Classical Mechanics 9861046007

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1. ·,

[For M.Sc. (Physics), B.Sc. (Honours), B.E., Net, GATE and Other Competitive Examinations]

Dr. J�.C. Upadhyaya MSc., PhD., F. Inst. P. (London) Formerly Director/Professor lncharge, Dau Dayal Institute,. Dr. B.R. Ambedkar University, Agra (India), Senior Reader in Physics, Agra Cot1ege, Agra {India)

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9iiinalaya GJluhlishingGflouse

MUMBAI• NEWDaHl• NAGPUR• BENGALURU•HYDERABAD• CHENNAI • PUNE LUCKNOW• AHMEDABAD• fRNAKULAM • BHUBANESWAR• IND0RE• KOLKA TA• GUWAHATI

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Preface. to th'e Second Edition The author feels pleasure that the present book, entitled 'Classical Mechanics" has been warmly welcomed and liked by the readers: Tiiat is why the first edition of the bbok was finished soon antl it was reprinted.four times, Now he is-presenting thoroughly revised and enlarged edition of the book in view of the latest syllabi of various Indian Universities and Model Syllabus, approved by the U,G.C., New Delhi. In the last few decades, a lot of workhas been done .in the field of non-linear dynamics and chaos. Looking at the importance and wide applications of this subject, several universities in India and abroad · have introduced this topic in the syllabi of Classical Mechanics. In the present edition a chapter on 'Non-:-linear Dynamics and Chaos' has been added to meet this requirement of the students. The author shall be grateful to the readers who would be kind enough to send their useful and constructive criticisms for the improvement of the subject matter. July 12; 2005

J.C. Upadhyaya

Preface to the First Edition Present book deals with an advanced course on mechanics, namely classical mechanics, for the . students of B.Sc. (Honours), M.Sc. (Physics) and B.E. classes. In addition tp a course book, it. has been written for the candidates, struggling to qulify competitive examinations at national and state levels such as NET, GATE;_SLET, I.A.S. etc. The concepts and formulatations involved in classical mechanics form the base to construct the entire building of physics. Of course, quantum mechanics plays the key role to study the phenomena at atomic scale, specially in the fields of atom~c.. and nuclear physics. The role of classical mechanics is of extreme importance on one hand in modem calculations involved in laµnching of satellites, motion. of rockets and relativily massive bodies, and on the other hand it makes essential background to switch over and move with couriosity and enthusiasm in the various branches of modem physics. In fact classical mechanics. provides an opportunity to a · 1• student to be famiiiar with and command many of the mathematical techniques needed in quantum mechanics. Classical mechanics had been developed over several centuries in particµlar by Newton, Lagrange, Hamilton and others. At lower level, the Newtonian mechanics and at higher level, the Lagra[!.gian and Hamiltonian dynamics, involving advanced topics, are taught. Here we mean by classical mechanics the mechanics of Lagrange and Hamilton. Classical mechanics was developed over long time on the basis of observations on moving bodies at relatively low speeds. Of course, the relativistic theory of Einstein deals with all particle-speeds, but it does not modify the classical ideas regarding the basic nature of matter and radiation and hence the relativistic theory is generally studied in classical mechanics. Therefore, in the present book, we also include the special theory of relativity and relevant advanced formulations, e.g:, four dimensional Mirikowski space and convariant fonnulation of electrodynamics. This course is conceptual in nature and involves intricate formulations, The course has been partly drifted to B.Sc. (Honours) classes in some universities and conventionally taught at M.Sc. level in the different universities. During teaching, the ·author had a feeling that the students need a textbook which deals the subject matter of classical mechanics with simplified treatments and good number of illustrations. Keeping this idea in mind, the author has made an effort to write a book on the subject in a simplified way with proper explanations so that.an average student may not feel 0 respectively . ..

1 12 Examples of central force are gravitational ·[F(r) = Gm;m2 ] and coulomb·[F(r) = q _; ] forces. 4 xe 0 , . r . . . .

Ex. 5. When a particle moves under a central force, show that (a) its angular momentum is conserved, (b) the motion takes place in a plane and {c) the ateal velocity remains constant. Solution i If a particle is moving under the influence of a central force F= F{r)rlr, so that the torque acting on it is given by

dJ r dt · r where J is the angular momentum about the origin. Therefore J = constant (vector). ... (i) Thus, when a particle moves under the action of a ·central force, its angular momentum ( J) is conserved, i.e., J remains the same in the magnitude and direction. But J = r x p, 't

= - = r x F = r x F(r)- = O [ •: r x r = OJ

where p is the linear momentum. Taking dot product with r of the both sides of this equation, we have r•J r•(rxp)=.(rxr)•p=O (since in a scalar triple product the positio11s of dot and cross may be interchanged and r x r = 0). Therefore r is perpenducular to the constant vector J i.e., the motion takes pface in a plane for .a central force. Now, let O be the centre of force. When the vector r. changes to r + Ar, the vector area AA swept by the radius vector in this time is

... (ii) I . AA= -r x dr 2

... (iii) .

This area is swept in At time, therefore dividing both the sides of this equation by At and taking the limit as At ➔ 0, . then

,J

dA 1 -·"" rxv dt'

Fig. 1.10 : Vector area swept by the radial vector

- ·..

2

2ni

... (iv)

which gives the areal velocity of. the narticle. Bu't J is constant for the .motion . under a central force [eq: (i)], I we mean that the areal velocity remains .., constant. I

'

fotroductory Ideas

19

Ex. 6. Harmonic Oscillator : The aifferential equations of a simple harmonic oscillator of mass m is given by d 2x m-7-+Cx

0

de

0 j

i"F-j

pc of or

'

(r- -r._·).= . (r- -r-)•d'(rJ ·. 1:Jc . lj

I

'' '

I

j

I

r 1-_)

=0

i,j

i >j

because. the internal force Eii is coll§icoordinate q1= 0 is sufficien\1~d two generalized coordinates

.

OJ

. r

w

Lagrangian Dynamics

35

q1= 8 , and qi==1

for a particle moving on the surface of a sphere. The generalized coordinates for a _system of N particles, constrained by m equations, are n = 3N - m. It is not necessary that these coordinates should be rectangular, spherical or cylindrical. In fact, the quantities like length, (length}2, angle, energy or a dimensionless quantity may be used as generalized coordinates but they should completely describe the state of the system. Further these n generalized coordinates are not restricted by any constraint. For a system of N particles, if xi, Yt , zi are the cartesian coordinates of the ith particle, then these coordinates in terms of the generalized coordinates qk can be expressed as

xi= x,(ql, qz,···,qk ,... ,qn, t) · ... (8 a)

Yi= Y1 (qi, qz, .. ,,qk ,... ,qn, t) zJq1, qz,···,qk ,. .. ,qn, t) or in general the positi-On vector ri (xi, yi' z1 ) of the ith particle is

ri= ri (qi' q2,... ,qk ,... ,qn, t) ... (8_ b) · Eqs. {8a) or (Sb) give the transformation equations. It may be mentioned that the .generalized coordinates may be the cartesian coordinates.

One should note that the system is said to be rheonomic, when there is an explicit time dependence in some or all ofthe functions defined by eq/ (8). If there is not the explicit time dependence, the system is called scleronomic and t is not written in the functional dependence, i. e., ...(9)

2.5. PRINCIPLE OF VIRTUAL WORK In order to investigate the properties of a system, we can imagine arbitrary instantaneous change in the position vectors of the particles of the system e.g., virtual displacements. An innnitesimal virtual displacement -~ of ith particle of a system of N partides is denoted by Sri' This is the displacement of position coordinates only and does not involve variation of time i.e.,

6ri or;(ql' qz, ... ,qn) Suppose the system is in equilibrium, then the total force on any particle· is zero i. e., i =l, 2, .... , N F.=O, l •

...(10)

The virtual work of the force F.l in-the virtual displadement 6r.l will also be zero i.e., . I

6W1

F1 • 6r; = 0

Similarly, the.sum of virtual work for all the particles.must vanish i.e., •

f

I

,

•

N

oW = 1:Fi •ori

0

... (11)

i=I

This result represents tlle principle of virtual work which states that the work done is zero in the case · of an arbitrary virtual displacement ofa system from a position of equilibrium .

-

~

The total force. Fi onthe ith particle can bf§xpressed as. . :

F.

t t

I

l

where

Ft is the applied .

l

F.a + fI I

force and .

Hence eq. (11) assumes the

Ll :s the force of constraip.t. . .

form

.Classical Mechanics

36

We restrict ourselves to the systems where the virtual work of the forces of constraints is zero, e.g., in case of a rigid body. Then N

~F~ I i=J ,t.,,

•or.I

=0

...(12)

i.e.Jor eq'f!,ilibrium of a system, the virtual work of applied forces is zero. We see that the principle of virtual work deals with the statics of a system of particles. However, we want a principle to deal with the general motion of the system and such a principle was developed by D' Alembert.

2.6. D' ALEMBERT'S PRINCIPLE According to Newton's second law of motion, the force acting on the ith particle is given by F,

= dpi = P·

I

I

dt

n

p

This can be written as i = 1, 2, ... , N

These equations mean that any particle in the system is in equilibrium under a force, which is equal to the actual force Fi plus a reversed effective force

Pi. Therefore, for virtual displacements ri,

But F; = F/' + f; , then N

N

i=l

i=I

I (F/' - p;) •or; + L f; • or; ~ O Again, we restrict ourselves to the systems for which the virtual work of the constraints is zero, i. e.,

Ifi • or; = 0. Then , ...(13)

This is known as.D'Alembert's principle. Since the forces of constraints do not appear in the equation and hence now we can drop the superscript. Therefore, the D'Alembert's principle may be written as N

I(F; -Ji;)•6r; =O i=l

...(14)

Ex. l. Two heavy particles of weights W1 a11j:fff2 are connected by a light inextensible string and hang over a fzxed smooth circular cylinder of radius· fj., the axis of which is horizontal [Fig. 2.3]. Find the ·\:ondition of equilibrium of the system by applying the· principle of virtual work. Solution : According to the principle ofvirtual work N

I Fi •6r; = 0 i=I

t

I t

I

I

Lagrangian Dynamics

37

Here, i = I, 2 and therefore

w1 sin0 or1+ w2 sinq, or2= o But

or1=R dB,

or2= R dq,

0

W1 sinB 08 + W2 sint/J oq, = O

But 8 + tp = constant Therefore

08 + oq, = 0 or oq, · = - 00

Fig. 2.3 ·

Thus The system is in equilibrium, hence the following condition is to be satisfied (08

.. 0 W1 sm

w.·

.

2 sm

,,. . 'I' ·=

0

t=

0) :

Wi simp -0r .-··-=-.- . ' Wz .

sine

Ex. 2. An inextensible string of negligible mass hanging over a smooth peg B [ Fig~ 2.4] connects one mass m1on a frictionless inclined plane of angle 6 w another mass m2• Using D'Alembert's principle, B

prove that the masses will be in equilibrium, if sin.0

Solution : According to D'Alembert's principle 2

:E(F; -pi) •Sr;= 0 i=l

Let r 1 and r 2 be the.position ve 1is independe,nt_. of v..\ , we . have/ .

I.'

'

! -

dA X dt

'~

d J!._ ( dt av_t

V•

A)

... (~1)

T : i:

51

Lagrangian Dynamics

~

r ~

Therefore

q [-~($

Fr

... (63)

•A)+, d {_q__($- v • A)}]

V

ax

avx

dt

We define a generalized potential U, given by

,)

U

q (~ -

V •

· ... (64)

A)

which i,s a velocity dependent potential in the sense of eq. (50). T~erefore, eq. (63) takes the form

Fx

au d au --+--

dt 8vx ·

Bx

...

(65)

The Lagrange's equations (48) in this case take the form (q k = x '

l,

CJ

k

= i = vX and Gk

!!_{ ar ) ~ ar · F dt·

ovx ' ax

FX)

·

... (66)

X

Substituting Fx from f66) in (65), we get the Lagrange's equation as

d(a

- -. (T-U) dt ox

1)

i

Bx

!!_(BL) iJL = O

or where

) -(T-U) a =O

dt .l

Bx

... (67)

Bx

= T U = T- q~ + q v • A

... (68)

Eq. (68) gives the Lagrangian for a charged particle moving in an electromagnetic field. Note: In Gaussian C.G.S. system Bis to be replaced by B/c in eqs. (52) and (53), where c is the speed of light. Therefore the expression for generalized potential isobtained to be (J

q~ _!!_ (v · A). C

Gyroscopic Forces ))

l)

All the velocity dependent (-0rces, which do not consume power, are called gyroscopic forces. If a charge q is moving wi-th velocity v in a magnetic field B, then the force acting on the particle . F = q (v x B) is gyroscopic in nature. For such a force #le power consumed happens to be zero, i.e., P = F • v q (v x BJ• v = qv • (v x B) q(v xv)• B = 0 because for a scalar triple·prnduct (Ax B) • C C •(Ax B) = (C x A)• B. Thus the vel-0city .dependent magnetic force, given in eq. (53) is an ex~mple of gyroscopic force. A gyroscopic force can be incorporated in a -generalised potential U similar to the one.due to magnetic force, given in-eq. (64) with the Lagrangian l {eq. (68)) and Lagrange's equation·(67).

2.11. Hamilton's Principle and .L~grange's Equatio.ns In Sec. 2.7, we have used the D'Atembert's principle t-0 deduce Lagrange's equations. This principie uses the idea of virtucil work and $fe~ns fromNewtop 's second law of motion. These Langrange 's equations · can be derived by an entirely diffo1~nt way; mitrieiy H;~milton'.s v~riational principle.

Hamilton's principle ~ ~ltis pril,i.c:iple states that Joi· a conservative holonomic system, its motion from time t 1 to time t2 is such tli;ij~f.11?, i17r~gral(known as action or action integral ) ·

line

Classi(::al Mecltanics

52

r I, r

... (69)

- r- V has stationm)' {extremum) value for the correct path of the motion.

ivi.th

The quantity Sis called as Hamilton's principal jimctio11. The principle may be expressed as 0

... (70)

where is the variation symbol.

Lagrange's equation from Hamilton's principle : The Lagrangian l is a function of generalized coordinates q1;s and generalized velocitiesqk 'sand time t, i.e., L (lf1,{/z,···,t/k, ... ,(/n,{i1 ,{h ,... ,qk ' ... ,iJ11, l)

ff the Lagrangian does not depend on time t explicitly, then the variation bL can be written as ·

oL_

11

aL

aL

II

I-oqk + L-'.-oiJ* k=18qk k ,1aqk

... (71)

0

Integrating both ~ides from t 12

J'1

ol dt

=

t 1 tot

=

t2, we get

=

f'1 L aqk 11

al oqk dt +

k

r~--I a~ oqkdt

JI

k aqk

But in view of the Hamilton's principle

... (72)

Therefore,

where

o, Solution : The constraint of motion is holonornic and the ~ ~ z - Rcos0 constraint equation is ? 2 2 2 x- + y + z - R = 0 'i We take 0 and lecule is

.~~

.

MM µ Hz = M+M

.

M

=2·

Ex. ·2. Show that the spectrallines of positronium are arranged in the same pattern as in the case of atpmic hydrogen spectrum but have nearly double the wavelengths. . Solution : Ac~ording to Bohr's are given by . theory; the frequencies of lines in the hydrdgen spectrum . . .

.

.

...(l)

where n andpai:e integers (p > n) and µ is the re.duced mass of the hydrogen atom.

. Classical Mechanics

106 Now, µH

= m [1- (1/1836)] =m, mass of the ~lectron

...(ii)

·the

Thus in the.·case of hydrogen spectrum, .frequencies are directly proportional to the mass of the electron m (v ex: µ or m). Positronium has the structure like hydrogen atom, the expression (i) will also provide the spectral lines, rediated ·by positronium, but the reduced mass of hydrogen atom µH m is to be replaced by the reduced mass of the positronium µp m/2. Hence in case of positronium, we will get the . . spectral lines like that of hydrogen but their frequencies will be nearly half, i.e., their wavelengths twill be nearly double because · ' ·

=

m/2

1 VH orv 2 P 2

--=m

...(iii)

...(iv)

Therefore, :

'.

. .

.

'

.

'

·.

.

.

.

.

.

.

where vH, vP and A..H, A..P are the frequencies and wavelengths of hydrogen and positronium respectively and c is the speed of light. Thus we see that the spectral lines of positronium are arranged in the same pattern as in case of hydrogen but have nearly double wavelengths.

,4.2. CENTRAL FORCE AND MOTION IN A PLANE · If a force acts on a particle in such a way that it is always directed towards·or away from a fixed centre and its magnitude depends only upon the distance (r) from, the centre, then this force is called central force. Thus a central force is represented by F =f (r) r ·= f {r) r/r .

... (9)

where f (r) is a function of distance r only and r r/r is a unit vector along r from the fixed centre. The force is attractive or repulsive, iff{r) < 0 or f (r) > 0 respectively. A central force is always a conservative force and if V(r) is the potential energy, then

f (r)

. av ~

ar

or F = -

av r

ar ;

... (10)

The potential energy for central force depends only on the distance r and hence the system possesses spherical symmetry. Thus any rotation aqout a fixed axis will not have any effect on the solution and .hence an angle coordinate for rotation about a fixed axis must be cyclic. This results in the conservation of angular momentum of the system i.e., · J = r x p =constant (vector) ...(11) where p is linear momentum. Taking dot product with r in eq. (1 ·l ), we have · . r. J r. (r x p) = (r x r). p = O since in a scalar triple product the position of dot and cross are interchangeable and r

...(12) x

r · 0.

Therefore, position vector r_is-always perpendicular to the constant J.vector. This means that the motion of the particle under central force takes place in a plane and we ~an describe the instantaneous position of the particle in plane polar coordinates r and 0.

'

.

Two-Body Central Force Problem

107

.

.

.

4.3. EQUATIONS OF MOTION UNDER CENTRAL FORCE AND FIRST INTEGRALS Consider a particle of mass m moving about a fixed centre of force 0. [Fig, 4.4] ·

p

The particle is moving under a central force r

av

r

r

ar

r

F=f(r)- = - - where V(r) is the potential energy. • Using polar coordinates (r, 8 ), the Lagrangian for the· system can be written as

0

Fig. 4.4 : Area swept out·by the radius v~ctor in tnfinitesimal small time dt

...(13)

In eq. (13), the Lagrangian L is independent of ecoor-dinate (i.e. iJL/a(J 0) and hence Bi~ the cyclic coordinate. The canonical momentump8 corresponding to the coordinate 8 is .given by

•Pa=-. oL =mr 29·

...(14)

ae

which is the angular momentum. Now, one of the equation of motion (Lagrange' s equation for B coordinate) is

d 2.9 ) = 0 -d(oL)· ---.- -oL = 0 or -(mr dt

ae

ae

dt ·

...(15)

..

Integration of this equation gives one of the first integral of motion. i.e, ·

mr2 e = J (constant)

...(16) '>,,

where J is the constant magnitude of the angular momentum and is conserved. Since m is. a constant , we obtain from eq. (15) d dt (

i r2 e) =0 or ½r2 e= h (constant)

The ,factor½ has been inserted above so that

...(17)

½r2 e represents the areal velocity (h),

i.e., the area swept

out by th~ radius vector per unit time. It can be seen from Fig. 4.4 that the differential area dA swept out in time dt is

dA

1·

.

12

=2 r (r dB) =2 r dB

and

dA

12d8

12

so that dt = 2 r dt =2 r

e is the· areal vel~city.

Thus from eq. (17) we see that the areal velocity is constant, when the motion is taking place under centra.l force. This is in accordance with the well known Kepler's second law of planetary motion. In other words the conservation of angular momentum is equivalent to say that the areal velocity is constant. Lagrange equation for. r coordinate is

. I,.

Classical .Mechanics

.108

aL · aL Bf =mf and a,.

Fr9m ( 13 ), . ·

.2

. d

-(mf)-mre 2 +

Therefore; ·

dt · ·

a;= -f

Putting from eq. {10),

\

·.

.-.

av

mre - a,. oV

... (18)

-= 0

8r

(r) for central force, eq. (18) takes the fonn.

. .

.

l

...(19)

mr-:-mre =f(r) .. . .~ ? J Also: from ( 16), e == · . mr

. ·. ·'.

=--r ,-ther~fore . .. . 12

..

mr--_3 =f(r)

...(20)

mr

This is the second order differential equation in r coordinate -0nly. Eq. (20) can also be writte~ as

.·

_.!.~[1 ·i _avar

2 2 1 _ av = . mr3 ar 2 ar mr 2

mr =

. .. a . mr ·= - or

or

[1

2

1 ] 2 mr2 + V

...(21) .

Multiplying both sides of eq. (21) by f, we get

mr ,.·'= ~'lq_fr~ + v] i , · ar [2mr 2

,,, .

. -

r

!Gmr'}=- :(i:, 1

or .

+v] d{l . 1

2

1 .J =O 2· or - -mr +--+V 2 dt- 2 2 mr

...(22)

Integrating it, we get · 1.. -2 . 12 V ·"- mr + - 2 + 2 2mr . .•

.

. ·._ .··

.

? .

Smee .from ( 16), _J = mr .

.

. .

E

, constant

'. ..(23)

.

e ,.therefore .

'

.

-_ }mr 2 + ½11tr 202 + V{r) = E ,° constant or ½m {r 2 + r2()2) +V(r) = E, constant

...(24)

where E represe11ts the total -energy.Thus, the sUm -of kinetic and potential energy i. conservation ofenergy.

e., total -energy E, is constant. This is the statement of .

Eq.,(23) or eq. (24) is known as another first integral of motion: .

I

4.4. DIFFERENTIAL EQUATION FOR AN ORBIT . ' .

.

.

.

· In case of a central for~, we want to deduce the equation of the orbit whose solution can giv~ us the radialdistatice (r) as function

ore:

.

. .

Twg-Body C~ntral Force Problem

1-09 .

.

The equation of motion for a particle of reduced mass m, tzj.oving under 'central force, can be written as [from eq. (20)] · · · · · · ' · ·· ·· · ·. · · · ·.

2

mr-~= f(r) 3 ·

.

"

mr

....(25).

·

.

.

.

.

.

Now,

dr dr de dr · . J · .dr ·•[. . J · . . . .· . . ] . r=-. = - . -=-_0=-·_--·. as-···-=0.fromeq. (16)· . · ·. · . · · . . .. · · dt _d0 dt d0 mr 2 dH . mr 2

and·

.. d [ .. J dr d[ · J dr Ji de · J d [ J dr ]· 2 ,. - dt mr d0 - d0 mr2 dH dt = mr 1 dR ·mr2 ·de

.

Let

Then

l

· .

.· u

1 • . •du. 1 .dr . therefore, .., . = - 2 de .. r . .. u0 . r

=- ,

.. ;: - . J2u2 2

·

m

[d_u]· - ·_

d.

2

de ·dB. ·

=

J2u2 d2u - J2u3 m d0 2 m

Hence eq, (25) is

J2u2

m

iu d0 d.

1(.!.) u ·

or

...(26)

This is the differential equation

of an

orbit, provided the force law / (r)

1(~ J= - :

or the

potential V is known.

4.5. lNVERSE SQUARE LAW OF FORCE · Gravitational and coulomb force between two particles .are the ·most important examples of central force. The force f (r), as usual, is expressed as f-(r)

-

Glfl1m2

r.2

.

(

•

Newton'-s law of Gravitation)

. ;.. (27)

.

... (28)

and .

.

.

.

.

The. general force law, g-oveming- eqs. (27) and (28), is the inverse square law of force, given .by K

f{r)

...(29)

,.2 .

If Vis the potential, then

av or

f(r)=--

K

and its integration gives V

K

r

where the constant of inte.gration is taken to be zero by assuming· V(r)

...(30) 0 at infinite separation (r · C>P).

liO

1

Classical Mechanics

4.6. KEPLER'S LAWS OF PLANETARY MOTION ANDTHEIR DEDUCTION .

.

The motion of planets has been a subject of much interest for astronomers from very early times. Kepler's laws of planetary motion are as follows : (1) The law of ~lliptical orbits : Every planet moves in an elliptical orbit around the sun, the sun being at one of the foci.

(2) The law of areas : The radius vector, drawn,.-from the sun to a .planet, sweep·s out equal areas in equal times i.e., the areal velocity of the radius vector is constant. (3) The harmonic law : The square of the period of revolution of the planet around the sun is proportional to the ...cube of the semi-major axis of the ellipse.

Kepler's laws have been enunciated pur~ly on the ba&i~ of observations,. taken for the motion of the planets. These laws give us a simple and accurate description of their motions, but do not offer any explanation.

1

The planets move around the sun under -the influence of gravitational force which is an inverse square law force. Henc~ we deduce the Kepler's laws of planetary motion around ·the sun on the basis of inverse square law of force.

4.6.1. Deduction of the Kepler's First Law For u = llr, the inverse square law force

!(~)

-Ku

ff (r) = - Klr2] is given by

2

Thus the differential equation (26) of the orbit can be exp,ressed as 2

d u

d0

m 2 +u=--Ku {·· f(r) 2 2 J u ·

or

d 2u mK d02 +u J2

Let

mK x=u--. J2

.

0

2

-Ku ]

-

*

...(31)

...(32)

Then which has the solution x == A cos ( B -8 ')

...(33)

where A and B I are the constants of integration.

. mK · Sincex = u. -12- and u

_!., we can write eq. {33) as r

1 mK r - 12 or

1

; =

A cos (B- 8 ')

mK +Acos(B-B') 12

... (34)

I

Two-Body Central Force Problem

111

1 2 /mK 12A · --=1+--c~~-e, r mK l

or

-:-r·= 1+ e cos ( 8 - 8)

where

12 12A · -=land - - = e mK mK

It is easy to identify e -

... (35)

~ with the help of the two first integrals of motion•. The cons'tanl 8'

appearing in eq. (33} is a consta11t of integration determined by initial ~onditions. Thus, when a particle is moving under inverse square law of force, its orbit is represented by eq: (35). This is the general equation of a .conic section with one focus at the origin and eccentricity e, given by 2

e=

·J··

.

2EJ l+-2

...(36)

·mK

The magnitude of e decides the nature of the orbit, represented by eq. (35), as following : Value oftotalenergy E

Value of.e

Conic

E>O E=O El e= 1 e 0 at r = r0 dr . Jr .

.

... (56)

.

If the potential energy. function for the central force is of the form V(r) = a r: n + .

.· .·

2

. and centrifugal energy Vcf(r)

..

.

.

1 ,

a being a constant,

br- , _(b > 0, a constant), then .

n+l

-2

Ve 0 or n > - 3 and the form

Classical· Mechanics

116

Further an orbit is said to be closed· if the particle eventually retraces its path. The stable and. closed orbits (circular and non-circular) are possible for n =l and n, = -2 and the corresponding force laws are as follows: l !• ...(59) For n = 1, f(r) =- Kr Hooke's law

. 1· (r) = - Kk,2I ...(60) Forn =-2, Inverse square l aw ihus ..the condition for bounded motion is that there is bounded region of r in which the total energy E 2: V/r), effective potential energy. The condition for stability of circular orbit is n > -3 with the force law f(r) = -Kr'. Further the orbits are closed only for the inverse square law of force (n = -2) of the· Coulombian or Newtonian type and for the linear law of force (n = ·1) of Hooke's type. From eq. (23 ), we have 2

I 2 [ E-V(r)--. J ) :=~/2

1

Vm

2tnr

tij

Now 4

Thus

dr= d8

~2mr - - [ E-V(r) ] -r or d0 2 2

J

.

t;' [

tt

dr

E - V(r)]- 1·

tf

"'

2

Its solution is ... (61)

If E. J and the form of the central force potential V(r) are given, the orbit is fixed. u0 and 00 refer to the starting point on the orbit. For V(r) a,fl+I, the above integral can be directly integrated for n = 1, -2, -3.

For n = 5, 3, 0,- 4, 5 and -7, the results can be obtained in terms of elliptical integrals. The equation of the orbit is not -obtainable in the closed form for other values of n.

Ex. 1. Use Hamilton's equation to find the differential equation for plpnetary motion and prove that the areal velocity is constant. (Agra 1992) Solution : Lagrangian L

T- V

~ m (r 2 + r 2tF) + ~

H= p,r+ p0 fJ

mr

Here, .

Therefore,

l

t r I

where u =l/r.

Hamiltonian ·

I'

EJL

,.

-. =mr8

and p 0

88

\

2 p2 . 2 :\ p2 K P + .:...:..!L _ Pr _ _li_ _ H H = _,. 2 2 or, m mr 2m 2mr r

\

2 2 K Pr + _f!L _ 2 2m 2mr r

I! I

two-Body Central Force Problem

117

Hamilton's equations are

r

aH

ap,. '

e-

P = 8H 0 = aH

ar '

r

ape

and _ Pa = oH

80

,, Pe K e· pr =---+mr3 r2 '

· =Prr

Therefore,

m '

Pe =--, mr-

From last two equations, p 8 =constant =mr 20 or { r 20

j:,0

and

0.

constant. This proves that the areal velocity is

constant in planetary motion. From the first two equations 2 - mr == - ~ + !_ 3 2

mr

2

·t

mr = - ( mr e3

or -

r

~2

+ !_ , whence ;: = r0 2 r2

mr

... (i)

mr

·

1

9 =hand u =-.

Suppose

,- 2

Then

0

r

' hu2 , r.. = h· 2u2 d-u dS (as done earlier)

h ,.2

. Hence the equation of planetary motion (i) is

lz2u2 d2u = _!_ h2u4 - K d0 2 u m

or d2u + u d0

u2

K mh 2

1 : K which is the equation of planetary motion with u =- and V = - - . r r Ex. 2. A particle of mass m moves under the action of central force whose potential is V(r) = Km,) (K> 0), then {i) For what kinetic energy and angular momentum will the orbit be a circle of radius R about

the origin ? (ii) Calculate the period of circular motion.

Solution :

V{r)

Therefore,

F

= Kmr

(Agra 1992)

3

av

2

3Kmr 2

(i) For circular motion F

Kinetic energy .:. mv 2 .

2

mv . - 3Kmr1 - --= r

!_ Kmr 3 2

.

·

.

·

3 ½

Angular momentum= mvr = mr (3Kr)

.

. . 2

3

[because v = 3Kr]

r-,

.. . 21tr . . . 2 · 3 (u)Smce v = rro = ~herefore from eq. v =•3Kr, we have

2rrr]2 3 47t2 . . . = 3/f.r or .r2 = 3Kr, whence r [T Ex. 3. The eccentricity of the eatth'sorbit speeds of the earth in its orbit.

21t

✓3Kr .

is 0.0167. C:alculate the ratio of maximum and mif!imum ·

·

(Agra 1991)

Classical Mechanics

118

n

Solution : From eqs. {38) and (39), the ratio of the minimum value of r, i.e., r1 (perihelion) and the maximum value of r, i.e., r2 (aphelion) is given by

to

·r2· = 1+ e

rJ

... (i)

1.-:e cc

However, the ~onstancy of angular momentum [mvr] at the two apsidal (turning) points gives Vi

mv r = mv r or11

22

V

2

rz = -r

... (ii)

I

Obviously when the radius vector has the minimum value r1, the speed of the planet ( earth) v1will be · maximum and when it is having maximum value r2, the speed of the earth v2 will be minimum. Thus th~ · desired ratio is

2 =2 = l+e = l+0:0167 =1.03. r1

v2

1-e

1-0.0167

Ex. 4. The maximum and minimum velocities ·

01a 'J

satellite are vmax · and 'vmm. respectively. Prove that ·

the eccentricity of the orbit of th~ satellite is

e= ,

V

-V · max . mm

Vrnax

(Agra 1998, 95, 93)

+ Vrnfn

Solution : As shownin Ex. 3, vrnax

= l+e

vrnax -vrnin

e=----

1- e'

vrnin

v rnax

+ ~rnin

Ex. 5. A particle of mass m is observed to move in a spiral oribit given by the equation r = C0 , where C is a constant. ls it moving in a central force field ? If it is so, find the force law.

Solution : The differential equation of the orbit [eq.(26) ] is given by

mf(~)

d 2u --+u=---d02

... (i)

l2u2

Since u =llr and r = C0 (given), hence

l u=-.

... (ii) ..

ce

Differentiating it, we get

1

du

-= --2

d0_

_

ce

2 d u 2 2 3 and --=--=2C u

d0 2

...(iii)

ce 2

Substituting in eq. (i) , we get

m

3

2Cu + u = -

Therefore,

1

(1)/

2u2 f -;; .

f(-1) =:--(u +2c u u· m · 1

2

3

· .--2 5

2

2

l

1 [ 1 . 2C )orf(r)=-- -_3 +_-5

m

r

r

Thus theparticle is moving under a central force fieldarid the force law is given by eq. (iv).

... (iv)

al

Two-Body Central Force Problem. ·

119

Ex. 6. A particle describes a circular orbit under the influence of an attractive ceniral force directed towards a point on the circle. Show that the force varies as the inverse fifth power of the distance. I

.

1

.

(Rohilkhand 1983; Meerut 83)

.

.

Solution : Let a be the radius of circle, described by the ·particle P [Fig. 4.7): If (r, 0) are the polar coordinates of P, then · r = 2a cos0 · ... (z) Differential equation of the orbit is

1)

2

d u · m . ( d0 2 + u == - 12u2 f -;; .1

1

· sec0

0 l'--'----.--------'1-x

u=-=·--- or u =--r . 2a cose 2a

Here,

2a

. . . du I . d 2u ·_ 1 · 3· .2 and hence-.= - sec8 tan0 and . .102 = - . .[sec 0 + sece tan 0] 2a . d0 ~ ~ .

Now,· J2u2

3

= - - - [sece + sec 0 + sece (sec20-1)] 2am

J 2u2 m

J 2 u 2 2sec 38

m f(r)

Thus .

--=---. 8a 2a

8J 2 a 2 .1· - - - . 5 orf(r) m

2 3.

u =

m

•5

-r

I oc -

r5

r

.

.

.

Thus the force varies as the inverse fifth power of the distance. Ex. 7. A particle, moving in a central force field located at r = 0, describes a spiral r = e-e_ Prove that the magnitude offorce is inversely proportional to ?. .

l

Solution : Here, r = e-e and therefore, u =- = e°

r

Differential equation of orbit is

{t2u .· - . m - +u d02 . - - J2u2

!(1) -. u

Substituting u = e°, we -get

t) ·( I) .

. m 20 f ( -_ e8, + e° = --e~ 2 . u

1

or

f(r)

2

J 39 orf - .=-:_2-e . u m

21 2 1 1 --·_-· 3i.e.,f(r) 0:3. m r r

Classical Mechanics

120

Ex. 8. A particle describes a conic r =

, where p and e involve constant quantities. Show p l+ec~0 · . that the force .under which the particle is moving is a central force. Deduce the force law.. (Agra 1992) Solution : Since the particle is moving on a curved path in a plane about a centre of force, its angular momentum remains constant i. e.,

J = mvr

m,.213

constant

J

r2e -m = a constant

Therefore,·

... (i)

The particle is moving on a .curved path, hence it is obvious from Newton's first law of motion that a force is acting on it. Consequently an acceleration is acting on the particle. Resolving this acceleration into its two components, along and perpendicular to the radius vector, we have · {I) Component/, along the radius vector, i.e., the radial acceleration of the planet is given by

(de ) rdt

2

2

d r f= dt 2

... (.ii)

(2) Component f ', perpendicular to the radius vector, i. e., the transverse acceleration of the planet is given by I

f

=_!_!{_[r2 01· d.

r dt•

dt

But from eq. (i), r 2 dS = a constant. dt Hence/'= 0 Thus the planet has no tra:nsverse acceleration and only the radial acceleration is acting on it i.e., the force on the planet is directed towards the -centre i. e., it is a central force.

From eq. (i) we have

dS dt

J

mr

J 2 -u m

2

...{iii)

where u = llr. Now

dr , dt

d l dt u

1 du u dt

-=--=-2

l du dB dr dS dt or dt

-7

l du Ju 2 dr -;J d0 · m or dt

J du

m·d8

J d 2u d8 m

Substituting the values

d0 dt

d2 r

and -

...(iv)

dB 2 dt

d0 2

from eqs. (iii) and (iv) in eq.(ii), we get

...(v) j

The equation of the conic is

p r = - - - - or 1+ e cos0 r

i+ e cose·or pu =1+ e cos8

... (vi)

Two-Body Central F~rce Problem .

121

Differentiating eq. (vi) twice with respect to 0 _, we have

d 2u P dS 2 =

.... (vii)

e cos0

Adding eq.(vi) and (vii), we get

2 2 P [u + d u2] = 1 hence u+ d u= J_ d8

d8

'

p

. eq. (v), we get . . th'1s va lue of u + d2u m Subst1tutmg

d8

J2 Let Then,

mp f=

k (a .constant). k

. I

1.e.,

l oc --)

...(viii)

,.-

· Thus, the acceleration and hence the force acting on the planet is inversely propohionai to the square of its distance from the sun. Negative sign indicates that the force is one of attraction.

4.8. ARTIFICIAL SATELLITES We have studied the motion of a planet and its orbit around the sun. In fact, a body which revolves constantly round a comparatively much larger body is said .to be satellite. We know that the earth and other planets revolve round the sun in their spe-cified orbits. The moon revolves round the earth and the planets Jupiter and Saturn have six and nine moons respectively revolving around them. All these are the ·examp-les of natural satellites. Each one of th-ese satellites is attracted by its primary with a force, given by Newton's law of gravitation. Scient~sts have also been abie to placed man-made sateUites, revolving round the -earth or sun. · They are caHed artificial satellites. The theory discussed above for the orbits and planetary motion is va~id for the discussion of satellites. p I'; · An a1~ificial sateHite of the earth is a body, place in a stable orbit around the earth with the help of muHistage rocket.In order to laurych a satellite .in a stable orbit, first it is necessary to take the sa-teliite to the altitude h, where at the point P by -some mechanism, it is given the necessary orbiuing velocity, called the insertion velocity vi (Fig. 4.8). The total energy of the satellite at P relative to the earth is given by 1

E=

,

2 11n'i~ -

GMm R+h

... (62)

·.,

· where mis the mass of the sa-tellite and M that of the earth, having radius R. The orbit will be an ellipse, a parabola or hyperbola, depending on whether E is negative, zero or positive. In each case, the centre of the .earth i's at one focus of the path. Therefore, the satellite will be moving in an elliptical orbit (Fig. 4.8), if

Fig. 4.8 : Elliptical path of a body projected horizontally from a height h above the earth's surface for· v;2 < 2GM /(R + h)

122

· Classical Mechanics 2GM. < R+h

2 Vj

(63)

Ii

The total energy E determines the size or semi-m·ajor axis - - - of the orbit. However the shape· or eccentricity e of the orbit is --determined by both total,energy E and angular momentum J by the relation : · · . e ~ J1+2EJ2 . mK 2

\

s

... (64)

'

f

with K = GMm. For elliptical orbits, layer the angular momentum; the less elongated is the orbit (Fig. 4.9). . For circularorbit, the ins~rtion velocity is found by equating the centripetal force mv 2/ r to the· grav,itational force GMmlr2 ·

(

c

Fig. 4.9 : Elliptical. orbits for 1,because (2Ep!zZe2) 2 is a positive quantity. Hence eq. (87) represents the path of the charged particle as hyperbola. ----cosec 2 dip 4E (41tt: 0 ) 2

... (96)

Substituting the value of p and dpldt/) from eqs. (95} and (96) in eq. {86), we get

cr((j>)=-

z 2 ]. · "' zZe2cot! [ 2 ze 't' .· - - - - cosec2 2 4 E (41teo) 2E (41te 0 ) 2sm!cos! 2 2

.[

1 cr(t/)) = -

or

4

l

.

]2 cosec

zZe 2 ----

(4ne 0 ) 2E

4

! 2

.... (97)

This is· the well known expression. for the Rutherford scattering cross-section. Thus the scattering cross~ section or the number of particles scattered per second along the direction ip are proportional to

l

(1) cosec

4

t,

(2) the square of the charge on. the nucleus (Ze), {3) the square of the charge on;the particle (ze), ano . (4) inversely proportional to the square of the initial kinetic energy E.

Thus if N 2 ccos f, where/"" [1- (KIE)] . 26. For a particle moving under the action of a central force, the effective potential energy is given by

100 50 U (r) = - - . +2 (MKS units) r

r

·

Sketch roughly. V as function of r and find the radius of' circular motion. Ans: r = 1 m.

.

1.

11-i=iiiil

A particle is describing a parn:bola about a centre of force which attract~ according to the inverse square of the distance. If the speed of the _particle is made one half without change of direction of motion when the particle is at one end of the latus rectum, pr-0ve that the new path is an ellipse with eccentricity e =

2.

.

(Mumbai.2001)

.Js /8 .

The eccentricity of the earth's orbit is e = 0.0167. If the orbit is divided into two by the minor axis, show that the tim~s spent in the tw.o halves of the ~rbit are (_!_±.:.)year. Also evaluate the difference . . '. 2 1C in hours. Ans: 93.2 hours.

3.

Show that the velocity of .a planet at any point of its orbit is the same asit would have been if it had fallen that point from rest at a distance from the sun equal to the length of the major axis.

4_. • • Calculate the time in which a particle moving under inverse square law force describes the area O ~ 0 . :::; a of elliptical orbit.

J

]

.- . J3 · . . . -i ( ~ 1+e a ~ · sina ·Ans: t= 2 tan -. tan-e iJl--:e~. · µK 2 (I-e2)312 · 1-e .2 · ..· l+e c.oso: ·

Classical Mechanics

136 5.

A particle moves under the action of a central force and describes a curve r,1= a/(1+0 )1, where a is constant: What is the force law ? If when tjJ = 0, the particle receives an impulse which reduces its radial velocity to zero and doubles its transverse velocity, show that its subsequent path is given by ·

3a 2r 6. 7.

{ ✓3

I l+-co -~)

2

2

A particle moves in a bounded orbit under an attractive inverse force. Prove that the time average of the kinetic energy is half the time average of the potential energy. Find the differential scattering cross-section for the scattering of particles. by the potential V (r), where V (r)

Ans:

(1 R1) for r < R and V(r) = 0 for r > R.·

== a ; -

R2 4

(l+y) .

(1 +y sin

2

2

,

where y

4

~R (a+ RE).

a .

~)

Objective Type QtAestions _________________ 1.

.A particle is moving under central force about a fixed centre of force. Choose the correct statement : (a) The motion of the particle is always on a circular path. (b) Its angular momentum is conserved. (c) Its kinetic energy remains constant. .. (d) Motion of the particle takes place in a plane. Ans : (b), (d).

2.

Two particles of masses m and 2m, interacting via gravitational force are • rotating about common centre of mass with angular velocity ro at a fixed distance r. If the particle of mass 2m is taken as the origin 0, 2

(a) the force between them can be represented as F == µro r. (b) in an inertial frame, fixed at ·the centre of mass, the origin is at rest. (c) in the inertial frame, the origin_ 0 is moving on a circular path of radius r/3. (d) in the inertial frame, the particle of mass m is m,oving on a circular path of radius r/3. Ans : (a), {c). 3.

·· 4.

A particle is moving on elliptical pa:.th under inverse square law fo;ce of the form F(r) = eccentricity of the orbit is (a) a function of totaLenergy. {b) independent of total energy.. (c) a function of a~gular momentum. (d) independent of angular momentum. Ans : (a), {c).

Kil The

Th~ maximum and minimum velocities of a satellite are v1 and v2 respectively. The eccentricity of th~ orbit ~f the satellite is given by

'.l

Two-pody Central . . Force Probl~ni

. v,

.

137

Vz Vi - Vz (a)e=- (b)e=- (c)e=-__;;_ (d) e v2 v1 v1 +v2

Vi

+ Vz

v1 -v2

Ans : (d).

5.

Rutheford's differential scattering cross-section (a) has the dimensions of area. (b) has the dimensions of solid angle. (c) is proportional to the square of the kinetic energy of the incident particle. 4 (d) is inversely proportional to cosec ( be the inclination ,of the inclined plane of length l with the horizontal. If a hoop of mass M and radius R is rolling ·dqwn an -inciined plane starting from a point O without' slipping, then x and e are two generalized-coordinates ah~· the' equatioii·af constraint is ·. .

'

x = R8 or dx

Rd0 or dx - Rde

0

...(i)

As.there is only ~ne,constraint equation, onlyone Lagrange's multiplier A will be required. Here a1:.:dx'+a 10 de = 0. .

Va

Classical Mechanics

152

n

[·: Iatkdqk +a/tdt::::o] k=l

.

. s:o that a. 1x = 1= ax (say) and a 10= -

R = a0 (say) ...(ii) Kinetic. energy of the hoop T == Kinetic energy of rriotion of centre of mass + Rotational kinetic energy about the centre of mass

Potential energy of the hoop V = Mg(l

x) sin· tf>

Lai cot

Fig. 5.8 : A hoop rolling down on inclined ,plane

...(iii)

Thus Equations of motion for two coordinates x and 0 are

...(iv)

and ... (v)

and

, aL aL . · . Mg sin,t, - .·=MR 20 - = O and also a =1 and a0= R.

aL . aL Here,-·= mx -

ax

'ax

't',

Therefore,

Mx

and

MR 0 =-KA

'

a0

as

'

·

x

... (vi)

Mg sin$= A 2 ..

.whe give

...(vii)

But from (i)_ x = RB . HenceJrom eq. (vii), we get ... (viii)

Mx=-1,., Substituting for 'A, in eq. (vi), we obtain

x = { g sin $

Mi - Mg sin $ = -: Mx or

... (it)

This is the acceleration of the hoop along the inclined plane. Note that it is one half of the acceleration it would have in slipping down a frictionless inclined plane. The force of constraint 'A, is [by using eq. (viii) and (ix)]

A = - ~2 Mg sin $

x

i

... (x)

This gives the frictional force due to .constraint which r.educes the.acceleration is only slipping without friction) to

I

x = g sin$ (when there

5.1

~ g sin$ (when the hoop is rolling without slipping).

Note : It is to be remarked that if we take constraint equation as R d0- dx =O, then the constraint force will

and

be obtained as A= { Mg sin~ which bears +ve ~ign .. Thus in such problems we obtain only the magnitude i

of the force of constraint.

(2) Simple pendulum : Find.the equation of motion and force of constraint in case of simple pendulum by using Lagrange's method of und~te~ined multipliers.

. A-va the I

a COi .the J

Variational Principles

153

Solution : Referring Fig. 5.9,the LagrangianL is given by

L ½mr 202 + mgrcos0

s

...(i)

where· V ~ - mgr cos0 with respect t