field with an arbitrary distribution of matter. After this, Newton's law began to be regarded as a fundamental law of nature.
At the same time, Newton's theory contained a number of difficulties. The main one is the inexplicable long-range action: the force of attraction was transmitted incomprehensibly through completely empty space, and infinitely quickly. Essentially, Newton's model was purely mathematical, without any physical content. In addition, if the Universe, as was then assumed, is Euclidean and infinite, and at the same time the average density of matter in it is non-zero, then a gravitational paradox arises. IN late XIX century, another problem was discovered: the discrepancy between the theoretical and observed displacement of the perihelion of Mercury.
Further development
General theory of relativity
For more than two hundred years after Newton, physicists proposed various ways to improve Newton's theory of gravity. These efforts were crowned with success in 1915, with the creation of Einstein's general theory of relativity, in which all these difficulties were overcome. Newton's theory, in full agreement with the correspondence principle, turned out to be an approximation of a more general theory, applicable when two conditions are met:
In weak stationary gravitational fields, the equations of motion become Newtonian (gravitational potential). To prove this, we show that the scalar gravitational potential in weak stationary gravitational fields satisfies the Poisson equation
Δ Φ = − 4 π G ρ (\displaystyle \Delta \Phi =-4\pi G\rho ).It is known (Gravitational potential) that in this case the gravitational potential has the form:
Φ = − 1 2 c 2 (g 44 + 1) (\displaystyle \Phi =-(\frac (1)(2))c^(2)(g_(44)+1)).Let us find the component of the energy-momentum tensor from the gravitational field equations of the general theory of relativity:
R i k = − ϰ (T i k − 1 2 g i k T) (\displaystyle R_(ik)=-\varkappa (T_(ik)-(\frac (1)(2))g_(ik)T)),Where R i k (\displaystyle R_(ik))- curvature tensor. For we can introduce the kinetic energy-momentum tensor ρ u i u k (\displaystyle \rho u_(i)u_(k)). Neglecting quantities of the order u/c (\displaystyle u/c), you can put all the components T i k (\displaystyle T_(ik)), except T 44 (\displaystyle T_(44)), equal to zero. Component T 44 (\displaystyle T_(44)) equal to T 44 = ρ c 2 (\displaystyle T_(44)=\rho c^(2)) and therefore T = g i k T i k = g 44 T 44 = − ρ c 2 (\displaystyle T=g^(ik)T_(ik)=g^(44)T_(44)=-\rho c^(2)). Thus, the gravitational field equations take the form R 44 = − 1 2 ϰ ρ c 2 (\displaystyle R_(44)=-(\frac (1)(2))\varkappa \rho c^(2)). Due to the formula
R i k = ∂ Γ i α α ∂ x k − ∂ Γ i k α ∂ x α + Γ i α β Γ k β α − Γ i k α Γ α β β (\displaystyle R_(ik)=(\frac (\partial \ Gamma _(i\alpha )^(\alpha ))(\partial x^(k)))-(\frac (\partial \Gamma _(ik)^(\alpha ))(\partial x^(\alpha )))+\Gamma _(i\alpha )^(\beta )\Gamma _(k\beta )^(\alpha )-\Gamma _(ik)^(\alpha )\Gamma _(\alpha \beta )^(\beta ))value of the curvature tensor component R 44 (\displaystyle R_(44)) can be taken equal R 44 = − ∂ Γ 44 α ∂ x α (\displaystyle R_(44)=-(\frac (\partial \Gamma _(44)^(\alpha ))(\partial x^(\alpha )))) and since Γ 44 α ≈ − 1 2 ∂ g 44 ∂ x α (\displaystyle \Gamma _(44)^(\alpha )\approx -(\frac (1)(2))(\frac (\partial g_(44) )(\partial x^(\alpha )))), R 44 = 1 2 ∑ α ∂ 2 g 44 ∂ x α 2 = 1 2 Δ g 44 = − Δ Φ c 2 (\displaystyle R_(44)=(\frac (1)(2))\sum _(\ alpha )(\frac (\partial ^(2)g_(44))(\partial x_(\alpha )^(2)))=(\frac (1)(2))\Delta g_(44)=- (\frac (\Delta \Phi )(c^(2)))). Thus, we arrive at the Poisson equation:
Δ Φ = 1 2 ϰ c 4 ρ (\displaystyle \Delta \Phi =(\frac (1)(2))\varkappa c^(4)\rho ), Where ϰ = − 8 π G c 4 (\displaystyle \varkappa =-(\frac (8\pi G)(c^(4))))Quantum gravity
However, the general theory of relativity is not the final theory of gravity, since it unsatisfactorily describes gravitational processes on a quantum scale (at distances on the order of the Planck distance, about 1.6⋅10 −35). The construction of a consistent quantum theory of gravity is one of the most important unsolved problems of modern physics.
From the point of view of quantum gravity, gravitational interaction occurs through the exchange of virtual gravitons between interacting bodies. According to the uncertainty principle, the energy of a virtual graviton is inversely proportional to the time of its existence from the moment of emission by one body to the moment of absorption by another body. The lifetime is proportional to the distance between the bodies. Thus, at short distances, interacting bodies can exchange virtual gravitons with short and long wavelengths, and at large distances only long-wave gravitons. From these considerations we can obtain the law inverse proportionality Newtonian potential versus distance. The analogy between Newton's law and Coulomb's law is explained by the fact that the graviton mass, like the mass
When he came to a great result: the same cause causes phenomena of an amazingly wide range - from the fall of a thrown stone to the Earth to the movement of huge cosmic bodies. Newton found this reason and was able to accurately express it in the form of one formula - the law of universal gravitation.
Since the force of universal gravitation imparts the same acceleration to all bodies regardless of their mass, it must be proportional to the mass of the body on which it acts:
But since, for example, the Earth acts on the Moon with a force proportional to the mass of the Moon, then the Moon, according to Newton’s third law, must act on the Earth with the same force. Moreover, this force must be proportional to the mass of the Earth. If the force of gravity is truly universal, then from the side of a given body a force must act on any other body proportional to the mass of this other body. Consequently, the force of universal gravity must be proportional to the product of the masses of interacting bodies. This leads to the formulation law of universal gravitation.
Definition of the law of universal gravitation
The force of mutual attraction between two bodies is directly proportional to the product of the masses of these bodies and inversely proportional to the square of the distance between them:
Proportionality factor G called gravitational constant.
The gravitational constant is numerically equal to the force of attraction between two material points weighing 1 kg each, if the distance between them is 1 m. After all, when m 1 =m 2=1 kg and R=1 m we get G=F(numerically).
It must be borne in mind that the law of universal gravitation (4.5) as a universal law is valid for material points. In this case, the forces of gravitational interaction are directed along the line connecting these points ( Fig.4.2). This kind of force is called central.
It can be shown that homogeneous bodies shaped like a ball (even if they cannot be considered material points) also interact with the force determined by formula (4.5). In this case R- the distance between the centers of the balls. The forces of mutual attraction lie on a straight line passing through the centers of the balls. (Such forces are called central.) The bodies that we usually consider falling on the Earth have dimensions much smaller than the Earth’s radius ( R≈6400 km). Such bodies can, regardless of their shape, be considered as material points and determine the force of their attraction to the Earth using the law (4.5), keeping in mind that R is the distance from a given body to the center of the Earth.
Determination of the gravitational constant
Now let's find out how to find the gravitational constant. First of all, we note that G has a specific name. This is due to the fact that the units (and, accordingly, the names) of all quantities included in the law of universal gravitation have already been established earlier. The law of gravitation provides a new connection between known quantities with certain names of units. That is why the coefficient turns out to be a named quantity. Using the formula of the law of universal gravitation, it is easy to find the name of the SI unit of gravitational constant:
N m 2 / kg 2 = m 3 / (kg s 2).
For quantification G it is necessary to independently determine all the quantities included in the law of universal gravitation: both masses, force and distance between bodies. It is impossible to use astronomical observations for this, since the masses of the planets, the Sun, and the Earth can only be determined on the basis of the law of universal gravitation itself, if the value of the gravitational constant is known. The experiment must be carried out on Earth with bodies whose masses can be measured on a scale.
The difficulty is that the gravitational forces between bodies of small masses are extremely small. It is for this reason that we do not notice the attraction of our body to surrounding objects and the mutual attraction of objects to each other, although gravitational forces are the most universal of all forces in nature. Two people with masses of 60 kg at a distance of 1 m from each other are attracted with a force of only about 10 -9 N. Therefore, to measure the gravitational constant, fairly subtle experiments are needed.
The gravitational constant was first measured by the English physicist G. Cavendish in 1798 using an instrument called a torsion balance. The diagram of the torsion balance is shown in Figure 4.3. A light rocker with two identical weights at the ends is suspended from a thin elastic thread. Two heavy balls are fixed motionless nearby. Gravitational forces act between the weights and the stationary balls. Under the influence of these forces, the rocker turns and twists the thread. By the angle of twist you can determine the force of attraction. To do this, you only need to know the elastic properties of the thread. The masses of the bodies are known, and the distance between the centers of interacting bodies can be directly measured.
From these experiments it was obtained next value for the gravitational constant:
Only in the case when bodies of enormous mass interact (or at least the mass of one of the bodies is very large) does the gravitational force reach a large value. For example, the Earth and the Moon are attracted to each other with a force F≈2 10 20 H.
Dependence of the acceleration of free falling bodies on geographic latitude
One of the reasons for the increase in the acceleration of gravity when the point where the body is located moves from the equator to the poles is that the globe is somewhat flattened at the poles and the distance from the center of the Earth to its surface at the poles is less than at the equator. Another, more significant reason is the rotation of the Earth.
Equality of inertial and gravitational masses
The most striking property of gravitational forces is that they impart the same acceleration to all bodies, regardless of their masses. What would you say about a football player whose kick would be equally accelerated by an ordinary leather ball and a two-pound weight? Everyone will say that this is impossible. But the Earth is just such an “extraordinary football player” with the only difference that its effect on bodies is not of the nature of a short-term blow, but continues continuously for billions of years.
The extraordinary property of gravitational forces, as we have already said, is explained by the fact that these forces are proportional to the masses of both interacting bodies. This fact cannot but cause surprise if you think about it carefully. After all, the mass of a body, which is included in Newton’s second law, determines the inertial properties of the body, that is, its ability to acquire a certain acceleration under the influence of a given force. It is natural to call this mass inert mass
and denote by m and.
It would seem, what relation can it have to the ability of bodies to attract each other? The mass that determines the ability of bodies to attract each other should be called gravitational mass m g.
It does not at all follow from Newtonian mechanics that the inertial and gravitational masses are the same, i.e. that
Equality (4.6) is a direct consequence of experiment. It means that we can simply talk about the mass of a body as a quantitative measure of both its inertial and gravitational properties.
The law of universal gravitation is one of the most universal laws of nature. It is valid for any bodies with mass.
The meaning of the law of universal gravitation
But if we approach this topic more radically, it turns out that the law of universal gravitation does not have the possibility of its application everywhere. This law has found its application for bodies that have the shape of a ball, it can be used for material points, and it is also acceptable for a ball having a large radius, where this ball can interact with bodies much smaller than its size.
As you may have guessed from the information provided in this lesson, the law of universal gravitation is the basis in the study of celestial mechanics. And as you know, celestial mechanics studies the movement of planets.
Thanks to this law of universal gravitation, it became possible to more precise definition the location of celestial bodies and the ability to calculate their trajectory.
But for a body and an infinite plane, as well as for the interaction of an infinite rod and a ball, this formula cannot be applied.
With the help of this law, Newton was able to explain not only how the planets move, but also why sea tides arise. Over time, thanks to the work of Newton, astronomers managed to discover such planets solar system, like Neptune and Pluto.
The importance of the discovery of the law of universal gravitation lies in the fact that with its help it became possible to make forecasts of solar and lunar eclipses and accurately calculate the movements of spacecraft.
The forces of universal gravity are the most universal of all the forces of nature. After all, their action extends to the interaction between any bodies that have mass. And as you know, any body has mass. The forces of gravity act through any body, since there are no barriers to the forces of gravity.
Task
And now, in order to consolidate knowledge about the law of universal gravitation, let's try to consider and solve an interesting problem. The rocket rose to a height h equal to 990 km. Determine how much the force of gravity acting on the rocket at a height h has decreased compared to the force of gravity mg acting on it at the surface of the Earth? The radius of the Earth is R = 6400 km. Let us denote by m the mass of the rocket, and by M the mass of the Earth.
At height h the force of gravity is:
From here we calculate:
Substituting the value will give the result:
The legend about how Newton discovered the law of universal gravitation after hitting the top of his head with an apple was invented by Voltaire. Moreover, Voltaire himself assured that this true story was told to him by Newton’s beloved niece Katherine Barton. It’s just strange that neither the niece herself nor her very close friend Jonathan Swift ever mentioned the fateful apple in their memoirs about Newton. By the way, Isaac Newton himself, writing in detail in his notebooks the results of experiments on the behavior of different bodies, noted only vessels filled with gold, silver, lead, sand, glass, water or wheat, not to mention an apple. However, this did not stop Newton’s descendants from taking tourists around the garden on the Woolstock estate and showing them that same apple tree before the storm destroyed it.
Yes, there was an apple tree, and apples probably fell from it, but how great was the merit of the apple in the discovery of the law of universal gravitation?
The debate about the apple has not subsided for 300 years, just like the debate about the law of universal gravitation itself or about who has the priority of discovery.uk
G.Ya.Myakishev, B.B.Bukhovtsev, N.N.Sotsky, Physics 10th grade
