Gravitation

CBSE-i

CLASS

XIUNIT-6

CENTRAL BOARD OF SECONDARY EDUCATIONShiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India

PHYSICS

GRAVITATION

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 inAncient India 4

• Kepler’sLawsofPlanetaryMotion 5

• Newton’sLawofGravitation 9

• ImportantCharacteristicsofGravitational force 10

• TheUniversalGravitationalConstant (G) 11

• VectorFormofNewton’sLawofGravitation 13

• PrincipleofSuperposition 13

• AccelerationDue toGravity (G) 16

• Variation inGDue toShape 19

• Variation inGwithHeight 19

• Variation inGwithDepth 22

• GravitationalField 25

• GravitationalPotentialEnergy 27

• GravitationalPotential (V) 32

• EscapeSpeed (Ve) 33

• MotionofSatellite 35

Post content studentworksheet 1 40





v

PrefaCe

TheCurriculuminitiatedbyCentralBoardofSecondaryEducation-International(CBSE-i)isaprogressivestepinmakingtheeducationalcontentandmethodologymore sensitive and responsive to global needs. It signifies the emergence ofa fresh thought process in imparting a curriculum which would restore theindependenceofthelearnertopursuethelearningprocessinharmonywiththeexistingpersonal, social andcultural ethos.

The Central Board of Secondary Education has been providing support to theacademicneedsof the learnersworldwide. Ithasabout11500schoolsaffiliatedto it and over 158 schools situated in more than 23 countries. The Board hasalwaysbeenconsciousofthevaryingneedsofthelearnersandhasbeenworkingtowardscontextualizingcertainelementsofthelearningprocesstothephysical,geographical, social and cultural environment inwhich they are engaged. TheCBSE-ihasbeenvisualizedanddevelopedwith these requirements inview.

Thenucleusof the entireprocessof constructing the curricular structure is thelearner. The objective of the curriculum is to nurture the independence of thelearner,giventhefactthateverylearnerisunique.Thelearnerhastounderstand,appreciate, protect and build on knowledge, values, beliefs and traditionalwisdom.Teachersneedtofacilitatetheleanertomakethenecessarymodifications,improvisationsandadditionswhereverandwhenevernecessary.

Therecentscientificandtechnologicaladvanceshavethrownopenthegatewaysof knowledge at an astonishing pace. The speed and methods of assimilatingknowledge have put forth many challenges to the educators, forcing them torethink their approaches for knowledge processing by their learners. In thiscontext, it has become imperative for them to incorporate those skills whichwillenable theyounglearners tobecome‘life longlearners’.Theability tostaycurrent,toupgradeskillswithemergingtechnologies,tounderstandthenuancesinvolvedinchangemanagementandtherelevantlifeskillshavetobeapartofthe learning domains of the global learners. The CBSE-i curriculum has taken cognizanceof these requirements.

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The CBSE-i aims to carry forward the basic strength of the Indian systemof education while promoting critical and creative thinking skills, effectivecommunicationskills,interpersonalandcollaborativeskillsalongwithinformationandmediaskills.There isan inbuiltflexibility in thecurriculum,as itprovidesa foundation and an extension curriculum, in all subject areas to cater to thedifferentpaceof learners.

The CBSE introduced the CBSE-i curriculum in schools affiliated to CBSE atthe international level in 2010 at primary and secondary level in classes I andIXand subsequently in the session2011-12 initiated the curriculumatClass II,VIandclassX.Thecurrent sessionwill take thecurriculumforward to classesIII,VII andXI.

An important feature of the Senior Secondary Curriculum is its emphasis onthespecialisationindifferentfieldsofstudyandpreparingastudentforhigherprofessional life and career at the work place. The CBSE-i, keeping in mind,thedemandsof thepresentGlobalopportunitiesandchallenges, isofferingthenew curriculum in the subject ofEnglish, Physics, Chemistry, Biology, Geography, Accountancy, Business Studies, Information and Communication Technology, and Mathematics at two levels, Mathematics-I for the students of pure sciences andMathematics-II for the studentsofCommerceandother subjects.

There are somenon-evaluative components in the curriculumwhichwouldbecommented upon by the teachers and the school. The objective of this part orthe core of the curriculum is to scaffold the learning experiences and to relatetacit knowledgewith formalknowledge.Thiswould involve trans-disciplinarylinkages that would form the core of the learning process.Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Researchwould bethe constituents of this ‘Core’. The Core skills are themost significant aspectsof a learner'sholisticgrowthand learning curve.

TheInternationalCurriculumhasbeendesignedkeepinginviewthefoundationsof theNational Curricular Framework (NCF 2005)NCERT and the experiencegatheredbytheBoardoverthelastsevendecadesinimpartingeffectivelearningtomillionsof learners,manyofwhomarenowglobal citizens.

TheBoarddoesnotinterpretthisdevelopmentasanalternativetoothercurriculaexisting at the international level, but as an exercise in providing the muchneededIndianleadershipforglobaleducationattheschoollevel.TheCurriculumenvisages pedagogy which would involve building on learning experiences

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inside the classroom over a period of time. The Board while addressing theissuesofempowermentandcapacitybuildingofteachersbelievesthatallschoolmust budget for and ensure teachers involved with CBSE-i are continuouslyupdated.

IappreciatethesincereeffortputinbyDr.SadhanaParashar,Director(Training)CBSE,Dr.SrijataDas,EducationOfficer,CBSEandtheteamofOfficersinvolvedin thedevelopmentand implementationof thismaterial.

The CBSE-i website enables all stakeholders to participate in this initiativethroughthediscussionforumsprovidedontheportal.Anyfurthersuggestionsarewelcome.

Vineet Joshi Chairman,CBSE

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aCknowledgements

ix

syllabus

learning outComes

At theendof thisunit, studentswouldbeable to:

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

• differentiate between the geocentric and the heliocentric model and thehistorical contextunderwhich these ideasprevailed

• know about the early astronomical observations and their significance andcontribution towards thedevelopmentof the lawofgravitation.

• state and interpret Kepler’s lawsofplanetarymotion

• recognize howKepler’slawsoriginatedfromtheanalysisandinterpretationofTychoBrahe’s astronomicaldata

• realize how measurements on planetary motion were in agreement withKepler’s lawof ‘periods’.

• State Newton’s lawofGravitation

• understand the conceptof central forces

• define G the universal gravitational constant and know about the variousexperimentsproposed for themeasurementofG

• draw thevectordiagram for thegravitational forcebetween twomasses

• understand how to compute the net gravitational force due to a collectionof masses

• recognize that gravitational force is one of the ‘basic forces of nature’ andit is aweak force that is alwaysattractive innature.

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

• recall the concept of acceleration anduse it toobtain the expression for g(theaccelerationdue togravity) fromNewton’s lawofGravitation

teaChers' manual

x

• recognize that the shape of the earth affects the value of g for differentpointson its surface

• comprehend the meaning and significance of the popular statement“Cavendishweighed theearth”

• realize thattheaccelerationduetothepulloftheearth(i.e.‘g’)isindependentof themassof theobject experiencing this acceleration.

• obtain the expression for variationof gwithheighth, (i.e. g(h)) above thesurfaceof earth, andunderstand the limitationsof this expression.

• recognize why the expression for g(h) cannot be used for negative valuesofh (i.e., forpointsbelow the surfaceof theearth)

• obtain the expression for variation of gwithdepthd , (i.e. g(d) below thesurface of earth

• recognize thatwork isdone, by or against, the gravitational force,when amass ismoved fromonepoint toanother

• recall the concept of potential energy and define Gravitational potentialenergy

• obtain thegeneralexpressionforthepotentialenergyofamassataheightfrom the surface of earth and get its usual (approximate) form for smallvaluesof thisheight.

• derive theformulaforthegravitationalpotentialenergyassociatedwithtwomasses separatedbyadistance

• differentiate between gravitational potential energy and gravitationalpotential

• know the conceptof escape speed

• obtain theexpression for theescapespeed fromthe lawofconservationofenergy

• appreciate the scientific efforts to place artificial earth satellites in orbitsaround theearthandknowabout theirdifferentuses.

• obtain theexpression for the timeperiodof anorbitingearth satellite

• list different types of artificial satellites and obtain the condition under whichanartificial satellitewouldbecomeageo-stationary satellite

• know the rangeof applicationsof ageo-stationary satellite.

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teaChing notes

Thediscoveryoftheconceptofgravitationalforce,followedbythediscoveryofthe quantitative law that determines themagnitude of this force between anytwoobjects isoneof themost significantdiscoveries in thehistoryofPhysics.It is useful to convey to the students thewayNewton used astronomical datarelatedtothemoontocometotheconclusionthatthegravitationalforcebetweentwo objectsmust vary as inverse of the square of distance between them. The‘inversesquare’natureofthislawgivesthegravitationalforceseveralinterestingpropertiesnotableamongwhicharethe(i)conservativenatureofthisforceand(ii)zerovalueofthegravitationalfieldinsideahollowobject.Theteachermayexplain that the other important’inverse square’ force in nature—the electricalforce—also has similar characteristics. One can therefore convey the ‘message’thatthesamephysicallawoftenhelpustounderstandavarietyofapparentlydiversephysicalphenomenon.

It may also be explained that the ‘inverse square’ law for gravitational force,proposed by Newton, was also consistent with Kepler’s laws of planetarymotion. These lawswere based on the painstaking and thorough astronomicalobservations of the great Danish astronomer Tycho Brahe. Newton’s law ofgravitation, therefore become the ‘core’ of our understanding of astronomicalphenonmenon in addition to its use in understanding a variety of terrestrialphenomenon.

The teacher may express this law in vector form and use ‘principle ofsuperpositioin’when thegravitational forceononeobject isdue toagroupoftwoormoreotherobjects.Itmayalsobestressedthatunliketheelectricalorthemagneticforce,thegravitationalforceisalwaysattractiveinnature.Thedirectionoftheforcebetweentwo(point)objectshastobealongtheline joiningthetwo(point) objects. This is so because in a region of spacewhere these two (point)objects alonemay be present, the only direction that can be uniquely definedisthelinejoiningthesetwopoints.Itmayalsobepointedoutthatthedirectionof the gravitational force on object 2 due to object 1 has to be opposite to thatof the position vector of object 2 with respect to object 1. The consistency ofthis statementwith the ‘alwaysattractive’natureof thegravitational forcehastobe clearlybroughtoutandemphasised.

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While discussing the universal nature of the law of gravitation, it needs to bestressed thatweordinarilydonotperceive the effectsof this forcebecauseofthevery smallvalueof the ‘universal constant’ ofgravitation ‘G’ .The teachermay explain how itwas thought thatwewould necessarily need to use verylarge,hugemassestomeasurethe(verysmall)effectsoftheweakgravitationalforce. These ‘intial attempts’need tobeappreciated in theirhistorical context.Itwouldbe interesting topoint out that inspite of the relatively crudenatureof thesemeasurements. The value ofG, obtained through these ‘experiments’thoughnotveryaccurate,was still of the correct’order ofmagnitude.

While discussing the measurement of G, the teacher may point out whyCavendish’s experiment—the first ‘in-house’ or laboratory experiment—on themeasurement of G, is regarded as one of themost important experiments inthe history of Physics.His novel idea ofmeasuring the (relatively) large effectof themoment of the couple due to a pair of (small) gravitational forces andhis accurate andprecise technique of carryingouthismeasurementsneeds tobe ‘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 studentsneed tobeclarified thata commonvalueof this ‘constant’ forallobjectsofdifferentmassesis aconsequenceoftheequalityofthe‘inertial’and‘gravitational’ mass on the other hand is the ratio of the (magnitudes) of thegravitationalforceontheobjecttoitsresulting(gravitational)acceleration.Theequality of this (gravitational) mass to the inertial mass of the object and thenatureofthegravitationalforce(proportionaltotheproductofthemassoftheobjectandthatoftheearth)thenimpliesthattheaccelerationduetothegravityfor theearth,wouldhave the samevalue forobjectsofdifferentmasses.

Whilediscussingthevariationofaccelerationduetogravitywithheightabovethe earth’s surface, it needs to be pointed out that we make calculations byassumingtheentiremassoftheearthtobeconcentratedat itscentre. Itshould

also be pointed out that we can use the formula g h/g = RR h h

R

2

2 21

1( )+

=

+

in

the form gh

g 1 2−

hR

only when hR << 1. In practice, for ordinary work,

hR<<1 canbe taken to imply that h

R

1100

or less.

The students generally find it difficult to appreciate the decrease in ‘g’withdepthbelowtheearth’ssurface.Theyneedtoberemindedthatsinceahollow

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spherical shalldoesnot exert anygravitational forceonanobject inside it, theeffectivemass of the earth, that exerts a gravitational force on anobject insideits core is only themassof that inner coreof the earth forwhich this object isan external object. This ’effective mass’ keeps on decreasing with increasingdepth.Hencetherewouldbeadecreaseinthegravitationalforceandtherefore,in the acceleration due to gravitywith increasing depth. The studentsmay be

toldaboutthedifferencebetweenthequantitativeformulaegd + g 1−

dR

for the

decrease in ‘g’withdepthandfor thedecrease in ’g’withheight. Agraphicalrepresentationofvariationof ‘g’with(small)heightsandwithdepthwouldbean interestingexercise for the students.

While introducing the concept of gravitational potential energy, the studentsneed to be reminded again that we can associate a definite value with thisenergy for a given position of the object, only because the gravitational forceis a conservative force. They also need to be clarified about the fact that it isonly the difference in gravitational potential energy between two points thathas a unique value. The value of the gravitational potential energy at a pointdependsonthe(somewhatarbitrary)choiceofthe‘zero’ofthispotentialenergy.If alsoneeds tobe emphasised that becauseof the alwaysattractivenatureofthe gravitational force, the gravitational potential energy is usually assigneda negative sign the maximum value of the gravitational potential energy isthereforezero.

Theconceptof ‘gravitationalpotential’ iscloselyrelatedto thatofgravitationalpotential energy. This is because the ‘gravitational potential’ at any point, isjust the gravitational potential energy of a ‘unit mass’ at that point. We can,therefore, thinkof theresults,associatedwithgravitationalpotentialenergy,asbeingalsoapplicable to ‘gravitationalpotential’.

While discussing the concept of the ‘escape speed’, the teacher must make itclear how a simple application of the law of conservation of energy enablesus to calculate this speed. It is important to emphasize that the escape speed,associatedwith theearth (orotherastronomicalobjects) isacharactericeof theearth(oroftheconcernedastronomicalobject)anditdoesnotdependuponthemassof theobjectbeing projected.

The concept of the satellite speed and its close relation with the radius ofthe satellite orbit and the ‘source’ of the gravitation force ( Earth, or someotherheavenlyobject)needs tobeclearlybroughtout.Theconceptof the ‘geo

xiv

stationary satellite’ and the need for having specifically located ‘launchingpads’ for such satellites,needs tobe clearly explained .

Averyinterestingconceptassociatedwiththatofsatellitesandtheirastronautsis that of weightlessness. The students must be made to understand how anobjectduringits‘freefall’canberegardedtobeinastateofweightlessness.Theymust be made to realise how an orbiting satellite (can the astronauts presentin it) canalsobe regarded tobe ina stateof ‘free fall’and therefore, ina stateof ‘weightlessness. It does not mean that the earth is no longer exerting agravitational force on it. Rather, the object is in a statewherewe are not ableto see ‘effects’of thisgravitational force thatwe refer toas ‘weight’ .

Anintroductiontotheconceptofsatellitesneedstobefollowedbyanintroductionto thedifferent typesof satellites and the role andapplicationof eachof thesedifferent types.Avisualdepictionof theprocessof satellite launchingand thebasicdetailsofdifferenttypesofsatellitesandtheroleandapplicationofeachof thesedifferent types.Avisualdepictionof theprocessof satellite launchingand the basicdetails ofdifferent types of satellites shouldbe of immensehelpand interest toall students.

xv

lesson Plan matrix

Content Skill learning outComeSEarly history of astronomy • Geocentric and

heliocentric models• Ability toappreciate theworksof scholarsandthinkers to enhance our poolofknowledge.

• appreciate the long historyofmankind’seffort to understand planetarymotion

• differentiate betweengeocentric and heliocentric models and the historical context underwhich these ideasprevailed

• know howearlyastronomical observations were takenandhowthey contributed to thedevelopmentof the lawofgravitation.

astronomy in anciEnt india• Contributionof Indian

mathematicians and thinkers

• Ability toappreciate thehighlydeveloped scientifictemperament inancientIndia

• Appreciate thecontributionof Indiain thedevelopmentofastronomy

KEplEr’s laws of planEtary motion• Kepler’sLawsofPlanetaryMotion

• Ability to stateandexplainKepler’s laws

• Understand the importanceof and interpretationofscientificdataanalysis

• Able to solveproblemsbasedonKepler’s thirdlaw.

• stateand interpret Kepler’s lawsofplanetarymotion

• recognize howKepler’slawsoriginated fromtheanalysis andinterpretationofTychoBrahe’s astronomicaldata

• realize howmeasurements on planetarymotionwere inagreementwithKepler’slawofperiods

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nEwton’s law of Gravitation• Newton’sLawofGravitation

• ImportantpropertiesofGravitational force

• DeterminationofvalueofG

• VectorFormofNewton’slawofGravitation

• PrincipleofSuperposition

• Appreciate thecontributionofNewtonin the formulation of theuniversal lawofGravitation

• ApplicationofNewton’slaw inpractical situations

• ApplyNewton’s lawto calculate the net gravitational force due to a collectionofmasses.

• Understand thecharacteristics of gravitational force.

• Ability toderive the laws

• stateNewton’s lawofGravitation

• know the conceptofcentral forces

• defineGandknowaboutthevariousexperimentsproposed for themeasurementofG

• comprehend thepopularstatement “cavendish weighed theearth”

• draw the vector diagram for the gravitational force between twomasses

• understand howto compute thenetgravitational forces due to a collection of masses

• recognize that gravitational force is oneof the ‘basic forcesof nature’ and it is a weak force that is alwaysattractive innature.

• obtainKepler’s lawsofplanetarymotionfromNewton’s lawofGravitation

accElEration dUE to Gravity• Accelerationdue toGravity

• Able toapply theNewton’s second law toobtain the value of g

• Able toappreciate the factthat g is not a constant

• recall the conceptofacceleration and use it to obtain theexpression forg fromNewton’s lawofGravitation

• recognize that the shapeof the earth affects the value of g for different pointson its surface

xvii

• The learnermustdeveloptheability toderive theexpressionsandalsoapplyit toproblems.

• Able todistinguishbetween thevariation ingwithheightanddepth

• meaning and significanceof the comprehendthepopular statement‘’Cavendishweighed theearth’’

• realize that the acceleration due to the gravitationalpullof theearth (i.e.g) is same forall masses

• derive the expressionforvariationofgwithheighth (i.e.g(h)) abovethe surface of earth and understand theexpressionlimitationsof this.

• recognizewhy theexpression forg(h) cannotbe used for negative h (pointsbelow the surface)

• obtain the expression forvariationofgwithdepthd (i.e.g(d)) below thesurface of earth

Gravitational potEntial EnErGy• Gravitationalfield

• GravitationalPotentialEnergy

• Able to comprehend theneed for thefieldpicturefor gravitational force

• Ability to co– relate the fact that gravitational potential energy is justoneof the formsofpotentialenergy

• Skill toderive theexpression forgravitationalpotential energyandbeable to solveproblemsbasedon the topic.

• recall the conceptofpotential energyanddefineGravitationalpotential energy

• obtain the general theexpression for thepotential energyof amass at a height from the surface of earth and obtain its usual approximate form.

• derive the formula for thegravitationalpotentialenergyassociatedwithtwomasses separatedbya distance

xviii

• GravitationalPotential

• Extrapolating the resultderived above to a multi particle system.

• Distinguishbetweengravitationalpotential andpotential energyand theneed for the conceptofgravitationalpotential.

• differentiatebetweengravitationalpotentialenergyandgravitationalpotential

EscapE spEEd • EscapeSpeed • Ability to recall the

conceptof escape speed

• Derive theexpressionandapply it toproblems

• know the conceptofescapeSpeed

• obtain the expression forescape speed from thelawof conservationofenergy

satEllitEs• Motionof satellite(Orbital speed,Timeperiodof a satellite)

• Able toderive theexpressionand solveproblemsbasedon thetopic

• appreciate the scientificefforts to launchartificialearth satellites and list theirfieldsofuse

• obtain the expressionfor the timeperiodof anorbiting earth satellite

• listdifferent typesofartificial satellites andobtain the condition for ‘ageo-stationary satellite’

• know the range of applicationsof ageo-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

http://www.schoolscienceguru.com

students'manual

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2 Unit 6 : Gravitation

6.1 IntroductIon

Long time back, even beforeman had learnt to read andwrite, hemust haveobservedtherisingandsettingofthesun,thephasesofthemoonandtheshiningstars at night. Hemust have wondered about the cause of these phenomena.Thein-builtcuriosityinthemindofmanmusthaveledtocenturiesofthoughtsand ideas. The subject ofGravitation, in the formwe study it today, is one ofthe consequencesof these thoughtsand ideas.

DidNewtondiscover ‘Gravitation’?

Theanswer is anemphaticNO!

TheforceofGravitationhadalwaysexistedanditsimportanceandsignificancewas realized by various philosophers and thinkers much before Newton.However, the credit for formulating a, quantitative law, which governs thenature of the force of gravitation,mustgo toNewton.

Early History of Astronomy

Earlier astronomy was based onthe geocentric model in whichthe Earth was regarded as thecenter of the universe, and allother heavenly objects weresupposed to orbit around it.This geocentric model was thepredominant cosmological systeminmanyancientcivilizationssuch

Figure 1. Geocentric system

EARTH

MOONVENUS

SUN

SATURN

Mercury

MarsJupiter

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Unit 6 : Gravitation 3

asancientGreece.

Two observations that must have made a significant contribution to theformulationofthegeocentricmodelwere(i)thestars,sun,andplanetsappeartorevolvearoundtheEartheachday,makingtheEarthappeartobethecenteroftheoverallsystem.(ii)TheEarthdoesnotseemtomovefromtheperspectiveofanEarthboundobserver,forwhomitappearsasasolid,stable,andunmovingobject.

It was not until the 16th century that a new astronomical model called theHeliocentric modelwas developed. In this model the Earth and planetswereimaginedtorevolvearoundastationarySunwhichwasconsideredasthecentreof the universe. This word comes from the Greek Helios (sun) and kentron(center).Thisnotion,thattheEarthrevolvesaroundtheSun,hadbeenproposedasearlyas the3rdcenturyBCEbyAristarchusofSamos,buthis ideareceivedno support from theotherancient astronomersof that time.

Figure 2. Heliocentric system

EARTH

MOON

VENUS

SUNSATURN

MERCURY

JUPITER

MARS

A fullypredictivemathematicalmodel of aheliocentric systemwaspresented,by the Renaissance mathematician, astronomer, and Catholic cleric, NicolausCopernicusofPoland, leading to the ‘CopernicanRevolution’. In the followingcentury, this model was elaborated and expanded by Johannes Kepler and

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Unit 6 : Gravitation4

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

WiththeobservationsofWilliamHerschel,Besselandothers,astronomerssoonrealized that the sunwas not the center of the universe. By the 1920s, EdwinHubble had shown that it was just a part of a galaxy (called theMilkyWaygalaxy) thathadbillionsof stars.TheMilkyWaygalaxy, in turn,was just oneof thebillionsof suchgalaxies in theuniverse.Theuniversewas,overall,veryveryvast indeed!

6.2 Astronomy In AncIEnt IndIA

Long before Kepler formulated his laws of planetary motion, the study ofplanetarymotionwasawelldevelopedbranchofScience in India.

Aryabhata (476–550?), in hismagnum opus Aryabhatiya (499?) propounded aplanetarymodel inwhich the Earthwas taken to be spinning on its axis andthe periods of the planets were given with respect to the Sun. He accuratelycalculatedmanyastronomicalconstants,suchastheperiodsoftheplanets,likelytimes of the occurrences of the solar and lunar eclipses, and the instantaneousmotionoftheMoon.EarlyfollowersofAryabhata’smodelincludedVarahamihira,Brahmagupta, andBhaskara II.

Nilakantha Somayaji (1444–1544?), in his Aryabhatiyabhasya, a commentaryonAryabhata’sAryabhatiya,developed a computational system for apartiallyheliocentricplanetarymodel, inwhich theplanetsorbit theSun,which in turnorbits theEarth, similar to theTychonicsystemlaterproposedbyTychoBrahein the late 16th century. In the Tantrasangraha (1500?), he further revised hisplanetary system, which was mathematically more accurate at predicting theheliocentricorbitsoftheinteriorplanetsthanboththeTychonicandCopernicanmodels. However, the Indian astronomy, in general, fell short of proposingmodels of the universe. Nilakantha’s planetary system also incorporated theEarth’srotationonitsaxis.MostastronomersoftheKeralaschoolofastronomyandmathematics seem tohaveacceptedhisplanetarymodel.

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6.3 KEplEr’s lAws of plAnEtAry motIon

The motion of the planets, as they revolve around theSun, and rise and set on the horizon, must have been apuzzle to the mankind. Johannes Kepler (1571 – 1630),afterstudyingthemotionofplanetsforalifetime,workedout the empirical laws governing planetarymotion. LaterNewton proved that Kepler’s laws can be derived fromNewton’s lawofgravitation.

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 ofthe fociof theellipse.

Theadjacentfigureshowsaplanetofmassm revolving around the sun having a mass M. (M>>m)

The orbit is characterized by twoparameters

(i) Semimajoraxis

(ii) eccentricity (e)

Forcircularorbits,eccentricityiszeroandsemimajoraxisisequaltotheradiusoftheorbit.Formostoftheplanetstheeccentricityoftheirorbitissosmallthatwecanassume them tobemoving in circularorbits.

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

PlanetSun

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Kepler’s second law — law of Areas

ThelinejoiningaplanettotheSunsweepsout equal areas in equal interval of time in theorbitalplaneof theplanet. i.e. the rate

dAdt

at which it sweeps out areaA (Arealvelocity) is a constant.

Kepler’s second law is themost importantof the three laws.

TheareasweptbytheplanetdA in time dt canbeexpressedas

dA = 12

r2 dq

(Areaof triangle= 12

base × height)

where r is the distance between Sun and planet, dq is the angle swept by the (radial line of the) planetin time dt, as shown in theaccompanyingfigure.

hence dAdt

= q= ω2 21 1

2 2ddt

r r

where ω= angular speedof theplanet

Themagnitudeof the angularmomentumof theplanet, about an axispassingthrough theSun, is L = rpn

wherepnisthecomponentofitslinearmomentum→p along the direction normal

to r L = r(mv)= r (mωr)=mr2ω

From theseequations,weget

dAdt

= 2Lm

Since, thegravitational force is a central force, it actsalong 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:

dqrdq

r

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

force,on theplanet, is zero.

In the absence of torque, the angular momentum→L remains conserved in

accordancewiththelawofconservationofangularmomentum.Hencethearealvelocity

dAdt

= a constant

One simple consequence of Kepler’s second law is that a planet moves fasterwhen it iscloser toSunandslowerwhen it is far fromit.Canyou thinkofanexplanation?

Kepler’s third law — law of periods

The square of the time period of revolution of a planet around the Sun, isdirectly proportional to the cube of the semi –major axis of the orbit. (radiusof theorbitwhen theorbit canbeassumed tobe circular.)Thus

T2 ∝ r3

whereT= timeperiodof revolutionof theplanet and r= radiusof itsorbit.

Alternately,we canalsowrite this lawas

2122

TT

= 3132

rr

Foran interactiveanimationvisit: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 butwas deficient in their theoretical interpretation.On theotherhand,JohannesKepler(1571-1630),aGerman,whowenttoPraguetobecomeBrahe’sassistant,hadastrongtheoretical intuition.

KeplerandBrahedidnotgetalongwell.BraheapparentlymistrustedKepler;He thusallowedonly limitedaccesstoKepler tohisvoluminousdata.

HesetKeplerthetaskofunderstandingtheorbitoftheplanetMars, which was particularly troublesome. It isbelievedthatpartofthemotivationforgivingtheMarsproblemtoKeplerwasthatitwasdifficult,andBrahehopeditwouldkeepKepleroccupiedwhileBraheworkedonhis theoryof theSolarSystem. Inasupreme irony, itwaspreciselytheMartiandatathatallowedKeplertoformulatethecorrectlawsofplanetarymotion,thuseventuallyachievingaplaceinthedevelopmentofastronomythatfar surpassed thatofBrahe.

It was left to to Kepler to provide an answer to the final piece of the puzzle.Afteralongstruggle,inwhichhetriedmightilytoavoidhiseventualconclusion,Kepler was finally forced to come to the conclusion that the orbits of theplanets were not the circles (demanded by Aristotle and assumed implicitlyby Copernicus), but were instead the “flattened circles” that geometers callellipses.

The irony, noted above, lies in the realization that the difficulties with theMartian orbit originate precisely from the fact that the orbit of Mars was themostellipticalofalltheplanetsforwhichBrahehadextensivedata.Thus,Brahehadunwittingly givenKepler the very part of his data that allowedKepler toeventually formulate the theory of the Solar System that surpassed Brahe’s own theory!

Figure 7: Tycho Brahe

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5.4 nEwton’s lAw of GrAvItAtIon

We all are familiarwith the famous legend ofNewtonobserving the free fall of an apple and formulating thelaw of gravitation. However, it will be appropriate tomention that Newton gave the law of gravitation afteralmost20yearsofhisfirstthoughtaboutitwhichisquiteoftenassociatedwith theepisodeof a fallingapple.

Newton’s law for gravitational force between twobodies isoneof themost far– reaching laws in thehistoryofhumanscientificendeavor.

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.

Ifmustbenotedthatthelawasstatedaboveapplies topointbodies.

For extended bodies, the distance mustbe taken between their centres of mass(geometrical centres if bodies are regularandhomogeneous) and thedirectionof theforce is along the line joining their centresofmass.

F ∝ m1m2

F ∝ 21

r

F ∝ 1 22

m mr

wherem1 and m2arethemassesofthetwobodiesandristhedistancebetweentheir centers.

Figure 8: Sir Issac Newton

m1 m2

r

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

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Notethatasthemassesofthebodiesincrease,theforcebetweenthemincreases.Also, as thedistancebetween their centres increases, the forcedecreases.

6.5 ImportAnt cHArActErIstIcs of GrAvItAtIonAl forcE

The forceofgravitation is associatedwith the followingcharacteristics:

(a) the gravitational force is a central force. Itactsalongtheline joiningthecentresof twobodies.

(b) it is a conservative force. This means that the work done by thegravitational force in displacing abody fromonepoint to another isonly dependent on the initial andfinal positions of the body and isindependentof thepath followed.

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

(d) Unlike electrostatic andmagnetic forces, the gravitational force is always attractive.

The entireUniverse 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)

TheproportionalitysignintheNewton’slawofgravitationcanbeeliminatedbyputtingaconstantofproportionality,denotedbyG. The equation then becomes

F = G 1 22

m mr

here Giscalledthegravitationalconstant.Itisauniversalconstant because the gravitational force between twobodies placed at a certain distance remains the same,wherever these bodies may be placed in the universe. The constantG is also independentof themedium inwhich the interactingbodiesareplaced.

The value of G,inSIunits,waslateron,foundtobeequalto6.67×10–11 nm2/kg2.

The units of Gareobtainedbyusingthefact thattheforcehastobeexpressedinnewtons.

Figure 11: Henry Cavendish

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The numerical value of G would change with a change in the units ofmeasurement formassanddistance.

The very small value ofG points to the fact that the gravitational force is anextremely 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 importantonlywhen themassesof thebodies involvedarevery large.That iswhythegravitationalforceplayssuchanimportantroleinthecaseofheavenlybodies.

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 GwasfirstmeasuredbyanEnglishPhysicist,HenryCavendishintheeighteenthcentury.Heachievedthisbymeasuringthesmall forcebetweenleadmasseswith an extremely sensitive torsion balance.A bettermethodwaslaterdevelopedbyPhilippvon Jolly.

As longas the sizesof theobjects are small compared to thedistancebetweenthem they can be treated as point objects which simplifies considerably themathematicsoftheirgravitationalinteraction.TheSunandSaturnarefarenough(in comparison to their sizes) for them tobe treatedaspointparticles.

If thedistancebetween twoobjects isvery large (in comparison to their sizes),wecantakethemaspointobjects.Whatabout thecaseofbodiesontheearth?For suchbodies, theearthdoesnot seem likeapointobject.

Theanswer to thisdilemma lies inNewton’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 ‘pointmass’, located atthe center of Earth andwithmass equal to that of Earth. We usually follow

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this theorem inallour calculations.

5.7 vEctor form of nEwton’s lAw of GrAvItAtIon

Since force is a vector quantity, it must beexpressed in a vector form. The gravitationalforce can also be expressed in a vector formbyattachingaunitvectortotheexpressionforthis force.

By convention, the direction of unit vector isalwaystakenasdirectedfromthebodyexperiencingtheforce(body1)towardsthebodyexerting the force (body2).Therefore,

F

= G 1 22

ˆm mr

r

IfwearecalculatingtheforceonbodyAduetobodyB, then r̂ willbetheunitvectordrawn fromA towardsB.

Itwillbeimperativetomentionherethatthisvectornotationisconsistentwiththebasicfactthatthegravitationalforceisalwaysanattractiveforce.Thisimpliesthatthegravitationalforceisacentralforce,andhencethedirectionofthisforcehas tobealong the line joining the centerof the twobodies.

5.8 prIncIplE of supErposItIon

Newton‘s law has been stated for two point bodies.How dowe calculate theforceonabody if therearemore than twobodies interactingwithoneother?

Figure 12: Diagram showing direction of the gravitational force acting along the line joining the center of two bodies.

1

A

2

B

r̂

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Thesolutiontothissituationliesinwhatiscalledtheprinciple of superposition.In a group of objects, the net gravitational force, on any one of the objects, isthe vector sumof the forces due to all the other objects. Theprinciple impliesthatwefirst calculate thegravitational force that actsonanobjectdue to eachof the other objects as if all other objects are absent. After doing this for allpossible pairs, the net force on the object under consideration is calculated bythevector sumofall the forcesactingon it.

→1F =

→ → → →+ + + +12 13 14 15 ...F F F F

here →1F is thenet forceonobject 1due toall theotherobjects 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 Adue toobjectB is

FAB = G 2A Bm mr

It is in thedirectionofAB

Putting thevaluesweget,

FAB 5×10–9 N

Similarly, the magnitude of the force onAdue tobodyC is

Fac 5×10–9 N

It is thedirectionofAC.

As per the principle of superposition, the net force on the bodyA is the

vector sum of forces →

ABF and →

ACF . Applyingthelawsofvectoraddition,

A

B C

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weget themagnitudeof the resultant as

Fa = + + °2 2 2 cos60AB AC AB ACF F F F

Fa = − − − × + × + ×

9 2 9 2 9 2 1

2(5 10 ) (5 10 ) 2(5 10 )N N N

= 5 3 ×10–9 N

Theresultant force isdirectedalong thebisectorof theanglebetween the

two forces→

ABF and →

ACF .

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 letusassume that theorbitof theplanet is circular.

Thegravitationalforceexertedbythesunontheplanetsprovidesthenecessarycentripetal force for the planet to remain in its orbit. So, applying the forceequation,

Centripetal force= Gravitational force

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weget2

pM vr

= 2s pGM M

r

putting v = p2 rT

p2

2

4pM r

T = 2

s pGM M

r

\ T2 = p

24

sGMr3

The quantity in the parentheses is a constant that depends only on the massMsof thecentralmassivebodywhich is theSun in thiscase. It canalsobe theEarth ifweare talkingof themotionof anartificial satellite.

\ T2 ∝ r3

This, asyouhave learnt above, isKepler’s third law forplanetarymotion.

6.9 AccElErAtIon duE to GrAvIty

WeassumethattheEarthisauniformsphereofmassM and radius R.Wecanassume then that themassof earth is concentratedat its center.

The gravitational force of Earth, on a particle ofmassm, at a distance r from the centerofEarth, isgivenbyNewton’sLawas follows:

F = 2GMm

r …(i)

whereM is themassof theearth.

This force gives rise to an acceleration in the particle which is called theacceleration due to gravity.

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ApplyingNewton’s second law, forceonabodycanbeexpressedas

F = ma …(ii)

Both the expressions (i) and (ii) aremeasuring the same force in twodifferentways.

Hence, from the twoequations (i) and (ii),weget

ma = 2GMm

r

\ a = 2GMr

We use the special symbol ‘g’ for this acceleration and call it as theaccelerationdue togravity.

Thus,theaccelerationduetogravity(g)istheaccelerationproducedonabody,on,ornear,theearth’ssurface,dueto thegravitationalpullof earth.

We have g = 2GMR

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,weputr = R, theradiusoftheearth.Wecan, therefore, expressaccelerationdue togravity (g) as

g = 2GMR

Forearth,takingM=6×1024 kg and R=6400km,wegetg=9.8m/s2.However,itmust be noted that the value of 9.8m/s2 is an average value and the value of g variesonearth fromoneplace toanotherbecauseofvarious factors.

On a given planet, the average value of acceleration due to gravity is thesame for all objects and is independent of themass 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 andradiusof thatplanet.

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. Inany suchquestion, thebestway is towrite twoequations:

Valueofg on earth

ge = 2

e

e

GMR

Valueofg onplanet

gp = 2p

p

GM

R Insertinggivenvalues,

p

e

gg

= 21 1 12 4 8

p e

e p

M RM R

= × =

We get gp = 18×9.8ms–2=1.225ms–2

≈ 1.2ms–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 9.8m/s2 for the accelerationdue to gravity onEarth is the averagevalueandisnotthesameatallplacesontheearth.Itvariesduetothefollowingreasons:

(i) Densityof earth isnotuniformat allplaces

(ii) Earth isnot aperfect sphere

(iii) Theearthhasa rotationalmotionabout itsownaxis.

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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 aswe go up above the surface ofearthandaswegodeepdown inside theearth.

Letusnowstudy thevariation in thevalueofg in somedetail.

6.9.1 due to the shape of the earth

The earth is not a perfect sphere. Infact, the shape of earth ismore like anoblate spheroid. It is bulgingout at theequators and a little compressed at thepoles.

Thus, since

Re>Rp,

wewouldhave

ge < gp

?Did You Know?A person weighs (a little) more at the poles than at the equator.

6.9.2 with height (h)

Thevalueofgat apointPon the surfaceof earth

g = 2GMR

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

North Pole

South PoleP

olar

dia

met

er12

,714

km

Equatorial diameter12,756 km

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Ifwegoup above the surface of earth,the value of gatapointP′ataheight‘h’ from the surface of earth,wouldbe

g′ = + 2( )

GMR h

\ ′gg

= 2

2( )R

R h+ …(i)

Wecanalsowrite the formulaas

This formula is valid for all values of ‘h’. However, it is usually used only when h is comparable to R.

′gg

= 2R

2R +

21 h

R

g′ = −

+

21 h

Rg

For h << R, the term−

+

21 h

R can be expanded by using binomial theorem.

Neglectinghigherorder terms,we canwrite

g′ = −1 2 h

Rg

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 changewillnotbepossibleatasmallheight.Henceitwillbe incorrect tousetheformulavalidonly for smallheights.

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

P′

P

R

h

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We use instead the formula

′gg

= +

2

2( )R

R h Putting g′=0.36g, and taking square rootofboth sides,

wegeth = 23R

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

solutIon. Inthisquestion,theheightatwhichgistobecalculatedissmallcomparedto R.Thus,we canuse the formulavalid forh < < R.

g′ = −1 2 h

Rg

Puttingvalues,weget

g′ = = − 32 6336

6400 64001 2g g

\ g′ = 99

100g=99%ofvalueofg on the surface of earth

Letusseewhathappenswhenthestudentusestheincorrectformulainagiven 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 toR; in fact it is equal to the radius of earth. Ignoring this fact, if a studentuses the formula

g′ = −1 2 h

Rg

which is actuallyvalidonly for smallheights,

the student gets

g′ = −1 2 R

Rg

g′ = –g

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The answer is completely wrong because a an invalid formula has beenused.

The rightapproach is touse the formula

′gg

= +

2

2( )R

R h Putting the value of h = R,weget

′gg

= +

2

2( )R

R R

′gg

= 2

24RR

g′ = 4g

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 aswe go deep inside the earth. It will be easier toderive the result, if we convert the expression for g in terms of the density,rather than themass,of theearth.

Weknow,

g = 2GMR

Assuming earth to be a sphere of uniform mass density ρ,we canwrite

M = 43pR3r

substituting in the formula for g,weget

g = 43pRGr

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Intermsoftheformuladerivedabovethevalueofgatadepthdcanbeexpressedas

g′ = 43pG(R –d)r

This is because, the point, at a depth d below the surface, is a ‘surface point’ona sphereof radius (R –d) and it isonly ‘this sphere’ that contributes to thegravitational force foraparticle locatedat thisdepth.

Thus,

′gg

= −1 d

Ror

g′ = g −1 d

R

Did you note that at the center of earth, i.e for d = R, the acceleration due togravitybecomeszero.

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.Valueofg, at aheightof 64km isgivenby

g′ = g 1 2 hR

−

Valueofgat adepthd isgivenby

g′ = g 1 2 dR

−

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Since thequestionsays thevalueofg at height = value of g atadepthd,wehave

1 2 hR

g

− = 1 dR

g

−

We get

2h = d

or d = 128km

?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 2GMm

Rmg

= = 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 = mω2R 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 2GMm

Rmg

= = directed along PO and a force F2 (= mω2R cos l) directed along

PN.

N

O Rl

Equator

PR cos l

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Simple calculations, based on the parallelogram law of vector addition, lead us to the formula

mgl = m (g2 + R2w4 – 2gRω2 cos2 l)1/2

gl = g 1/22 2 4

22

21 cosR Rg g

ω ω − l+

Since Rω2 is a small quantity, we can neglect R2w4 and higher terms. Then

gl = g2

21 cosRg

ωl

−

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

Therearesomeforcesinnaturewhichcanactevenwhentheinteractingbodiesare not in direct contact with each other. Such forces are called action-at-a-distanceforces.Thegravitational,electrostaticandmagneticforcesarewellknowexamplesofsuchforces.Ontheotherhand,thereareforceswhichactonlywhenbodies are in direct contact. Frictional, viscous and muscular forces are someexamplesof such ‘contact forces’

The‘action-at-a-distance’,ornon-contactforces,forceswhichcanactevenwhenthe bodies are not in direct contact, are usually discussed on the basis of theconceptoftheirassociated‘field’.Forthegravitationalforce,weusetheconceptof the ‘gravitationalfield’.

Body (source)

The gravitational field, of a body is defined as the space around this body inwhich itsgravitational force canbeexperiencedbyotherbodies.

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An alternativeway to explain the gravitational attraction between two bodiesis to imagine that each body creates around it a force field that causes otherbodies tobeattracted to it.

It is important to realize that to verify the existence of a gravitational field, atestbodyhas tobebrought into the region.

Thebodywhichcreates thefield iscalled the ‘sourcebody’and theonewhichisbrought to ‘test’ theexistenceof thisfield is called the testbody.

Thevalueofgravitationalfieldatapoint(alsocalledgravitationalfieldintensity)isdefinedas the force experiencedbyaunitmassat thatpoint.

We have F→

= 2ˆGMm

rr

Thegravitationalfield

E→

= 2

ˆF GMm r

r→

=

It is a vector quantity. The direction of the gravitational field is same as thedirection of the gravitational force and since the gravitational force is alwaysattractive, the gravitational field is always directed towards the center of thesourcebodycreating thefield.

S. I.unitofgravitationfield isN/kg.

The ‘test body’ has to be a body whose mass is very small as compared to that of the ‘source body’. This eliminates the possibility of the disturbance of the gravitational field of ‘source body’ by the field produced by the ‘test body’.

On the surface of earth, r = R. Hence themagnitude of the gravitational fieldof theearthon its surfacewillbe

E = 2GMR

Notice, that thisexpression issimilar to theexpression for theaccelerationduetogravityon the surfaceof theearth.

Acceleration due to gravity is actually a measure of the gravitational field of earth. It is always directed towards the center of Earth. The direction to the centre of the earth is called the vertical direction.

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IllustratIon 7. the distance between earth and moon is 3.8 × 105 km and the mass of earth is 81 times the mass of the moon. Find the position of a point, on the line joining the earth to moon, where the net gravitational field, due to the two, equals zero. Given Radius of earth = 6400 km.

solutIon. Asweknow,thegravitationalforcebeingacentralforce,thegravitationalfield is directed along the line joining the center of the twomass bodies.In this case, the twobodiesare theMoonandEarth.

Thepoint,wherethenetgravitationalfieldiszero,liesonthislinejoiningtheEarthandMoon.Atthispoint,thegravitationalfieldoftheEarthwillbe equal andopposite to thegravitationalfieldof theMoon.

Aswealreadyknow,thedirectionsofthetwofieldbeingoppositetoeachother,weneedequateonly themagnitudesof the twofields.

Let thegravitationalfieldbezeroat apointP at a distance x fromEarth.The distance of this point, from theMoon, is (r – x), where r = distance betweenEarthandMoon.

\ 2GMex

= 2( )

GMmr x−

Substituting thevaluesgiven in thequestion,

281Mm

x =

2( )Mm

r x−

hence x = 9(r –x)

\ x = 0.9r = 3.42 × 105 m from the earth

6.11 GrAvItAtIonAl potEntIAl EnErGy

WearealreadyfamiliarwiththeconceptofPotentialEnergy.Thepresenttopicis anextensionof the samediscussion.

In general, the term Potential Energy (U) is defined as the energy associatedwith an arrangement, or configuration, of a system of objects, exerting forcesoneachother.

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Forpotentialenergytoexist, theremustbeaninteractionbetweentheparticlesof the system. i.e. theparticlesmust exert forceson eachother. The conceptofpotential energy canalsobeassociatedwithafield.

One of the most common examples of potential energy is the Gravitationalpotentialenergy.Inthiscasetheassociatedfieldisthegravitationalfield.Whena body is moved away or towards earth (or any other massive body), in thegravitational field of the earth, the configurationof the earth – body system isbeing changed.To facilitate this change, someworkhas to bedonewhich canbe viewed as having got converted into the gravitational potential energy ofthe system.

Fortheearth–bodysystem,wedefine the gravitational potential energy, at a given point, as the work done in bringing the body, of mass m, from infinity to that point, in the gravitational field without any acceleration.

dx

r

Figure 15: A body of mass m is being brought from infinity to a point P at distance r from the center of Earth

Suppose a body of mass m is being brought from infinity to a point P at a distance r from the center of the earth of mass M.

let A be a point in the pathwhen the body is at a distancex from the centre of earth.

Gravitational forceonm at thepointA, isgivenby

F = 2mGM

xThisistheforceactingonthebodyduetothegravitationalpullofearth.Since,our definition says that the bodymustmovewithout acceleration,we need tothinkof anequal andopposite (external) forcebeingexertedon thebody.

Workdone, ina smalldisplacementdx,wouldbe

dW = F dx

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\ Theworkdone,by thegravitational force, inmoving thebody, fromx = ∝ to x = r, is

W = r r

dW F dx∞ ∞

=∫ ∫

W = 2

r GMmx∞

∫ dx

W = GMm 2r

x dx−

∞∫

W= –GMm 1 1r

− ∞

W= – GMmr

…(a)

ThisworkWcanberegardedasstoredinthesystemofbodiesasthegravitationalpotential energyU.

\ U= – GMmr

Gravitational potential energy, being work done, is a scalar quantity. Its S.I.unit is joule (J)

The maximum value of gravitational potential energy is zero for the point at infinity. As it comes closer to earth, it loses some of its potential energy and attains a lower value of this energy.

What happens to the energy that a particle loses when it falls in a gravitational field? This energy is radiated away. When a large number of particles fall in a very strong gravitational field of a black hole, they radiate energy in the form of X-rays .

It is the detection of this X-ray signal that gives a clue to the presence of a black hole at that location. For more information carryout a search on the internet for the detection of black holes.

Thenegative sign in thegravitationalpotential energydenotes that the systemis losing someof itspotential energyas it comes close to earth.

Negative sign also implies that the force responsible, for the gravitationalpotential energy, is anattractive force.

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Negativesignalsodenotesthatthesystemis inaboundstatei.e. itcan’tbreakaparton itsown.

The gravitational potential energy is a property of the system of two particlesand not of either particle alone. However, for systems like earth – body, weoftensay that it is thepotentialenergyof thebodybecauseanychange, in thepotential energy of the system, almost entirely appears as the change in thekinetic energyof thebody.

IllustratIon 8. Find the change in the gravitational potential energy of a body of mass 50 kg, when it is moved from the surface of the earth to a point at a height twice the radius of earth. given radius of earth = 6.4 × 106 m and g = 9.8 m/s2.

solutIon. Initialpotential energy

Ui = –GMm

R

andfinalpotential energy,

Uf = –GMmR h+

Putting h=2R,weget

Uf = – 3GMm

R

\ Change inpotential energy

DU = Uf –Ui = – 3GMm GMm

R R − −

DU = 23

GMmR

Herewecanmakea substitution:

g = 2GMR

\ GMR

= gR

\ DU = 23

gRm

Putting thevalues,weget change inpotential energy=2.09× 109 J

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?Did You Know?Gravitational Red Shift

The light, from a massive star, can be said to undergo a ‘gravitational red shift’. This implies a (very slight) increase in the wavelength of the light emitted (by the star) due to the gravitational force exerted on the emitted photons by the star itself.

A formula for the gravitational red shift can be obtained by assuming a photon of frequency

n to have an energy h n and a mass h 2cn

.

When the photon is on the surface of the star, it has, in addition to its intrinsic energy h

n, a gravitational potential energy equal to – 2GM h

R cn

.

At a far off point, the photon energy would be h n′ (n′ = frequency of photon when it is very far away from the star).

Hence, by principle of energy conservation,

h n + 2.GM hR c

n −

= hn′

\ n′ = n 21 GMRc

−

or cl′

= 21c GMRc

l −

\ l′ = l 1

21 GMRc

− −

= l 21 GMRc

+ 2 1GM

Rc

<<

\ l′ − ll

= 2GMRc

This formula gives the fractional gravitational red shift, produced in a line of wavelength l, due to the gravitational field of the star emitting this line.

The gravitational red shift is usually a very small quantity. If becomes important in objects like the neutron stars and black holes because of their strong gravitational fields. In fact, the observation of gravitational red shift is an indication that the source of radiation is a very compact object having a strong gravitational field.

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6.12 GraviTaTional PoTenTial (v)

Gravitational potential energy has been defined as the energy of a particle ina gravitational field. Another way to express the potential energy at a pointis by using the concept of the Gravitation potential (V) at that point in thegravitationalfield.

Thegravitationalpotential(V),atapoint,isdefinedastheworkdoneinbringingabodyofunitmassfrominfinitytothatpointinthegravitationalfieldwithoutanychange in itskinetic energy.

As discussed above, the work (W) done in bringing a body of massm, frominfinity toadistance r from the (heavier)body,ofmassM, isgivenby

W= – GMmr

\ V= – W GMm r

= −

TheSIunitofpotential is J/kg. It is a scalarquantity.

Unlike gravitational potential energy, which is a property of the particle,gravitationalpotential canbeviewedas thepropertyof the space surroundingthe sourcebody.

Bycalculatingthegravitationalpotential,wecanassignevaluestodifferentpointin the space.Eachof thesevaluesmeasures theworkdone inbringingabodyofunitmass from infinity to thepointunder consideration.

Thus if gravitational potential at a point is say –10 J/kg, its implies that –10 Jworkneeds tobedone inbringingabodyof 1kg from infinity to thatpoint.

Hence, if a bodyof 2 kg is brought from infinity to that point, theworkdonewouldequal –20 J and soon.

Hence, if we know the gravitational potential at a point, we can calculate thegravitationalpotential energyat thatpointby the formula

U = mV

wherem=massof thebodycoming from infinity to thatpoint.

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6.13 esCaPe sPeed (Ve)

If you throw a ball up, it reaches a height and comes back to your hands.Asthe initial speed is increased, themaximumheight attainedalso increases.Theminimumspeedwithwhichthebodyisthrownsothatitnevercomesbackandreaches apointwhere the earth’sgravitationalfield ceases is called theescape speed.

Wecanobtainanexpression, for theescape speed, as follows:

If abodyofmassm, is thrownwitha speedve,

the total Initial energyof thebody

Ei = K.E. + P.E.

Ei = 212 e

GMmR

mv + −

Let thisbodycome to restonlyafter reaching thepoint at infinity

We thenhave thefinal energyof thebodyat infinity,

Ef= 0+0

becauseat infinitybothK.E. andP.E. arezero.

Applying the lawof conservationof energy,

We have Ei = Ef

\ 212 e

GMmR

mv − = 0

\ 212 emv = GMm

R

\ ve = 2GMR

The escape speed, for a planet, is thus seen todependupon themass and theradius of the planet. It is independent of the mass of the body thrown. The

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equationcanbeusedtoobtainescapespeedforanyastronomicalbodyprovidedwe substitute themassand radiusof the relevantastronomicalbody.

Substitutingthevaluesformassandradius,theescapespeedforearthis=11.2km/s

since g = 2GMR

,wehave

GM = gR2

\ ve = 2gR

The escape speed v does not depend on the direction in which the object (orprojectile) is thrown, orfired.However, it is helpful if theprojectile isfired inthedirectionof the rotationof theplanet.

?Did You Know?Satish Dhawan Space Centre SHAR, located at Sriharikota, a spindle shaped island on the East Coast of Andhra Pradesh, is the spaceport of India. This island was chosen in 1969 for setting up of a satellite launching station. Rockets are launched eastward to take advantage of the eastward speed due to Earth’s rotation.

Our universe has certain bodies called black holes. A black hole is formed when a star dies. At the end of its life, the star collapses into a highly dense sphere of large mass. The escape velocity on a black hole is so high that even light cannot escape from it and thus, it appears to be black and is invisible.

IllustratIon 9. earth has a mass 9 times and a radius twice that of mars. Calculate the escape velocity on Mars. (Given escape velocity on earth = 11.2 km/s.)

solutIon. Escape speedonearth

ve = 2GMR

Escape speedonMars

vm = 2GMR

′′

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\ vm = 29

2

GM

R

vm = 2 23

GMR

\ vm = 23

ve

vm = 5.28m/s

ConCept probe

1. Earth’s atmosphere does not contain lighter gases. Why?

2. Moon doesn’t have an atmosphere. Why?

6.14 motIon of sAtEllItE

A satellite can be viewed as a smaller mass body revolving around a moremassive body. Thus, the moon is a satellite of earth and earth itself can beregardedasa satelliteofSun.

orbital speed (Vo)

If a satellite ofmassm is revolving around a planet ofmassM in an orbit ofradius r, then for its equilibrium,wehave

20mv

r = 2

GMmr

\ v0 = GMr

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But r = R + h

whereR=radiusoftheplanetandh = height of satellite above the surfaceof theplanet

\ v0 = GMR h+

For smallheights,R + h ≈ R

\ v0 = GMr

gR=

wherewehave substitutedGM = GR2.

It is thus seen that fornear earth satellites

v0 = 11.2 km/s2 2ev =

=7.92km/s

Time Period (T)

We have T = 0

2 rvp

\ T= 2pr r

GM

T= 2p33 ( )2 R hr

GM GM+

= p

Forh << R,R + h ≈ R

\ T= 2p3R

GM

T= 2p3

2RgR

T= 2p Rg

Substituting the relevant values, we get T ≈ 84 minutes for the near earthsatellites.

Figure 16: Diagram showing a satellite of mass m revolving around a planet of mass M at a distance r from its centre

M

r

m

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Notethatthetimeperiodandorbitalspeedforasatellite is independentofthemassof thesatellite. In fact, itdependsonlyon themassof theplanetand theradiusof theorbitof the satellite.

Energy of a satellite

Asatellite in its orbit, possessesK.E. due to its rotational motion and P.E. due to itspositionwith respect to theEarth.

KineticEnergy

K.E. = 20

12

mv

Substituting thevalueof theorbital speedv0weget

K.E. = 12

GMmr

Potential energy

P.E. = – GMmr

total energy

E = K.E. + P.E.

E = 12

GMm GMmr r

+ −

E = –2

GMmr

Remember the relation

E = –K.E. = . .2

P E

The negative sign in the total energy indicates that the satellite – Earth system is a bound system. i.e. the satellite is bound to Earth and will not leave the orbit unless energy E is given to it.

The maximum value of the total energy is at infinity and this Max. Energy = 0

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types of satellites

Satellitesareusuallyclassifiedaccording to the typeoforbit theyare in.Thereare four types of orbit associated with satellites. It is the type of orbit thatdictatesa satellite’suse.

low Earth orbits

These satellites revolve in loworbits andhaveveryhighorbital speeds.Thesesatellites complete one revolution in about 90minutes. These orbits have veryshort lifetimesof theorderofweeks,The correspondingperiod isof theorderof decades for geostationary satellites. These satellites are generally used forspyingand formappingpurposes.

sun-synchronous orbits

Meteorologicalsatellitesareoftenplacedinasun-synchronousorheliosynchronousorbit. These satellites are in polar orbits. The orbits are designed so that thesatellite’s orientation is fixed relative to the Sun throughout the year, allowingvery accurate weather predictions to be made. Most meteorological satellitesorbit theEarth15 to16 timesperday.

Geosynchronous satellites

Earth-synchronous or geosynchronous satellites are placed into orbit so thattheirperiodof rotationexactlymatches theperiodofEarth’s rotationabout itsaxis. They take (nearly) 24 hours tomake one rotation. The plane of orbit forthesesatellites isgenerallynot theequatorialplane.Thesatellitesareplaced inhighly elliptical orbitswhich enable them to appear to hover above one pointon the Earth formost of the day. These satellites are used for communicationandGPS (GlobalPositioningSystem)purposes.

Geostationary satellites

The majority of communications satellites are in fact geostationary satellites.Geostationarysatellites likegeosynchronoussatellites, take (nearly)24hours tocomplete a rotation. However, geostationary satellites are positioned directlyover theequatorand theirpath follows theequatorialplaneof theEarth.Asa

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result geostationary satellites don’t move North or South during the day andappear tobepermanentlyfixedaboveonepointon theequatorof theEarth.

Most video or T.V. communications systems use geostationary satellites.Geosynchronousandgeostationarysatellitesare typicallyorbitingat35,788km(22,238miles) above the surfaceof theearth (42,000km from its centre).

Modern satellites have a mass of several thousand kilograms, compared withjust180kilogramsforthefirstmanmadesatellite,theSputnik.Modernsatellitesareplaced in spaceusing launchvehicles like theArianneRocketor theSpaceShuttle.Onceinspace,mostsatellitesobtaintheirpowerrequirementsfromtheSunusing solarpanels.

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Q.1 Whatwillbethenatureofgraphofthechangeintheaccelerationduetogravityanddepthbelow the surfaceof theearth?

Q.2 An object is taken from the surface of earth to a height equal to the radius oftheearth.Whatwillbe the change in itspotential energy?

Q.3 The time period of a satellite in a circular orbit of radiusR is T.Whatwill bethe timeperiodof another satellite ina circularorbitof radius4R?

Q.4 Howwill theorbitalvelocityofasatellitechangeif itweretobebroughtcloserto the surfaceof theplanet?

Q.5 Whatwillbe the timeperiodofapendulum ina satellite revolvingaround theearth?

Q.6 A man can jump six times as high on moon as on earth. Give reason for thesame.

Q.7 If thereisaforceofattractionbetweenallobjects,whydowenotfeelourselvessticking tomassivebuildings?

Q.8 A clock, controlled by a pendulum,when taken from the planes tomountains,slowsdown.Why?

Q.9 The sun is continuously attracting earthwith a strong gravitational force.Whydoes theearthnot fall towards the sun?

Q.10 An astronaut, while revolving in a circular orbit at a heightH (<<R) from thesurfaceoftheEarthhappenstoreleaseasmallpacketoutside.Calculatethetimethepacketwill take to reach theearth.

post contentstudent worksheet 1

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41Unit 6 : Gravitation

Q.1 Calculatetheforceofattractionbetweentwoballseachofmass1kg,whentheircentersare separatedby10 cm.G=6.67× 10–11 nm2/kg2.

Q.2 Threeequalmasses,ofmkgeach,arefixedattheverticesofanequilateraltriangleABCof side1m, as shown in thefigure.

(a) What is the force acting on a mass 2 m placedat the centroidG of the triangle?

(b) What would be this force if the mass atthe vertex A isdoubled?

Q.3 IftheradiusoftheEarthshrinksby2%,keepingthe mass constant, find the percentage changein thevalueof accelerationdue togravity.

Q.4 HowfarfromthesurfaceoftheEarthwouldtheaccelerationdue togravitybecome8%of itsvalueon the surfaceof theEarth?

Q.5 Findthevalueofaccelerationduetogravityinamineatadepthof80kmfromthe surfaceof earth. (Radiusof earth= 6400km)

Q.6 Abodyisprojectedverticallyupwardsfromthesurfaceoftheearthsoastojustreachaheightequaltotheradiusoftheearth.Neglectingfrictionofair,calculatethe speedwithwhich it was projected.Mass of earth = 6 × 1024 kg, Radius ofearth=6400km. (7.8 km/s)

Q.7 A rocket isfiredvertically from the surfaceofMarswith a speedof 2 km/s. If20% of its initial kinetic energy is lost due to Martian atmospheric resistance,howfarwilltherocketgofromthesurfaceofMarsbeforereturningback.Massofmars=6.4× 1023kgRadiusofmars=4000km,G=6.67× 10-11 nm2/kg2

Q.8 ASaturnyear is 29.5 times that ofEarth.How far is Saturn fromearth if earthis 1.5× 108kmaway fromsun?

Q.9 ThedistanceoftheplanetJupiterfromthesunis5.2timesthatoftheearth.Findtheperiodof the jupiter’s revolutionaround the sun. (11.85 years)


m

2 m

C mB m

G

A

Three equal masses are placed at the three vertices of the DABC. A mass 2

m is placed at the centroid G.

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Q.1 Assuming the earth to be sphere of radiusR (= 6360 km), calculate its averagedensity.

(given: G=6.67×10–11 nm2 kg–2 and g=9.8ms–2).

Q.2 Atwhatpoint,betweentheearthandthesun,wouldthenetgravitational forceonanyobject,due to these two,becomezero?

given:Radiusof earth’sorbit=1.5×1011m,

massof sum=2×1030 kg

andmassof earth=6×1024kg)

Q.3 Show that thevalueofaccelerationdue togravityatapointataheightnearly,equal to43%of theradiusof theearth,becomes50%of itsvalueonthesurfaceof theearth.

Q.4 Compare the time periods of a given simple pendulum on a planet X, and theearth.Given that themassandradiusof theplanetX,bothhavevalues3 timesthoseof theearth.

Q.5 Two particles, of masses m1 and m2, are placed a distance d apart. If thenet gravitational field due to these two masses becomes zero at a point p on the line goining them, show that the gravitational potential at p equal

– 1 2 1 22Gd

m m m m + − .

Q.6 Aparticle is imaginedtofalldown,fromrest, tothesurfaceof theearthfromapoint that is

(i) veryvery far away

(ii) at aheight equal to10R (R=Radiusof earth).

Calculatetheratioofthevelocitiesoftheparticle,onreachingthesurfaceoftheearth, in the twocases.

Q.7 Asatelliteisorbitingtheearthina‘closetotheearth’orbit.Whatislikelytohappen

tothissatelliteifitsspeedweretosuddenlyincreasebyafactorof 2 ?

Q.8 An astronaut, inside a man-made satellite, is regarded as being in a ‘state ofweightlessens.’Whycan’twesaythesameforanastronautonthemoonwhichis alsoa (natural) satelliteof theearth?


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Q.9 The most abundant gas in nature–hydrogen–is almost absent in the earth’satmosphere.Why?

Q.10 Wouldthe ‘escapespeed’,neededbyanobject toescapefromtheearth,changewitha change in the

(i) locationof thepointofprojection?

(ii) heightof thepointofprojection?

Explainyouranswers.

Q.11 The radii of two planets are r1 and r2 and their (surface) acceleration due togravity,havevaluesg1 and g2.Findtheratioofthe‘escapespeeds’forthesetwoplanets.

Q.12 The total energy, kinetic energy andpotential energy, of a satellite orbiting theearth, equal E, K and U respectively. State the relation between E, K and U. (U=2E=–2K)

Q.13 Ifasatellite,orbitingaroundtheearth,weretosuddenly‘loseheight’,howwouldthis affect the

(i) speedof the satellite?

(ii) timeperiodof the satellite?

Justifyyouranswers.

14. Twopoints,A and B, are atdistancesd1 and d2 (d2>d1) from the centre of theearth. The difference, in the gravitational potential energy (DU) of a mass m,when it ismoved frompointA topointB,withoutacceleration,wascalculatedby

(i) studentX as:DU = GMm1 2

1 1d d

−

(ii) studentY as:DU = mg (d2 –d1)

whereM is themass of the earth, ‘g’ is the acceleration due to gravity on thesurface of the earth and G is theuniversalgravitational constant.

comment on the range of value of d1 and d2 where

(a) both studentsare likely tobe correct

(b) onlyoneof the students is likely tobe correct.

Supportyouranswerwithadequate reasoning.

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Q.1 Thegravitationalfieldinaregionofspaceisgivenby ˆ ˆ(4 6 )E i j→= + N/kg.Show

thattheworkdone,bythegravitatationalfield,whenaparticleismovedontheline (6x+2y=10), is zero.

Q.2 Asatellite,ofmass1000kg,issupposedtoorbittheearthataheightof2000kmabove the earth’s surface. Find its (a) orbital speed (b)K.E. and (c) timeperiod.Massof earth=6× 1024kg.

Q.3 Ifearth,withmass6×1024kg,weretoreducetoablackholewithescapespeedequal to the speedof light,find the radiusof thatblackhole.

Q.4 A body is projected from the surface of earth with a speed equal to half theescape speedof earth.How farabove the surfaceof earthwill it reach?

Q.5 A particle is fired vertically upwardswith a speed of 15 km/s. Find the speedwithwhichitwilltravelinspace.Assumeonlyearth’sgravitationalinfluencetobeactingon theparticle.

Q.6 Twoparticles,ofmassesM and m,arekeptadistancexapart.Findthepositionof thepoint, if any,where thenet

(i) gravitationalfield

(ii) gravitationalpotential,

due to these twoparticles,becomeszero.


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multiPle CHoiCe QueStionS

Q.1 If the acceleration due to gravity on the surface of the earth (of radiusR) is g,thegain inpotentialenergyofabody,ofmassm,whenraisedfromthesurfaceto a height Rwouldbe

(a)mg4R (b)mg

2R (c)mgR (d)2 mgR

(e) 4mgR

Q.2 Accelerationdue togravity ‘g’ at the center of the earth is

(a) 9.8m/s2 (b) 4.9m/s2 (c) infinite (d)zero

Q.3 If the ratioof the radii of theorbits ofmars andearth around theSun is 1.526,the timeperiodofmars (in earthyears)wouldbe (nearly)

(a) 1.89years (b) 32years (c)45years (d)48years

Q.4 If a satellite of mass m is revolving around the earth at a distance r from its centre, its total energy is

(a) 2GMmr

− (b) GMmr

− (c)2

GMmr

− (d)2

GMmr

Q.5 Asatellite isorbitingaround theearth (andclose to it)witha totalenergyE. Ifthe satellite’skinetic energy ismade2E,

(a) itsperiodof revolutionwouldgetdoubled

(b) radiusof itsorbitwouldgethalved

(c) radiusof itsorbitwouldgetdoubled

(d) the satellitewouldescapeaway from theearth.

Q.6 Kepler’s second law isbasedupon the

(a) Newton’s second law

(b) Lawof conservationof energy

(c) Lawof conservationofmomentum

(d) lawof conservationof angularmomentum


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Q.7 The average density of a hypothetical planet is twice that of the earth. If theacceleration due to gravity on the surface of the planet is equal to that of theearthand if the radiusof theearth isR, the radiusof theplanetmustbe

(a)4R (b)

2R (c)2R (d)4R

Q.8 For a satellitemoving in an orbit around the earth, themagnitude of the ratioof itskinetic energy to itspotential energy is

(a) 12 (b) 2 (c) 1

2 (d)2

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