KEPLER LAWS AND CENTRAL ORBIT, Lecture notes of Mathematics

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B.Sc. Mathematics – 2

nd

Semester

MTB 202 – Statics and Dynamics

by

Dr. Krishnendu Bhattacharyya

Department of Mathematics,

Institute of Science, Banaras Hindu University

Part – V

Central Orbit and Kepler’s Laws

Equation of Motion (for a Central Orbit):

Let O be the centre of force. Let OX be a fixed line through O. Let P be

the position of the particle at any time t. With O as pole and OX as initial

line let (^)  r ,^  be the coordinates of P.

Let F be the central force per unit mass

acting along the line PO.

There is no force acting along the cross

radial direction. Hence the equation of

motion of the particle along radial and transverse directions are

2 2

2

d r d m r mF dt dt

 ^    

 ^  

, i.e.,

2 2

2

d r d r F dt dt

 , (1)

d d m r r dt dt

, i.e.,

2 0

d d r dt dt

where m is the mass of the particle.

Integrating (2) we have

2 constant

d r h dt

  (say) (3)

2 2 2 2 2 2 5 2

d r dr h r h r h r F d  d 

Case-II: Equation in u and 

Let

u r

 , i.e.,

r u

Then from (3) we get

2

  hu (5)

Now 2

dr d d 1 d 1 du d r d dt d u dt u d dt

2 2

1 du du hu h

u d  d 

d du d du d r h h dt d d d dt



  

2 2 2 2 2 2 2

d u d u h hu h u

d  d 

Then from (1) we get

2 2 2 2 4 2

d u 1 h u h u F

d  u

2 2 2 2 3  2  

d u h u h u F

d 

2 2 2 2

d u F u u

d  h

, Or,

2

2 2 2

d u F u d  h u

This equation of motion of central orbit is used in practical purpose.

We know

r u

 then 2

dr 1 du

d  u d 

Then we have

2 2 2

1 du u

p d 

Now, differentiating both sides with respect to  we get

2

3 2

dp dr du du d u u

p dr d  d  d  d 

2

3 2 2

dp du du d u u p dr u d  d  d 

  ^ 

2 2 3 2

1 dp d u u u

p dr d 

3 2

1 dp F

p dr h

2

3

h dp F p dr

Linear velocity at a position:

We know that the angular velocity of a particle is given by 2

vp

r

 

2 2

h vp h 1 v v r r p p

     . We also have

2 2 2 2 du v h u

d 

 ^  

in time  t. As   t 0,    0 and  r  0 and arc PQ  chord PQ.

Then the rate of description of the sectorial area

0

area of the sector OPQ lim   t (^) t

^0

area of the triangle OPQ lim   t (^) t

 

0

sin 2 lim t

r r r

t

 

  0

1 sin lim t 2

r r r t

 ^  

 ^  

  0 0 0

1 sin lim lim lim 2 t^ t

r r r  t

 

d r r r h dt

The rate of describing the sectorial area by the radius vector to the particle

moving under a centre force is called areal velocity.

Therefore, h  2  rate of describing the sectorial area

 2  areal velocity

---

Angular momentum of the particle is the moment of momentum of the

particle about the pole (or origin).

The radial and transverse components of momentum are mr and mr .

Therefore, angular momentum = moment of momentum

2

 r mr.   mr   mh

Thus, h  Angular momentum per unit mass

(II) Law of velocity is given

Suppose

2 2 6 3

1 du v u u r d 

Differentiate w.r.t.

2 5 2

d u u u d 

5 5 2 7

F 1

u F u F u r

Also,

3 3 p  r  p  ar.

(III) Pedal Equation is given

Suppose

2 3 3 3 a p  r  p  r  v  u and so on.

Polar Equation of the central orbit is given:

1) Elliptic Orbit:

Let the central orbit be an ellipse whose equation is

1 cos

l e r

where

2 b l a

 is the semi-latus rectum, a is the semi-major axis, b is the

semi-minor axis and  

2 2 2

b  a 1  e , i.e.,  

2 2 1 2 , 1

b e e a

   is

eccentricity.

We have

2 2 2 2 du v h u

d 

 ^  

2 2 2 1 e^ cos^ e sin v l l l

  

 ^ ^  

2 2 2 2 2

1 2 cos cos sin

l e e e l

  ^      

2 1 2 cos e e l

 ^    

____________________

2

1 2 cos (^1 )

b e l a

 

2

2

2 1 e cos b

l a l

r a

v r

  and the linear velocity varies inversely to square root of radius

vector of that position.

We have

2 2 2

h v p r a

2

l 2 1

p r a