Gravitation

CBSE-i

CLASS

XI

UNIT

-

6

CENTRAL BOARD OF SECONDARY EDUCATION

Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India

PHYSICS

Contents

Preface

v

Acknowledgements

viii

Teachers’ manual

learning outcomes

ix

Teaching notes

xi

lesson Plan matrix

xv

Pre-requisites

xix

Weblinks/vediolinks/other references

xix

sTudenTs’ manual

• Introduction

2

• Astronomy in Ancient India

4

• Kepler’s Laws of Planetary Motion

5

• Newton’s Law of Gravitation

9

• Important Characteristics of Gravitational force

10

• The Universal Gravitational Constant (G)

11

• Vector Form of Newton’s Law of Gravitation

13

• Principle of Superposition

13

• Acceleration Due to Gravity (G)

16

• Variation in G Due to Shape

19

• Variation in G with Height

19

• Variation in G with Depth

22

• Gravitational Field

25

• Gravitational Potential Energy

27

• Gravitational Potential (V)

32

• Escape Speed (V

_e

)

33

• Motion of Satellite

35

Post content student worksheet 1

40

Post content student worksheet 2

41

Post content student worksheet 3

42

Post content student worksheet 4

44

Post content student worksheet 5

45

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PrefaCe

The Curriculum initiated by Central Board of Secondary Education -International (CBSE-i) is a progressive step in making the educational content and methodology more sensitive and responsive to global needs. It signifies the emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning process in harmony with the existing personal, social and cultural ethos.

The Central Board of Secondary Education has been providing support to the academic needs of the learners worldwide. It has about 11500 schools affiliated to it and over 158 schools situated in more than 23 countries. The Board has always been conscious of the varying needs of the learners and has been working towards contextualizing certain elements of the learning process to the physical, geographical, social and cultural environment in which they are engaged. The CBSE-i has been visualized and developed with these requirements in view. The nucleus of the entire process of constructing the curricular structure is the learner. The objective of the curriculum is to nurture the independence of the learner, given the fact that every learner is unique. The learner has to understand, appreciate, protect and build on knowledge, values, beliefs and traditional wisdom. Teachers need to facilitate the leaner to make the necessary modifications, improvisations and additions wherever and whenever necessary.

The recent scientific and technological advances have thrown open the gateways of knowledge at an astonishing pace. The speed and methods of assimilating knowledge have put forth many challenges to the educators, forcing them to rethink their approaches for knowledge processing by their learners. In this context, it has become imperative for them to incorporate those skills which will enable the young learners to become ‘life long learners’. The ability to stay current, to upgrade skills with emerging technologies, to understand the nuances involved in change management and the relevant life skills have to be a part of the learning domains of the global learners. The CBSE-i curriculum has taken cognizance of these requirements.

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The CBSE-i aims to carry forward the basic strength of the Indian system of education while promoting critical and creative thinking skills, effective communication skills, interpersonal and collaborative skills along with information and media skills. There is an inbuilt flexibility in the curriculum, as it provides a foundation and an extension curriculum, in all subject areas to cater to the different pace of learners.

The CBSE introduced the CBSE-i curriculum in schools affiliated to CBSE at the international level in 2010 at primary and secondary level in classes I and IX and subsequently in the session 2011-12 initiated the curriculum at Class II, VI and class X. The current session will take the curriculum forward to classes III, VII and XI.

An important feature of the Senior Secondary Curriculum is its emphasis on the specialisation in different fields of study and preparing a student for higher professional life and career at the work place. The CBSE-i, keeping in mind, the demands of the present Global opportunities and challenges, is offering the new curriculum in the subject of English, Physics, Chemistry, Biology, Geography,

Accountancy, Business Studies, Information and Communication Technology, and Mathematics at two levels, Mathematics-I for the students of pure sciences and Mathematics-II for the students of Commerce and other subjects.

There are some non-evaluative components in the curriculum which would be commented upon by the teachers and the school. The objective of this part or the core of the curriculum is to scaffold the learning experiences and to relate tacit knowledge with formal knowledge. This would involve trans-disciplinary linkages that would form the core of the learning process. Perspectives, SEWA

(Social Empowerment through Work and Action), Life Skills and Research would be

the constituents of this ‘Core’. The Core skills are the most significant aspects of a learner's holistic growth and learning curve.

The International Curriculum has been designed keeping in view the foundations of the National Curricular Framework (NCF 2005) NCERT and the experience gathered by the Board over the last seven decades in imparting effective learning to millions of learners, many of whom are now global citizens.

The Board does not interpret this development as an alternative to other curricula existing at the international level, but as an exercise in providing the much needed Indian leadership for global education at the school level. The Curriculum envisages pedagogy which would involve building on learning experiences

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inside the classroom over a period of time. The Board while addressing the issues of empowerment and capacity building of teachers believes that all school must budget for and ensure teachers involved with CBSE-i are continuously updated.

I appreciate the sincere effort put in by Dr. Sadhana Parashar, Director (Training) CBSE, Dr. Srijata Das, Education Officer, CBSE and the team of Officers involved in the development and implementation of this material.

The CBSE-i website enables all stakeholders to participate in this initiative through the discussion forums provided on the portal. Any further suggestions are welcome.

Vineet Joshi

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ix

syllabus

learning outComes

At the end of this unit, students would be able to:

• know the long history of mankind’s efforts to understand planetary

motion

• differentiate between the geocentric and the heliocentric model and the

historical context under which these ideas prevailed

• know about the early astronomical observations and their significance and

contribution towards the development of the law of gravitation. • state and interpret Kepler’s laws of planetary motion

• recognize how Kepler’s laws originated from the analysis and interpretation

of Tycho Brahe’s astronomical data

• realize how measurements on planetary motion were in agreement with

Kepler’s law of ‘periods’.

• State Newton’s law of Gravitation

• understand the concept of central forces

• define G the universal gravitational constant and know about the various

experiments proposed for the measurement of G

• draw the vector diagram for the gravitational force between two masses

• understand how to compute the net gravitational force due to a collection

of masses

• recognize that gravitational force is one of the ‘basic forces of nature’ and

it is a weak force that is always attractive in nature.

• obtain Kepler’s laws of planetary motion from Newton’s law of

Gravitation

• recall the concept of acceleration and use it to obtain the expression for g

(the acceleration due to gravity) from Newton’s law of Gravitation

x

• recognize that the shape of the earth affects the value of g for different

points on its surface

• comprehend the meaning and significance of the popular statement

“Cavendish weighed the earth”

• realize that the acceleration due to the pull of the earth (i.e. ‘g’) is independent

of the mass of the object experiencing this acceleration.

• obtain the expression for variation of g with height h, (i.e. g(h)) above the

surface of earth, and understand the limitations of this expression.

• recognize why the expression for g(h) cannot be used for negative values

of h (i.e., for points below the surface of the earth)

• obtain the expression for variation of g with depth d , (i.e. g(d) below the

surface of earth

• recognize that work is done, by or against, the gravitational force, when a

mass is moved from one point to another

• recall the concept of potential energy and define Gravitational potential

energy

• obtain the general expression for the potential energy of a mass at a height

from the surface of earth and get its usual (approximate) form for small values of this height.

• derive the formula for the gravitational potential energy associated with two

masses separated by a distance

• differentiate between gravitational potential energy and gravitational

potential

• know the concept of escape speed

• obtain the expression for the escape speed from the law of conservation of

energy

• appreciate the scientific efforts to place artificial earth satellites in orbits

around the earth and know about their different uses.

• obtain the expression for the time period of an orbiting earth satellite

• list different types of artificial satellites and obtain the condition under

which an artificial satellite would become a geo-stationary satellite • know the range of applications of a geo-stationary satellite.

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teaChing notes

The discovery of the concept of gravitational force, followed by the discovery of the quantitative law that determines the magnitude of this force between any two objects is one of the most significant discoveries in the history of Physics. It is useful to convey to the students the way Newton used astronomical data related to the moon to come to the conclusion that the gravitational force between two objects must vary as inverse of the square of distance between them. The ‘inverse square’ nature of this law gives the gravitational force several interesting properties notable among which are the (i) conservative nature of this force and (ii) zero value of the gravitational field inside a hollow object. The teacher may explain that the other important’inverse square’ force in nature—the electrical force—also has similar characteristics. One can therefore convey the ‘message’ that the same physical law often help us to understand a variety of apparently diverse physical phenomenon.

It may also be explained that the ‘inverse square’ law for gravitational force, proposed by Newton, was also consistent with Kepler’s laws of planetary motion. These laws were based on the painstaking and thorough astronomical observations of the great Danish astronomer Tycho Brahe. Newton’s law of gravitation, therefore become the ‘core’ of our understanding of astronomical phenonmenon in addition to its use in understanding a variety of terrestrial phenomenon.

The teacher may express this law in vector form and use ‘principle of superpositioin’ when the gravitational force on one object is due to a group of two or more other objects. It may also be stressed that unlike the electrical or the magnetic force, the gravitational force is always attractive in nature. The direction of the force between two (point) objects has to be along the line joining the two (point) objects. This is so because in a region of space where these two (point) objects alone may be present, the only direction that can be uniquely defined is the line joining these two points. It may also be pointed out that the direction of the gravitational force on object 2 due to object 1 has to be opposite to that of the position vector of object 2 with respect to object 1. The consistency of this statement with the ‘always attractive’ nature of the gravitational force has to be clearly brought out and emphasised.

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While discussing the universal nature of the law of gravitation, it needs to be stressed that we ordinarily do not perceive the effects of this force because of the very small value of the ‘universal constant’ of gravitation ‘G’ . The teacher may explain how it was thought that we would necessarily need to use very large, huge masses to measure the (very small) effects of the weak gravitational force. These ‘intial attempts’ need to be appreciated in their historical context. It would be interesting to point out that inspite of the relatively crude nature of these measurements. The value of G, obtained through these ‘experiments’ though not very accurate, was still of the correct’order of magnitude.

While discussing the measurement of G, the teacher may point out why Cavendish’s experiment—the first ‘in-house’ or laboratory experiment—on the measurement of G, is regarded as one of the most important experiments in the history of Physics. His novel idea of measuring the (relatively) large effect of the moment of the couple due to a pair of (small) gravitational forces and his accurate and precise technique of carrying out his measurements needs to be ‘shown’ to the students.

The concept of ‘acceleration due to gravity (g) --- a characteristic ‘constant’ for the earth (and other astronomical objects)- needs to be discussed carefully. The students need to be clarified that a common value of this ‘constant’ for all objects of different masses is a consequence of the equality of the ‘inertial’ and ‘gravitational’ mass on the other hand is the ratio of the (magnitudes) of the gravitational force on the object to its resulting (gravitational) acceleration. The equality of this (gravitational) mass to the inertial mass of the object and the nature of the gravitational force (proportional to the product of the mass of the object and that of the earth) then implies that the acceleration due to the gravity for the earth, would have the same value for objects of different masses.

While discussing the variation of acceleration due to gravity with height above the earth’s surface, it needs to be pointed out that we make calculations by assuming the entire mass of the earth to be concentrated at its centre. It should also be pointed out that we can use the formula g _h/g = R

R h h R 2 2 2 1 1 ( + ) = +       in the form g_{h  g} 1 2−      h

R only when _Rh << 1. In practice, for ordinary work, h R << 1 can be taken to imply that h R  1 100 or less.

The students generally find it difficult to appreciate the decrease in ‘g’ with depth below the earth’s surface. They need to be reminded that since a hollow

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spherical shall does not exert any gravitational force on an object inside it, the effective mass of the earth, that exerts a gravitational force on an object inside its core is only the mass of that inner core of the earth for which this object is an external object. This ’effective mass’ keeps on decreasing with increasing depth. Hence there would be a decrease in the gravitational force and therefore, in the acceleration due to gravity with increasing depth. The students may be told about the difference between the quantitative formulae g_d + g1 −   _Rd_ for the decrease in ‘g’ with depth and for the decrease in ’g’ with height. A graphical representation of variation of ‘g’ with (small) heights and with depth would be an interesting exercise for the students.

While introducing the concept of gravitational potential energy, the students need to be reminded again that we can associate a definite value with this energy for a given position of the object, only because the gravitational force is a conservative force. They also need to be clarified about the fact that it is only the difference in gravitational potential energy between two points that has a unique value. The value of the gravitational potential energy at a point depends on the (somewhat arbitrary) choice of the ‘zero’ of this potential energy. If also needs to be emphasised that because of the always attractive nature of the gravitational force, the gravitational potential energy is usually assigned a negative sign the maximum value of the gravitational potential energy is therefore zero.

The concept of ‘gravitational potential’ is closely related to that of gravitational potential energy. This is because the ‘gravitational potential’ at any point, is just the gravitational potential energy of a ‘unit mass’ at that point. We can, therefore, think of the results, associated with gravitational potential energy, as being also applicable to ‘gravitational potential’.

While discussing the concept of the ‘escape speed’, the teacher must make it clear how a simple application of the law of conservation of energy enables us to calculate this speed. It is important to emphasize that the escape speed, associated with the earth (or other astronomical objects) is a characterice of the earth (or of the concerned astronomical object) and it does not depend upon the mass of the object being projected.

The concept of the satellite speed and its close relation with the radius of the satellite orbit and the ‘source’ of the gravitation force ( Earth, or some other heavenly object) needs to be clearly brought out. The concept of the ‘geo

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stationary satellite’ and the need for having specifically located ‘launching pads’ for such satellites, needs to be clearly explained .

A very interesting concept associated with that of satellites and their astronauts is that of weightlessness. The students must be made to understand how an object during its ‘freefall’ can be regarded to be in a state of weightlessness. They must be made to realise how an orbiting satellite (can the astronauts present in it) can also be regarded to be in a state of ‘free fall’ and therefore, in a state of ‘weightlessness. It does not mean that the earth is no longer exerting a gravitational force on it. Rather, the object is in a state where we are not able to see ‘effects’ of this gravitational force that we refer to as ‘weight’ . An introduction to the concept of satellites needs to be followed by an introduction to the different types of satellites and the role and application of each of these different types. A visual depiction of the process of satellite launching and the basic details of different types of satellites and the role and application of each of these different types. A visual depiction of the process of satellite launching and the basic details of different types of satellites should be of immense help and interest to all students.

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lesson Plan matrix

Content Skill learning outComeS

Early history of astronomy

• Geocentric and

heliocentric models • Ability to appreciate the works of scholars and thinkers to enhance our pool of knowledge.

• appreciate the long history of mankind’s effort to understand planetary motion • differentiate between

geocentric and

heliocentric models and the historical context under which these ideas prevailed • know how early astronomical observations were taken and how they contributed to the development of the law of gravitation.

astronomy in anciEnt india

• Contribution of Indian mathematicians and thinkers • Ability to appreciate the highly developed scientific temperament in ancient India • Appreciate the contribution of India in the development of astronomy

KEplEr’s laws of planEtary motion

• Kepler’s Laws of

Planetary Motion • Ability to state and explain Kepler’s laws • Understand the importance of and interpretation of scientific data analysis • Able to solve problems based on Kepler’s third law. • state and interpret Kepler’s laws of planetary motion • recognize how Kepler’s laws originated from the analysis and interpretation of Tycho Brahe’s astronomical data • realize how measurements on planetary motion were in agreement with Kepler’s law of periods

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nEwton’s law of Gravitation

• Newton’s Law of Gravitation • Important properties of Gravitational force • Determination of value of G • Vector Form of Newton’s law of Gravitation • Principle of Superposition • Appreciate the contribution of Newton in the formulation of the universal law of Gravitation • Application of Newton’s law in practical situations • Apply Newton’s law

to calculate the net

gravitational force due to a collection of masses. • Understand the characteristics of gravitational force. • Ability to derive the laws • state Newton’s law of Gravitation • know the concept of central forces • define G and know about the various experiments proposed for the measurement of G • comprehend the popular statement “cavendish weighed the earth” • draw the vector diagram

for the gravitational force between two masses • understand how

to compute the net

gravitational forces due to a collection of masses • recognize that gravitational force is one of the ‘basic forces of nature’ and it is a weak force that is always attractive in nature. • obtain Kepler’s laws of planetary motion from Newton’s law of Gravitation

accElEration dUE to Gravity

• Acceleration due to

Gravity • Able to apply the Newton’s second law to obtain the value of g

• Able to appreciate the fact that g is not a constant

• recall the concept of acceleration and use it to obtain the expression for g from Newton’s law of Gravitation

• recognize that the shape of the earth affects the value of g for different points on its surface

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• The learner must develop the ability to derive the expressions and also apply it to problems. • Able to distinguish between the variation in g with height and depth • meaning and significance of the comprehend the popular statement ‘’Cavendish weighed the earth’’

• realize that the

acceleration due to the gravitational pull of the earth (i.e. g) is same for all masses • derive the expression for variation of g with height h (i.e. g(h)) above the surface of earth and understand the expression limitations of this.

• recognize why the

expression for g(h) cannot be used for negative h (points below the surface) • obtain the expression for variation of g with depth d (i.e. g(d)) below the surface of earth

Gravitational potEntial EnErGy

• Gravitational field

• Gravitational Potential Energy

• Able to comprehend the need for the field picture for gravitational force • Ability to co – relate the

fact that gravitational potential energy is just one of the forms of potential energy • Skill to derive the expression for gravitational potential energy and be able to solve problems based on the topic. • recall the concept of potential energy and define Gravitational potential energy • obtain the general

the expression for the potential energy of a mass at a height from the surface of earth and obtain its usual approximate form. • derive the formula for

the gravitational potential energy associated with two masses separated by a distance

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• Gravitational Potential

• Extrapolating the result derived above to a multi particle system. • Distinguish between gravitational potential and potential energy and the need for the concept of gravitational potential. • differentiate between gravitational potential energy and gravitational potential EscapE spEEd • Escape Speed • Ability to recall the concept of escape speed • Derive the expression and apply it to problems • know the concept of escape Speed • obtain the expression for escape speed from the law of conservation of energy satEllitEs • Motion of satellite (Orbital speed, Time period of a satellite) • Able to derive the expression and solve problems based on the topic • appreciate the scientific efforts to launch artificial earth satellites and list their fields of use • obtain the expression

for the time period of an orbiting earth satellite • list different types of artificial satellites and obtain the condition for ‘a geo-stationary satellite’ • know the range of

applications of a geo-stationary satellite

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Pre-requisites

web-links/vedios/other referenCes

Recall the following concepts already learnt in previous classes

• Newton's law of gravitation

• Equations of motion under free fall. • Concept of potantial energy.

• Circular motion

http://www.drennon.org/science/kepler.htm

http://www.physicsclassroom.com/class/circles/u6l3c.cfm http://www.physicsclassroom.com

students'

manual

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Unit 6 : Gravitation

6.1 IntroductIon

Long time back, even before man had learnt to read and write, he must have observed the rising and setting of the sun, the phases of the moon and the shining stars at night. He must have wondered about the cause of these phenomena. The in-built curiosity in the mind of man must have led to centuries of thoughts and ideas. The subject of Gravitation, in the form we study it today, is one of the consequences of these thoughts and ideas.

Did Newton discover ‘Gravitation’? The answer is an emphatic NO!

The force of Gravitation had always existed and its importance and significance was realized by various philosophers and thinkers much before Newton. However, the credit for formulating a, quantitative law, which governs the nature of the force of gravitation,

must go to Newton.

Early History of Astronomy

Earlier astronomy was based on the geocentric model in which the Earth was regarded as the center of the universe, and all other heavenly objects were supposed to orbit around it. This geocentric model was the predominant cosmological system in many ancient civilizations such

Figure 1. Geocentric system EARTH MOON VENUS SUN SATURN Mercury Mars Jupiter

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as ancient Greece.

Two observations that must have made a significant contribution to the formulation of the geocentric model were (i) the stars, sun, and planets appear to revolve around the Earth each day, making the Earth appear to be the center of the overall system. (ii) The Earth does not seem to move from the perspective of an Earth bound observer, for whom it appears as a solid, stable, and unmoving object.

It was not until the 16th _{century that a new astronomical model called the}

Heliocentric model was developed. In this model the Earth and planets were

imagined to revolve around a stationary Sun which was considered as the centre of the universe. This word comes from the Greek Helios (sun) and kentron (center). This notion, that the Earth revolves around the Sun, had been proposed as early as the 3rd century BCE by Aristarchus of Samos, but his idea received no support from the other ancient astronomers of that time.

Figure 2. Heliocentric system EARTH MOON VENUS SUN SATURN MERCURY JUPITER MARS

A fully predictive mathematical model of a heliocentric system was presented, by the Renaissance mathematician, astronomer, and Catholic cleric, Nicolaus Copernicus of Poland, leading to the ‘Copernican Revolution’. In the following century, this model was elaborated and expanded by Johannes Kepler and

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observations, supporting it, made by using a telescope, were presented by Galileo Galilei.

With the observations of William Herschel, Bessel and others, astronomers soon realized that the sun was not the center of the universe. By the 1920s, Edwin Hubble had shown that it was just a part of a galaxy (called the Milky Way galaxy) that had billions of stars. The Milky Way galaxy, in turn, was just one of the billions of such galaxies in the universe. The universe was, overall, very very vast indeed!

6.2 Astronomy In AncIEnt IndIA

Long before Kepler formulated his laws of planetary motion, the study of planetary motion was a well developed branch of Science in India.

Aryabhata (476–550?), in his magnum opus Aryabhatiya (499?) propounded a planetary model in which the Earth was taken to be spinning on its axis and the periods of the planets were given with respect to the Sun. He accurately calculated many astronomical constants, such as the periods of the planets, likely times of the occurrences of the solar and lunar eclipses, and the instantaneous motion of the Moon. Early followers of Aryabhata’s model included Varahamihira, Brahmagupta, and Bhaskara II.

Nilakantha Somayaji (1444–1544?), in his Aryabhatiyabhasya, a commentary on Aryabhata’s Aryabhatiya, developed a computational system for a partially heliocentric planetary model, in which the planets orbit the Sun, which in turn orbits the Earth, similar to the Tychonic system later proposed by Tycho Brahe in the late 16th century. In the Tantrasangraha (1500?), he further revised his planetary system, which was mathematically more accurate at predicting the heliocentric orbits of the interior planets than both the Tychonic and Copernican models. However, the Indian astronomy, in general, fell short of proposing models of the universe. Nilakantha’s planetary system also incorporated the Earth’s rotation on its axis. Most astronomers of the Kerala school of astronomy and mathematics seem to have accepted his planetary model.

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6.3 KEplEr’s lAws of plAnEtAry motIon

The motion of the planets, as they revolve around the Sun, and rise and set on the horizon, must have been a puzzle to the mankind. Johannes Kepler (1571 – 1630), after studying the motion of planets for a lifetime, worked out the empirical laws governing planetary motion. Later Newton proved that Kepler’s laws can be derived from Newton’s law of gravitation.

Kepler’s laws were deduced for planets revolving around the Sun. But, do you realize that these laws are equally valid for artificial satellites, and also for any other body revolving around a massive central body. What do you think is the reason?

Kepler’s first law — law of orbits

All planets revolve around the Sun in elliptical orbits with the Sun at one of the foci of the ellipse.

The adjacent figure shows a planet of mass m revolving around the sun having a mass M. (M >> m)

The orbit is characterized by two parameters (i) Semi major axis (ii) eccentricity (e) For circular orbits, eccentricity is zero and semi major axis is equal to the radius of the orbit. For most of the planets the eccentricity of their orbit is so small that we can assume them to be moving in circular orbits.

For the sake of simplicity, we, therefore, generally assume all orbits to be circular unless mentioned otherwise.

Figure 3: Kepler

Figure 4: Diagram showing elliptical path of a planet with Sun at one of the foci.

A B

focl

Planet Sun

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Kepler’s second law — law of Areas

The line joining a planet to the Sun sweeps out equal areas in equal interval of time in the orbital plane of the planet. i.e. the rate

dA

dt at which it sweeps out area A (Areal

velocity) is a constant. Kepler’s second law is the most important of the three laws. The area swept by the planet dA in time dt can be expressed as dA = 1 2r2 dq (Area of triangle = 1 2base × height) where r is the distance between Sun and planet, dq

is the angle swept by the (radial line of the) planet in time dt, as shown in the accompanying figure. hence dA_dt = 1 2 q=1 2ω 2 2 d dt r r where ω = angular speed of the planet The magnitude of the angular momentum of the planet, about an axis passing through the Sun, is L = rp_n

where p_n is the component of its linear momentum →p along the direction normal to r L = r(mv) = r (mωr) = mr2_ω From these equations, we get dA dt = 2 L m Since, the gravitational force is a central force, it acts along the line joining the

Figure 5: Diagram showing area covered by a planet at two different positions on its orbit.

dt A1 A2 A1=A2 planet sun dt Figure 6: dq rdq r

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Sun to planet i.e. parallel to →r . Hence the torque, exerted by the gravitational force, on the planet, is zero.

In the absence of torque, the angular momentum →L remains conserved in accordance with the law of conservation of angular momentum. Hence the areal velocity

dA

dt = a constant

One simple consequence of Kepler’s second law is that a planet moves faster when it is closer to Sun and slower when it is far from it. Can you think of an explanation?

Kepler’s third law — law of periods

The square of the time period of revolution of a planet around the Sun, is directly proportional to the cube of the semi – major axis of the orbit. (radius of the orbit when the orbit can be assumed to be circular.) Thus T2 _{∝ r}3 where T = time period of revolution of the planet and r = radius of its orbit. Alternately, we can also write this law as 2 1 2 2 T T = 3 1 3 2 r r For an interactive animation visit: http://www.drennon.org/science/kepler.htm

ConCept probe

1. Consult a table of planetary data. Calculate T2_/r3 _{for each planet. Verify that this quantity} is almost constant for all planets.

2. You are given that the period of rotation of the moon around the earth is approximately 30 days and its distance from the earth is approximately 64 earth radii. Can you calculate the height of geosynchronous satellite?

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Brahe was the master of quantitative observations but was deficient in their theoretical interpretation. On the other hand, Johannes Kepler (1571-1630), a German, who went to Prague to become Brahe’s assistant, had a strong theoretical intuition. Kepler and Brahe did not get along well. Brahe apparently mistrusted Kepler; He thus allowed only limited access to Kepler to his voluminous data. He set Kepler the task of understanding the orbit of the planet Mars, which was particularly troublesome. It is

believed that part of the motivation for giving the Mars problem to Kepler was that it was difficult, and Brahe hoped it would keep Kepler occupied while Brahe worked on his theory of the Solar System. In a supreme irony, it was precisely the Martian data that allowed Kepler to formulate the correct laws of planetary motion, thus eventually achieving a place in the development of astronomy that far surpassed that of Brahe.

It was left to to Kepler to provide an answer to the final piece of the puzzle. After a long struggle, in which he tried mightily to avoid his eventual conclusion, Kepler was finally forced to come to the conclusion that the orbits of the planets were not the circles (demanded by Aristotle and assumed implicitly by Copernicus), but were instead the “flattened circles” that geometers call ellipses.

The irony, noted above, lies in the realization that the difficulties with the Martian orbit originate precisely from the fact that the orbit of Mars was the most elliptical of all the planets for which Brahe had extensive data. Thus, Brahe had unwittingly given Kepler the very part of his data that allowed Kepler to eventually formulate the theory of the Solar System that surpassed Brahe’s

own theory!

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5.4 nEwton’s lAw of GrAvItAtIon

We all are familiar with the famous legend of Newton observing the free fall of an apple and formulating the law of gravitation. However, it will be appropriate to mention that Newton gave the law of gravitation after almost 20 years of his first thought about it which is quite often associated with the episode of a falling apple. Newton’s law for gravitational force between two

bodies is one of the most far – reaching laws in the history of human scientific endeavor.

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the two bodies.

If must be noted that the law as stated above applies to point bodies.

For extended bodies, the distance must be taken between their centres of mass (geometrical centres if bodies are regular and homogeneous) and the direction of the force is along the line joining their centres of mass. F ∝ m₁m₂ F ∝ 1₂ r F ∝ 1 2 2 m m r

where m₁ and m₂ are the masses of the two bodies and r is the distance between their centers.

Figure 8: Sir Issac Newton

m₁ m₂

r

Figure 9: Two masses m₁ and m₂ separated by distance r.

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Note that as the masses of the bodies increase, the force between them increases. Also, as the distance between their centres increases, the force decreases.

6.5 ImportAnt cHArActErIstIcs of GrAvItAtIonAl

forcE

The force of gravitation is associated with the following characteristics:

(a) the gravitational force is a central force. It acts along the line joining the centres of two bodies.

(b) it is a conservative force. This means that the work done by the gravitational force in displacing a body from one point to another is only dependent on the initial and final positions of the body and is independent of the path followed.

(c) it is a long range force. The gravitational force is effective even at large distances.

(d) Unlike electrostatic and magnetic forces, the gravitational force is always

attractive.

The entire Universe is held together by the gravitational force. It is, therefore, the most important force in nature. However, it is the weakest of all the

fundamental forces in nature.

Note that earlier mass was regarded only as a measure of the inertia of the body. This led to the concept of inertial mass. Newton’s law of gravitation, however, bestows mass with another property. Mass is also a measure of the gravitational force. This leads us to the concept of the gravitational mass of an object. However, all the precise experiments done so far, show that both inertial and gravitational masses are equal.

1

2

3

Figure 10: Work done by gravitation force along path 1 is the same as that along path 2 or path 3.

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ConCept probe

1. What happens to the gravitational force between two bodies if the mass of one of the bodies is halved and the distance between them is doubled?

2. The gravitational force acts on all bodies. Why, then does an apple fall towards earth but the earth does not move towards the apple?

The formulation of Newton’s law of gravitation is a story of human determination and quest for scientific enquiry. In the first instance, Newton‘s calculations did not work and he put aside his papers, in a drawer, for almost 20 years. It was during the advent of a comet in 1680, and at the prodding of Sir Edmund Halley, his friend, that he again worked on his calculations and, subsequently, obtained excellent results.

For an interesting explanation, visit the link:

http://www.physicsclassroom.com/class/circles/u6l3c.cfm

6.6 The Universal GraviTaTional ConsTanT(G)

The proportionality sign in the Newton’s law of gravitation can be eliminated by putting a constant of proportionality, denoted by G. The equation then becomes

F = G 1 2 2

m m r

here G is called the gravitational constant. It is a universal constant because the gravitational force between two bodies placed at a certain distance remains the same,

wherever these bodies may be placed in the universe. The constant G is also independent of the medium in which the interacting bodies are placed.

The value of G, in SI units, was later on, found to be equal to 6.67 × 10–11 _nm2_/kg2_.

The units of G are obtained by using the fact that the force has to be expressed in newtons.

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The numerical value of G would change with a change in the units of measurement for mass and distance.

The very small value of G points to the fact that the gravitational force is an extremely weak force. In fact, it is the weakest of all the fundamental forces (Can you find out, which are the other fundamental forces?) and it becomes important only when the masses of the bodies involved are very large. That is why the gravitational force plays such an important role in the case of heavenly bodies.

http://www.physicsclassroom.com

ConCept probe

1. If there is a gravitational attractive force between all objects, why do we not feel ourselves attracted towards massive structures in our surroundings?

2. Why is the gravitational force an important force for heavenly (or astronomical) objects?

The value of G was first measured by an English Physicist, Henry Cavendish in the eighteenth century. He achieved this by measuring the small force between lead masses with an extremely sensitive torsion balance. A better method was later developed by Philipp von Jolly.

As long as the sizes of the objects are small compared to the distance between them they can be treated as point objects which simplifies considerably the mathematics of their gravitational interaction. The Sun and Saturn are far enough (in comparison to their sizes) for them to be treated as point particles. If the distance between two objects is very large (in comparison to their sizes), we can take them as point objects. What about the case of bodies on the earth? For such bodies, the earth does not seem like a point object. The answer to this dilemma lies in Newton’s shell theorem:

A uniform spherical shell of matter attracts a body that is outside the shell as if all the mass of the shell were concentrated at its center.

According to this theorem, Earth can be regarded as a ‘point mass’, located at the center of Earth and with mass equal to that of Earth. We usually follow

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this theorem in all our calculations.

5.7 vEctor form of nEwton’s lAw of GrAvItAtIon

Since force is a vector quantity, it must be expressed in a vector form. The gravitational force can also be expressed in a vector form by attaching a unit vector to the expression for this force.

By convention, the direction of unit vector is

always taken as directed from the body experiencing the force (body 1) towards the body exerting the force (body 2). Therefore, F  = G 1 2 2 ˆ m m r r If we are calculating the force on body A due to body B, thenˆr will be the unit vector drawn from A towards B. It will be imperative to mention here that this vector notation is consistent with the basic fact that the gravitational force is always an attractive force. This implies that the gravitational force is a central force, and hence the direction of this force has to be along the line joining the center of the two bodies.

5.8 prIncIplE of supErposItIon

Newton‘s law has been stated for two point bodies. How do we calculate the force on a body if there are more than two bodies interacting with one other?

Figure 12: Diagram showing direction of the gravitational force acting along the

line joining the center of two bodies.

1 A 2 B ^ r

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The solution to this situation lies in what is called the principle of superposition. In a group of objects, the net gravitational force, on any one of the objects, is the vector sum of the forces due to all the other objects. The principle implies that we first calculate the gravitational force that acts on an object due to each of the other objects as if all other objects are absent. After doing this for all possible pairs, the net force on the object under consideration is calculated by the vector sum of all the forces acting on it.

→ 1

F = →F12 +→F13+→F14+→F15+...

here →F1 is the net force on object 1 due to all the other objects 2, 3, 4, 5, …

The Principle of superposition is based on the fact that the gravitational interaction between two bodies is independent of the presence of other bodies in the neighborhood.

The same concept is applicable to electrostatic force which will be studied in class XII.

IllustratIon. three objects of masses 5 kg, 3 kg and 3 kg are placed at the corners of an equilateral triangle of side 20 cm. Calculate the net gravitational force on the object of 5 kg.

solutIon. The magnitude of the force on object A due to object B is F_AB = Gm mA B₂ r It is in the direction of AB Putting the values we get, F_AB  5 × 10–9 _N

Similarly, the magnitude of the force on A due to body C is

F_ac  5 × 10–9 _N It is the direction of AC.

As per the principle of superposition, the net force on the body A is the vector sum of forces →FAB and

→ AC

F . Applying the laws of vector addition, A

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we get the magnitude of the resultant as F_a = 2 + 2 +_{2 cos60}° AB AC AB AC F F F F F_a = × − + × − + × −  _{ }   9 2 9 2 9 2 1 2 (5 10 N) (5 10 N) 2(5 10 N) = 5 3 × 10–9 _N The resultant force is directed along the bisector of the angle between the two forces →FAB and → AC F .

Four equal masses are placed at the corners of a square of side 2cm as shown in the figure. Another mass is placed at the centre of the square. Find the magnitude and direction of the net force on the body kept at the center of the square due to all the other masses.

practice problem

?

Did You Know?

There are two high tides per day in oceans due to the gravitational pull of the moon on the earth. Surprisingly, although the gravitational pull of Sun is 180 times greater than the pull of the Moon, the effect of Sun is much less on ocean tides. Search for the possible reason for this?

derivation of Kepler’s law of time periods from newton’s law of

gravitation

For simplicity let us assume that the orbit of the planet is circular.

The gravitational force exerted by the sun on the planets provides the necessary centripetal force for the planet to remain in its orbit. So, applying the force equation,

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we get M vp 2 r = 2 s p GM M r putting v = 2 rp T p2 2 4 p M r T = 2 s p GM M r \ T2 ₌ _ p _     2 4 s GM r3

The quantity in the parentheses is a constant that depends only on the mass

M_s of the central massive body which is the Sun in this case. It can also be the Earth if we are talking of the motion of an artificial satellite.

\ T2 _{∝ r}3

This, as you have learnt above, is Kepler’s third law for planetary motion.

6.9 AccElErAtIon duE to GrAvIty

We assume that the Earth is a uniform sphere of mass M and radius R. We can assume then that the mass of earth is concentrated at its center.

The gravitational force of Earth, on a particle of mass m, at a distance r from the center of Earth, is given by Newton’s Law as follows:

F = GMm₂

r …(i)

where M is the mass of the earth.

This force gives rise to an acceleration in the particle which is called the

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Applying Newton’s second law, force on a body can be expressed as

F = ma …(ii)

Both the expressions (i) and (ii) are measuring the same force in two different ways. Hence, from the two equations (i) and (ii), we get ma = GMm₂ r \ a = GM₂ r

We use the special symbol ‘g’ for this acceleration and call it as the acceleration due to gravity. Thus, the acceleration due to gravity (g) is the acceleration produced on a body, on, or near, the earth’s surface, due to the gravitational pull of earth. We have g = GM₂ R

Aristotle taught that heavy objects fall faster than light objects. Galileo argued that all objects, irrespective of their mass, should take the same time to fall to the ground from a given height. Do you agree with Galileo? If so, why?

Search the internet for Galileo’s Pisa experiment. It is said that Galileo went to the top of the tower of Pisa and dropped bodies of various masses and showed that they all take the same time to fall to the ground.

For points lying very close to the surface of earth, we put r = R, the radius of

the earth. We can, therefore, express acceleration due to gravity (g) as

g = GM₂

R

For earth, taking M = 6 × 1024 _{kg and R = 6400 km, we get g = 9.8 m/s}2_{. However,}

it must be noted that the value of 9.8 m/s2 _{is an average value and the value}

of g varies on earth from one place to another because of various factors. On a given planet, the average value of acceleration due to gravity is the same for all objects and is independent of the mass of the object. In fact, it is

r m

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a characteristic constant for that planet as it depends only on the mass and radius of that planet.

IllustratIon. Find the value of g on a planet whose mass is half that of earth and

radius twice that of earth. given g on earth = 9.8 m/s2_.

soltuIon. In any such question, the best way is to write two equations: Value of g on earth g_e = ₂e e GM R Value of g on planet g_p = ₂p p GM R Inserting given values, p e g g = 2 1 1 1 2 4 8 p _e e p M _R M R     = × =     We get g_p = 1 8 × 9.8 ms–2 = 1.225 ms–2 ≈ 1.2 ms–2

At the time when Cavendish determined the value of G, there was lot of excitement the world over. In fact, Cavendish’s determination of G was publicized by the popular statement “Cavendish weighed the earth”. Can you explain how knowing the value of G, it is possible to calculate the mass of the Earth?

variation in the value of g

The value 9.8 m/s2 _{for the acceleration due to gravity on Earth is the average}

value and is not the same at all places on the earth. It varies due to the following reasons:

(i) Density of earth is not uniform at all places (ii) Earth is not a perfect sphere

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Can you explain how variation of gravity can be caused by variation of density in the earth’s interior? In fact, variation of gravity inside the earth is a tool for exploration of mineral and other deposits inside the earth. Search for ‘gravity measurement and exploration of mineral deposits’ on the internet for more information on this topic.

It is easy to realize that ‘g’ would also vary as we go up above the surface of earth and as we go deep down inside the earth.

Let us now study the variation in the value of g in some detail.

6.9.1 due to the shape of the earth

The earth is not a perfect sphere. In fact, the shape of earth is more like an oblate spheroid. It is bulging out at the equators and a little compressed at the poles.

Thus, since

R

_e

> R

_p

,

we would have

g

_e

< g

_p

?

Did You Know?

A person weighs (a little) more at the poles than at the equator.

6.9.2 with height (h)

The value of g at a point P on the surface of earth

g = GM₂

R

Figure 13: Picture showing difference in the radius of Earth at equator and the poles.

North Pole

South Pole

Polar diameter 12,714 km

Equatorial diameter 12,756 km

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If we go up above the surface of earth, the value of g at a point P′ at a height ‘h’ from the surface of earth, would be g′ = + 2 ( ) GM R h \ g′ g = 2 2 ( ) R R h+ …(i) We can also write the formula as

This formula is valid for all values of ‘h’. However, it is usually used only when h is comparable to R. ′ g g = 2 R 2 R _ + _   2 1 h R g′ = _ _−  +  2 1 h R g

For h << R, the term _ _−

 + 

2

1 h

R can be expanded by using binomial theorem.

Neglecting higher order terms, we can write g′ =   1 2−  h R g

This formula is valid only when the value of h is small compared to R, the radius of earth. If this condition is not satisfied, we use equation (i) above.

IllustratIon 3. at what height above the surface of the earth, will the acceleration due to gravity be 36% of its value on the surface of earth? Radius of earth = 6400km.

solutIon. note the fact that the change in g is of about 64%. Such a large change will not be possible at a small height. Hence it will be incorrect to use the formula valid only for small heights.

Figure 14: Diagram showing point P on earth and P’ at a height h from the surface of earth.

P′

P R

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We use instead the formula ′ g g = + 2 2 ( ) R R h Putting g′ = 0.36 g, and taking square root of both sides, we get h = 2 3 R

IllustratIon 4. Find the value of g at a height of 32 km above the surface of

earth. solutIon. In this question, the height at which g is to be calculated is small compared to R. Thus, we can use the formula valid for h < < R. g′ ₌ _ _ 1 2−  h R g Putting values, we get g′ = _ _= _ _  −    32 6336 6400 6400 1 2 g g \ g′ = 99

100g = 99% of value of g on the surface of earth Let us see what happens when the student uses the incorrect formula in a

given case

IllustratIon 5. Find the value of g at a height equal to the radius of earth.

solutIon. Note that in this question, the value of height is not small compared to

R; in fact it is equal to the radius of earth. Ignoring this fact, if a student

uses the formula

g′ =   1 2−  h R g which is actually valid only for small heights, the student gets

g′ = g_1 2− R_R_

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The answer is completely wrong because a an invalid formula has been used. The right approach is to use the formula ′ g g = + 2 2 ( ) R R h

Putting the value of h = R, we get ′ g g = + 2 2 ( ) R R R ′ g g = 2 2 4 R R g′ = 4 g

At what height above the surface of the earth, will the value of g be 5% of its value on the surface of earth? Given R = 6400 km.

try this

6.9.3 with depth

The value of g also varies as we go deep inside the earth. It will be easier to derive the result, if we convert the expression for g in terms of the density, rather than the mass, of the earth.

We know,

g = GM₂

R

Assuming earth to be a sphere of uniform mass density ρ, we can write M = 4₃pR3r

substituting in the formula for g, we get

g = 4

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In terms of the formula derived above the value of g at a depth d can be expressed as

g′ = 4₃pG(R – d)r

This is because, the point, at a depth d below the surface, is a ‘surface point’ on a sphere of radius (R – d) and it is only ‘this sphere’ that contributes to the gravitational force for a particle located at this depth. Thus, ′ g g =     1−  d R or g′ = g_1−_Rd_

Did you note that at the center of earth, i.e for d = R, the acceleration due to gravity becomes zero.

ConCept probe

1. A body of weight mg is taken to the centre of the earth. What would be its mass there?

2. Draw a graph showing variation of the value of g with distance from the centre of earth.

For a description of how gravity actually varies as we go deep into the earth

http://www.schoolscienceguru.com

IllustratIon 6. at what depth, below the surface of the earth, the value of g is same as that at a height of 64 km above the surface of earth?

solutIon. Value of g, at a height of 64 km is given by g′ = g 1 2h R      −  Value of g at a depth d is given by g′ = g 1 2d R      − 

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Since the question says the value of g at height = value of g at a depth d, we have 1 2_Rh g   −  = 1 d R g   −  We get 2h = d or d = 128 km

?

Did You Know?

Variation of ‘g’ on the surface of the Earth

It is well know that the acceleration due to gravity (g) for the earth, does not have the same value for all points on its surface. ‘g’ has its maximum value of the ‘poles’ and its minimum value at the ‘equator’. How can we understand this variation in ‘g’?

A simple explanation for this variation can be given by saying that the earth is not a ‘perfect sphere’ and its ‘polar radius’ is (slightly) less than its ‘equatorial radius’. A particle, on the poles, would, therefore, be (slightly) nearer to the centre of the earth than, when it is at the equator. If would, therefore, experience a (slightly) greater gravitational force due to the earth at the poles. This implies a (slightly) higher value for ‘g’ at the poles compared to its value at the equator.

The difference between the equatorial and the polar radii, however, cannot alone explain the observed difference between the values of ‘g’ at the two places. There has to be ‘another cause’ for the ‘observed difference’.

The ‘other cause’ is associated with the rotation of the earth around its axis. A particle P, at a latitude l, moves in a circular path (of radius r = R cos l) around this axis.

It would, therefore, need a centripetal force (= mrω2_{) to}

‘stay in its circular path’. Where from does it get this required ‘centripetal force’. The earth exerts a gravitational force GMm₂

R mg

 

 

= =  on this particle (of mass m) and this

force is directed radially in wards along the direction PO. The particle needs a centripetal force (= mrω2 _{= mω}2_{R cos l) directed along PN. We can now say that a part of the earth’s} gravitational force, on the particle, gets ‘used up’ in providing this centripetal force. The effective gravitational force, experienced by the particle, is, therefore, the difference of a force F₁ GMm₂

R mg

 

 

= =  directed along PO and a force F2 (= mω

2_{R cos l) directed along} PN. N O R l Equator P R cos l

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Simple calculations, based on the parallelogram law of vector addition, lead us to the formula mg_l = m (g2 _{+ R}2w4 _{– 2gRω}2 _cos2 _l)1/2 g_l = g 1/2 2 2 4 2 2 2 1 R cos R g _g  _{ω ω}       − l + 

Since Rω2 _{is a small quantity, we can neglect R}2w4 _{and higher terms. Then}

g_{l = g} 1 R 2 cos2 g  _ω  l      − 

The effective value of g, i.e. g_l is, therefore, dependent on l, the latitude of the place. ‘g_l’, clearly, has its minimum value at the equator where l = 0. As the poles (l = 90°), g_l = g and this is the maximum value of g.

It is worth mentioning here that this ‘cause’—the rotation of the earth about its own axis—is the main cause for the observed difference between the values of ‘g’ at the poles and at the equator.

6.10 GrAvItAtIonAl fIEld

There are some forces in nature which can act even when the interacting bodies are not in direct contact with each other. Such forces are called action-at-a-distance forces. The gravitational, electrostatic and magnetic forces are well know examples of such forces. On the other hand, there are forces which act only when bodies are in direct contact. Frictional, viscous and muscular forces are some examples of such ‘contact forces’

The ‘action-at-a-distance’, or non-contact forces, forces which can act even when the bodies are not in direct contact, are usually discussed on the basis of the concept of their associated ‘field’. For the gravitational force, we use the concept of the ‘gravitational field’.

Body (source)

The gravitational field, of a body is defined as the space around this body in which its gravitational force can be experienced by other bodies.