1. FORCES AND NEWTON’S
LAW OF GRAVITATION
Standard Competency:
Analyze the nature phenomenon and its regularity within the
scope of particle’s Mechanics
Base Competency:
Analyze the regularity of planetary motion within the universe
base on Newton’s Law
Learning Objectives:
After completing this chapter, students should be able to
[1] Analyze the relation between gravitational force and
object’s masses with their distance
[2] Calculate the gravitational forces resultant on particles
within a system
[3] Compare the gravitational acceleration at different
positions
[4] Formulate the quantitiy of potential gravity at a point due
to several object’s masses
[5] Analyze the planetary motion within a universe base on
Keppler’s Law
[6] Calculate the speed of satellite and its terminal velocity
References:
[1] John D Cutnell and Kenneth W. Johnson (2002). Physics 5th Ed
with Compliments. John Wiley and Sons, Inc. pp
[2] Douglas C. Giancolli (1985). Physics: Principles with
Applications, 2nd Edition. Prentice Hall, Inc. pp
[3] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA
untuk SMA/MA Kelas XI. CV Yrama Widya pp
[4] http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l3d.html

2. WHAT IS GRAVITY
Gravity is a force that exist between the earth
and any object surround it. The force of gravity
tends to attract objects. It attracts the objects
with same force.
The name of gravity is proposed by both Galileo
Galileo and Newton. Newton showed that the same
force exists between all objects.
GRAVITATIONAL FIELD
DEFINITION: A space around an object where the
gravitational force exist and has a certain value
in its every point
Gravitational field is a concept introduced
by Newton to describe a force acting on
objects throught out a distance in a
space.
According Newton, there seems to be out
of mind if there are two object that can interact each other
without being physically interact.
Field, a distance space, is then “to be created” to bridge how
remote objects (distance separated objects) could interacting
each other.
Base on Newton, FIELD is a quantification of
distance acting force. Recall that F = m a
(Newton’s II Law) is a type of contact force.
Also FIELD is a vector quantity so it could be
visualized by an arrow which shows the
magnitude and direction of grafivitational
field.
Representation of fields as arrangement arrows is well known
as force line

3. GRAVITATION PHENOMENON
Base on Newton’s Observations of how an apple could fall
from its tree:
[1] All objects which are on
certain height will
always fall freely
toward the earth
surface
[2] A cannon ball will move
in archery trajectory
while shot in a certain
angle and initial force
[3] Moon is always stay in
its trajectory while
evolves the Earth and
Earth is always stay in
its trajectory while
evolves the Sun
Newton Proposed:
[1] There is a force whose kept all
the remote objects downward to
the earth surface
[2] Such force works on two
different objects which is
separated in a certain distance
[3] Such force has a property of
being attractive
[4] Such force works without any
physical contact between those
objects
[5] Such force apply to all objects in
universe, hence it is universal

4. Gravitational Force Formula
Base on Newton’s Proposes, there is an attractive force which
works on two different object separated in a certain distance.
The force is propotional to two masses
F ≈ m1 m2
The force is propotional to inverse square of the distance
between those masses
1
F ≈
r2
m1 m2 m1 m2
F ≈ → F =G
r2 r2
F = the attractive force G = a constant,
m1 = mass of first particle UNIVERSAL GRAVITATIONAL
m2 = mass of second particle CONSTANT
r = linier distance of both particles = 6.67 x 10−11 N.m2/kg2
(measured by Henry Cavendish)
GRAVITATION FIELD FORCE
Gravitation field force is a force experience by an object due
to gravitational attraction per unit mass
F m
g= → g=G
m r2
Direction of gravitational field is goes to the
center of object’s mass.
If an object is under gravitational force
influenced by several objects, the gravitational
field force on the object is the sum of each
gravitational field another objects
g = (g1 )2 + (g1 )2 + 2(g1 )(g2 ) cos θ

5. NEWTON’S LAW OF UNIVERSAL GRAVITATION
In 1666 Isaac Newton determined that the same force that
kept the planets in motion must also exist between every
object.
Newton’s law of gravitation is an empirical
physical law which is describing the gravitional
attraction between bodies with their masses.
He stated that every object in the universe
attracts every other object in the universe.
Newton’s law gravitation resembles Coulomb’s
S Isaac N ton
ir ew law of electrical forces, which is used to
calculate the magnitude of electrical force between two
charged bodies. Both are inverse-square laws, in which forse
is inversely proportional te the square of the distance between
the bodies. Coulomb’s law has the product of two charges in
place of the product of the masses, and the electrostatic
constant in place of the gravitational constant.
What was Newton stated is now become a Universal Law of
Gravitation. It is universal because it works for all kinds,
types and sizes of object. No matter how small or big the
objects are, attraction between those objects is apply.
Newton’s Law of Universal Gravitation
For two particles, which have masses m1 and m2 and are
separated by a distance r, the force that exerts on the other is
directed along the line joining the particles

6. The law of gravitation is universal and very fundamental. It
can be used to understand the motions of planets and moons,
determine the surface gravity of planets, and the orbital
motion of artificial satellites around the Earth
Some consequences on Gravitation Formula:
masses distance Force (F)
(m1 and m2) (r)
)
bigger constant stronger
Condition smaller constant weaker
constant smaller stronger
constant bigger weaker
The very small value of G is affectly effective for massive mass

7. PROBLEM SOLVED
What is the magnitude of the gravitational force between
the earth (m = 5.98 x 1024 kg) and a 60-kg man whose
stand on 6.38 x 106 m away
SOLUTION
m1 m2
F =G
r2
(5.98 x 1024 kg)(60 kg)
= (6.67 x 10 −11 N.m2 /kg2 )
(6.38 x 106 m)2
= 587.9 N

8. PROBLEM SOLVED
What is the magnitude of the gravitational force that acts
on each bicycle’s tyres where m1 = 12 kg and m2 = 25 kg
and r = 1.2 m?
(approximately the mass of a bicycle)
SOLUTION
m1 m2
F =G
r2
(12 kg)(25 kg)
= (6.67 x 10 −11 N.m2 /kg2 ) = 1.4 x 10 − 8 N
(1.2 m)2
(this force is extremely small and could be neglected and this is
why your bicycle will not bended by itself)
PROBLEM SOLVED
What is the magnitude of the gravitational force due to
moon (m = 7.35 x 1022 kg) on a 50-kg man where moon-
earth is 3.84 x 108 m away
SOLUTION
m1 m2
F =G
r2
−11 2 2 (7.35 x 1022 kg)(50 kg)
= (6.67 x 10 N.m /kg )
(3.84 x 108 m)2
= 1.67 x 10 −3 N = 0.00167 N
Gravitational force due to moon on earth could be ignored

9. Cavendish and the Value of G
The value of G was not experimentally determined until nearly
a century later (1798) by Lord Henry Cavendish using a
torsion balance.
Cavendish's apparatus
involved a light, rigid rod
about 2-feet long. Two small
lead spheres were attached
to the ends of the rod and
the rod was suspended by a
thin wire.
When the rod becomes
twisted, the torsion of the
wire begins to exert a
torsional force which is
proportional to the angle of
rotation of the rod. The more
twist of the wire, the more the system pushes backwards to
restore itself towards the original position. Cavendish had
calibrated his instrument to determine the relationship
between the angle of rotation and the amount of torsional
force.
Cavendish then brought two large lead spheres near the
smaller spheres attached to the rod. Since all masses attract,
the large spheres exerted a gravitational force upon the
smaller spheres and twisted the rod a measurable amount.
Once the torsional force balanced the gravitational force, the
rod and spheres came to rest and Cavendish was able to
determine the gravitational force of attraction between the
masses.
By measuring m1, m2, d and Fgrav, the value of G could be
determined. Cavendish's measurements resulted in an
experimentally determined value of 6.75 x 10-11 N m2/kg2.
Today, the currently accepted value is
6.67259 x 10-11 N m2/kg2.

10. GRAVITATIONAL ACCELERATION
Symbolized by letter g, gravitational acceleration is the
magnitude of gravitational field
F Mm
F =mg → g= ; F =G
m r2
Mm
G
g= r2 = G M
m r2
where
F = gravitational force
m = mass of test object
M = mass of source object
g = gravitational field strength
(gravitational acceleration)
r = distance of a point to the object source
The value of gravitational
acceleration g is depend
on the location of the
object.
If M is mass of earth,
then r is the distance of
object from the center of
the earth (the location of
object)

11. WEIGHT: Acceleration Due to Gravity
WEIGHT is a measurement of the force on a object caused by
gravity trying to pull the object down. The weigth of an
object of on the earth is the gravitational force that earth
exerts on the object.
- The weight always acts downward,
toward the center of the earth
- On another astronomical body, the
weight is the gravitational force
exerted on the object by that body
(appear as the consequence of
Newton’s III law)
- Weight is opposite to Gravitational
Force
An object has weight whether or not it is resting on the
earth’s surface
Gravitational force is acting even when the distance r is much
bigger than the radius of the earth R
Mearth mobject
W =G
r2
Mearth
W =G mobject
r2
W =mg

12. PROBLEM SOLVED
The mass of Hubble Space
Telescope is 11600 kg.
Determine the weight of the
telescope (a) when it was
resting on the earth and (b) as
it is in its orbit 598 km above
the earth (R-earth = 6380 km)
SOLUTION
ME m
(a) W =G
r2
(5.98 x 1024 kg)(11600 kg)
= (6.67 x 10 −11 N.m2 /kg2 ) 6 2
= 1.14 x 105 N
(6.38 x 10 m)
(b) as in (a) but with r = 6.38 x 106 m + 598 x 103 m = 6.98 x 106 m
W = 0.950 x 105 N
Potential Energy and Potential Gravities
Potential Energy Gravity (Ep) is a work that needed to
displace an object from one position to infinity.
Mearth mobject
Ep = − G
r
Potential Energy Gravity
Potential Gravity (V) =
object' s mass
Mearth
V = −G
r

13. The Law of Energy Conservation on Gravity
The total energy is
E = E p + Ek F = ma
Mm v2 M
Mm G =m → v2 = G
= −G + mv 2
2
1
2
r r r
r Mm 1 Mm
E = −G 2 + G
Mm r 2 r
= −G Mm
2r = −G
2r
The conservation energy is
E p (1) + E k (1) = E p (2) + E k (2)
Mm 1 2 Mm 1 2
−G + 2 mv1 = − G + 2 mv2
r1 r2
Escape (Drift) velocity (vesc) E = E p + Ek
Mm 1 2
0 = −G + 2 mv esc
M R
v esc = 2 G → v esc = 2 gR Mm
R 1 mv 2
2 esc = G
R
2 M
v esc = 2 G
R
M
v esc = 2 G
R

14. KEPPLER’S LAW AND NEWTON’S SYNTHESIS
F = m as
Mm
G = m as → v = ωR
R2
M v2 2π 4π 2
G 2 = = R → v 2 = 2 R2
R R T T
M M 4π 2
G = v2 → G = 2 R2
R R T
M 1
G = 2 R3
4π 2
T
T2 4π 2 T2
= → =k
R3 GM R3
Keppler’s law state that ratio of square any planet’s revolution
period revolves the sun and triple rank of average distance
between planet and sun is always constant
KEPPLER’S LAWS OF PLANETARY MOTION
First Law: The path of the planets are ellips
with the center of the sun at one focus
(The Law of Ellips)
Second Law: An imaginary line from the
sun sweeps out equal areas
in equal time intervals. Thus,
planets move fastest when
closest to the sun, slowest when farthest away
(The Law of Equals Areas)
Third Law: The ratio of the squares of the periods of any two
planets revolving about the sun is equal to the
ratio of the cubes of their average distance from
the sun
(The Law of Harmonies)

15. Exercises
[1] Determines the gravitational force that works on a
satellite (mass ms) while is orbiting the earth (mass me)
in the position of height earth radius (re) from the surface
[2] Three masses of objects (each has mass of m) are on the
corner of a triangle which side s. Determines the
gravitational force on each mass.
[3] Calculate the gravitational field force on earth’s surface
[4] On height h from earth’s surface the gravitational field
force is known to be equal to half the gravitational field
force of earth’s surface. Define the value of h in the
term of earth radius (re)
[5] On a point beyond earth’s surface it is known that the
potential gravity is −5.12 x 107 J/kg and gravitational
earth acceleration is 6.4 m/s2. If the earth radius is
6400 km, calculate the height of such point from earth
surface.
[6] By what minimal velocity required is in order a bullet
(mass m) which fired from earth surface could reach a
height of R? (R = earth radius)
[7] A satellite is orbiting earth in a circle orbital. Determine
the satellite periode when is
(a) orbiting exactly on earth’s surface.
(b) orbiting on the height of h above the earth surface