Acknowledgements

Rof. R.N. Mathur and Dr. S.C. Gurg for co~.;mentson the unitt.
1

4
UNIT 6 MOTION UNDER CENTRAL
CONSERVATIVE FORCES
Structure

Objectives

6.2 Centrd Conservative Force

Properties of Motion under Central Conservative Forces

6.3 Inverse Square Central Conservative Forces .

6.4 Summary
6.5 ~ e r h i n aQuestions
l
6.6 Answers

6.1 INTRODUCTION

In Block 1 you have studied the basic concepts of mechanics, In Unit 5 of Block 1 we have
discussed gravitation. You know that the planets move under the influence of the
gravitational field of the sun. How do we solve the equation of motion of a planet ? In this
unit we will try to answer this and similar questions.
In fact one of the most important problems of mechanics is to understand the motion of a
particle moving under the influence of a force field: The force may be due to another particle
or a system of particles, as in the Solar System or'a system of fixed charged particles. It
could even be due to an electromagnetic field. In this unit we will restrict ourselves to what
we call central conservative forces. You will first learn what a central conservative force is.
The motion of particles under the influence of such forces has special properties which
simplify its description. So you will also study these properties.
There are many examples of such motion. We have mentioned the motion of planets around
the sun. Other examples are the motion of satellites around the earth, of spacecrafts sent out
to probe the universe and that of two charged particles with respect to each other. The forces
associated with these systems, namely the gravitational and electrostatic, obey the inverse
square law. We shall see that the inverse square central conservative forces are of special
importance. So we shall concentrate chefly on inverse squire central conservative forces. We
shall solve the equation of motion of a particle inoving under the influence of such forces,
We shall then apply the results to determine the possible orbits of a body moving around the
sun. This provides the theoretical basis for Kepler's empirical laws. We shall also determine
the trajectory of an alpha particle approaching the nucleus. Such a calculation led to the
nucle .rmodel of the atom.
So far we have studied single particle motion. In the next unit, we shall turn our attention ta
many-particle systems. In this unit we shall refer to the contents of Units 3 . 4 and 5 of
bloc^ 1 very often. So we suggest that you go through these units once again before
studying this unit. You may also go through Appendix A on conic sections before'studying
Sec. 6.3. It is given after Unit 10.

Objectives
After studying this unit you should be able to
e identify a central conservativeforce
e solve problems by applying the properties of motion under a central conservative force
e determine the possible orbits under a given inverse square central conservative force.
 Systems of Particles
6.2 CENTRAL CONSERVATIVE FORCE
Ln nature we come across many forces which are either directed towards or away from a fixed
point. For example, the gravitational force experienced by a mass due to a fixed point mass
is directed towards the point mass. Again the force experienced by a positive charge due to
another fixed positive charge is dlrected away from the latter charge. The force on a particle
of mass m attached to a string and moving in a circle in a horizontal plane is also directed
toward the centre of the circle (see Fig. 4.17 of Ull;r 4). Such forces are examples of central
forces. We define a central force as one that is everywhere directed towards or away from a
fixed point. This fixed point is called the centre of force. Mathematically, we can express
a central force acting on a particle as
Fig. 6.1 : A particle moving
under a cenbal force.
A
0 =Centre of force where r is a unit vector pointing from the centre of force to the particle (see Fig. 6.1). For
P = Panicle the above mentionedFkst three examples of central forces, F depends only on the separation
APE =Trajectory
between the centre ofTorce and the particle. For such forces Eq. 6.1 can be1written as

We can show that the central forces given by Eq. 6.2 are also conservative. For this recall
the definition of a conservative force from Sec. 3.3 of Unit 3. Let us compute the work done
by the force on particle P as it moves from point A to B (Fig. 6.2).
Let N b e the work done by the central force on the particle astit undergoes a displacement
dl along the path, It is given by
A
dW = F. dl = f ( r ) r.dl = f ( r ) dl cos a, (6.3)
A
where a is the angle between r and dl . Since dl is infinitesimal, from Fig. 6.2 you can see
that

where dr is the change in the particle's separation from 0,as it undergoes displacement dl.
P So Eq. 6.3 becomes
Fig. 6.2 : Work done on a particle my = f (r) dr.
moving from A to B. Since dl is
infinitesimal, the angle indicated by The work done by the force on the particle as it moves from the point A to B is ghen as
double stripes can be considered
alternate and hence equal to a.
Have you wondered about the use of w = J flr)dr. (6.4)
the term 'conservative' force field7
You h o w that for a conservative r,
€om, the work done in taking a NOW,the value of this integral depends on its limits only. So the work done depends only
system mund a closed path is zero. on the end points and not on the path being followed by the particle. Thus, the central force
If this were not so, we could fmd a
closed path, haversing which would given by Eq. 6.2 is conservative. We term the forces represented by Eq. 6.2 as central
yield negative work, i.e, energy to conservative forces.
1 us. Thus we could recover any
amount of energy going around the You may like to use this concept to identify some central conservative forces in the
loop. That this does not happen is following SAQ.
related to the consewationof energy.
Thus path independenceof work in a
SAQ 1
'conservative' force field is related to Which forces among the following are central ? Also identify the central conservative force.
'energy comrvationf.

(a) (b, (c)

Fig. 6.3 : (a) Ideal spring-mass system; (b) real spring-mass system: (c) a current-carrying conductor.
 a) The force acting on a particle of mass m in the spring-mass system shown in ~ig.';6.3a Motion Under Central
for which F = - k x . Conservative Forces

b) The force acting on a particle of a real spring-mass system kept inside water '(shown in
Fig. 6.3b). Its vibration is subject to damping due to water. For such a system

where k2 and k3 are constants.

C) The force acting on a charge P due to an element (fl of a current-carrying conductor

shown in Fig. 6 . 3 for
~ which

F = k , (v . f ) d l - (V . d l ) P.
r2
where v is the velocity of the charge and kl is a constant depending on the magnitude of
the current and the nature of the medium.

Now that you know what a central conservative force is, let us find out the equation of
motion for a particle of mass m moving under its influence. From Newton's second law it is
given as
A
ma =f (r) r (6.5)

We find that the study of motlion under central conservative forces is much simplified
because it has certain general properties. Let us first discuss these properties.

6.2.1 Properties of Mation under Central Conservative Forces

A
.
The central force is directed alon,g r So, the torque on the particle about the centre of force
is

Angular momentum is constant

5
You know from Unit 4 that r =-. So for zero net torque, L is a constant. This means
dt
that for motion under central force, the magnitude and direction of angular.momentum is
constant. We shall now see that another interesting property arises only from the fact that
the direction of angular momentur,p is constant.

Motion is restricted to a plane

We know from Unit 4 that L = r x * p= r X m v. SOLis a vector perpendicular to r. In
other words, the vector r always rernains in a plane perpendicular to L.Since the direction of
L is fixed, this plane is also fixed (Fig. 6.4).
Since the motion is restricted to a plane, we can use a two-dimensional coordinate system to Fig. 6.4 : A particle having
describe the particle's motion. Sinoe r will be occurring very often in the mathematical - constant angular momentum L
moves qn a fixed plane perpendicular
treatment, it will be convenient to use plane polar courdinates which you have studied in to L.
Unit 4.

We have already determined the maignitude of L in Unit 4. Fro-mEq. 4.25

L.= mr2 0 (6.6)

which is constant for a central force.

The property that angular mornenturn is constant for cpntral force motion gives mse to the:
following law.

Law of equal areas

Refer to Fig. 6.5. Let r be the radius vector of a particle at a time t , executing cental forck
S v s i e i z s of Particles . motion. Let its radius vector be r + Ar at time I + A I. .The polar coordinates of the particle a:
\ t and t + At ark (r, 0 ) and ( r + Ar, 0 + AO), respectively. The area AA swept out by the radius
Q vector during the time interval At is shown shaded in the figure. For small values of A8,the
a r e a d is approximately equal to the area of the triangle OPQ, i.e.

1
2 2
'
AA = - r (r + Ar )sin A0 = ( r + hr )A0, ('.. sin A8 A8, for small A8)
M 1 A0
Fig. 6.5 : Area swept out by the Ignoring the term b A 0 , we get - = - r2 -
:irii;::y vector. OX - polar axis,
Af 2 At
Op=r,OQ=r+Ar.
Therefore, the rate at which area is swept out is given by
'The area of a *gle can be
expressed as half of the product of the
length of my two sides and the sine
of the angle contained between them.
dA
From Eq. 6.6. we understand that 3-28= a constant for a particle of given mass m . So - is
dt
a constant, which gives the law of equal areas. It states thatfor any central force the
radius vector of a particle sweeps out equal areas in equal times. Kepler's second law of
planetary motion is precisely this law applied to the central force of gravitation. You will
understand the physical meaning of this law better after deriving Kepler's fist law.
The property that angular momentum is a constant vector holds for all central forces. Motion
under central conservative fotces has another property that the total mechanical energy is
constant.
Total mechanical energy is constant
From Eq. 3.21 of Unit 3, you know that the total mechanical energy E for a conservative
force is constant, is.,
1 .,
E = - mv 2 + U (r)= constant.
2
The potential energy U(r)is given by

where YO is some arbitrary reference position. Both these equations 6.7a and 6.7b apply to
those central forces which we conservative.
Let us now apply the concepts that angular momentum and total mechanical energy of a
particle moving under a central conservative force are constants of motion.

Fxample : 1
A spacecraft is launched from the point A of the surface of a spherical planet of mass M
having no atmosphere with a speed vo at an angle of 30°from the radial direction. It goes
into an orbit, where its maximum distance OB from the centre of the planet is twice its
radius R. Find vain terms of G, R and M.
Refer to Pig. 6.6. Let the mass of the spacecraft be m. When the spacecraft is at position r
GMm
with respect to the centre of the planet then the force of gravitation on it is F = - 7 r.
A

On comparing with Eq.6.2we realise that this is a central conservativeforce. This means
Fig. 6.6 that the spacecraft is moving under the influence of a central conservative force. Hence, its
angular momentum and the total mechanicaI energy E are constant. We know that
6 = K,B + P.E. From Eq. 5.16 we also know that the P.E. of a mass m at a point at a
distance r from the centre of a spherical mass M is -OM".Therefore, the total mechanical
r
energy of the spacecraft at point A on the surface of the planet is
 The total mechanical energy of the spacecraft at point B corresponding to the maximum Motion Under C
clistance.2~'is Conservative 5 .

EB - -1 ,,,2 -G
- Mm 9

-2 2R
We know that EA = Ee , since the total mechanical energy is constant. The conservation of
angular momentum gives us another relation. Recalling that L = m r x v,we get for the
magnitudes of angular momentum at points A and B,

LA = mRvO sin 300 =

2 '
LB = m v (2R ) sin 90° = 2mRv.
Since LA = LB,we get

"0
Setting EA= EBand putting v = - in the equation, we get
4

After simplification, we get

- - -

So far we have studied some general properties of motion under central conservative forces.
We shall now use these properties to determine the path of a particle moving under inverse
square central conservative forces. Examples of such forces are the familiar gravitational and
electrostatic forces.

6.3 INVERSE SQUARE CENTRAL CONSERVATIVE

FORCES

For any general inverse square central conservative force, Eq. 6.2 is expressed as

If k is positive, then the force is repulsive and if it is negative, the force is attractive. For
example, you k n ~ w that the force between two like charges is repuIsive and that between
two unlike charges is attractive. Similarly, gravitation is an attractive inverse square force.
Let us now solve the equation of motion to determine the orbit of a body moving under the
I influence of gravitational force of the sun. We will regard the sun to be stationary. In order
to determine the orbit, we need to know r (t) and 8 (t), or r a s a function of 8.We will now
use a simple method to obtain r (8).

Refer to Fig. 6.7. As has been pointed out in Sec. 6.2,.1, we shall be using plane polar
coordinates. Let the sun be at the origin located at the centre of force, The equation of
motion of the body under the influence of the gravitational attraction of the sun is given by
I
Fig. 6.7 : Motion of B boay
moving under the gravitational forcc
of the sun (S ). P is the position of,
the body at t = 0.

1 where m and M are the masses of the body and the sun, respectively. 9
 , Systems of Particles
"i .

Let us first solve this equation to obtain v. Then we will use Eq. 4.13a to obtain r from the
expression of v.

Since the force is central, we have fiom,Eq. 6.6 that L = mr2 8 = a coostant.

d6
We also know from Eq. 4.10 that - = - 0; . Using Eqs. 4.10 and 6.6 we can write
dt
Eq. 6.9b as

-dv- - GM -
d6 - -G-M m d6 - -
A d6 \,.

dt - 20 dt - L dt - L dt'
where A = GMm = a constant.

On integrating, we get

where C is a constant vector of integration. We sh

C. Let us choose the origin of time (t = 0) at the i 31usent when
the initial conditions determine
thebody
to
is closest to
the
sun, i.e.. r is a minimum. Thus &-= 0 at t = 0. Again v (0) (i.e. v at t = 0) is in the s
m
dr
A
6 6
d i i t i o n as (0) (Le. at t = 0). Let 0 (0) = 8. Hence, fmm Eq.6.10a we get

where e --
-A
v(0)-1 =awnstant.

Hence, from Eqs. 6.10a and b, we get

Now that we have obtained an expression fot v, we can find r as a function of 6 in a simple
A
manner. Taking the scalar product of EQ.6.10~with 0 , we get

LA v . 4 = b.4 A A
+en .e= l + e c o s e . (6.11)

- A
We know from Eq.4.13a that v = ;i?+ r 0 0 .
. 1
~ i n c e ~= d0 a n d 4 . 6 = l . w e g e t v . d = r e = - r 2 0
.
=mrLhmEq.6.6.
r
So we get from Eq. 6.11,
Fig

Comparing Eq.6.12 and Eq.A.3 of Appendix A, we get

So we can say that the orbit of the body is a conic with its pole inside. e is called the
8 eccentricity, of the conic. Now this conic can be either a parabola, a hyperbola or W
10 , ellipse depending on whether e is equal to, greater thh or less than 1.

k k
In the special case when e = 0, the conic is a circle. Motion Under C ~ n t r a l
Conservative Forces
The solution that we have obtained for the path of a body moving under the sun's
gravitational field is based on some simplifying assumptions. W e have assumed that the sun
is stationary and that the only force acting on the body is the gravitational attraction of ;he
sun. We know that both these assumptions are not exactly true in the real Solar System.
The sun is not stationary and all other members of the Solar System also exert gravitational
forces on the body. However, these forces are negligible in comparison with the gravitational
attraction of the massive sun. For our Solar Systein containing one huge sun and a small
number of little planets (called a Keplerian system) these assulnptions are reasonable.

Let us now analyse what kinds of orbits (elliptical, parabolic or hyperbolic) are followed by
the various bodies in the Solar System. For this w e shall relate the eccentricity e to the total
mechanical energy of the moving body.

Energy and*eccentricity
We know that E = K.E. + P.E. (6.14a)

I 1
K.E. = - nnv2 = - nl v.v
2 2

To calculate K.E. in polar coordinates we use Eq.6. I Oc to get

K.E. = - - ( 8 + e n ) . ( e + e n ) = A2m
, A 2 A

2 L"
A A A
-
2L'
( 1 + 2c cos 9 + e2). (6.14b)

GMI?I- A .
Similarly, from Eq.5.16 we know that P.E. = -- -- I'
- I'

From Eq. 6.12. P.E. = -& (1 + e cos 8).

L~

From Eqs.6.14a, 6.14b and 6 . 1 4 ~we get

A2nt
E =-(LJ~- I),
2 ~ '

Eqs. 6.13 and 6.15b give the values of 17 and e, which together determine the orbit of the
body moving under the sun's gravitation. These can be calculated if we know the values of
E, L and A. Altl~oughwe have determined the orbit of a body moving under the sun's
gravitation, these results can be applied more generally. These equations hold for every
particle of mass n7 moving under the influence of an attractive inverse square force given by

Before proceeding further you may like to solve an SAQ to get some practice on-qs, 6.12
to 6.15.

S%AQ2

The elliptical orbit of a 2000 kg satellite about the earth is given by the equation

Find the (a) eccentricity of the orbit; (b) angular momentu,m and (c) total mechanical energy
of the satellite. (Note that for this problem m and M are the masses of the satellite and the
earth, respectively.) /I