# Newton Basic Laws of Mechanics Content Newtonian mechanics

Dynamics is the study of changing the motion of a body and statics is the study of bodies in undisturbed or uniform motion. From the law of inertia follows that each body has a certain resistance to change its motion and an interaction with the environment of the body is needed for this change to occur.

## Newton's Laws

Newton stated in his Principia in 1687, the following three laws of motion, although in a more modern form.

### First Law: Inertial motion

Every body remains being at rest or being in uniform rectilinear motion unless it is compelled to change that state by forces impressed on it.

### Mass

The property of matter to resist a change of motion is called mass. The mass ratio between two bodies in an inertial reference frame is defined as the negative of the reverse acceleration ratio's of these two bodies in an interaction between the center of mass of the two bodies: $\frac{m_2}{m_1}= - \frac{\vec{a}_1}{\vec{a}_2}$ The mass of a specific object has been selected as a standard reference for measurement purposes and called the standard kilogram (kg).

In classical mechanics mass has the following properties:

• Mass is a scalar positive real number;
• The mass of a body is constant in time and space and independent of velocity.

Also notice that this first law excludes force laws of the following forms: $F=m \dot{\vec{r}}$ because velocity does not require a force and $F=m\dddot{r}$ because in that case a particle would move with constant acceleration without a force instead of with constant velocity. If we want to describe the amount of motion it seems that velocity is not enough, but that we also need to take into account the mass of the object. Therefore we define the concept of momentum: $p=m\dot{\vec{r}}$ which combines both the mass and velocity. Momentum is a vector quantity.

From the the first law follows that no force is needed for maintaining the speed of a uniformly rectilinear moving body. Mathematically a change in velocity results from a change in acceleration. Newton stated that the change in acceleration of a body is proportional to a force and inversely proportional to the mass of the body.

### Second law: Force law

The rate of change of the momentum, $p=m\dot{\vec{r}}$, of an object is proportional to the applied force, where $m$ is a constant mass: $F=\dot{\vec{p}}=m\frac{\mathrm{d}\dot{\vec{r}}}{\mathrm{dt}}=m\vec{a}$

The unit of force is the Newton, $1N=1kg.m.s^{-2}$. Note that $F = m\vec{a}$ is a vector equation, so it is really three equations in one. In Cartesian coordinates, it says that $F_x = m\vec{a}_x$, $F_y = m\vec{a}_y$, and $F_z = m\vec{a}_z$.

A force is anything that changes the momentum of a body. We observe different kind of forces, contact-forces where there is a visible push or pull and non-contact forces or field forces which seem to act at a distance. The 3 fundamental forces of physics are all field forces:

• gravitation, responsible for the attraction between large bodies;
• electromagnetic, responsible for the interaction between atoms;
• strong nuclear force, responsible for the stability of the nucleus of an atom.

Sometimes the weak nuclear force is defined as a separate fundamental force, but is has been shown that the electromagnetic and weak nuclear force are essentially the same.

The force could be a function of $r$,$\dot{\vec{r}}$,$t$ or a combination of these. The trajectory of a system is the solution of the following second order differential equation with initial conditions $r(0),\dot{\vec{r}}(0)$: $F(r(t),\dot{\vec{r}}(t),t)=m\vec{a}(t)$ Notice that the initial conditions a system consisting of one point particle is describes by 3 dimensions for position, and 3 dimensions for velocity. The total number of dimensions is called the degrees of freedom of the system.

Notice that the fist law $\dot{\vec{p}} = 0 \leftrightarrow F=0$ directly follows from the second law.

The force that matters in the second law is the net resulting force acting on the system $\sum F_i$. ### Third law: Action-Reaction

If body A and Body B are in constant contact and body A exerts a force F on a body B, then body B will exert a force -F on body A: $F_{AB} = - F_{BA}$

This law is valid only when the force between $A$ and $B$ is independent of the force with other objects e.g $C$.

In an isolated system (no net external forces), with n-bodies, all interacting with each other we have: $(1) \: \sum_{i=1}^{n}\sum_{j=1}^{n}F_{ij} =0$ For the rate of change of momentum we have: from which follows the conservation of linear momentum: $(2) \: \dot{\vec{p}}=\sum_{i=1}^n {m_i\vec{a}_i}=\sum_{i=1}^{n}\sum_{j=1}^{n}F_{ij}$ Combining with $(1)$ gives the conservation of momentum: $\dot{\vec{p}}=0$ The conservation of momentum only holds for forces of the “pushing” and “pulling” type. It fails for the magnetic force, for example. In that case, momentum is carried off in the electromagnetic field (so the total momentum of the particles and the field is conserved).

The interacting of the bodies within the system causes only to flow the momentum between the bodies, without changing the total amount of momentum. Therefore motion never stops, it is only exchanged between bodies in a system. Motion often disappears from visible motion on a macroscopic scale into the microscopic scale at the molecular level by a process we call friction. Transforming visible motion into friction results in heat. Heat is the disorganized motion of the microscopic constituents of every material.

## Forces

In classical mechanics we study the impact of forces applied to a selected body.

The forces in nature can be reduced to the fundamental forces. In a classical mechanics problem this fundamental force is normally gravitation.

### Gravitational force

The gravitational force $\vec{F}_g$ is the force that the gravitation of the earth exerts on each body in its neighborhood. This force gives each body the same acceleration depending on its distance from the center of the earth. Near the earth's surface this acceleration is approximately $g=9.8 m/s^2$. The direction of this force is towards the center of the earth. $\vec{F}_g=m\vec{a}_g=m\vec{g}$

### Weight

The weight of a body is a force exerted by the body on its surface interface due to the gravitational force in the direction of the support surface: $\vec{W}=m\vec{g}$

Next to the fundamental force of gravitation there are also forces which are consequences of other fundamental forces which play a role in classical mechanics problems.

### Normal force

The normal force, $\vec{F}_N$, is the force exerted by the interface surface on the body due to the repulsive forces of interaction between atoms at close contact. Their direction is normal to the surface interface between two objects. When the incline between the interface surface and the earth is $0^\circ$, we notice four situations:

• $a_y>g$: the body is accelerated upward, $\vec{F}_N>-\vec{W}$
• $a_y=g$: free fall,$\vec{F}_N=0$
• $a_y\less g$: the body is accelerated downward with a drag,$\vec{F}_N\less-\vec{W}$
• $a_y=0$: the body is at rest, $\vec{F}_N=-\vec{W}$

The resulting normal force pushing on the body is sometimes called its apparent weight. Humans experience this force as their weight because the spines and bones act as a spring compressed by this normal force. However the true weight, $\vec{W}$, as defined above, is unchanged.

### Friction

Friction force, $\vec{F}_f$, is a surface force that opposes motion along the surface. The frictional force is directly related to the normal force which acts to keep two solid objects separated at the point of contact. $\vec{F}_f=\mu \vec{F}_N$ where $\mu$ is the coefficient of friction which depends on the specific material of the surface. There are two types of friction coefficients:

• static friction,$\mu_s$: applied as long as there is no motion along the surface, such that $0\leq F_f \leq \mu_s\vec{F}_N$
• kineticc friction, $\mu_k$: applied when objects slide, $F_f=\mu_k\vec{F}_N$.
For most surface $\mu_k \less \mu_s$.

The weight of an object at rest is balanced due to Newton's Third Law, against the contact force of the body it rests on. This contact force consists of two components. The part that lies within the plane of contact, the frictional force and the part that is normal to the plane of contact, the normal force. We have for a body at rest: $\vec{F}_g+\vec{F}_N+\vec{F}_f=0$

### Tension forces

Tension, $\vec{F}_T$, is the pulling force exerted by a string, cable, robe or similar solid object on another object. It results from the net electrostatic attraction between the particles in a solid when it is deformed so that the particles are further apart from each other than when at equilibrium. It is the pull exerted by a solid trying to restore its original, more compressed shape.

Tension forces are normally modeled using ideal strings which are massless, frictionless, unbreakable, and unstretchable. They can be combined with ideal pulleys which allow ideal strings to switch physical direction. Ideal strings transmit tension forces instantaneously in action-reaction pairs. So if two objects are connected by an ideal string, the force directed along the string by the first object is transferred along the string to the second object.

### Elastic force

When an elastic material is deformed due to an external force, it experiences internal forces that oppose the deformation and restore it to its original state if the external force is no longer applied. An elastic force acts to return a spring to its natural length. An ideal spring is taken to be massless, frictionless, unbreakable, and infinitely stretchable. Such springs exert forces that push when contracted, or pull when extended, in proportion to the displacement of the spring from its equilibrium position.

This linear relationship was described by Robert Hooke in 1676, for whom Hooke's law is named. If $\Delta \vec{x}$ is the displacement, the force exerted by an ideal spring equals: $\vec{F}=-k \Delta x$ where $k$ is the spring constant (or force constant), which is particular to the spring. The minus sign accounts for the tendency of the force to act in opposition to the applied load.

## Application of Newton's laws

In classical mechanics we study the impact of forces applied to a selected body. In generel solving such a problem consists of a number of steps.

• Step 1: Identify the body of interest, the relevant bodies in the environment
• Step 2: Identify the forces applied by these bodies to the body of interest
• Step 3: Choose a convenient inertial reference frame
• Step 4: Draw a free body diagram, which shows only the body and all forces working on the body in the choosen reference frame
• Step 5: Apply Newton's second law to each component of force and acceleration and solve the system of equations

## Work and Energy

We defined the 'amount of motion' as the momentum $\vec{p}=m\vec{v}$. This is a vector quantity and not a scalar. A scalar quantity that measures the amount of motion must be based on speed. In the case of two bodies with a different mass but the same momentum, the body with the highest speed must have the highest value for the scalar amount of motion. We will call this scalar amount kinetic energy and define it by:

### Kinetic energy

The kinetic energy is a function of the velocity, $\vec{v}$, defined by: $K(\vec{v})=1/2 \vec{p} \cdot \vec{v}=1/2m \vec{v} \cdot \vec{v}=1/2m \norm{\vec{v}}^2$

We defined kinetic energy in this particular way because then its time derivative equals the impact of applying a force, $\vec{F}(\vec{r},\vec{v},t)$ , along the line of motion of a body. The force will change the speed of an object. If the force acts along the line of motion of the body it will increase its speed and if its acts in the opposite direction it will decrease its speed.

Let us do the mathematical derivation. To obtain the derivative of a dot product $\vec{a} \cdot \vec{b}$ we use the following rule: $\frac{\mathrm{d}(\vec{a}\cdot\vec{b})}{\mathrm{dt}}=\vec{a} \cdot \frac{\mathrm{d}\vec{b}}{\mathrm{dt}}+\vec{b} \cdot \frac{\mathrm{d}\vec{a}}{\mathrm{dt}}$ $\begin{equation} \frac{\mathrm{dK}}{\mathrm{dt}}=\frac{\mathrm{d}(\tfrac{1}{2}m \vec{v}\cdot \vec{v})}{\mathrm{dt}}=m\frac{\mathrm{d}\vec{v}}{\mathrm{dt}}\cdot \vec{v} = \vec{F} \cdot \vec{v} = \vec{F} \cdot \frac{\mathrm{d}\vec{r}}{\mathrm{dt}}=\frac{\mathrm{dW}}{\mathrm{dt}}=P \end{equation}$

The quantity on the left is the rate of change in kinetic energy and the quantity is called the power delivered by the force to the body upon which it acts. If we multiply both sides by a small amount of time $\mathrm{dt}$ we find that the differential change in kinetic energy equals the force times the differential distance moved:

$\mathrm{dK}=\vec{F} \cdot \mathrm{d}\vec{r} = \mathrm{dW}$

The quantity $\mathrm{dW}$ is called the increment of work done by the force $F$ in the displacement $\mathrm{d\vec{r}}$. The quantity $\mathrm{dK}$ is the differential of the kinetic energy. Both quantities have the same dimension $N.m$ and $1Nm=1J$ where $J$ stands for Joules.

This can also be showed as follows: \begin{align} \vec{F} \cdot \mathrm{d\vec{r}} & = \vec{F} \cdot \vec{v}\mathrm{dt} \\ &= m\vec{a} \cdot \vec{v}\mathrm{dt} \\ &=m \left(\frac{\vec{v}\cdot\vec{v}}{2}\right)^\prime \mathrm{dt}\\ &=\mathrm{dt} \frac{\mathrm{d}}{\mathrm{dt}}\left(1/2m\norm{\vec{v}}^2\right) \\ &=\mathrm{dK} \end{align}

If we integrate this equation we get: $W=\Delta K = \int_{x_1}^{x_2} \vec{F} \cdot \mathrm{d}\vec{r}$

$W$ is called the work done along the curve between $x_1$ and $x_2$.

We now summarize these important definitions:

### Work

The work done by a force along a curve is defined by the line integral: $W=\int_C{\vec{F} \cdot \mathrm{d\vec{r}}}=\int_{t_1}^{t_2}{\vec{F} \cdot \vec{v}\mathrm{dt}}$

### Power

Power is the amount of work done per unit of time and measured in Watt $(1\mathrm{W}=1J.s^{-1})$: $P=\frac{\mathrm{d}W}{\mathrm{dt}}=F.\vec{v}$

### Work Energy Theorem

The work done by a force along a curve is equivalent to the change in the kinetic energy: $W=\int_C{\vec{F} \cdot \mathrm{d\vec{r}}}=\int_{t_1}^{t_2}{\vec{F} \cdot \vec{v}\mathrm{dt}}=1/2m\norm{\vec{v}_2}^2-1/2m\norm{\vec{v}_1}^2=\Delta K$

Now that we have established the work-energy theorem we could assume that $F$ is a function of $\vec{r}$ alone, $\vec{F}(\vec{r})$. If the integral $\int_{r_1}^{r_2} \vec{F}(\vec{r})\cdot \mathrm{d}\vec{r}$ is path independent and only depends on the positions of $\vec{r}_1$ and $\vec{r}_2$ the force is called a conservative force. Examples of such foreces are gravity and the spring force. We then have the following result for the total amount of work done between the begin and end point:

$\Delta K = K_2-K_1 = \int_{\vec{r}_1}^{\vec{r}_2} \vec{F}(\vec{r})\cdot \mathrm{d}\vec{r} = G(\vec{r}_2) - G(\vec{r}_1)=G_2-G_1$ where $F(\vec{r})=\frac{\mathrm{d}G(\vec{r})}{\mathrm{d}\vec{r}}$ From this directly follows: $K_2-G_2=K_1-G_1$ For convenience physicist define the potential energy function $U(\vec{r})$ as: $U(\vec{r})=-\int_{\vec{r}_1}^{\vec{r}_2}\vec{F}(\vec{r})\cdot \mathrm{d}\vec{r}+C$ with this convention $-G_2=U_2$ and $-G_1=U_1$ and thus $K_2+U_2=K_1+U_1$ From this we can formulate the following important conservation principle.

### Conservation of mechanical energy

If only conservative forces acts on a body the total mechanical energy $E$ of the body which is the sum of the kinetic energy, $K$, plus the potential energy, $U$, remains constant: $K+U=E \text{ with } \Delta E=0$

In a mechanical system there is often a frictional force. Frictional forces are non-conservative forces. Suppose we have a system with a frictional force, $F_f$, and a number of other conservative forces, $F_c$. then based on the work-energy theorem we have for this system

$W_f+\sum W_c = \Delta K$ We also have $\sum W_c=-\sum \Delta U_p$ so following results: $W_f=\Delta K + \sum \Delta U_p$ So it is obvious that the law of conservation of mechanical energy does not hold in this case: $\Delta E = E_2 - E_1 = W_f<0$

However friction transforms mechanical energy into internal energy (heat) $U_i$ which results in a temperature rise. The internal energy produced is equal to the work done by the frictional force. We define $W_f=-U_i$ and get:

$\Delta E + U_i = 0 \to \Delta K + \sum \Delta U_p + U_i=0$

So the principle of conservation of energy still holds.

The general principle of conservation of energy is much more general. Energy may be transformed from one kind to another, but it cannot be created or destroyed, the total amount of energy remains constant.

## Torque

If an object rotates around an axis the motion we observe is rotational motion. Rotational motion is caused by a twisting force called torque. The torque, $\tau$, on a particle $P$ depends on the force $F$ and the vector $r$ between the point of rotation $O$ of an inertial reference frame and $P$. The torque is proportional to the force and position vector, both are vector quantities, therefore it seems natural to assume that the torque is also a vector. We define it as the cross product. It is good to know that although the torque mathematically behaves like a vector, it is in fact a pseudo-vector, invented for our purpose. A pseudo-vector differentiates from a polar vector, like position and force, in that its direction is not something we can measure in nature, but is based on a rule of agreement. In physics we use the so called right-hand screw rule. Everytime we refer to this rule we have a pseudo-vector.

If the force acts on a particle that is part of a rigid body which is confined to rotate about an axis that is fixed in an inertial reference frame then the torque acts on the rigid body as a whole.

### Torque

$\vec{\tau} = \vec{r} \times \vec{F}$ Torque is a vector quantity with magnitude given by: $\tau = rF \sin \phi$

and direction normal to the plane formed by $\vec{r}$ and $\vec{F}$. The sense is given by the right-hand screw rule. Torque had the dimensions forces times distance, these are the same as the dimensions of work. However torque and work are very different physical quantities. Work is a scalar for example. The unit of torque is $N.m$.

We can also write the magnitude of $\tau$ as $\tau = (r\sin \phi) F=r_\perp F$ or as $\tau = r(F\sin \phi)=rF_t$

$r_\perp$, the moment arm (or lever arm), is the component of $\vec{r}$ at right angles to the line of action of $\vec{F}$ (The line of action of a force is an imaginary line extending out both ends of the vector representing the force), and $F_t$ is the component of $\vec{F}$ at right angles to $\vec{r}$. The horizontal component $F\cos \phi$, because its line of action passes through $O$, has no tendency to produce rotation about an axis passing through $O$.

If two or more forces are acting on a rigid object each tends to produce rotation about the axis at $O$. We use the convention that the sign of the torque resulting from a force is positive if the turning tendency of the force is counterclockwise and is negative if the turning tendency is clockwise. The net torque is given by: $\sum\limits_i^n{\tau_i}=F_ir_{i_\perp}$

### Angular momentum

Consider a particle of mass $m$ and linear momentum $\vec{p}$ at a position $\vec{r}$ relative to the origin $O$ of an inertial reference frame. We define angular momentum of $\vec{l}$ of the particle with respect to the origin $O$ to be

#### Angular momentum

$\vec{I}=\vec{r} \times \vec{p}$ Angular momentum is a vector quantity with magnitude given by: $l = rp \sin \phi$

and direction normal to the plane formed by $\vec{r}$ and $\vec{p}$. The sense is given by the right-hand screw rule. Torque had the dimensions momentum times distance. The unit of torque is $kg.m^2.s^{-1}$.

We can also write the magnitude of $l$ as $l = (r\sin \phi)p=r_\perp p$ or as $l = r(p\sin \phi)=rp_\perp$

$r_\perp$, the moment arm (or lever arm), is the component of $\vec{r}$ at right angles to the line of action of $\vec{p}$ (The line of action of a momentum is an imaginary line extending out both ends of the vector representing the momentum), and $p_\perp$ is the component of $\vec{p}$ at right angles to $\vec{r}$. The horizontal component $p\cos \phi$, because its line of action passes through $O$, has no tendency to produce momentum about an axis passing through $O$.

For translational motion we have: $\vec{F}=\frac{\mathrm{d}\vec{p}}{\mathrm{dt}}$ this implies for rotational motion: $\vec{\tau}=\vec{r} \times \vec{F} = \vec{r} \times \frac{\mathrm{d}\vec{p}}{\mathrm{dt}} \tag{1}$

We now differentiate the rotational momentum with respect to $t$: \begin{align} \frac{\mathrm{d}\vec{l}}{\mathrm{dt}}&=\frac{\mathrm{d}}{\mathrm{dt}}\left( \vec{r} \times \vec{p} \right) \\ &=\frac{\mathrm{d}\vec{r}}{\mathrm{dt}}\times \vec{p}+\vec{r} \times \frac{\mathrm{d}\vec{p}}{\mathrm{dt}} \\ &=\vec{v} \times m\vec{v}+\vec{r} \times \frac{\mathrm{d}\vec{p}}{\mathrm{dt}} \\ &=\vec{r} \times \frac{\mathrm{d}\vec{p}}{\mathrm{dt}} \tag{2}\\ \end{align} Notice $\vec{v} \times m\vec{v}=0$, because the vector product of two parallel vectors is zero.

From (1 and (2 follows:

#### Time rate of change of angular momentum

$\vec{\tau}=\frac{\mathrm{d}\vec{l}}{\mathrm{dt}}$
This result states that the time rate of change of the angular momentum of a particle is equal to the torque acting on the particle. This result is the rotational analog of the fact that the time rate of change of the linear momentum of a particle is equal to the force acting on it.

The total angular momentum of a system of particles about a given point is the vectorial sum of the angular momenta of all individual particles about this same point. $\vec{L}=\sum\limits_i^n{\vec{l}_i}$

Let us differentiate this sum with respect to time. $\frac{d\vec{L}}{dt}=\sum\limits_i^n{\frac{d\vec{l}_i}{dt}}=\sum\limits_i^n{\vec{\tau_i}}$

The torques acting on the particles of the system are those associated with internal forces between particles and those associated with external forces. However, the net torque associated with all internal forces is zero. Recall that Newton’s third law tells us that internal forces between particles of the system are equal in magnitude and opposite in direction. If we assume that these forces lie along the line of separation of each pair of particles, then the total torque around some axis passing through an origin O due to each action–reaction force pair is zero. That is, the moment arm from O to the line of action of the forces is equal for both particles and the forces are in opposite directions. In the summation, therefore, we see that the net internal torque vanishes. We conclude that the total angular momentum of a system can vary with time only if a net external torque is acting on the system.

$\vec{\tau_{ext}}=\frac{d\vec{L}}{dt}$

This is the rotational analog of $\vec{F_{ext}}=\frac{d\vec{p}}{dt}$ for translational motion.

The resultant torque acting on a system about an axis through the center of mass equals the time rate of change of angular momentum of the system regardless of the motion of the center of mass.

### Angular momentum of a rotating rigid object

Consider a rigid object rotating about a fixed axis that coincides with the $z$ axis of a coordinate system. Each particle of the object rotates in the $xy$ plane in circular moption about the $z$ axis with an angular speed $\omega$. The velocity of each particle is perpendicular to the radius, and therefore the momentum arm is $r_i$ for the linear momentum $\vec{p}$. The magnitude of the angular momentum of a particle of mass $m_i$ about the $z$ axis is then: $l_i=m_iv_ir_i=m_ir_i^2\omega$ The direction is along the positive $z$-axis.

The angular momentum of the whole object is the sum of the $l_i$ of the individual particles. $L=\sum\limits_i^n{m_ir_i^2\omega}=\omega\sum\limits_i^n{m_ir_i^2}=I\omega$ $I$ is called the rotational inertia or the moment of inertia.

#### Angular momentum

$L=I\omega$

###### In general, the vector expression $\vec{L}=I\vec{\omega}$ is not always valid. If a rigid object rotates about an arbitrary axis, $\vec{L}$ and $\omega$ may point in different directions. The vector $\vec{\omega}$ always points along the (fixed)axis of rotation of a spinning body, but in general the vector $\vec{L}$ does not. In this case, the moment of inertia cannot be treated as a scalar. Strictly speaking, $\vec{L}=I\vec{\omega}$ applies only to rigid objects of any shape that rotate about one of three mutually perpendicular axes (called principal axes) through the center of mass.

Differentiating the total angular momentum with respect to time (assuming $I$ is constant) gives. $\frac{dL}{dt}=I\frac{d\omega}{dt}=I\alpha$

Using $\tau_{ext}=\frac{dL}{dt}$ we have

#### Force law of rotational motion

$\tau_{ext}=I\alpha$

## Gravitation

The great discovery of Newton was the accurate description of the heavenly bodies with a simple and unified law compared to the laws derived before by Ptolemaeus in his work the Almagest.

Ptolemaeus assumed that all motion is circular around the earth (based on the idea of Plato) and to explain the observed motion accurately he used so called epicylces.Epicycles are circles a planet orbits, and the center of this circle is orbitting the earth. So with a very complicated system Ptolemaeus managed to descibre the observed motions of planets.

Based on the observations of his mentor, Tycho Brahe, Johannes Kepler, proposed three new laws of planetary motion in the early 1600s.

• Law of Ellipses: The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus.
• Law of Equal Areas: An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time.
• Law of periods: The ratio of the squares of the periods, $T$, of any two planets is equal to the ratio of the cubes of their average distances $S$ from the sun: $\frac{T^2}{S^3}$
These laws are still applicable, but replaced by the laws of Newton.

It was Newton's great discovery that the interplanetary force holding celestial bodies in their orbits is of the same kind as the force of gravity that causes apples, and other things, to fall downward near the surface of the Earth. The law of universal gravitation formulated by Newton states:

Every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
In mathematical terms this law is formulated as follows for the gravitation between two particles of mass $m_1$ and $m_2$: $\vec{F}=-G\frac{m_1m_2}{r^2}\frac{\vec{r}}{r}=-G\frac{m_1m_2}{r^2}\hat{\vec{r}}$ $G$ is called the universal constant and is experimental determined as: $G =6.6720 \times 10^{-11} Nm^2/kg^2$

The gravitational forces are an action-reaction pair, the forces are equal and opposite in direction. This law implies a force between two particles along the line of joining them. This force in independent of forces applies by any other particles. For two bodies, which consists of multiple particles, the resulting net force between the two bodies is the vector sum of the individual forces between all the particles making up the bodies. The gravitational force obeys the principal of superposition. The net gravitational force between two spherical bodies acts just as though the mass of each body were concentrated at the center of its respective sphere.