3.2 KEPLER'S LAW

CHAPTER 3
GRAVITATION

3.2 KEPLER’S LAW

TEXT BOOK : PAGE 96 - 102

SUE ROSE

3.2 KEPLER’S LAW

LEARNING STANDARD
Student is able to:
3.2.1 Explain Kepler’s law I, II and III
3.2.2 Express Kepler’s Third Law T2 α r3
3.2.3 Solve problems using Kepler’s

Third law.

LEARNING STANDARD 1

3.2.1 Explain Kepler’s Law I, II dan III

• A German astronomist,
mathematician and astrologist
who formulated three laws that
describe the movement of planets
around the Sun

1 All planets move in elliptical orbits with
the Sun at one focus (Law of Orbit)







Ellipse

The distance is further
at the major axis.

The shape of the orbit is
almost round.

The planets in the Solar System have Planets can be assumed to make
elliptical shaped orbits circular motion around the Sun

Sun always stays on
a focus of the ellipse

The major axis is longer
than the minor axis

As such, the shape of The radius of orbit is the average value
the elliptical orbit of the of the distance between the planet and
planets in the Solar the Sun.
System is almost round

2 A line that connect a planet to the Sun
sweeps out equal areas in equal times
(Law of Areas)

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

The planet is moving
at a higher linear speed
from A to B than from
C to D



3 The square of the orbital period of any
planet is directly proportional to the cube
of the radius of its orbit (Law of Period)

- A planet which orbits
with a larger radius has
a longer orbital period.

- Planets which are
further from the Sun
take a longer time to
complete one orbit
around the Sun.

- For example, the Earth
takes 1 year to make
one complete orbit
while Saturn takes 29.5
years.



Planet and orbital period

PLANET ORBITAL PERIOD
Mercury 0.2 years
Venus 0.6 years
1.0 years
Earth 1.9 years
Mars 11.9 years
Jupiter 29.5 years
Saturn 84.0 years
Uranus
Neptune 164.8 years

LEARNING STANDARD 2

3.2.2 Express Kepler’s Third Law
T2 α r3

ACTIVITY 2 Aim: Formulating Kepler’s Third Law

➢ Kepler’s third law can be formulated using Newton’s Universal Law of
Gravitation and concept of circular motion.

➢ Planets make circular motions around the Sun.

The centripetal = the gravitational force

force between the Sun and the planet.

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

ACTIVITY 2 Aim: Formulating Kepler’s Third Law

Derive the relationship between the orbital period
of the planet and the radius of the orbit.

Centripetal Force = Gravitational Force Mass of the Sun = M
Mass of the planet = m
mv2 = GMm Radius of orbit = r
r Gravitational force = F
r2 Linear speed of planet = v
v2 = GM Orbital period = T
r
Linear speed = distance travelled in one complete orbit v = 2πr
T
of planet, v orbital period

v2 = GM = (2πr)2 T2 = (4π2 )r3 T2 = kr3 T2 α r3
r T2 GM
KEPLER’S THIRD LAW
k = 4π2 k = constant
GM

LEARNING STANDARD 3
3.2.3 Solve problems

using Kepler’s Third Law
Formula

ACTIVITY 3 Aim: Solving Problems using Kepler’s Third Law Formula

From Kepler’s Third Law, T2 = (4π2 )r3
relationship between period, T GM
and radius is

Compare two planets

Planet 1 Planet 2

T12 = (4π2 )r13 T22 = (4π2 )r23
GM GM

k = constant

T12 = kr13 k = T12 k = T22 T12 = T22
T22 = kr23 r13 r23 r13 r23

T12 = T22
r13 r23



All planets move in elliptical
orbits with the Sun at one
focus (Law of Orbit)

A line that connect a
planet to the Sun
sweeps out equal

areas in equal times
(Law of Areas)

Speed at Z > X > Y

The square of the orbital period of any planet is directly r12 = T12r23
proportional to the cube of the radius of its orbit (Law of Period) T22

r1 = 6.37 x 106 + h

T12 = T22 r2 = 3.83 x 108 (6.37 x 106 + h)3 = (24)2(3.83 x 108)3
r13 r23 (655.2)2
T1 = 24 hr

T2 = 27.3 x 24 = 655.2 hr 2.58 x 1020 + h3 = 3.23 x 1028
2.81 x 108

2.58 x 1020 + h3 = 7.5 x 1022 h3 = 7.47 x 1022
h3 = 7.5 x 1022 - 2.58 x 1020 h = 4.21 x 107 m