Kepler Law .pptx

1. Inter- Collegiate Model and Seminar Competition
To Celebrate
“National Mathematics Day”
Nikhil V. Jaipurkar
B.Sc. – II yr
Sardar Patel Mahavidyalaya,
Chandrapur

2. The Founder of the Modern Dynamical Astronomy is No
Body Other than Mr. Johannes Kepler who was a German
Astronomer and Mathematician.
Kepler Three Laws are as:-
1) The Orbit of Each Planet is an Ellipse with the Sun
at One focus….
2) The Radius vector drawn from the Planet to the Sun
Sweeps out equal Areas in Equal times….
3) The Square of the Periodic Time of the Planet is
proportional to the Semi-Major Axis of its Elliptic
Orbit….
The First Two laws were published in 1609 and the Third in
1619…

3.  Kepler’s First Law :-
Consider a Planet moving in a Central Attractive Force
field with Centre at the Point S.
We Consider generalized Polar Co-ordinates r & θ of
the Planet at any time t.
For the Inverse Square Law of Force,
We Have
F= – k / r2 k>0
And then V = – k / r = –ku
……(1.1)
Then Equation of The Path of The Planet becomes
θ = θ0 – ∫ du / √ α+βu–u2
where α=2mE/h2 & β=2mk/h2

4. θ = θ0 – {1/√–(–1)} {cos-1 (–β–2u / √β2+4α) }
θ = θ0 – cos-1 {u–(mk/h2) / (mk/h2) √1+2(Eh2/mk2) }
After Solving This Equation We get,
u = mk /h2 {1+ e cos(θ – θ0)} …... (1.2)
where e = √{1+ (2Eh2 / mk2)}
1/r= mk/h2 {1+e cos(θ – θ0)} OR
(h2/mk)/r = 1+e cos(θ – θ0)
Above represents a Conic Section with one Focus at the Origin,
Eccentricity ‘e’ and Latus Rectum ‘(h2/mk)’.

5. The classification of the Orbit is as follows:-
Hyperbola: e>1 1+(2Eh2/mk2)>1 i.e. E>0
Parabola: e=1 E=0
Ellipse: e<1 E<0
Circle: e=0 E= – mk2/2h2
Hence, the path of the Planet is a Conic Section.
Since an Ellipse is a Particular case of the General Conic,
Kepler’s First Law is Proved….

6.  Kepler’s Second Law :-
Let a planet trace an Elliptic orbit around the Sun.
It takes time t12 to move from position 1 to 2 and time t34 to
move from 3 to 4 As Shown in Fig.
During these times Suppose that the Radius Vector
sweeps out Areas A12 and A34 respt.
Kepler’s Second Law states that,
t12 = t34 A12 =A34 ……(2.1)

7.  Kepler’s Third Law:-
The Areal Velocity of a Planet is given by
dA/dt =1/2 r2θ
dA =1/2 r2 θ dt = h/2m dt (Here mr2θ=h)
If the periodic time (the Time for One Revolution) of the
Planet is ‘ ’ A
∫dA = ∫h/2m dt
0 0
OR A = (h/2m)
The area of an Ellipse having Major and Minor
Axes 2a and 2b respt. is A = ab. Then Above implies that
ab = (h/2m)
= 2 abm/h ….. (3.1)

8. In case of an Ellipse, if r1 and r2 denote the Maximum and Minimun
values of r, then r1 + r2 = 2a
From First Law Equation, the Quantity u is maximum when cos(θ–θ0)=+1
and is Minimum when cos(θ–θ0)= –1.
Hence u max =mk/h2 (1+e) And u min =mk/h2 (1–e)
r1 + r2 = 1/ u max + 1/ u min ( Bcoz r = 1/u )
= h2 /mk(1–e) + h2 /mk(1+e)
2a = 2h2 /mk(1–e2) …… (3.2)
amk (1–e2) = h2
amk (b2 / a2 )= h2 ( Bcoz b2 = a2(1–e2) )
b/h = √a/mk …...(3.3)
From eqn (3.1) and (3.3)
= 2 a3/2 √m/k ……(3.4)

9. For Planetary Motion, the quantity m in eqn is to be replaced
by the reduced mass,
m = (ms•mp) / (ms+ mp) ……(3.5)
where ms = Mass of the Sun
mp = Mass of the Planet
The Force of Attraction, Here Gravitation Force, between
the Sun and Planet is given by,
F = – G (ms•mp / r2 )
Comparing it with Eqn (1.1)
k = G ms•mp ……(3.6)
Then eqn (3.4 – 3.6)
= 2 a3/2 √1/G(ms+ mp)

10. As ms>>mp , We have ms+ mp ≈ ms
= 2 a3/2 √1/Gms = {2 / √Gms } a3/2
……(3.7)
The quantity in the Curly Bracket is Same for All Planets
And Hence is Constant for Planetary System.
= (Constant) a3/2
i.e. a3/2
OR 2 a3
This is Kepler’s Third Law….

11. To All Honorable Members of
Mathematics Department
And
My Dear Friends