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

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 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

Galileo

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

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 The force is propotional to inverse square of the distance between those masses GRAVITATION FIELD FORCE Gravitation field force is a force experience by an object due to gravitational attraction per unit mass 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

21 mmF ≈

2

1r

F ≈

221

221

rmm

GFrmm

F =→≈

F = the attractive force m1 = mass of first particle m2 = mass of second particle r = linier distance of both particles

G = a constant, UNIVERSAL GRAVITATIONAL CONSTANT = 6.67 x 10−11 N.m2/kg2

(measured by Henry Cavendish)

2rm

GgmF

g =→=

θcos))((2)()( 212

12

1 ggggg ++=

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 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.

Sir Isaac Newton

NNeewwttoonn’’ss LLaaww ooff UUnniivveerrssaall GGrraavviittaattiioonn 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

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

SSoommee ccoonnsseeqquueenncceess oonn GGrraavviittaattiioonn FFoorrmmuullaa::

The very small value of G is affectly effective for massive mass

weaker bigger constant

stronger smaller constant

weaker

stronger

FFoorrccee ((FF))

constant smaller

constant bigger

CCoonnddiittiioonn

ddiissttaannccee ((r))

mmaasssseess ((mm11 aanndd mm22))

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

PROBLEM SOLVED

N9.587)m10x38.6(

)kg60)(kg10x98.5()/kgN.m10x67.6( 26

242211

221

=

=

=

−

rmm

GF

(this force is extremely small and could be neglected and this is why your bicycle will not bended by itself)

N10x4.1)m2.1(

)kg25)(kg12()/kgN.m10x67.6( 8

22211

221

−− ==

=rmm

GF

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

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

PROBLEM SOLVED

N00167.0 N10x67.1

)m10x84.3()kg50)(kg10x35.7(

)/kgN.m10x67.6(

3

28

222211

221

==

=

=

−

−

rmm

GF

Gravitational force due to moon on earth could be ignored

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.

GRAVITATIONAL ACCELERATION Symbolized by letter g, gravitational acceleration is the magnitude of gravitational field 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)

2

2

2;

rM

Gmr

MmG

g

rMm

GFmF

ggmF

==

==→=

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

2r

mMGW objectearth=

objectearth mr

MGW 2=

gmW =

Potential Energy and Potential Gravities Potential Energy Gravity (Ep) is a work that needed to displace an object from one position to infinity.

r

mMGEp objectearth−=

mass sobject'Gravity Energy Potential

(V) Gravity Potential =

r

MGV earth−=

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

N10x14.1)m10x38.6(

)kg11600)(kg10x98.5()/kgN.m10x67.6(

)(

526

242211

2

==

=

−

rmM

GWa E

N10x950.0

m10x98.6 m10x598m10x38.6 with but (a) in as)(5

636

=

=+=

W

rb

The Law of Energy Conservation on Gravity The total energy is

rmM

G

mvrmM

G

EEE kp

2

22

1

−=

+−=

+=

The conservation energy is

222

1

2

212

1

1

)2()2()1()1(

mvrmM

GmvrmM

G

EEEE kpkp

+−=+−

+=+

Escape (Drift) velocity (vesc)

gRvRM

Gv escesc 22 =→=

rMm

G

rMm

Gr

MmGE

rM

Gvr

vm

rMm

G

maF

2

21

2

22

2

−=

+−=

=→=

=

RM

Gv

RM

Gv

RmM

Gmv

mvRmM

G

EEE

esc

esc

esc

esc

kp

2

2

0

2

22

1

22

1

=

=

=

+−=

+=

KEPPLER’S LAW AND NEWTON’S SYNTHESIS

kRT

MGRT

RT

MG

RTR

MGv

RM

G

RT

vRTR

vRM

G

RvamRmM

G

amF

s

s

=→=

=

=→=

=→==

=→=

=

3

22

3

2

322

22

22

22

22

2

2

2

4

14

4

42

π

π

π

ππ

ω

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)

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