COMPARISON OF TRANSLATIONAL MOTION AND ROTATIONAL MOTION

1          Linear displacement                      → d                    Angular displacement                 θ

2          Linear velocity                              → v                    Angular velocity                          ω

3          Linear acceleration                       → a = Dv / Dt     Angular acceleration                  α = ω / t

4          Mass                                           → m                    Moment of inertia                     I

5          Linear momentum                         → p = m v          Angular momentum                  L = Iω

6          Force                                           → F = m a          Torque                                    t = I α

7          Newton’s Second Law of motion  → F = DP / Dt    Result similar to Newton’s       t = DL / Dt

Second Law

8          Translational kinetic energy           → k = ½ mv²     Rotational kinetic energy           K = ½ Iω²

9          Work                                           → W = F d        Work                                       W = t θ

10        Power                                          → P = F v          Power                                      P = t ω

11        Equations of linear motion             → v = v₀ + at     Equations of rotational              ω = ω_i + αt

with constant linear                     S = v₀t + ½ at²      motion with constant                θ = ω_i t + ½αt²

            acceleration                                     2aS = v_f ² + v_i ² angular acceleration               2αθ = ω_f² - ω_i²

RADIUS OF GYRATION

Suppose the rigid body of mass M consists of n particles each of mass m.

Therefore             M = mn.

The moment of inertia of the body about a given axis,

I = m r₁ ² + m r₂ ²  +.………….+ m r_n ²    = m [ r₁ ² + r₂ ²  +.………….+ r_n ² ] = mK²  

where, K is called the radius of gyration corresponding to the given axis and is the mean of the squares of perpendicular distances of the particles of the body from the given axis.

Moment of inertia and radius of gyration for some symmetric bodies

	Body	Axis	I	K
1	Thin rod of length L	Passing through its center and Perpendicular to its length	1 ML2 12	L 2 Ö3
2	Ring of radius R                               → Thin-walled hollow cylinder of radius R	Passing through its center and Perpendicular to its plane Geometric axis	MR 2	R
3	Ring of radius R                               → Circular disc of radius R                  → Solid cylinder of radius R                 →	Any diameter Passing through its center and Perpendicular to its plane Geometric axis	1 MR 2 2	R Ö2
4	Circular disc of radius R                  →	Any diameter	1 MR 2 4	R 2
5	Thin-walled hollow sphere of radius R	Any diameter	2 MR 2 3	Ö2/3 R
6	Solid sphere of radius R                  →	Any diameter	2 MR 2 5	Ö2/5 R
7	Solid right circular cone of radius R→	Geometric axis	3 R 2 10	Ö3/10 R

MOMENT OF INERTIA

If m₁, m₂ , ..., m_n are the masses of the particles of a rigid body and r₁, r₂ , ..., r_n are their perpendicular distances from a given axis respectively, then the moment of inertia of the body corresponding to the given axis is given by

                                                            _n

I = m ₁ r ₁ ² + m ₂ r ₂ ² + ... + m _n r _n ² =Σ mi r_i²  

                                                                      ⁱ⁼¹

The magnitude of moment of inertia depends on the selection of the axis and the distribution of mass about it. Its S I unit is kg m2 and dimensional formula is M ¹ L ² T ⁰.

The equations L = Iω and t = I α are analogous to the equations of linear motion P = mv and F = ma respectively which shows that the moment of inertia plays the same role in rotational motion as the mass plays in linear motion. The moment of inertia is the inertia for rotational motion just as the mass is the inertia for linear motion.

TWO THEOREMS REGARDING MOMENT OF INERTIA

( i ) Parallel axes theorem:

The moment of inertia ( I ) of a body about a given axis is equal to the sum of its moment of inertia I c about a parallel axis passing through its center of mass and the product of its mass and square of perpendicular distance ( d ) between the two axes.

I = I _c + M d ²

( ii ) Perpendicular axes theorem:

( a ) For laminar bodies:

For laminar bodies, the moment of inertia I _z about Z-axis normal to its plane is equal to the sum of its moments of inertia about X-axis, I _x and Y-axis, I _y.

I _z = I _x + I _y

( b ) For three-dimensional bodies:

The sum of moments of inertia of a three dimensional body about any three mutually perpendicular axes drawn through the same point is equal to twice the moment of inertia of the body about that point.

I _x + I _y + I _z = 2 I ₀

CALCULATION OF MOMENT OF INERTIA OF CERTAIN SYMMETRIC OBJECTS

There are many calculations in which the following are very important.

( a ) Moment of inertia of a thin uniform rod about an axis, passing through its centre and perpendicular to its length:

To calculate moment of inertia of a thin rod of length l and mass M about an axis yy’ passing through its center O and perpendicular to its length, consider O as origin and X-axis along the length of the rod. A small element of length dx of the rod is at a distance x from O. 

The moment of inertia of this element about yy’ is

d I = M dx · x² because moment of inertia of the rod,

         ℓ

    +ℓ/2

I =   ∫  [ M dx · x²]

  -ℓ/2         ℓ

   =  M [ℓ³ + ℓ³]

       3ℓ [8  +  8 ]

    = Mℓ²

       12

 ( b ) Moment of inertia of a thin ring or a thin walled hollow cylinder or a thin walled hollow sphere:

As the entire mass, M, of a thin ring is at the same distance, equal to the radius R of the ring from its centre, the moment of inertia of a thin ring about an axis passing through its centre and perpendicular to its plane is MR2. Similarly, the moments of inertia of a thin walled cylinder about its geometric axis or of a thin walled hollow sphere about its center are also given by MR2, where M represents their mass and R their radii.

( c ) Moment of inertia of a disc or a solid cylinder:

To calculate moment of inertia of a disc of uniform thickness t, radius R and mass M about an axis passing through its centre and perpendicular to its plane, consider an element of the disc in the form of a thin ring of thickness dx at a distance x from its centre. Mass of this ring is 2 π x dx ⋅ t ⋅ ρ, where ρ is the density of the material of the ring. Therefore, the moment of inertia of the ring about an axis passing through the centre, O, of the disc and perpendicular to its plane is 

d I = ( 2 π x dx . t . ρ ) x² = ( 2 π t ρ ) x³ dx

Therefore moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is

     R

Ι = ∫( 2 t ρ ) x3 dx = ( 2 π t ρ ) [x⁴ /4]  = 2 π t ρR⁴ / 4

     0

   = 1/2 M R²

so therefore moment of inertia of a solid cylinder about its axis is also  1/2 M R² 

 ( d ) Moment of inertia of a thin walled hollow sphere about its diameter:

Moment of inertia of a thin walled hollow sphere about its center is I ₀ = MR².

By perpendicular axes theorem for three dimensional bodies, 2 I ₀ = I _x + I _y + I _z 

Now I _x = I _y = I _z  = moment of inertia, I , of the hollow sphere about its diameter.

Therefore I = 2 I ₀ = 2 MR²

                      ^{3           3}

ROTATIONAL MOTION

Introduction

In this chapter, rotational motion of a rigid body about a fixed axis of rotation is discussed. A rigid body is a system of particles in which inter particle distances do not change and the body cannot be deformed no matter how large a force is applied to it. Although a solid body is not a rigid body, it can be so considered for most of the practical applications.

Rotational Kinematics and Dynamics

In rotational motion of a body, all its particles move on circular paths having centers on a definite straight line, called the axis of rotation. Kinematics deals with motion without considering its cause, whereas dynamics deals with motion along with its cause and properties of the body.

Relations between variables of rotational and linear motion

( a ) Angular displacement:

The figure shows a rigid body rotating about a fixed axis OZ normal to the plane of the figure. P and P’ are the positions of a particle of the body at time t and t + Δt. Angle θ made by the line joining the particle to the centre of its rotation with a reference line OX shows its angular position at time t. Similarly, angle θ + Δθ is its angular position at time t + Δt. The change in angular position, of a particle is called its angular displacement. The angular displacement of the particle P is Δθ in time Δt. As the inter particle  distances do not change in a rigid body, all its  particles will have the same angular displacement in a given time. Hence, the angular displacement, Δθ, of the particle P can be considered as the angular displacement of

the rigid body. 

( b ) Angular speed and angular velocity:

The average angular speed of a particle or of the rigid body is defined as

 < ω > = time interval / angular displacement = Δ θ / Δ t

The instantaneous angular speed of a particle or of the rigid body is given by

 ω = limit       Δ θ / Δ t

        Δ t → 0

The unit of ω is radian / s or rotation / s.

The direction of angular velocity is given by the right-handed screw rule. When a right-handed screw is kept parallel to the axis of rotation and rotated in the direction of rotation of the body, the direction of advancement of screw gives the direction of angular velocity.

( c ) Scalar relation between angular velocity and linear velocity:

As shown in the figure, the particle P covers a linear distance equal to the arc length PP’ in time Δ t. Hence, average linear speed,

< v > = arc length PP' / Δ t  = r dθ / d t = r ω

Linear velocity is a vector quantity and its direction at any point on the path of motion is tangential to the path at that point. In the above equation, v, r and ω are the magnitudes of the vector quantities  v , r and ω.

( d ) Vector relation between angular velocity and linear velocity:

The position vector r w. r. t. the centre of the circular path of a particle, angular velocity ω and linear velocity V are shown in the figure.  As v is perpendicular to the plane formed by ω and r , the scalar relation v = r ω can be written in the vector form as 

v = ω × r

( e ) Angular acceleration:

The average angular acceleration in the time interval Δ t is

< α > = change in angular velocity / time interval = Δ ω / Δ t and the instantaneous angular acceleration at time t is given by 

α  =   limit        Δ ω / Δ t = d ω / d t = = rate of change of angular velocity

        Δ t → 0

α  is in the direction of Δ ω and in the case of fixed axis of rotation, both are parallel to the axis. The unit of α is rad / s 2 or rotation / s 2. 

( f ) Relation between linear acceleration and angular acceleration :

Differentiating the equation v = ω × r with respect to time gives the linear acceleration

a = d v  = d  (ω × r) = ω × dr    +  dω  × r

      dt       dt                       dt         dt

From the figure on the previous page and using right-handed screw rule it can be found that the direction of

ω × v is radial towards the centre. Hence, it is called the radial component, ar , of the linear acceleration a.

Similarly, the direction of α × r is tangential to the circular path at the position of the particle. Hence it is called the tangential component, aT , of the linear acceleration a.

Therefore

a = ar  +  aT

or a = (ar²  +  aT²)^1/2  because ar is perpendicular to aT

Even if α = 0, that is the angular velocity is constant, ar is not zero. As the angular displacement, θ, angular velocity, ω and angular acceleration, α are the same for all the particles of a rigid body, they are known as variables of the rotational kinematics.

NUCLEAR CHAIN REACTION AND NUCLEAR REACTOR

From the fission of one uranium nucleus by a single neutron, on an average, 2 to 3 neutrons are released which induce fission in more uranium nuclei. This sets up a self-sustaining chain of reaction called nuclear chain reaction. Following precautions have to be taken in carrying out such a reaction.

( 1 ) The neutrons released during the fission are quite fast having kinetic energy of about 2 MeV. They have to be slowed down to the level of thermal neutrons ( kinetic energy 0.04 MeV ) so that they do not escape the fission material and induce further fission. To slow down the neutrons, reflectors and moderators like heavy water ( D2O ), carbon in the form of graphite, Beryllium and ordinary water are used in nuclear reactors. Moreover, the core region of the reactor is kept large to prevent the neutrons leakage
from the surface.

( 2 ) Large amount of energy released in nuclear reaction can raise the temperature to about 105 K. Hence to cool the fission material and moderators, coolants like suitable gases, water, liquid sodium, etc. are used.

( 3 ) The ratio of number of neutrons produced to the number of neutrons incident at any stage of the chain reaction is called the multiplication factor ( k ). For controlled chain reaction, this ratio should be kept nearly one. With higher value of k, chain reaction can go out of control and with lower value of k, it may tend to stop. For this, some controlling rods of neutron absorbing materials like Boron and Cadmium are inserted in the fission material with automatic control device. Rods move further inside the fission material if k increases beyond one and come out if k reduces below one.

NUCLEAR TEACTOR POWER PLANT

A schematic diagram of a specially designed nuclear power plant is shown in the figure. 

 92U235 is used as fuel. But the ore of uranium contains 99.3 % of 92U238 and only 0.7 % of 92U235. Hence ore is enriched to contain 3 % of U235. When 92U238 absorbs neutron, it converts into plutonium though the following reactions:

92U238 + 0n1 ® 92U239 ® 93Np239 + e- + n (antineutrino)

93Np239 ® 94Pu239 + e- + n (antineutrino)

Plutonium is highly radioactive and can be fission with slow neutron.

In the reactor shown, ordinary water, used as a moderator and coolant, is circulated through the core of the reactor by means of a pump. Outlet water, at 150 atm and temperature of about 600 K is passed through a steam generator. The steam produced drives the turbine connected to electric generator and the low-pressure outlet steam from the turbine is condensed, cooled and pumped back to the steam generator. Inside the reactor, some safety rods are used in addition to the controlling rods to quickly reduce the multiplication factor, k, below 1 in case of crisis.

In some specially designed reactors, uranium oxide pallets are used as fuel, which are filled in long tubes from one end to another. The tubes through which liquid moderators are circulated surround the tubes. This device forms the core of the reactor.

DIELECTRIC SUBSTANCES AND THEIR POLARIZATION

There are many dielectric substances but following are best dielectric substances.
· A non-conducting material is called a dielectric.


· The dielectric does not possess free electrons like a conducting material.

· Introduction of a dielectric between the two parallel plates of a capacitor considerably increases the capacitance of a capacitor.

· Dielectric materials are of two types: ( 1 ) Polar and ( 2 ) Non-polar.

· The atoms of a dielectric ( like HCl, H2O ) that have permanent dipole moment is called a polar dielectric.

· The atoms of a dielectric ( like H2, O2 ) that do not have permanent dipole moment is called a non-polar dielectric.

CONDUCTORS AND ELECTRIC FIELDS

When a conducting material is placed in a uniform
electric field as shown in the figure, free electrons
migrate in a direction opposite to the electric field
and get deposited on one side of the metal surface
while the positive charge gets deposited on the other
side of the conductor. This produces an electric field
inside the conductor and the migration of charges stops when the internal electric field becomes equal to the external field.

If we draw a Gaussian surface inside the conductor as shown in the figure, then since the electric field on it is zero, the net electric charge enclosed by it is also zero.
The important conclusions are:
( 1 ) Stationary electric charge distribution is induced on the surface of the conductor.
( 2 ) Both the electric field and the net electric charge inside the conductor are zero.
( 3 ) At every point on the outer surface of the conductor, the electric field is perpendicular to the surface. This is so because the electric charge on the surface is stationary which means that no tangential force acts on it, thus proving that the electric field on the surface has no tangential component.

Consider another example of a hollow conductor placed in an external electric field. Here also, the electric charges deposit on the outer surfaces and the electric field inside is zero as there is no charge inside. This phenomenon is called Electro-static Shielding. When a car is struck by lightning, the person sitting inside is saved from lightning as the car is hollow and acts like an electrostatic shield. Electric field inside a charged conductor which is NOT in an electric field is also zero. Consider a Gaussian surface close to the surface of the conductor as shown by broken line in the figure. The line integration of the electric field along the Gaussian surface being zero, the net electric charge enclosed by it is also zero. This shows that in a charged conductor, the electric charge gets distributed on the outer surface of the conductor. As the electric charges are stationary, the direction of the electric field will be perpendicular to the surface of the conductor as shown in the figure and its magnitude will be
 σ / Є0 

To explain it, consider a pillbox shaped Gaussian surface on the surface of the conductor as shown in the above figure.

The charge enclosed by the Gaussian surface = A .σ
The total flux passing through this surface = A E
Therefore by Gauss’s law, A E = σ / Є0 
If σ  is not uniform along the surface, its proper value at the point should be used to calculate value of E at that point.
If a positive electric charge is placed in the cavity of the conductor as shown in the adjoining figure, it induces
charges on the inner and outer surfaces of the conductor in such a way that the field will be zero in the interior portion of the conductor. 

ELECTRIC POTENTIAL ENERGY AND POTENTIAL DIFFERENCE

A stationary electric charge at infinity has no energy ( kinetic or potential ) associated with it. If a unit positive charge is brought from infinity to an arbitrary point P in the electric field such that it has no velocity at that point then, the field being conservative, work done on it is stored with it in the form of potential energy and is called the electric potential of the point P and is given by

                P

 V (P) = - ∫ q E . dl

               ∞

If the electric charge is of magnitude q instead of unity, then the work done is called the potential energy of the charge q at point P and is given by

                               P

  U (P) = qV (P) = - ∫ q E . dl

                              ∞

The original electric field or the arrangement of charges in the field should remain unaffected by bringing the electric charge q or the unit charge from an infinite distance to the point in the electric field.

Generally, one needs to calculate the potential difference or the difference in potential energy of charge q when it is moved from P to Q, which is given by

                            Q

 U (Q) - U (P) = - ∫ q E . dl

                            P

It should be noted that this potential energy or the potential energy change is associated not only with the charge q but also with the entire charge distribution which gives rise to the electric field.

ELECTROSTATIC POTENTIAL AND CAPACITANCE

LINE INTEGRAL OF ELECTRIC FIELD: -


If a unit positive charge is displaced by dl in an electric field of intensity E, work done is given by dW = E. dl Line integration of this equation gives the work done in displacing a unit positive charge from P to Q as 

       Q

W = ∫ E . dl

       P

This work depends only on the initial and final positions of the unit charge and not on the path followed by it. Hence, work done in moving a charge along a closed path is equal to zero. Thus electric field like gravitational field is a conservative field. 

ELECTROSTATIC POTENTIAL: -

The work done by the electric field in moving a unit positive electric charge from an arbitrarily selected reference point q, which may be inside or outside the field, to point P is given by

        p

W = ∫ E . dl

       q

For the selected reference point, the value of WP depends only on the position of point P and not on the path followed in going from reference point to point P.

Let q be at infinity. The electric field at infinite distance due to finite charge distribution will be zero. The electric field due to an infinitely long charged plane at infinite distance will not be zero. However, in practice, one cannot have such a charge distribution.

The work done in a direction, opposing the electric field in bringing a unit positive charge from an infinite position to any point in the electric field is called the static electric potential ( V ) at that point.

Its sign is taken as negative as the work done is in a direction opposite to the electric field. Thus, work done in bringing a unit positive charge from infinity to points P and Q will be

                  P                                   Q

V ( P ) = - ∫ E . dl  and V ( Q ) = -  ∫ E . dl

                 ∞               ∞

So therefore

                                Q            P

V ( Q ) - V ( P ) = - ∫ E . dl + ∫ E . dl

                               ∞     ∞

                             Q

                        = - ∫ E . dl

                             P

This equation gives the electric potential of point Q with respect to point P. Its unit is volt ( joule / coulomb ) denoted by V and its dimensional formula is M1L(exp2)T(exp – 3)A(exp – 1).

Science Introduction: CHEMICAL EFFECTS OF ELECTRIC CURRENT

Science Introduction: CHEMICAL EFFECTS OF ELECTRIC CURRENT: INTRODUCTION: - Solid and molten metals are good conductors of electricity due to free electrons. When current flows through metals, only h...

THERMONUCLEAR FUSION IN SUN AND OTHER STARS

Just as energy is released in nuclear fission, it is also released when light nuclei like proton and deuteron fuse together at a very high temperature to form helium nucleus. Such a process is called thermonuclear fusion.

Hydrogen works as fuel and helium is the end product of the process, called proton-proton cycle, occurring in the Sun. The reactions are given by
1H1 + 1H1 ® 1H2 + e  + n + 0.42 MeV                                   ( 1 )

e+ + e- ® 2 g + 1.02 MeV                                                   ( 2 )

1H2 1H1 ® 2He3 + g + 5.49 MeV                                          ( 3 )

2He3 + 2He3 ® 2He4 + 1H1 + 1H1 + 12.86 MeV                     ( 4 )

The first three reactions should occur twice so that the fourth reaction becomes possible. The 

total energy released in this process is = 2 ´ ( 0.42 + 1.02 + 5.49 ) + 12.86 = 26.7 MeV. 

Carbon-nitrogen cycle is also proposed in case of the Sun in which 25 MeV energy is

released and the net reaction is fusion of 4 protons forming 1 nucleus of helium, 2He4.

Choosing a dog
Choosing a new dog is exciting and
emotional. We are all prone to choosing
what looks familiar and attractive,
whether it suits us or not. Making the
right choice is more likely, however, if
you take time to consider temperament
and characteristics, and think carefully
about what kind of dog will complement your
own personality. This section will help you to
do that, explaining why different breeds
have different behavior traits
as well as physical forms, so that
you can find a breed that will fit
into your lifestyle. It will also
help you decide whether a puppy
or an adult may be more suitable,
and will enable you to find a dog
that is just right for you.
ACTIVE DOGS
If you choose an energetic dog,
you need to be prepared to be
active yourself if you are to
have a happy life together

CHEMICAL EFFECTS OF ELECTRIC CURRENT

INTRODUCTION: -

• Solid and molten metals are good conductors of electricity due to free electrons. When current flows through metals, only heating occurs and no chemical effect is observed.
• Most liquids have no free electrons and hence do not conduct electricity, e.g., water.
• When an acid, base or an inorganic salt is added to water it dissociates into positive and negative ions which conduct electricity. The solutions which conduct electric current are called electrolytes and the vessel containing it along with electrodes is called an electrolytic cell.
• Inorganic salts like NaCl and KCl conduct electricity in molten form.
• Silver iodide ( AgI ) conduct electric current even in solid form.
• Normally, the solutions of organic compounds are non-conductors.
• In NaCl crystal, Na+ and Cl- ions are bound to each other due to electrostatic attraction. About 7.9 eV [ 1 eV = 1.6 × 10exp(- 19) joule ] energy is required to separate them. Only 0.03 eV thermal energy is available at room temperature which is insufficient to break NaCl crystal. When NaCl is added to water, polar water molecules get arranged in space between the ions which reduce attraction between them. Also, due to a specific distribution of charge inside them, some water molecules stick to each other forming a cluster around the ions. Each cluster is electrically polarized which reduces the strength of electric field between the ions. For this reason, dielectric constant, K, of water is very high which reduces the electric field between Na+ and Cl- ions to 1 / K times and they get dissociated. Due to its high dielectric constant, water acts as a good solvent. Ions dissociated this way participate in conduction of electric current. 
• At room temperature, electrical conductivity of electrolytes is 10exp(- 5) to 10exp(- 6) times that of metals because ( i ) the number density of ions is less as compared to the number density of electrons in metals, ( ii ) viscosity of solution increases electrical resistance and ( iii ) drift velocity of ions is less compared to electrons due to their larger mass.

NATURAL RADIOACTIVITY

Heavy elements like uranium are unstable and emit invisible radiations spontaneously to gain stability. This phenomenon is called “natural radioactivity”. This was accidentally discovered by Becquerel in 1896 while studying the relation between X-rays discovered by Rontgen in 1895 and the phenomenon of fluorescence. He called them Becquerel rays.

Madame Curie and her husband Pierre Curie isolated radium and polonium from an ore of uranium called pitch-blend which showed much larger radioactivity than uranium. Later thorium and actinium possessing radioactivity were also discovered. Radiations from radioactive elements are called radioactive radiations.

Radioactive radiations are spontaneous and instantaneous and are not affected by pressure, temperature, electric and magnetic fields etc. Their emission rates also cannot be changed by any means not even by combining radioactive elements chemically with other elements to form different compounds.

CONCEPT OF THERMODYNAMIC SYSTEM AND ENVIRONMENT

Thermodynamic system is a part of the universe under thermodynamic study which can be one, two or three dimensional and may consist of one or more objects called components of the system. A system may be made up of radiation or radiation may be one of the components of the system. Remaining part of the universe surrounding the system and interacting with it is known as its environment. The boundary separating the system and its environment is called wall of the system. The type of interactions between the system and its
environment depends on the nature of its wall.

The macroscopic description of any system in physics is done in terms of its measurable properties. For example, in kinematics of rotational motion, position and velocity are macroscopic properties which are called mechanical coordinates. Potential and kinetic energy of a rigid body are determined using these mechanical coordinates with respect to some system of coordinate axes. Similarly, in thermodynamics, macroscopic quantities are thermodynamic coordinates which help in determining internal state of the system. Thermodynamic state of a system is determined from values of its mechanical and thermal properties like pressure, volume, Temperature, internal energy, etc.

Interaction between a system and its environment is called a thermodynamic process. If no interaction occurs, the system is called an isolated system. Thermal and mechanical properties of an isolated system remain constant and the system is said to be in a definite thermodynamic equilibrium state. During interaction with the environment, thermal and mechanical properties of the system change continuously at the end of which the system attains an equilibrium state. The amount of heat energy exchanged during the interaction, known as heat, is denoted by Q and the mechanical energy exchanged, called work, is denoted by W.

The thermodynamic equilibrium state of a gas is decided by its pressure, volume, temperature and quantity, not all of which are independent. These are called thermodynamic or state variables. The mathematical equation of state gives the relations between these variables, e.g., PV = μRT is the equation of state of an ideal gas. Thermodynamic state variables are of two types:
( i ) Extensive variables which depend upon the dimensions of the system, e.g., mass, volume, internal energy,  ( ii ) Intensive variables which are independent of the dimensions of the system, e.g., temperature.

KEPLER'S LAWS

First Law:
“The orbits of planets are elliptical with the Sun at one of their two foci.”

Second Law:
“The area swept by a line, joining the Sun to a planet, per unit time ( known as areal velocity of the planet ) is constant.”

Third Law:
“The square of the periodic time ( T ) of any planet is directly proportional to the cube of the semi-major axis
(a) of its elliptical orbit.