Gravity- this is the force acting on the body from the side of the Earth and informing the body of the acceleration of free fall:

\(~\vec F_T = m \vec g.\)

Any body located on the Earth (or near it), together with the Earth, rotates around its axis, i.e., the body moves in a circle with a radius r with a constant modulo speed (Fig. 1).

A body on the surface of the Earth is affected by the gravitational force \(~\vec F\) and the force from the earth's surface \(~\vec N_p\).

Their resultant

\(~\vec F_1 = \vec F + \vec N_p \qquad (1)\)

imparts centripetal acceleration to the body

\(~a_c = \frac(\upsilon^2)(r).\)

Let us decompose the gravitational force \(~\vec F\) into two components, one of which will be \(~\vec F_1\), i.e.

\(~\vec F = \vec F_1 + \vec F_T. \qquad (2)\)

From equations (1) and (2) we see that

\(~\vec F_T = - \vec N_p.\)

Thus, the force of gravity \(~\vec F_T\) is one of the components of the force of gravity \(~\vec F\). The second component \(~\vec F_1\) tells the body centripetal acceleration.

At the point Μ at geographic latitude φ gravity is not directed along the radius of the Earth, but at some angle α to him. The force of gravity is directed along the so-called sheer line (vertically down).

The force of gravity is equal in magnitude and direction to the force of gravity only at the poles. At the equator, they coincide in direction, and the absolute difference is greatest.

\(~F_T = F - F_1 = F - m \omega^2 R,\)

where ω is the angular velocity of the Earth's rotation, R is the radius of the earth.

\(~\omega = \frac(2 \pi)(T) = \frac(2 \cdot 2.34)(24 \cdot 3600)\) rad/s = 0.727 10 -4 rad/s.

Because ω very small, then F T≈ F. Consequently, the force of gravity differs little in modulus from the force of gravity, so this difference can often be neglected.

Then F T≈ F, \(~mg = \frac(GMm)((h + R)^2) \Rightarrow g = \frac(GM)((h + R)^2)\) .

This formula shows that the free fall acceleration g does not depend on the mass of the falling body, but depends on the height.

Literature

Aksenovich L. A. Physics in high school: Theory. Tasks. Tests: Proc. allowance for institutions providing general. environments, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsia i vykhavanne, 2004. - C. 39-40.

In this paragraph, we will remind you about gravity, centripetal acceleration and body weight.

Every body on the planet is affected by the Earth's gravity. The force with which the Earth attracts each body is determined by the formula

The point of application is at the center of gravity of the body. Gravity always pointing vertically down.

The force with which a body is attracted to the Earth under the influence of the Earth's gravitational field is called gravity. According to the law of universal gravitation, on the surface of the Earth (or near this surface), a body of mass m is affected by the force of gravity

F t \u003d GMm / R 2

where M is the mass of the Earth; R is the radius of the Earth.
If only gravity acts on the body, and all other forces are mutually balanced, the body is in free fall. According to Newton's second law and the formula F t \u003d GMm / R 2 free fall acceleration modulus g is found by the formula

g=F t /m=GM/R 2 .

From formula (2.29) it follows that the free fall acceleration does not depend on the mass m of the falling body, i.e. for all bodies in a given place on the Earth it is the same. From formula (2.29) it follows that Fт = mg. In vector form

F t \u003d mg

In § 5 it was noted that since the Earth is not a sphere, but an ellipsoid of revolution, its polar radius is less than the equatorial one. From the formula F t \u003d GMm / R 2 it can be seen that for this reason the force of gravity and the acceleration of free fall caused by it is greater at the pole than at the equator.

The force of gravity acts on all bodies in the Earth's gravitational field, but not all bodies fall to the Earth. This is due to the fact that the movement of many bodies is hindered by other bodies, such as supports, suspension threads, etc. Bodies that restrict the movement of other bodies are called connections. Under the action of gravity, the bonds are deformed and the reaction force of the deformed bond, according to Newton's third law, balances the force of gravity.

The acceleration of free fall is affected by the rotation of the Earth. This influence is explained as follows. The frames of reference associated with the surface of the Earth (except for the two associated with the poles of the Earth) are not, strictly speaking, inertial frames of reference - the Earth rotates around its axis, and with it move along circles with centripetal acceleration and such frames of reference. This non-inertiality of reference systems is manifested, in particular, in the fact that the value of the acceleration of free fall turns out to be different in different places on the Earth and depends on the geographical latitude of the place where the reference frame associated with the Earth is located, relative to which the acceleration of gravity is determined.

Measurements carried out at different latitudes showed that the numerical values ​​of the gravitational acceleration differ little from each other. Therefore, with not very accurate calculations, one can neglect the non-inertiality of reference systems associated with the Earth's surface, as well as the difference in the shape of the Earth from spherical, and assume that the acceleration of free fall in any place on the Earth is the same and equal to 9.8 m / s 2.

From the law of universal gravitation it follows that the force of gravity and the acceleration of free fall caused by it decrease with increasing distance from the Earth. At a height h from the Earth's surface, the gravitational acceleration module is determined by the formula

g=GM/(R+h) 2.

It has been established that at a height of 300 km above the Earth's surface, the free fall acceleration is less than at the Earth's surface by 1 m/s2.
Consequently, near the Earth (up to heights of several kilometers), the force of gravity practically does not change, and therefore the free fall of bodies near the Earth is a uniformly accelerated motion.

Body weight. Weightlessness and overload

The force in which, due to attraction to the Earth, the body acts on its support or suspension, is called body weight. Unlike gravity, which is a gravitational force applied to a body, weight is an elastic force applied to a support or suspension (i.e., to a connection).

Observations show that the weight of the body P, determined on a spring balance, is equal to the force of gravity F t acting on the body only if the balance with the body relative to the Earth is at rest or moving uniformly and rectilinearly; In this case

P \u003d F t \u003d mg.

If the body is moving with acceleration, then its weight depends on the value of this acceleration and on its direction relative to the direction of free fall acceleration.

When a body is suspended on a spring balance, two forces act on it: the force of gravity F t =mg and the elastic force F yp of the spring. If at the same time the body moves vertically up or down relative to the direction of free fall acceleration, then the vector sum of the forces F t and F yn gives the resultant, causing the acceleration of the body, i.e.

F t + F pack \u003d ma.

According to the above definition of the concept of "weight", we can write that P=-F yp. From the formula: F t + F pack \u003d ma. taking into account the fact that F t =mg, it follows that mg-ma=-F yp . Therefore, P \u003d m (g-a).

The forces F t and F yn are directed along one vertical straight line. Therefore, if the acceleration of the body a is directed downward (i.e., it coincides in direction with the acceleration of free fall g), then modulo

P=m(g-a)

If the acceleration of the body is directed upwards (i.e., opposite to the direction of free fall acceleration), then

P \u003d m \u003d m (g + a).

Consequently, the weight of a body whose acceleration coincides in direction with the acceleration of free fall is less than the weight of a body at rest, and the weight of a body whose acceleration is opposite to the direction of acceleration of free fall is greater than the weight of a body at rest. The increase in body weight caused by its accelerated movement is called overload.

In free fall a=g. From the formula: P=m(g-a)

it follows that in this case P=0, i.e., there is no weight. Therefore, if bodies move only under the influence of gravity (i.e., fall freely), they are in a state weightlessness. A characteristic feature of this state is the absence of deformations and internal stresses in freely falling bodies, which are caused in resting bodies by gravity. The reason for the weightlessness of bodies is that the force of gravity imparts the same accelerations to a freely falling body and its support (or suspension).

A private, but extremely important for us, type of universal gravitational force is force of attraction of bodies to the earth. This force is called gravity . According to the law of universal gravitation, it is expressed by the formula

\(~F_T = G \frac(mM)((R+h)^2)\) , (1)

where m- body mass, M is the mass of the earth, R is the radius of the earth, h is the height of the body above the earth's surface. The force of gravity is directed vertically downward towards the center of the earth.

• More precisely, in addition to this force, in the reference frame associated with the Earth, the body is affected by the centrifugal force of inertia \(~\vec F_c\) , which arises due to the daily rotation of the Earth, and is equal to \(~F_c = m \cdot \ omega^2 \cdot r\) , where m- body mass; r is the distance between the body and the earth's axis. If the height of the body above the Earth's surface is small compared to its radius, then \(~r = R \cos \varphi\) , where R is the radius of the earth, φ is the geographic latitude at which the body is located (Fig. 1). With that in mind, \(~F_c = m \cdot \omega^2 \cdot R \cos \varphi\) .

gravity called the force acting on any body near the earth's surface.

It is defined as the geometric sum of the force of gravitational attraction to the Earth \(~\vec F_g\) acting on the body and the centrifugal force of inertia \(~\vec F_c\) , which takes into account the effect of the daily rotation of the Earth around its own axis, i.e. \(~\vec F_T = \vec F_g + \vec F_c\) . The direction of gravity is the direction of the vertical at a given point on the earth's surface.

BUT the magnitude of the centrifugal force of inertia is very small compared to the force of the Earth's attraction (their ratio is approximately 3∙10 -3), then usually the force \(~\vec F_c\) is neglected. Then \(~\vec F_T \approx \vec F_g\) .

Acceleration of gravity

The force of gravity imparts an acceleration to the body, called the acceleration of free fall. According to Newton's second law

\(~\vec g = \frac(\vec F_T)(m)\) .

Taking into account expression (1), for the free fall acceleration module we will have

\(~g_h = G \frac(M)((R+h)^2)\) . (2)

On the surface of the Earth (h = 0), the free fall acceleration modulus is

\(~g = G \frac(M)(R^2)\) ,

and the force of gravity is

\(~\vec F_T = m \vec g\) .

The gravitational acceleration modulus included in the formulas is approximately 9.8 m/s 2 .

From formula (2) it can be seen that the acceleration of free fall does not depend on the mass of the body. It decreases as the body rises above the Earth's surface: free fall acceleration is inversely proportional to the square of the body's distance from the center of the earth.

However, if the height h body above the Earth's surface does not exceed 100 km, then in calculations that allow an error of ≈ 1.5%, this height can be neglected in comparison with the radius of the Earth (R = 6370 km). Free fall acceleration at altitudes up to 100 km can be considered constant and equal to 9.8 m/s 2 .

But still At the surface of the Earth, the acceleration of free fall is not the same everywhere. It depends on geographic latitude: more at the poles of the Earth than at the equator. The fact is that the globe is somewhat flattened at the poles. The equatorial radius of the Earth is greater than the polar one by 21 km.

Another, more significant reason for the dependence of the acceleration of free fall on geographic latitude is the rotation of the Earth. Newton's second law is valid in an inertial frame of reference. Such a system is, for example, the heliocentric system. The frame of reference associated with the Earth, strictly speaking, cannot be considered inertial. The earth rotates on its axis and moves in a closed orbit around the sun.

The rotation of the Earth and its oblateness at the poles leads to the fact that the acceleration of free fall relative to the geocentric reference system is different at different latitudes: at the poles g floor ≈ 9.83 m / s 2, at the equator g eq ≈ 9.78 m / s 2, at a latitude of 45 ° g≈ 9.81 m / s 2. However, in our calculations we will consider the free fall acceleration to be approximately equal to 9.8 m/s 2 .

Due to the rotation of the Earth on its axis, the acceleration of gravity in all places except the equator and the poles is not directed exactly towards the center of the Earth.

In addition, the acceleration of free fall depends on the density of rocks occurring in the bowels of the Earth. In areas where rocks occur, the density of which is greater than the average density of the Earth (for example, iron ore), g more. And where there are oil deposits, g less. This is used by geologists in the search for minerals.

Body weight

Body weight- this is the force with which the body, due to its attraction to the Earth, acts on the support or suspension.

Consider, for example, a body suspended from a spring, the other end of which is fixed (Fig. 2). The force of gravity \(~\vec F_T = m \vec g\) acting downwards acts on the body. It therefore begins to fall, dragging the lower end of the spring with it. The spring will be deformed because of this, and the elastic force \(~\vec F_(ynp)\) of the spring will appear. It is attached to the upper edge of the body and directed upwards. The upper edge of the body will therefore "lag behind" in its fall from its other parts, to which the spring force is not applied. As a result, the body is deformed. There is another force of elasticity - the force of elasticity of the deformed body. It is attached to the spring and directed downward. This force is the weight of the body.

According to Newton's third law, both of these elastic forces are equal in absolute value and directed in opposite directions. After several oscillations, the body on the spring is at rest. This means that the force of gravity \(~m \vec g\) is equal in modulus to the force of elasticity F spring control. But the same force is equal to the weight of the body.

Thus, in our example, the weight of the body, which we will denote by the letter \(~\vec P\) , is modulo equal to the force of gravity:

\(~P = m g\) .

Second example. Let the body BUT is on a horizontal support AT(Fig. 3). On the body BUT the force of gravity \(~m \vec g\) and the reaction force of the support \(~\vec N\) act. But if the support acts on the body with the force \(~\vec N\) then the body also acts on the support with the force \(~\vec P\) , which, in accordance with Newton's third law, is equal in absolute value and opposite in direction \(~ \vec N\) \[~\vec P = -\vec N\] . The force \(~\vec P\) is the weight of the body.

If the body and the support are stationary or move uniformly and rectilinearly, i.e. without acceleration, then, according to Newton's second law,

\(~\vec N + m \vec g = 0\) .

\(~\vec N = -\vec P\) , then \(~-\vec P + m \vec g = 0\) .

Consequently,

\(~\vec P = m \vec g\) .

Means, if acceleration a = 0, then the weight of the body is equal to the force of gravity.

But this does not mean that the weight of a body and the force of gravity applied to it are one and the same. The force of gravity is applied to the body, and the weight is applied to the support or suspension. The nature of gravity and weight is also different. If gravity is the result of the interaction of the body and the Earth (gravitational force), then the weight appears as a result of a completely different interaction: the interaction of the body BUT and supports AT. Support AT and body BUT at the same time, they are deformed, which leads to the appearance of elastic forces. In this way, body weight(as well as the reaction force of the support) is a special type of elastic force.

Weight has features that significantly distinguish it from gravity.

Firstly, the weight is determined by the totality of the forces acting on the body, and not only by gravity (for example, the weight of a body in a liquid or air is less than in a vacuum due to the appearance of a buoyant (Archimedean) force). Secondly, the weight of the body depends significantly on the acceleration with which the support (suspension) moves.

Body weight when the support or suspension moves with acceleration

Is it possible to increase or decrease body weight without changing the body itself? It turns out yes. Let the body be in the elevator cabin moving with acceleration \(~\vec a\) (Fig. 4 a, b).

Rice. four

According to Newton's second law

\(~\vec N + m \vec g = m \vec a\) , (3)

where N is the reaction force of the support (elevator floor), m- body mass.

According to Newton's third law, the weight of the body is \(~\vec P = -\vec N\) . Therefore, taking into account (3), we obtain

\(~\vec P = m (\vec g - \vec a)\) .

Let's direct the coordinate axis Y reference system associated with the Earth, vertically down. Then the projection of the body weight on this axis will be equal to

\(~P_y = m (g_y - a_y)\) .

Since the vectors \(~\vec P\) and \(~\vec g\) are co-directed with the coordinate axis Y, then R y= R and g y= g. If the acceleration \(~\vec a\) is directed downward (see Fig. 4, a), then a y= a, and the equality takes the following form:

\(~P = m (g - a)\) .

It follows from the formula that only a= 0 body weight is equal to gravity. At a≠ 0 body weight is different from gravity. When the elevator moves with an acceleration directed downward (for example, at the beginning of the descent of the elevator or in the process of stopping it when moving up) and in absolute value less than the acceleration of free fall, the weight of the body is less than the force of gravity. Therefore, in this case, the weight of the body is less than the weight of the same body if it is on a resting or uniformly moving support (suspension). For the same reason, the weight of the body at the equator is less than at the poles of the Earth, since due to the daily rotation of the Earth, the body at the equator moves with centripetal acceleration.

Let us now consider what happens if the body moves with an acceleration \(~\vec a\) directed vertically upwards (see Fig. 4, b). In this case, we get

\(~P = m (g + a)\) .

The weight of a body in an elevator moving with acceleration directed vertically upwards is greater than the weight of a body at rest. The increase in body weight caused by the accelerated movement of the support (or suspension) is called overload. The overload can be estimated by finding the ratio of the weight of the rapidly moving body to the weight of the body at rest:

\(~k = \frac(m (g + a))(m g) = 1 + \frac(a)(g)\) .

A trained person is able to briefly withstand about six times the overload. This means that the acceleration of the spacecraft, according to the obtained formula, should not exceed five times the value of the free fall acceleration.

Weightlessness

Let's pick up a spring with a load suspended from it, or rather a spring balance. On the scale of spring scales, you can read the weight of the body. If the hand holding the scales is at rest relative to the Earth, the scales will show that the weight of the body is modulo equal to the force of gravity mg. Let's release the scales from our hands, they, together with the load, will begin to fall freely. In this case, the arrow of the scales is set to zero, showing that the weight of the body has become equal to zero. And this is understandable. In free fall, both the scale and the load move with the same acceleration equal to g. The lower end of the spring is not carried away by the load, but follows it, and the spring is not deformed. Therefore, there is no elastic force that would act on the load. This means that the load is not deformed and does not act on the spring. The weight is gone! The cargo is said to have become weightless.

Weightlessness is explained by the fact that the force of universal gravitation, and hence the force of gravity, informs all bodies (in our case, the load and the spring) the same acceleration g. Therefore, any body that is affected by only gravity or in general the force of universal gravitation, is in a state of weightlessness. Under such conditions, there are freely falling bodies, for example, bodies in a spaceship. After all, both the spacecraft and the bodies in it are also in a state of long free fall. However, each of you is in a state of weightlessness, albeit for a short time, jumping from a chair to the floor or jumping up.

The same can be proved mathematically. When a body is in free fall, \(~\vec a = \vec g\) and \(~P = m (g - g) = 0\) .

Literature

1. Kikoin I.K., Kikoin A.K. Physics: Proc. for 9 cells. avg. school - M .: Pro-sveshchenie, 1992. - 191 p.
2. Lutsevich A.A., Yakovenko S.V. Physics: Proc. allowance. – Mn.: Vysh. school, 2000. - 495 p.
3. Physics: Mechanics. Grade 10: Proc. for in-depth study of physics / M.M. Balashov, A.I. Gomonova, A.B. Dolitsky and others; Ed. G.Ya. Myakishev. – M.: Bustard, 2002. – 496 p.

Definition 1

The force of gravity is considered to be applied to the body's center of gravity, determined by suspending the body from a thread at its various points. In this case, the point of intersection of all directions that are marked by a thread will be considered the center of gravity of the body.

The concept of gravity

Gravity in physics is the force acting on any physical body that is near the earth's surface or another astronomical body. The force of gravity on the planet's surface, by definition, will be the sum of the gravitational attraction of the planet, as well as the centrifugal force of inertia, provoked by the daily rotation of the planet.

Other forces (for example, the attraction of the Sun and the Moon), due to their smallness, are not taken into account or are studied separately in the format of temporal changes in the Earth's gravitational field. Gravity imparts equal acceleration to all bodies, regardless of their mass, while representing a conservative force. It is calculated based on the formula:

$\vec(P) = m\vec(g)$,

where $\vec(g)$ is the acceleration imparted to the body by gravity, denoted as the free fall acceleration.

In addition to gravity, bodies moving relative to the Earth's surface are also directly affected by the Coriolis force, which is the force used in studying the motion of a material point with respect to a rotating frame of reference. The addition of the Coriolis force to the physical forces acting on a material point will make it possible to take into account the effect of the rotation of the frame of reference on such a movement.

Important formulas for calculation

According to the law of universal gravitation, the force of gravitational attraction acting on a material point with its mass $m$ on the surface of an astronomical spherically symmetric body with mass $M$ will be determined by the relation:

$F=(G)\frac(Mm)(R^2)$, where:

• $G$ is the gravitational constant,
• $R$ - body radius.

This relation turns out to be valid if we assume a spherically symmetric mass distribution over the volume of the body. Then the force of gravitational attraction is directed directly to the center of the body.

The modulus of the centrifugal force of inertia $Q$ acting on a material particle is expressed by the formula:

$Q = maw^2$ where:

• $a$ is the distance between the particle and the axis of rotation of the astronomical body that is being considered,
• $w$ is the angular velocity of its rotation. In this case, the centrifugal force of inertia becomes perpendicular to the axis of rotation and directed away from it.

In vector format, the expression for the centrifugal force of inertia is written as follows:

$\vec(Q) = (mw^2\vec(R_0))$, where:

$\vec (R_0)$ is a vector perpendicular to the axis of rotation, which is drawn from it to the specified material point located near the Earth's surface.

In this case, the force of gravity $\vec (P)$ will be equivalent to the sum of $\vec (F)$ and $\vec (Q)$:

$\vec(P) = \vec(F) = \vec(Q)$

law of attraction

Without the presence of gravity, the origin of many things that now seem natural to us would be impossible: thus, there would be no avalanches coming down from the mountains, no rivers, no rains. The Earth's atmosphere can only be maintained by the force of gravity. Planets with less mass, such as the Moon or Mercury, lost their entire atmosphere at a rather rapid pace and became defenseless against aggressive cosmic radiation.

The atmosphere of the Earth played a decisive role in the process of formation of life on Earth, her. In addition to gravity, Earth is also affected by the moon's gravity. Due to its close proximity (on a cosmic scale), the existence of ebb and flow is possible on Earth, and many biological rhythms coincide with the lunar calendar. Gravity, therefore, must be viewed in terms of a useful and important law of nature.

Remark 2

The law of attraction is considered universal and can be applied to any two bodies that have a certain mass.

In a situation where the mass of one interacting body turns out to be much greater than the mass of the second, one speaks of a special case of the gravitational force, for which there is a special term, such as "gravity". It is applicable to tasks focused on determining the force of attraction on the Earth or other celestial bodies. When substituting the value of gravity into the formula of Newton's second law, we get:

Here $a$ is the acceleration of gravity, forcing the bodies to tend towards each other. In problems involving the use of free fall acceleration, this acceleration is denoted by the letter $g$. Using his own integral calculus, Newton mathematically managed to prove the constant concentration of gravity in the center of a larger body.

Absolutely all bodies in the Universe are affected by a magical force that somehow attracts them to the Earth (more precisely, to its core). There is nowhere to escape, nowhere to hide from the all-encompassing magical gravity: the planets of our solar system are attracted not only to the huge Sun, but also to each other, all objects, molecules and the smallest atoms are also mutually attracted. known even to small children, having devoted his life to studying this phenomenon, he established one of the greatest laws - the law of universal gravitation.

What is gravity?

The definition and formula have long been known to many. Recall that gravity is a certain quantity, one of the natural manifestations of universal gravitation, namely: the force with which any body is invariably attracted to the Earth.

The force of gravity is denoted by the Latin letter F heavy.

Gravity: Formula

How to calculate directed to a certain body? What other quantities do you need to know in order to do this? The formula for calculating gravity is quite simple, it is studied in the 7th grade of a comprehensive school, at the beginning of a physics course. In order not only to learn it, but also to understand it, one should proceed from the fact that the force of gravity, invariably acting on a body, is directly proportional to its quantitative value (mass).

The unit of gravity is named after the great scientist Newton.

It is always directed strictly down to the center of the earth's core, due to its influence all bodies fall down with uniform acceleration. We observe the phenomena of gravity in everyday life everywhere and constantly:

• objects, accidentally or specially released from the hands, necessarily fall down to the Earth (or to any surface preventing free fall);
• a satellite launched into space does not fly away from our planet for an indefinite distance perpendicularly upwards, but remains in orbit;
• all rivers flow from mountains and cannot be reversed;
• it happens that a person falls and is injured;
• the smallest dust particles sit on all surfaces;
• air is concentrated at the surface of the earth;
• hard to carry bags;
• rain falls from clouds and clouds, snow falls, hail.

Along with the concept of "gravity", the term "body weight" is used. If the body is placed on a flat horizontal surface, then its weight and gravity are numerically equal, so these two concepts are often replaced, which is not at all correct.

Acceleration of gravity

The concept of "acceleration of free fall" (in other words, is associated with the term "gravity." The formula shows: in order to calculate the force of gravity, you need to multiply the mass by g (acceleration of St. p.).

"g" = 9.8 N/kg, this is a constant value. However, more accurate measurements show that due to the rotation of the Earth, the value of the acceleration of St. p. is not the same and depends on latitude: at the North Pole it is = 9.832 N / kg, and at the sultry equator = 9.78 N / kg. It turns out that in different places of the planet a different force of gravity is directed to bodies with equal mass (the formula mg still remains unchanged). For practical calculations, it was decided to allow for minor errors in this value and use the average value of 9.8 N/kg.

The proportionality of such a quantity as gravity (the formula proves this) allows you to measure the weight of an object with a dynamometer (similar to ordinary household business). Please note that the instrument only displays force, as the local "g" value must be known to determine the exact body weight.

Does gravity act at any (both close and far) distance from the earth's center? Newton hypothesized that it acts on the body even at a considerable distance from the Earth, but its value decreases inversely with the square of the distance from the object to the Earth's core.

Gravity in the solar system

Is there a Definition and formula regarding other planets retain their relevance. With only one difference in the meaning of "g":

• on the Moon = 1.62 N/kg (six times less than on Earth);
• on Neptune = 13.5 N/kg (almost one and a half times higher than on Earth);
• on Mars = 3.73 N/kg (more than two and a half times less than on our planet);
• on Saturn = 10.44 N/kg;
• on Mercury = 3.7 N/kg;
• on Venus = 8.8 N/kg;
• on Uranus = 9.8 N/kg (practically the same as ours);
• on Jupiter = 24 N/kg (almost two and a half times higher).