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General properties of Central Force
1.
Central Force is Confined to a Plane
If p is the linear momentum of a particle of mass µ,
the torque τ about an axis passing through the
center of force is :
 If the angular momentum L of mass µ is constant,
its magnitude and direction are fixed in space.
Hence, by definition of the cross product, if the
direction of L is fixed in space, vectors r and p
must lie in a plane perpendicular to L. That is, the
motion of particle of mass µ is confined to a plane
that is perpendicular to L.
 As we the force acting at body is central force, three
dimensional problem can be reduced into two
dimensional. Using polar coordinate system :
2. Angular Momentum and Energy are Constant
The angular momentum of a particle of mass µ at a
distance r from the force center is :
Total Energy of System
 Since there are no dissipative systems and central
forces are conservative, the total energy is constant :
 But
Orbits in an Inverse Square Field
 The differential equation of orbit under central forces is
d u
 1
 u   2 2 F   ....(1)
2
d
L u u
2
 The inverse square force is given by
F (r )  k / r 2
where k  Gm1m2 or k  q1q2 / 4πε0
Put r  1 / u
F (1 / u )  ku2 ....(2)
Eq. (1) 
d 2u

k
2
 u   2 2 (  ku )  2
2
d
Lu
L
d 2u
k

 u  2  0 ....(3)
2
d
L
k
put x  u  2 ....(4)
L
dx
du
d 2x
d 2u


and

2
d
d
d
d 2
d 2x
Eq (3) 
 x  0 ....(5)
2
d
which is equation of simple harmonic motion.
The solution of equation (5) is
x  A cos (θ  θ0 ) ....(6)
from eq. (4) and (6)
k
u  2  A cos (θ  θ0 )
L
k
 u  2  A cos (θ  θ0 )
L
1 k
  2  A cos (θ  θ0 )
r
L
1 k 
L2
  2 1 
A cos (θ  θ0 )
r
L 
k
put
k
2
 l and



L2 A
  (eccentric ity)
k
L
1
  l 1   cos (θ  θ0 ) 
r
The nature of orbit will depend on value of  .
Hyperbola, >1
Parabola, 1
Ellipse, 0<<1
Directrix
For parabola
Circle, =0
Turning Points of Motion
 The total energy of the system of mass μ under central forces
is given by
1 2
L2
E  r 
 U ....(1)
2
2
2 r
Potential energy wil l be given by
U
k
F U    Fdr  
r
r
1 2
L2
k

 E  r 

2
2
2 r
r
1 2
k
r represents Kinetic energy,  represents Potential energy &
2
r
L2
represents Centripeta l Potential Energy
2
2 r
Turning Points of Motion
 Effective potential energy of the system
k
L2
 
 U (Effective Potential Energy) ....(2)
2
r 2r
 U` is fictitious potential that combines the real or actual
potential U(r) with the energy term associated with the
angular motion about the center of force.
 Therefore total energy
1 2
E  r  U  ...(3)
2
Turning Points
 As written, this is actually a correct equation for any two-body central force
problem. For the specific case of a gravitational potential energy, we have
U eff
l2
 U (r ) 
.
2 r 2
 Notice that one term of this is negative, while the
other is positive.
U eff
Gm1m2
l2


.
2
r
2 r
U
~ l2/r2
r
Ueff
~ 1/r
Turning Points of Motion
 The point at which total energy of the system is equal to its
effective potential energy is called the Turning Point.
E U
 eq (3) 
1 2
r  0
2
i.e., Radial kinetic energy of the system  0
1
but   0
 r 2  0
2
 Radial velocity of the system  0
Relation between Eccentricity and Energy
• At turning points,
total energy of the system is
E U
k
L2
E 
r 2 r 2
L2
k


E 0
2
2 r
r
Put r  1/u
2
2
Lu

 ku  E  0 ...(1)
2
which is quardtic equation in ' u'.
Its solution w ill be
4 EL2
k k 
2
u
L2
2.
2
2
u 
u 
u 
k
2
L
k
2
L
k
2
L




4 EL2
k 
2
2
L2
2
2
EL
4

2
 4.
k
.
4
L 2
L
2
 2k 2
L4
2 E
 2
L
Relation between Eccentricity and Energy
since u  1/r
1 k
 2 k 2 2 E
  2 
 2
4
r L
L
L
differenti ating w.r.t. ' t' , we get
1
 2 r   A sin (θ  θ0 )
r
2 2
2
1
k
 k
2 E


r


r
A sin (θ  θ0 )

 2 

...(2)
rmin
L
L4
L2
since at turning points r  0
2
1
k
 2 k 2 2 E
 sin (θ  θ )  0

r
A

&
 2 

...(3)
0
4
2
rmax
L
L
L
but equation of conic is given by
 sin (θ  θ0 )  0

1 k 
L2
 sin (θ  θ0 )  sin n
 2 1 
A cos (θ  θ0 ) 
r L  k

 θ  θ0  n ...(5)
1  k

   2  A cos (θ  θ0 )  ...(4) where n  0,1,2,3,...
r L

Relation between Eccentricity and Energy
• Thus turning points occur at various
values of  given by eq. (5)
Comparing eq.(2) and (6)
A
 2k 2
1.when n  0 , θ  θ0  0
L4
k

2 E
L2
2 EL2
1
k 2
from eq. (4)
A
 k
 k
  2  A cos ( 0 )   2  A ...(6)
rmin  L
 L
2.when n  1, θ  θ0  
L2 A
2 EL2

 1
....(8)
2
k
k
1
from eq. (4)
1
rmax
 k
 k
  2  A cos ()   2  A ...(7)
L
 L
L2
L2 A
Now  
k
2 EL2
   1
...(9)
2
k
The Equivalent 1-D Problem
rmin
U
rmax
r

rmin
E>0
 As before, if the total energy is less than zero
then the object (planet, star, comet…) has two
E<0
rmin
rmax
turning points, rmin and rmax. However, the actual
motion is not purely radial, but rather the object
~ 1/r
is moving in an ellipse. The “force” that turns the
object at the inner point is the irreducible angular momentum of the object.
 The object can get no closer to the center of force than rmin, because to do so would
require losing some angular momentum.
r
Graphical Discussion of the Nature of
motion under Central Forces
 We know total energy of the system of mass μ under
central force is given by
1 2
L2
E  r 
 U ....(1)
2
2
2 r
1 2
r represents Kinetic energy, U represents Potential energy &
2
L2
represents Centripeta l Potential Energy
2
2 r
 L2

The term is called 
 U  Effective Potential Energy (U' )
2
 2 r

2
L
U' 
U
2
2 r
first term is always positive, second term can
be positive or negative depending on value of U.
k
For attractive inverse square force U  
r
L2
k
 U' 

2
2 r
r
1 2
 E  r  U' ....(2)
2
1 2
or E  U '  r ...(3)
2
 As the particle coming from infinity, approaches the
centre of force O
i.e., as r varies from infinity to zero
L2
first term
: varies from zero to +∞
2 r 2
& second term

k
r
: varies from zero to -∞
U
Since the total energy is
constant under central
force motion, therefore
graph between E & r is a
always a straight line
parallel to r-axis.
~ l2/r2
E
r
Ueff
~ 1/r
WHEN E=E1 > 0
 The minimum distance up to which particle can approach is r1.
On reaching point A, particle will go back.
 Point A is called Turning point for E > 0
 If r < r1 , then E-U’< 0, i.e., K.E.<0
which is not possible
 Motion of the particle is unbounded
 Since E>o, hence ε>1,
 the trajectory of the particle is hyperbolic
WHEN E=E2=0
 The minimum distance up to which particle can approach is r2.
(Point B)
 On reaching point B, particle returns
 Point B is called Turning point for E=0
 Motion of the particle is unbounded
 Since E=o, hence ε=1,
 the trajectory of the particle is parabolic
Energy
U’(r)
Turning point(s)
(apsidal distances)
A
r1
bound
unbound
B
r
2
r5
1 2
r&
2
r4
D
S
•
•
If ε>1, conic is Hyperbola (E>0)
If ε=1, conic is Parabola (E=0)
(UNBOUNDED MOTION)
E2=0
r
E3<0
E4<0
r3
C
E1>0
•
•
If ε<1, conic is Ellipse (E<0)
If ε=0, conic is Circle E   2Lk
(BOUNDED MOTION)
2
2
WHEN E=E3 < 0
 In this case energy E cuts the curve at two points C & D,
where C & D are turning points.
 Hence motion of the particle is bounded between points C
& D.
 Since E < o , hence ε < 1,
 hence the trajectory of the particle is elliptical
WHEN E=E4=U’min < 0
 In this case, radial K.E. is zero at point S
 So S is turning point
 Hence motion of the particle is confined to single value of
r.
 the trajectory of the particle is circular.
Minimum energy for circular orbit
2
L
k
We know
U' 

2
2 r
r
dU '
'
For U '  U min ,
0
dr
d  L2
k
 
   0
2
dr  2 r
r
L2
k
 3  2 0
r r
2
L
r
....(3)
k
L2
k
 3 2
r
r
Total energy of the system is
L2
k
E

2 r 2 r
L2
from eq. (3), r 
k
L  k 
k k 2  k 2 
k 2
E
 2  2
 2   k. 2 
2
2  L 
L
2L
L
2L
We know, relation b/w E and  is
2
2
2 EL2
  1
k 2
2 L2
k 2
   1
. 2  0
2
k
2L
Hence particle moves in a circular orbit.
KEPLER’S LAWS
Kepler's First Law
1. The orbits of the planets are ellipses with the sun at one
focus.
=closest to the Sun
=farthest from the Sun
Kepler's Second Law
 Kepler noticed that the
planets sweep out equal areas
in their orbit over equal times
 this means the planet must
speed up and slow down at
different points
 If it takes the same amount of
time to go through A as it does
C, at what point is it moving
faster?
 C, when it is closest to the Sun
Kepler's 2nd Law: An imaginary line
connecting the Sun to any planet
sweeps out equal areas of the ellipse
over equal intervals of time.
Kepler's Third Law
 Finally, Kepler noticed that
the period of planet's orbit
squared is proportional to
the cube of its semi major
axis
Kepler's 3rd Law
T a
2
 This law allowed the orbits
of all the planets to be
calculated
 It also allowed for the
prediction of the location of
other possible planets
3
NOTE: In order to use the equation
as shown, you must be talking
about a planet in the Solar System,
T must be in years, and a must be
in A.U. !!!
Kepler's First Law
 For bounded motion, total energy of the system must be less
than zero. Hence particle must move in an ELLIPITICAL
ORBIT.
 Since the motion of planet around sun is bounded, so every
planet must move in ellipitical orbit.
 The general equation of the conic is given by
1
 l 1   cos (θ  θ0 ) 
r
1
r
...(1)
l 1   cos (θ  θ0 ) 
For elliptical orbit
(i) When (θ  θ0 )  0; r  rmin (Perihelio n)
1
rmin 
...(2)
l 1   
(ii ) When (θ  θ0 )   ; r  rmax (Aphelion)
rmax
1

l 1  

...(3)
(closest to the Sun)
(farthest from the Sun)
 The semi-major axis of the ellipse is given by
rmin  rmax
a
2
1
1

l 1    l 1   

2
1  1
1 
 


2l  1    1    
 2 
1
 1  2   l 1  2


1
a
...(4)
2
l 1 
1

2l






k
2 EL2
Since l  2 &   1 
L
k 2
L2
a 
k
1
2 EL2
1  (1 
)
2
k
1
L2  k 2

2
2 EL
k 2 EL2

k 2
k
a
2E
k
or E  
....(5)
2a
L2

k
Eq. (5) shows that major axis depends on total energy of the
system/particle moving in orbit.
Semi - minor axis is given by
b 2  a 2 (1   2 )
2
2




2
EL
2
EL
2
2
2
  a  

 b  a 1  1 
2 
2 
k 

 k 
 2 2 EL
 b    a . 2
k

but a  k / 2 E
2
1/ 2



  k  2 EL
 b  

  2 E  k 2

L
 b
....(6)
2 E
2
2
 Eq. (6) shows that
minor axis depends on
total energy and
momentum of the
system/particle
moving in orbit.
1/ 2




 Hence shape & size of
orbit is determined by
ENERGY & ANGULAR
MOMENTUM
Kepler's Second Law
 The line connecting the Sun to any planet sweeps out
equal areas of the ellipse over equal intervals of time i.e.,
AREAL VELOCITY remains constant.
 Consider a mass µ at a distance r(θ) at time t from the
force center O. Let vector r rotates through angle dθ in
time dt, then area swept by the vector r in time dt is
 Substituting
dA 1 2 L
 r  2
dt 2
r
dA L


...(7)
dt 2
but for motion under central force,
total angular momentum remains constant
dA

 constant
dt
which is Kepler’s second law.
Kepler's Third Law
 Let ‘a’ and ‘b’ be the semi-major and semi-minor axis of
the ellipse.
 Area of ellipse= πab
 Time-period of the planet is given by
Area of the ellipse
T
Areal Velocity
from eq. (7), areal velocity  L / 2
ab
T 
...(8)
L / 2
b  a (1   2 )
For ellipse
2 EL2
but   1 
k 2
2 EL2
 b  a (1  1 
)
2
k
but E  -k/ 2a
2 L2  k
ba  2 
k 2a
 b  a1 / 2
L2
k
....(9)
2 EL2
ba 
k 2
T 
a
L / 2
 T  a
3/ 2
2
a
1/ 2
L
k
2
k
 2  3
 T   4
a
k

2
3
T  a
2
 Which is Kepler’s third law.