- Johannes Kepler modified the heliocentric model and formulated Kepler's laws of planetary motion. - Kepler's laws state that planets move in elliptical orbits with the Sun at one focus, planets sweep out equal areas in equal times, and the square of the orbital period is directly proportional to the cube of the radius of the orbit. - Kepler's laws can be derived from Newton's law of universal gravitation and the concept of circular motion, providing a mathematical relationship between orbital period and radius.

1. Chapter 3 Gravitation
3.2 Kepler's laws
Physics Form 4

2. Johannes Kepler, a German astronomist, mathematician and astrologist who
modified the heliocentric model according to Kepler’s Law.
Kepler’s Laws
Physics Form 4
Chapter 3 Gravitation

3. All planets move in elliptical orbits with the Sun at one focus (Law of Orbits)
Minor axis
Major
axis
Sun
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Chapter 3 Gravitation

4. Physics Form 4
Chapter 3 Gravitation
Ellipse (string and optical properties) - YouTube Video

5. The planets in the Solar System have elliptical shaped orbits.
The Sun always stays on a focus of the ellipse.
The major axis is longer than the minor axis.
Most orbits of the planets in the Solar System have major axis and minor axis of
almost the same length.
As such, the shape of the elliptical orbit of the planets in the Solar System is
almost round.
Planets can be assumed to make circular motion around the Sun.
The radius of orbit is the average value of the distance between the planet and
the Sun.
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Chapter 3 Gravitation

6. A line that connects a planet to the Sun sweeps out equal areas in equal times
(Law of Areas).
Sun
Elliptical
orbit
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Chapter 3 Gravitation

7. If a planet takes the same amount of time to move from A to B and from C to D, the
area AFB is the same as the area CFD.
Distance AB is longer than distance CD.
This means the planet is moving at a higher linear speed from A to B than from C
to D.
Physics Form 4
Chapter 3 Gravitation

8. The square of the orbital period of any planet is directly proportional to the cube of the
radius of its orbit (Law of Periods).
Mathematically,
T = orbital period of a planet
r = radius of orbit
2 3
T r

A planet which orbits with a larger radius has a longer orbital period. As such, planets
which are further from the Sun take a longer time to complete one orbit around the Sun.
Physics Form 4
Chapter 3 Gravitation

10. For example, the Earth takes 1 year to make one complete orbit while Saturn takes 29.5 years.
0.2 tahun
0.6 tahun
1.0 tahun
1.9 tahun
11.9 tahun
29.5 tahun
84.0 tahun
164.8 tahun
Utarid
Zuhrah
Bumi
Marikh
Musytari
Zuhal
Uranus
Neptun
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Chapter 3 Gravitation

11. The Kepler’s Third Law can be formulated using Newton's Law of Gravitation and the
concept of circular motion.
Planets make circular motions around
the Sun. The centripetal force is the
same as the gravitational force
between the Sun and the planet.
Assuming that the orbit of the planet
around the Sun is circular, we can
derive the relationship between the
orbital period of the planet and the
radius of the orbit as in Kepler’s third
law.
Physics Form 4
Chapter 3 Gravitation

12. Gravitational force acting on the planet,
Sun
Planet
m
F
v
M
r
The gravitational force acts as the centripetal force for the planet to create a circular
motion around the Sun.
Centripetal force,
Mass of the Sun = M
Mass of the planet = m
Radius of orbit = r
Gravitational force = F
Linear speed of planet = v
Orbital period = T
2
GMm
F
r

2
mv
F
r

Physics Form 4
Chapter 3 Gravitation

14.  
 
   
2
2
2
2
2
2 3
1
2
,
2
2
2 into
4
1
Centripetal force Gravitational force
Distance travelled in one complete orbit
Linear speed of planet v
Orbital period
r
v
T
Substit
mv GMm
r r
GM
v
u
r
r GM
T r
T r
te
GM





 

 
 
 
  





As GM is constant,
is Kepler’s Third Law
2 3
T r

2 3
T r

Physics Form 4
Chapter 3 Gravitation

15. Solving Problems Using Kepler’s Third Law Formula
From Kepler’s third law, relationship between orbital period, T and radius
of orbit, r is 2
2 3
4
T r
GM

 
  
 
Compare two planets:
For planet 1,
For planet 2,
(1) ÷ (2) gives
 
2
2 3
2 2
4
2
T r
GM

 
  
 
 
2
2 3
1 1
4
1
T r
GM

 
  
 
2 3
1 1
2 3
2 2
T r
T r

Physics Form 4
Chapter 3 Gravitation
Planet 1
Planet 2
2
2

16. Solving Problems Using Kepler’s Third Law Formula
Physics Form 4
Chapter 3 Gravitation

17. Solving Problems Using Kepler’s Third Law Formula
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Chapter 3 Gravitation

18. Solving Problems Using Kepler’s Third Law Formula
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Chapter 3 Gravitation

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Chapter 3 Gravitation

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Chapter 3 Gravitation

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Chapter 3 Gravitation