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7. Central Force and Kepler’s System
7.1 The Motion in Two Dimension in Polar Coordinate System
The drawing shows the unit vectors iˆ, ˆj and r̂ , ˆ at a point in the
x-y plane. We see that the orthogonality of r̂ and ˆ plus the
fact that they are unit vectors,
y
ĵ
ˆ2
iˆ
r2
rˆ  1, ˆ  1 ,
r̂2
ˆ1
ĵ
r̂1
iˆ
r1
rˆ  iˆ cos ˆ  ˆj sin  and ˆ  iˆ sin   ˆj cos  which is shown.
x
The transformation can be shown by rotational Matrix
 rˆ   cos 
ˆ    sin 
  
sin    iˆ 
 
cos    ˆj 
ĵ
ˆ
drˆ
drˆ  ˆ
 iˆ sin   ˆj cos  
 
dt
dt
r̂
cos
sin 
r
dˆ
dˆ
 iˆ cos   ˆj sin  
 rˆ
dt
dt
iˆ
r
cos

7.1.1 The Position Vector in Polar Coordinate


r  x iˆ  y ˆj , r  r cos iˆ  sin  ˆj


rˆ  r cos iˆ  r sin  ˆj


r  r cos iˆ  sin  ˆj  r  r rˆ


r  r rˆ is sometimes confusing, because the equation as written seems to make no
reference to the angle  . We know that two parameters needed to specify a position in
two dimensional space (in Cartesian coordinates they are x and y ), but the equation
r  r rˆ seems to contain only the quantity r . The answer is that r̂ is not a fixed vector
and we need to know the value of  to tell how r̂ is origin. Although  does not occur
explicitly in r r̂ , its value must be known to fix the direction of r̂ . This would be
apparent if we wrote r  r rˆ   to emphasize the dependence of r̂ on  . However, by
common conversation r̂ is understood to stand for r̂   .
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1
sin 
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7.1.2 Velocity Vector in Polar Coordinate
v  r  ˆ
v  r rˆ
r
case 1
r
case 2
 d
 d (rrˆ) dr
drˆ
drˆ

ˆ
 ˆ  r
v  rr̂   v 
 .rˆ  r
 rrˆ  r
 v  rr
dt
dt
dt
dt
dt
where r is radial velocity in r̂ direction and r is tangential velocity in ˆ direction as
shown in figure and the magnitude to velocity vector v  r 2  r 2 2
7.1.3 Acceleration Vector in Polar Coordinate

dv dr
drˆ dr  ˆ
d ˆ
dˆ
 rˆ  r  
r
  r
dt dt
dt dt
dt
dt

dv
 ˆ  r
 ˆ  r
 ˆ  r()rˆ
 
rrˆ  r
dt


a  (
r  r2 )rˆ  r  2r ˆ  a  ar rˆ  a ˆ


ar  r  r 2 is radial acceleration and a  r  2r is tangential acceleration .
So, Newton’s law in polar coordinate can be written as
f r  mar  m(
r  r2 ) , where f r is force in radial direction .
f  ma  m(r  2r) , where f is force in tangential direction.
7.1.4 Circular Motion
For circular motion r  r0 , then 
r  0 . So, f r  mar   mr02 , where f r is force in radial
direction and f  ma  mr0 , where f is force in tangential direction.
If there is not any force in tangential direction i.e., f  0 is condition, then motion is
uniform circular motion i.e.,    is constant known as angular speed and tangential
speed is given by v  r0
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For non-uniform circular motion radial acceleration is ar  
acceleration is given by at 
v2
r
and tangential
dv
dt
 a  ar2  at2
7.2 Central Force
In classical mechanics, the central-force problem is to determine the motion of a
particle under the influence of a single central force. A central force is a force that points
from the particle directly towards (or directly away from) a fixed point in space, the
center, and whose magnitude only depends on the distance of the object to the center.
In central force potential V is only function of r and only central force is always
a conservative force; the magnitude F of a central force can always be expressed as the
derivative of a time-independent potential energy
 
 F 
1
r sin 
 F  ˆ 1  F  ˆ

  
  0
r   
  


V
And the force F is defined as F  
rˆ (force is only in radial direction)
r
7.2.1 Angular Momentum and Areal Velocity
The equation of motion in polar coordinate is given by m(r  r2 )  Fr and
m( r  2r)  F but for central force,
  
  r  Fr
V

   rrˆ  
rˆ ,
r

External torque   0 , so angular momentum is conserved
d ( mr 2)
m( r  2r)  F but for central force F  0 , so m( r  2r)  0 
 0 means
dt
 
J
angular momentum of mr 2  J  r  p   
mr 2
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    
 

r  J  r  r  p   0  r  J , hence position vector r is perpendicular to angular


momentum vector J and hence J is conserved, its magnitude and direction both are

fixed so direction of r is also fixed.

So, motion due to central force is confined into a plane and angular momentum J is
perpendicular to that plane.
rd
Central force problem. Prove that Areal velocity is constant.
For the central force problem  
Now, Areal velocity =
1
r  rd
2
d  1 2 d 1 2 
 r
 r
dt 2 dt 2
d
d 1 2 
J
d
J
 r  . It is given that  
, so

2
dt 2
mr
dt 2m
which means equal area will swept in equal time.
7.2.2 Total Energy of the System
Total energy is not explicitly function of time t , so
E
 0 . So, one can conclude that
t
total energy in central potential is constant.
E

1
mv 2  V r  and Velocity v  rrˆ  rˆ
2
So, total energy, E 



1 2 1 2 2
J
mr  mr   V ( r ) , where  
2
2
mr 2

1 2
J2
mr 
V r , r  0
2
2mr 2
E
where
1
m r 2  r 2 2  V r 
2
1 2
mr  Veff ,
2
J2
 V r  is identified as effective potential Veffective .
2mr 2
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7.2.3 Condition for Circular Orbit
From equation of motion in radial part m(r  r2 )  f (r )
J2
For circular orbit of radius r0 , r  r0 and 
r  0  f  r   m 0  r2   3 at r  r0
mr
Veffective
which can be also derived by
 0 and   0 is identified as angular frequency
r
r r


0
of circular orbit.
Radius r  r0 of circular orbit is also identified as stable equilibrium point
so
 2Veffective
r 2
 0 . If somehow particle of mass m changes its orbit without changing its
r  r0
angular momentum and new orbit is bounded then new orbit is identified as elliptical
orbit. The angular frequency in new elliptical orbit is
 2Veffective
r 2

r  r0
m
7.2.4 Equation of Motion and Differential Equation of Orbit
So, Lagrangian can be reduced to
L
Since,
1
 r 2  r 2 2  V  r 
2


L
 0 . So,  is cyclic co-ordinate. Hence, angular momentum is conserved

during the motion.
d  L 
J
2 


  0  mr   J or  
dt   
mr 2
Equation of Motion
The Lagrangian equation of motion is given by
d  L  L
0
 
dt  r  r
V
V
 mr  mr 2 
 0 or mr  mr 2  
r
r
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mr  mr 2  f  r 
From equation of motion in radial part,
d 2r J 2
m(r  r2 )  f (r )  m 2  3  f (r ) ………..(1)
dt
mr
where, J  mr 2  d 
J
d
J d
dt 

2
dt mr 2 d
mr
d 2  d  d   J
     
dt 2  dt  dt   mr 2
 d  J

 2
 d  mr
 d

 d
Substituting in (1)
J 2 1 d  1 dr  J 2
J 2 1 d  d  1 / r   J 2


 f r 
 f r 




m r 2 d  r 2 d  mr 3
m r 2 d  d  mr 3
 J 2 1 d 2 1 / r  J 2 
J2



 


f
r


2
d 2
mr 3 
mr 2
m r
Let,
1
J 2u 2
u 
r
m
 d 2u

 2  u  
 d

 d 2 1 / r  1 

   f r 
2
r
 d
1
f   (differential equation of an orbit)
u 
kr 2
Example: Consider the motion of a particle of mass m in the potential field V (r ) 
.
2
If l is angular momentum then
(a) What is effective potential ( Veff ) of the system? Plot Veff vs r
(b) Find value of energy such that motion is circular in nature.
(c) If particle is slightly disturbed from circular orbit such that its angular momentum
remains constant. What will be the nature of new orbit? Find the angular frequency of
new orbit in term of m, l , k .
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J2
1 2
V

Solution: (a) eff 2mr 2  2 kr
(b)
 J2 
J2

  3  kr  0 at r  r0 , so r0  
dr
mr
mk


dVeff
1/ 4
and J  m0 r02
Veff
J2
2mr 2
Veff
1 2
kr
2

r
r  r0
for circular motion m02 r0  kr0 , where r0 is radius of circle 0 
total energy E 
r
k
m
J2
1 2 mkr04 1 2

kr 
 kr0  E  kr02 ,
2
2
2
2mr
2mr0 2
1/ 4
 J2 
on putting r0  

 mk 
gives E  J
k
m
(c) orbit is elliptical in nature
d 2Veff
dr 2
3J 2

k 
mr 4
r  r0
3J 2
 k  4k
 J2 

m
 mk 
d 2Veff
dr 2

m
r  r0
 
4k
k
  2
 2 0
m
m
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Example: A particle of mass m moves under the influence of an attractive central
force f ( r ) .
(a) What is condition that orbit is circular in nature if J is the angular momentum of
particle
(b) If force is in form of f (r ) 
k
determine the maximum value of n for which the
rn
circular orbit can be stable.
Solution: (a) If Veff 
Veff
 2Veff
J2

V
(
r
)
,
for
circular
stable
orbit

0,
0
r
2mr 2
r 2
(b) f (r ) 
Veff
J 2 V
k
,
for
circular
motion

0



0
rn
r
mr 3 r
It is given
V
k
V
k
  f ( r ) , if f ( r )  n 
 n
r
r
r r

J2
k
k
J2


0


mr 3 r n
r n mr 3
 2Veff
r 2
3 J 2 nk
3J 2 n J 2
 0  4  n1  0 
 .
 0 , so n  3
mr
r
mr 4 r mr 3
Example: A particle of mass m and angular momentum l is moving under the action of a
central force f  r  along a circular path of radius a as shown in the figure.
y
The force centre O lies on the orbit.
(a) Given the orbit equation in a central field motion.
2
d u
m
1
 u   2 2 f , where u  .
2
r
d
l u
O
r

C (a,0)
Determine the form of force in terms of l , m, a and r .
(b) Calculate the total energy of the particle assuming that the potential energy V ( r )  0
as r   .
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Solution: (a) from the figure, r  2a cos  
J 2u 2

m
1 sec 

r
2a
 d 2u

1
 2  u  f  
u
 d

J 2 sec 2   1
sec  
1

sec  tan 2   sec3  
 f 
2


4a
2a 
 2a
u


y
J 2 sec 2 
4a 2

r

C (a,0)
O
x
1

1
2
3
 2a sec  tan   sec   sec    f  u 


J 2 sec3 
2 J 2 sec5 
1
1
1
2
2



tan   sec   1  f    
 f    f (r )  5
3
3

8a m
8a m
r
u
u
(b)
E
mr 2
J2
J2


V
(
r
)
,
as
r


,

0
V
(
r
)

0
2
2mr 2
2mr 2
J2
mr 2



and r  2a cos and r  2a sin  ,  
as r  
E
2
mr 2
J2

hence,  
 0 , so r  0 , so E  0
mr 2
7.3 Two Body Problem
Reduction of two body central force problem to the equivalent one body problem: A
system of two particles of mass m1 and m2 whose instantaneous position vectors of
inertial frame with origin O are r1 and r2 respectively.
  
Position vector m2 relative to m1 is r  r2  r1
m1
r1
The potential energy V is only function of distance between the particles.
 
So, V  V ( r2  r1 )
R
  
r  r2  r1
C.M .
r2
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Total energy of the system is given in lab frame by
E
1  2 1  2
 
m1r1  m2 r2  V  r2  r1 
2
2
Let the position vectors of m1 and m2 be r1 and r2 . The position vector
of the center of mass, measured from the same origin, is
 m r  m2 r2
R 11
m1  m2
m2

r2
The center of mass lies on the line joining m1 and m2 . To show this,

suppose first that the tip of R does not lie on the line, and consider the

 
vectors r1 , r2 from the tip of R to m1 and m2 . From the sketch we see that
  
r1  r1  R
  
r2  r2  R


 
m1r1
m2 r2
m2
 
r1  r1 


 r1  r2 
m1  m2 m1  m2 m1  m2
 
r1  r2

r2 
 r1
R

r1


 m1   
 
m1r1
m2 r2
r2  r2 

 
  r1  r2 
m1  m2 m1  m2
 m1  m2 


 


r1 and r2 are proportional to r1  r2 , the vector from m1 to m2 . Hence r1 and r2 lie along
the line joining m1 and m2 as shown. Furthermore,

r1 
m2
 
m2 

m1
 
m1 
r1  r2 
r and r2 
r1  r2 
r
m1  m2
m1  m2
m1  m2
m1  m2
E
 2 1
 2
1
 
m1 r1'  m2 r2 '  V  r2  r1  . The total energy is transformed
2
2
E
 2 1 mm  2
1
 m1  m2  R   1 2  r  V  r 
2
2  m1  m2 
Centre of mass moving with constant momentum and equation of motion for three
generalized co-ordinates or will not terms in R and R . While discussing the motion of
the system, one can ignore
1
 m1  m2  R 2 .
2
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So, energy in centre of mass reference frame is reduced to E 
where,  
1 m1m2  2
r  V  r 
2 m1  m2
m1m2
is reduced to one body system in centre of mass reference frame.
m1  m2
7.3.1 Kepler’s Problem
Kepler discuss the orbital motion of the sun and Earth system under the potential
V (r )  
k
where, k  Gms me . Here ms and me is mass of Sun and Earth respectively.
r
Although Kepler discuss Sun and Earth system but method can be used for any system
which is interacting with potential V ( r )  
k
r
The reduced mass for Sun and Earth system is  
me ms
me

 me , ms  me
me
me  ms
1
ms
Let us assume mass of Earth me  m
7.3.2 Kepler’s First Law
Every planet (earth) moves in an elliptical orbit around the sun, the sun is being at one of
k
the foci. Where sun and earth interact each other with potential V ( r )   , we solve
r
equation of motion in center of mass reference frame with reduced mass   me  m
7.3.3 Equation of Motion
Lagrangian can be reduced to
L
1
 r 2  r 2 2  V  r  , put   m
2


L
 0 , so  is cyclic co-ordinate. Hence angular momentum is conserved during the

motion.
d  L 
J
2 

    0  mr   J or   2
dt   
mr
k
J
mr  mr 2   2 , put  
mr 2
r
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mr 
l2
k
 2
3
mr
r
Equation of orbit is given by
f r  

J 2u 2  d 2u
1
 2  u   f  
m  d
u 


d 2u
ku 2 m
k
J 2u 2  d 2u
1
2
2

f


ku


u



u


ku
 


r2
m  d 2
d 2
J 2u 2
u 

d 2u 
km 
d 2u d 2 y
km

u


0
,
put
u


y
,
so



d 2 d 2
d 2 
J2 
J2
The equation reduces to,
d2y
y0
d 2
The solution of equation reduces to y  A cos  u 
km
km
 A cos   u  2  A cos 
2
J
J
 AJ 2 
1 km
J 2 / km

 A cos 
 1 
 cos 
r J2
r
 km 
J2
AJ 2
l
Put
 l and e 
the equation reduce to  1  e cos which is equation of
km
km
r
conic, where l is latus rectum and e is eccentricity.
In a central force potential given by potential V ( r )  
k
, the trajectory can be any conic
r
section depending on eccentricity e .
Now, we shall discuss the case specially of elliptical orbit as
Veff
Kepler discuss for planetary motion.
Total
energy
Veffective
1
J2
k
E  mr2 

2
2
r
2mr
rmin rmax
,
where
E ellipse
J2
k

 with constant angular momentum J .
2
r
2mr
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If one will plot Veffective vs r , it is clear that for negative
r
energy the orbit is elliptical which is shown in figure.
Earth is orbiting in elliptical path with Sun at one of its foci
as shown in figure.
Let equation of this ellipse is
r
m
r

B  ae,0  0,0 
s ae,0 
x2 y2


r

0


 1 , where
a2 b2
b  a 1  e 2 . Minimum value of r is a  ae  and maximum value of r is  a  ae  and
rmax  rmin  2a , from plot of effective potential it is identified that rmax and rmin is the
turning point so at these points radial velocity is zero.
E
J2
k
  2mEr 2  2mkr  J 2  0 , given equation is quadratic in terms of r for
2
r
2mr
their root at rmax and rmin . Using theory of quadratic equation, sum of roots,
rmax  rmin  
2mk
k
 E   , which is negative.
2mE
2a
7.3.4 Relationship between Energy and Eccentricity
For central potential V ( r )  
The energy is given by E 
So,
J2
k
l
the solution of orbit is  1  e cos with l 
r
r
km
1 2
J2
k
mr 

2
2
r
2mr
l
eJ sin 
J
r  e sin   r 
, where   2
2
ml
mr
r
after putting the value of
J2
l
eJ sin 
 1  e cos and r 
with l 
in equation of
r
ml
km
energy,
2 EJ 2
one will get, e  1 
.
mk 2
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A
r  0
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The condition on energy for possible nature of orbit for potential
E 0 ;
e  1 Hyperbola
E 0 ;
e  1 Parabola
E0 ;
e 1
mk 2
E 2 ;
2J
e  0 circle
Ellipse
7.3.5 Kepler’s Second Law
Equal Area will swept in equal time or Areal velocity is constant.
dA
J

( which is derived earlier )
dt 2m
7.3.6 Kepler’s Third Law
The square of time period ( T ) of revolution in elliptical orbit is proportional to cube of
semi major axis a ie T 2  a 3
dA
J
J
J

  dA 
dt   ab 
T ( ab is the area of ellipse)

dt 2m
2m
2m
a  a 1  e 2 
e2  1 
2 EJ 2
J
k
T . It is given e  1 
and E  
2
2m
2a
mk
2kJ 2
2kJ 2
2

1

e

2amk 2
2amk 2
T2 
4m 2 2 2 2
2kJ 2
J2
2
2

a

a
(
1

e
)
put
value
of
1

e


J2
2amk 2 amk
T2 
4m 2 2 4
4m 2 2 4 J 2
4 2 ma 3
2
2

a
(
1

e
)

T


a
.

mak
k
J2
J2
T2 
4 2 ma 3
4 2 a 3
if k  Gms m  T 2 
, where ms is mass of the sun.
k
Gm s
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Example: Given a classical model of tritium atom with nucleus of charge +1 and a single
electron in a circular orbit of radius r0 . Suddenly the nucleus emits a negatron and
changes to charge +2 (the emitted negatron escapes rapidly and we can forget about it)
the orbit suddenly has a new situation.
(a) Find the ratio of the electron’s energy after to before the emission of the negatron
(b) Describe qualitatively the new orbit
(c) Find the distance of closest and the farthest approach for the new orbits in units of r0
Solution: (a) As the negatron leaves the system rapidly, we can assume that its leaving
has no effect on the position and kinetic energy of the orbiting electron.
From the force relation for the electron,
mv02
mv02
e2
e2

, and we find its kinetic energy

r0
4 0 r02
2
8 0 r0
and its total mechanical energy
E1 
mv02
e2
e2


2
4 0 r0
8 0 r0
before the emission of the negatron. After the emission the kinetic energy of the electron
is still
e2
2e2
e 2
, while its potential energy suddenly changes to

8 0 r0
4 0 r0 2 0 r0
Thus after the emission the total mechanical energy of the orbiting electron is
E2 
mv02
2e 2
3e 2
E


, giving 2  3 .
2
4 0 r0 8 0 r0
E1
In other words, the total energy of the orbiting electron after the emission is three times
as large as that before the emission.
(b) As E2 
3e 2
, the condition equation (i) for circular motion is no longer satisfied
8 0 r0
and the new orbit is an ellipse.
(c) Conservation of energy gives
m r 2  r 2 2
3e 2
e 2


8 0 r0 2 0 r0
2


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At positions where the orbiting electron is at the distance of closest or farthest approach
to the atom, we have r  0 , for which
3e 2
mr 22
e2
J2
e2




8 0 r0
2
2 0 r 2mr 2 2 0 r
Then, with J 2  m 2 v02 r02 
me 2 r0
4 0
From above equations
3r 2  4r0 r  r02  0
the solutions are r 
r0
, r  r0
3
Hence, the distances of closest and farthest approach in the new orbit are respectively,
1
rmin  , rmax  1
3
Example: A satellite of mass m  2000 kg is in elliptical orbit about earth. At perigee it
has an altitude of 1,100 km and at apogee it has altitude 4,100 km. assume radius of the
earth is Re  6, 400 km. it is given GmM e  8 1017 J .m
(a) What is major axis of the orbit?
(b) What is eccentricity of the orbit?
(c) What is angular momentum of the satellite?
(d) How much energy is needed to fix satellite in an orbit from
4100
6400
6400
surface of the earth?
Solution: rmax  4100  6400  10500 km
rmin  1100  6400  7500 km
(a) rmax  rmin  2a  18000  2a  a  9000 km
(b) e 
rmax  rmin
10500  7500 3000 1
1
e

 e
10500  7500 18000 6
6
rmax  rmin
(c) It is given, k  8 1017 J  m
E
k
8  1017
8

, E f    1011 J  4.5  1010 J
3
2a
18
18000  10
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2
e  1
2 EJ 2
2 EJ 2
1


1

 
mk 2
mk 2
 6
1
2 EJ 2
1 
 140  10 26  J 2  J  140  1013  1.2  1014 kgm / sec 2
36
mk 2
(d) When satellite is at surface of the earth, R  6400 Km
Ei  
GMm  8  1017
 1017
 1012



 12.5  1010
3
3
R
8
6400  10
800  10
Ef  
GMm
 4.5  1010 J  E  E f  Ei  8  1010 J
2a
Example: For circular and parabolic orbits in an attractive 1 / r potential having the same
angular momentum, show that perihelion distance of the parabola is one-half the radius of
the circle.
Solution: For Kepler’s problem,
parabola e  1 ,
l
l
 1  e cos  , for circular orbit e  0   1 and for
r
rc
l
 1  cos  , then rp is minimum when cos is maximum.
r
rp 1
l
l
 2 and  1 

rp
lc
rc 2
Example: A planet of mass m moves in the inverse square central force field of the Sun
of mass M . If the semi-major and semi-minor axes of the orbit are a and b respectively,
then find total energy of the planet by assuming Sun is at the center of ellipse.
v2
Solution: Assume Sun is at the centre of elliptical orbit.
b
1 2 GMm 1 2 GMm
Conservation of energy, mv1 
 mv2 
s
2
a
2
b
a
v1
Conservation of momentum, L  mv1a  mv2b
 a
v2  v1  
 b
1 2 1 2 GMm GMm
1 
a2 
ba
mv1  mv2 

 m  v12  v12 2   GMm 

2
2
a
b
2 
b 
 ab 
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1 2  b2  a2 
1
1
 b  a
b
 mv12  GMm   
mv1 
 GMm 

2

 ab 
2
2
 b

 a  b  a 
E

1 2 GMm
b 1
GMm
mv1 
 GMm

2
a
a b  a
a
 GMm  b  b  a 
GMm  b
GMm
 1 


  

a  b  a  
a  b  a  
b  a 
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