PHYSICS VOL.1

154∴ g = 43G π R ρ... (1)The acceleration due to gravity at at a depth from the surfaceQdof the Earth isg d= 22( Rd) −d GMwhere M d is the mass of the inner sphere of the Earth of radius (R d)−.M d = 43 ( πR d−)3 ρ∴ g d = 43G π (R– d)ρ... (2)dividing (2) by (1), d gR-d=gRg = gd1dR⎛⎞ −⎜⎟ ⎝⎠The value of acceleration due to gravity decreases with increase ofdepth.(iii) Variation of g with latitude (Non sphericity of the Earth)−The Earth is not a perfect sphere. It isan ellipsoid as shown in the Fig. 4.6. It isflattened at the poles where the latitude is 90oand bulged at the equator where the latitudeis 0 .oThe radius of the Earth at equatorialplane R e is greater than the radius along thepoles R p by about 21 km.We know that = g2 GMR∴ gα 2 1RThe value of varies inversely as the square of radius of thegEarth. The radius at the equator is the greatest. Hence the value of gR eR pFig.4.6 Non sphericity−of the Earth

155is minimum at the equator. The radius at poles is the least. Hence, thevalue of is maximum at the poles. The value of increases from theggequator to the poles.(iv) Variation of g with latitude (Rotation of the Earth)Let us consider the Earth as a homogeneous sphere of mass Mand radius . The Earth rotates about an axis passing through its northRand south poles. The Earth rotates fromwest to east in 24 hours. Its angularvelocity is 7.3 × 10 rad s .− 5− 1Consider a body of mass m onthe surface of the Earth at at aPlatitude . Let be the angular velocity.θωThe force (weight) F = mg acts alongPO. It could be resolved into tworectangular components (i) mg cos alongθPB and (ii) mg sin along θPA (Fig. 4.7).From the ∆OPB, it is found thatBP = cos . The particle describes aRθcircle with B as centre and radiusBP = R cos . θThe body at experiences a centrifugal force (outward force) PF Cdue to the rotation of the Earth.(i.e)F C = mRω 2cos θ .The net force along PC = mg cos θ − mRω 2 cos θ∴The body is acted upon by two forces along PA and PC .The resultant of these two forces isF= (mg sin √θ )2+(mg cosθ −mRω 2 cos )θ 2F = mg 2224222RωcosθRωcosθ1-+ggsince 242 Rgω is very small, the term 2422cosRg ωθ can be neglected.The force, F = mg222Rωcosθ1-g... (1)WNESODABθθPF cFig. 4.7 Rotation ofthe Earth

156If g ′ is the acceleration of the body at due to this force ,PFwe have, F = mg′... (2)by equating (2) and (1)mg′ = mg222Rcos 1ωθ−gg ′ = g22cos1ωθ⎛⎞ −⎜⎟ ⎜⎟ ⎝⎠RgCase (i)At the poles, = 90 ; θocos θ = 0∴ g ′ = gCase (ii)At the equator, = 0 ; θcos = 1 θ∴ ′ = gg21ω⎛⎞ −⎜⎟ ⎜⎟ ⎝⎠RgSo, the value of acceleration due to gravity is maximum at thepoles.4.5Gravitational fieldTwo masses separated by a distance exert gravitational forces onone another. This is called action at–a–distance. They interact eventhough they are not in contact. This interaction can also be explainedwith the field concept. A particle or a body placed at a point modifies aspace around it which is called gravitational field. When another particleis brought in this field, it experiences gravitational force of attraction.The gravitational field is defined as the space around a mass in which itcan exert gravitational force on other mass.4.5.1 Gravitational field intensityGravitational field intensity orstrength at a point is defined as the forceexperienced by a unit mass placed atthat point. It is denoted by E. It is avector quantity. Its unit is N kg .–1Consider a body of mass placedMat a point and another body of mass Qm placed at at a distance Prfrom .QFig. 4.8 Gravitational fieldMmrQP

157The mass M develops a field at and this field exerts a forceEPF = mE .The gravitational force of attraction between the masses m andM is F=2GM mrThe gravitational field intensity at P is = EFm∴E = 2 GMrGravitational field intensity is the measure of gravitational field.4.5.2 Gravitational potential differenceGravitational potential difference between two points is defined asthe amount of work done in moving unit massfrom one point to another point against thegravitational force of attraction.Consider two points and separatedABby a distance dr in the gravitational field.The work done in moving unit mass fromA to is Bdv = WA → BGravitational potential difference dv = E dr−Here negative sign indicates that work is done against thegravitational field.4.5.3 Gravitational potentialGravitational potential at a point is defined as the amount of workdone in moving unit mass from the point to infinity against the gravitationalfield. It is a scalar quantity. Its unit is N m kg .− 14.5.4 Expression for gravitational potential at a pointConsider a body of mass M at thepoint . Let be a point at a distance CPrfrom C. To calculate the gravitationalpotential at consider two points andPAB. The point A, where the unit mass isplaced is at a distance from .xCABdrFig. 4.9 Gravitationalpotential differenceABdxxrPC MFig. 4.10 Gravitational potential

158The gravitational field at is = AE2 GMxThe work done in moving the unit mass from to through aABsmall distance dx is dw = dv = −E.dxNegative sign indicates that work is done against the gravitationalfield.dv = −2 GMxdxThe work done in moving the unit mass from the point P toinfinity is ∞=−∫∫rdv2 GMx dxv = – GMrThe gravitational potential is negative, since the work is doneagainst the field. (i.e) the gravitational force is always attractive.4.5.5 Gravitational potential energyConsider a body of mass m placed at at a distance from thePrcentre of the Earth. Let the mass of the Earth be M .When the mass m is at A at adistance from , the gravitationalxQforce of attraction on it due to mass isMgiven by = F2GM mxThe work done in moving the massm through a small distance dx from to along the line joining theABtwo centres of masses m and M is dw = –F.dxNegative sign indicates that work is done against the gravitationalfield.∴dw = – 2GM mx.dxThe gravitational potential energy of a mass m at a distance r fromanother mass M is defined as the amount of work done in moving themass m from a distance r to infinity.The total work done in moving the mass m from a distance torFig. 4.11 Gravitationalpotential energyABdxxrPQEarth

159infinity isr = -dw∞∫∫ 2GM mdx xW = – GMm2 1 ∞∫rdxx*U = – GM mrGravitational potential energy is zero at infinity and decreases asthe distance decreases. This is due to the fact that the gravitationalforce exerted on the body by the Earth is attractive. Hence thegravitational potential energy is negative.U4.5.6Gravitational potential energy near the surface of the EarthLet the mass of the Earth be M and its radius be . Consider aRpoint on the surface of the Earth and another point at a height ABhabove the surface of the Earth. The work done inmoving the mass m from to is ABU = U UB−AU = −GMm11 - (Rh)R⎡⎤ ⎢⎥ +⎣⎦ U = GMm11 -(R h)R⎡⎤ ⎢⎥+⎣⎦ U = GMmhR(R +h)If the body is near the surface of the Earth,h is very small when compared with . Hence (RR+h)could be taken as .R∴U = 2GM mhRU = mgh2 GM = gR⎛⎞ ⎜⎟ ⎝⎠ ∵4.6Inertial massAccording to Newton’s second law of motion ( = Fma), the mass ofa body can be determined by measuring the acceleration produced in it* Potential energy is represented by U (Upsilon).BAhOREarthFig. 4.12 Gravitationalpotential energynear the surface ofthe Earth

160by a constant force. (i.e) = / . mF a Intertial mass of a body is a measureof the ability of a body to oppose the production of acceleration in it byan external force.If a constant force acts on two masses m A and m B and producesaccelerations a A and a B respectively, then, = Fm a = m aA AB B= ABBAmama∴The ratio of two masses is independent of the constant force. If thesame force is applied on two different bodies, the inertial mass of thebody is more in which the acceleration produced is less.If one of the two masses is a standard kilogram, the unknownmass can be determined by comparing their accelerations.4.7Gravitational massAccording to Newton’s law of gravitation, the gravitational force ona body is proportional to its mass. We can measure the mass of a bodyby measuring the gravitational force exerted on it by a massive body likeEarth. Gravitational mass is the mass of a body which determines themagnitude of gravitational pull between the body and the Earth. This isdetermined with the help of a beam balance.If F A and F B are the gravitational forces of attraction on the twobodies of masses m A and m B due to the Earth, thenF A = 2 AGm MR and F B = 2 BGm MRwhere is mass of the Earth, is the radius of the Earth and is theMRGgravitational constant.∴ = AABBmFmFIf one of the two masses is a standard kilogram, the unknownmass can be determined by comparing the gravitational forces.4.8Escape speedIf we throw a body upwards, it reaches a certain height and thenfalls back. This is due to the gravitational attraction of the Earth. If wethrow the body with a greater speed, it rises to a greater height. If the

161body is projected with a speed of 11.2 km/s, it escapes from the Earthand never comes back. The escape speed is the minimum speed withwhich a body must be projected in order that it may escape from thegravitational pull of the planet.Consider a body of mass m placed on the Earth’s surface. Thegravitational potential energy is E = – PGM mRwhere M is the mass of the Earth and is its radius.RIf the body is projected up with a speed ve, the kinetic energy isE K = 2e1mv2∴ the initial total energy of the body isEi = 2e1mv2 – GM mR... (1)If the body reaches a height above the Earth’s surface, thehgravitational potential energy isE P = – (Rh) GM m+Let the speed of the body at the height is , then its kinetic energy is,vE K = 212mv.Hence, the final total energy of the body at the height isEf = 212mv – (Rh) GM m+... (2)We know that the gravitational force is a conservative force andhence the total mechanical energy must be conserved.∴E = Eif(i.e) 22e mvGMmmvGMm - = - 2R2(R +h)The body will escape from the Earth’s gravity at a height wherethe gravitational field ceases out. (i.e) = h∞. At the height = h∞, thespeed of the body is zero.v

162Thus2e mvGMm = 02R−ve = 2GMRFrom the relation = g2 GMR , we get GM = gR2Thus, the escape speed is v = e2gRThe escape speed for Earth is 11.2 km/s, for the planet Mercuryit is 4 km/s and for Jupiter it is 60 km/s. The escape speed for themoon is about 2.5 km/s.4.8.1An interesting consequence of escape speed with theatmosphere of a planetWe know that the escape speed is independent of the mass of thebody. Thus, molecules of a gas and very massive rockets will require thesame initial speed to escape from the Earth or any other planet ormoon.The molecules of a gas move with certain average velocity, whichdepends on the nature and temperature of the gas. At moderatetemperatures, the average velocity of oxygen, nitrogen andcarbon–di–oxide is in the order of 0.5 km/s to 1 km/s and for lightergases hydrogen and helium it is in the order of 2 to 3 km/s. It is clearthat the lighter gases whose average velocities are in the order of theescape speed, will escape from the moon. The gravitational pull of themoon is too weak to hold these gases. The presence of lighter gases inthe atmosphere of the Sun should not surprise us, since the gravitationalattraction of the sun is very much stronger and the escape speed is veryhigh about 620 km/s.4.9 SatellitesA body moving in an orbit around a planet is called satellite. Themoon is the natural satellite of the Earth. It moves around the Earth oncein 27.3 days in an approximate circular orbit of radius 3.85 × 10 km.5The first artificial satellite Sputnik was launched in 1956. India launchedits first satellite Aryabhatta on April 19, 1975.

1634.9.1 Orbital velocityArtificial satellites are made to revolve in an orbit at a height offew hundred kilometres. At this altitude, the friction due to air isnegligible. The satellite is carried by a rocket to the desired height andreleased horizontally with a high velocity, so that it remains moving ina nearly circular orbit.The horizontal velocity that has to be imparted to a satellite at thedetermined height so that it makes a circular orbit around the planet iscalled orbital velocity.Let us assume that a satellite of mass moves around the Earthmin a circular orbit of radius with uniform speed . Let the satellite bervoat a height from the surface of the Earth. Hence, hr = R+h, where Ris the radius of the Earth.The centripetal force required to keep the satellite in circularorbit is = F22oomvmv = rR +hThe gravitational force between the Earth and the satellite isF = 22GMmGMm = r(R + h)For the stable orbital motion,2o2mvGMm = R+ h(R +h)vo = GMR+hSince the acceleration due togravity on Earth’s surface is = g2 GMR,vo = 2gRR+hIf the satellite is at a height of few hundred kilometres(say 200 km), (R+h) could be replaced by .R∴ orbital velocity, vo = gRIf the horizontal velocity (injection velocity) is not equal to thecalculated value, then the orbit of the satellite will not be circular. If thehRrEarthv oFig. 4.13 Orbital Velocity

164injection velocity is greater than the calculated value but not greaterthan the escape speed ( =ve2 vo), the satellite will move along an ellipticalorbit. If the injection velocity exceeds the escape speed, the satellite willnot revolve around the Earth and will escape into the space. If theinjection velocity is less than the calculated value, the satellite will fallback to the Earth.4.9.2 Time period of a satelliteTime taken by the satellite to complete one revolution round theEarth is called time period.Time period, = Tcircumference of the orbitorbital velocityT = oo2 r π2 (R +h)π= vv where is the radius of the orbit which is equalrto (R+h).T = 2 π(R+h)R+ hGMoGM v=R+h⎡⎤ ⎢⎥ ⎣⎦ ∵T = 2π 3(R+h)GMAs GM = gR , T2 = 2π 32 (R+h)gRIf the satellite orbits very close to the Earth, then h << R∴ = 2 TπRg4.9.3 Energy of an orbiting satelliteA satellite revolving in a circular orbit round the Earth possessesboth potential energy and kinetic energy. If is the height of the satellitehabove the Earth’s surface and is the radius of the Earth, then theRradius of the orbit of satellite is r = R+h.If m is the mass of the satellite, its potential energy is,E P = -GMm-GMm = r(R + h)where is the mass of the Earth. The satellite moves with an orbitalMvelocity ofvo = GM(R+h)

165Hence, its kinetic energy is, E = K212o mvE = KGMm2(R+h)The total energy of the satellite is, E = E + EPKE = −GMm2(R+h)The negative value of the total energy indicates that the satelliteis bound to the Earth.4.9.4Geo–stationary satellitesA geo-stationary satellite is a particular type used in television andtelephone communications. A number of communication satellites whichappear to remain in fixed positions at a specified height above the equatorare called synchronous satellites or geo-stationary satellites. Sometelevision programmes or events occuring in other countries are oftentransmitted ‘live’ with the help of these satellites.For a satellite to appear fixed at a position above a certain placeon the Earth, its orbital period around the Earth must be exactly equalto the rotational period of the Earth about its axis.Consider a satellite of mass m moving in a circular orbit around theEarth at a distance from the centre of the Earth. For synchronisation, itsrperiod of revolution around the Earth must be equal to the period of rotationof the Earth (ie) 1 day = 24 hr = 86400 seconds.The speed of the satellite in its orbit isv = Circumference of orbitTime periodv = 2 rπTThe centripetal force is = F2mvr∴ = F22 4mπrTThe gravitational force on the satellite due to the Earth is2GMmF=rFor the stable orbital motion 22 4mπrT = 2GMmr (or) r 3= 22 GMT4 π

166We know that, g = 2 GMR∴ r 3 = 222 gR T4 πThe orbital radius of the geo- stationary satellite is, = r1/3222 gR T4 π⎛⎞ ⎜⎟ ⎜⎟ ⎝⎠ This orbit is called parking orbit of the satellite.Substituting = 86400 s, = 6400 km and = 9.8 m/s , theTRg2radius of the orbit of geo-stationary satellite is calculated as 42400 km.∴ The height of the geo-stationary satellite above the surface ofthe Earth is = hr−R = 36000 km.If a satellite is parked at this height, it appears to be stationary.Three satellites spaced at 120 intervals each above Atlantic, Pacific andoIndian oceans provide a worldwide communication network.4.9.5 Polar satellitesThe polar satellites revolve around the Earth in a north south−orbit passing over the poles as the Earth spins about its north south−axis.The polar satellites positioned nearly 500 to 800 km above theEarth travels pole to pole in 102 minutes. The polar orbit remains fixedin space as the Earth rotates inside the orbit. As a result, most of theearth’s surface crosses the satellite in a polar orbit. Excellent coverageof the Earth is possible with this polar orbit. The polar satellites areused for mapping and surveying.4.9.6 Uses of satellites(i) Satellite communicationCommunication satellites are used to send radio, television andtelephone signals over long distances. These satellites are fitted withdevices which can receive signals from an Earth – station and transmitthem in different directions.(ii) Weather monitoringWeather satellites are used to photograph clouds from space andmeasure the amount of heat reradiated from the Earth. With thisinformation scientists can make better forecasts about weather. You

167might have seen the aerial picture of our country taken by the satellites,which is shown daily in the news bulletin on the television and in thenews papers.(iii) Remote sensingCollecting of information about an object without physical contactwith the object is known as remote sensing. Data collected by theremote sensing satellities can be used in agriculture, forestry, droughtassessment, estimation of crop yields, detection of potential fishing zones,mapping and surveying.(iv) Navigation satellitesThese satellites help navigators to guide their ships or planes inall kinds of weather.4.9.7 Indian space programmeIndia recognised the importance of space science and technologyfor the socio-economic development of the society soon after the launchof Sputnik by erstwhile USSR in 1957. The Indian space efforts startedin 1960 with the establishment of Thumba Equatorial Rocket LaunchingStation near Thiruvananthapuram for the investigation of ionosphere.The foundation of space research in India was laid by Dr. VikramSarabai, father of the Indian space programme. Initially, the spaceprogramme was carried out by the Department of Atomic Energy. Aseparate Department of Space (DOS) was established in June 1972.Indian Space Research Organisation (ISRO) under DOS executes spaceprogramme through its establishments located at different places inIndia (Mahendragiri in Tamil Nadu, Sriharikota in Andhra Pradesh,Thiruvananthapuram in Kerala, Bangalore in Karnataka, Ahmedabad inGujarat, etc...). India is the sixth nation in the world to have the capabilityof designing, constructing and launching a satellite in an Earth orbit.The main events in the history of space research in India are given below:Indian satellites1. Aryabhatta - The first Indian satellite was launched on April 19,1975.2. Bhaskara - 1

1683. Rohini4. APPLE - It is the abbreviation of Ariane Passenger Pay LoadExperiment. APPLE was the first Indian communication satellite put ingeo - stationary orbit.5. Bhaskara - 26. INSAT - 1A, 1B, 1C, 1D, 2A, 2B, 2C, 2D, 3A, 3B, 3C, 3D, 3E(Indian National Satellite). Indian National Satellite System is a jointventure of Department of Space, Department of Telecommunications,Indian Meteoro-logical Department and All India Radio and Doordarshan.7. SROSS - A, B, C and D (Stretched Rohini Satellite Series)8. IRS - 1A, 1B, 1C, 1D, P2, P3, P4, P5, P6 (Indian Remote SensingSatellite)Data from IRS is used for various applications like droughtmonitoring, flood damage assessment, flood risk zone mapping, urbanplanning, mineral prospecting, forest survey etc.9. METSAT (Kalpana - I) - METSAT is the first exclusivemeteorological satellite.10. GSAT-1, GSAT-2 (Geo-stationary Satellites)Indian Launch Vehicles (Rockets)1. SLV - 3 - This was India’s first experimental Satellite LaunchVehicle. SLV - 3 was a 22 m long, four stage vehicle weighing 17 tonne.All its stages used solid propellant.

1692. ASLV - Augmented Satellite Launch Vehicle. It was a five stagesolid propellant vehicle, weighing about 40 tonnes and of about 23.8 mlong.3. PSLV - The Polar Satellite Launch Vehicle has four stages usingsolid and liquid propellant systems alternately. It is 44.4 m tall weighingabout 294 tonnes.4. GSLV - The Geosynchronous Satellite Launch Vehicle is a 49m tall, three stage vehicle weighing about 414 tonnes capable of placingsatellite of 1800 kg.India’s first mission to moonISRO has a plan to send an unmanned spacecraft to moon in theyear 2008. The spacecraft is named as CHANDRAYAAN-1. This programmewill be much useful in expanding scientific knowledge about the moon,upgrading India’s technological capability and providing challengingopportunities for planetory research for the younger generation. Thisjourney to moon will take 5½ days.CHANDRAYAAN - 1 will probe the moon by orbiting it at the lunarorbit of altitude 100 km. This mission to moon will be carried by PSLVrocket.4.9.8 WeightlessnessTelevision pictures and photographs show astronauts and objectsfloating in satellites orbiting the Earth. This apparent weightlessness issometimes explained wrongly as zero–gravity condition. Then, what shouldbe the reason?Consider the astronaut standing on the ground. He exerts a force(his weight) on the ground. At the same time, the ground exerts anequal and opposite force of reaction on the astronaut. Due to this forceof reaction, he has a feeling of weight.When the astronaut is in an orbiting satellite, both the satelliteand astronaut have the same acceleration towards the centre of theEarth. Hence, the astronaut does not exert any force on the floor of thesatellite. So, the floor of the satellite also does not exert any force ofreaction on the astronaut. As there is no reaction, the astronaut has afeeling of weightlessness.

1704.9.9 Rockets principle− − − − −A rocket is a vehicle which propels itself by ejecting apart of its mass. Rockets are used to carry the payloads(satellites). We have heard of the PSLV and GSLV rockets.All of them are based on Newton’s third law of motion.Consider a hollow cylindrical vessel closed on bothends with a small hole at one end, containing a mixture ofcombustible fuels (Fig. 4.14). If the fuel is ignited, it isconverted into a gas under high pressure. This high pressurepushes the gas through the hole with an enormous force.This force represents the action A. Hence an opposite force,which is the reaction R, will act on the vessel and make itto move forward.The force (F m) on the escaping mass of gases andhence the rocket is proportional to the product of the massof the gases discharged per unit time dmdt⎛⎞ ⎜⎟ ⎝⎠ and the velocitywith which they are expelled ( )v(i.e) F mαdmdtv(v) α⎡⎤ ⎢⎥ ⎣⎦ ∵d FmdtThis force is known as momentum thrust. If thepressure ( ) of the escaping gases differs from the pressureP e( ) in the region outside the rocket, there is an additional thrust calledP othe velocity thrust ( ) acts. It is given by F vF = A Pv ( e−P o) where is theAarea of the nozzle through which the gases escape. Hence, the totalthrust on the rocket is F = F + Fmv4.9.10 Types of fuelsThe hot gases which are produced by the combustion of a mixture ofsubstances are called propellants. The mixture contains a fuel which burnsand an oxidizer which supplies the oxygen necessary for the burning ofthe fuel. The propellants may be in the form of a solid or liquid.Fig. 4.14Principleof RocketFFAR

1714.9.11 Launching a satelliteTo place a satellite at a height of 300 km, the launching velocityshould atleast be about 8.5 km s or 30600 kmph. If this high velocity–1is given to the rocket at the surface of the Earth, the rocket will beburnt due to air friction. Moreover, such high velocities cannot bedeveloped by single rocket. Hence, multistage rockets are used.To be placed in an orbit, a satellite must be raised to the desiredheight and given the correct speed and direction by the launching rocket(Fig. 4.15).At lift off, the rocket, with a manned or unmanned satellite on top,is held down by clamps on the launching pad. Now the exhaust gasesbuilt up an upward thrust which exceeds the rocket’s weight. The clamps−are then removed by remote control and the rocket accelerates upwards.4.15 Launching a satelliteSatelliteSatelliteThird StageSecond StageSecond StageCombustionchamberFirst StageFirst StageCombustionchamber

172To penetrate the dense lower part of the atmosphere, initially therocket rises vertically and then tilted by a guidance system. The first stagerocket, which may burn for about 2 minutes producing a speed of3 km s , lifts the vehicle to a height of about 60 km and then separates–1and falls back to the Earth.The vehicle now goes to its orbital height, say 160 km, where itmoves horizontally for a moment. Then the second stage of the rocketfires and increases the speed that is necessary for a circular orbit. Byfiring small rockets with remote control system, the satellite is separatedfrom the second stage and made to revolve in its orbit.4.10 The UniverseThe science which deals with the study of heavenly bodies inrespect of their motions, positions and compositions is known asastronomy. The Sun around which the planets revolve is a star. It is oneof the hundred billion stars that comprise our galaxy called the MilkyWay. A vast collection of stars held together by mutual gravitation iscalled a galaxy. The billions of such galaxies form the universe. Hence,the Solar system, stars and galaxies are the constituents of the universe.4.10.1 The Solar systemThe part of the universe in which the Sun occupies the centralposition of the system holding together all the heavenly bodies such asplanets, moons, asteroids, comets ... etc., is called Solar system. Thegravitational attraction of the Sun primarily governs the motion of theplanets and other heavenly bodies around it. Mercury, Venus, Earth,Mars, Jupiter, Saturn, Uranus, Neptune and Pluto are the nine planetsthat revolve around the Sun. We can see the planet Venus in the earlymorning in the eastern sky or in the early evening in the western sky.The planet Mercury can also be seen sometimes after the sunset in theWest or just before sunrise in the East. From the Earth, the planet Marswas visibly seen on 27 August 2003. The planet Mars came closer toththe Earth after 60,000 years from a distance of 380 × 10 km to a6nearby distance of 55.7 × 10 km. It would appear again in the year62287.Some of the well known facts about the solar system have beensummarised in the Table 4.1.

Table 4.1 Physical properties of the objects in the Solar system(NOR FOR EXAMINATION)gE = 9.8 m s–2, 1 year = 365.257 days ; 1 AU = 1.496 × 108 km ; RE = 6378 km ; ME = 5.98 × 1024 kgMercury0.0560.3870.24158.6 days5,4000.380.3674Nil0.060Venus0.8150.7230.615243 days51000.960.88610.5CO20.850(E → W)Earth1.0001.0001.00023 hours 56.1 minutes55201.001.00011.2N2 O20.401Mars0.1071.5241.88124 hours 27.4 minutes39700.530.3835C20.152Ceres (Asteroid)0.00012.7674.60390 hours33400.0550.18–––Jupiter317.95.20311.8649 hours 50.5 minutes133011.232.52260He, CH4, NH30.4538Saturn95.29.54029.4610 hours 14 minutes7009.411.07437He, CH40.6130 + 3 ringsUranus14.619.1884.0110 hours 49 minutes13303.980.92221H2, He,0.3524 (E → W)CH4Neptune17.230.07164.115 hours16603.881.43522.5H2, He, CH40.352Pluto0.00239.442476.39 days20300.1790.0511.1–0.140Moon0.0123––27.32 days33400.270.1702.5Nil0.07–ObjectsMass in Earth unitSemi-major axis oforbit (AU)Period of revolutionin yearsRotation periodMean density(kg m )–3Radius in Earth unitg in Earth unitEscape speed(km/s)AtmosphereAlbedoNumber of satellites173

1744.10.2 Planetary motionThe ancient astronomers contributed a great deal by identifyingthe planets in the solar system and carefully plotting the variations intheir positions of the sky over the periods of many years. These dataeventually led to models and theories of the solar system.The first major theory, called the Geo-centric theory was developedby a Greek astronomer, Ptolemy. The Earth is considered to be thecentre of the universe, around which all the planets, the moons and thestars revolve in various orbits. The great Indian Mathematician andastronomer Aryabhat of the 5th century AD stated that the Earth rotatesabout its axis. Due to lack of communication between the scientists ofthe East and those of West, his observations did not reach thephilosophers of the West.Nicolaus Copernicus, a Polish astronomer proposed a new theorycalled Helio-centric theory. According to this theory, the Sun is at restand all the planets move around the Sun in circular orbits. A Danishastronomer Tycho Brahe made very accurate observations of the motionof planets and a German astronomer Johannes Kepler analysed Brahe’sobservations carefully and proposed the empirical laws of planetarymotion.Kepler’s laws of planetary motion(i)The law of orbitsEach planet moves in an elliptical orbit with the Sun at one focus.A is a planet revolving roundthe Sun. The position P of the planetwhere it is very close to the Sun is Pknown as perigee and the positionQ of the planet where it is farthestfrom the Sun is known as apogee.(ii)The law of areasThe line joining the Sun and the planet (i.e radius vector) sweepsout equal areas in equal interval of times.The orbit of the planet around the Sun is as shown in Fig. 4.17.The areas A and A are swept by the radius vector in equal times. The12planet covers unequal distances S and S in equal time. This is due to12Fig. 4.16 Law of orbitsAQSunMajor axis

175the variable speed of the planet.When the planet is closest to theSun, it covers greater distancein a given time. Hence, the speedis maximum at the closestposition. When the planet is faraway from the Sun, it coverslesser distance in the same time. Hence the speed is minimum at thefarthest position.Proof for the law of areasConsider a planet moving from to AB. The radius vector OAsweeps a small angle d θ at the centre in a small interval of time dt.From the Fig. 4.18, AB = rd θ. The small area dA swept by theradius is,dA = 12 × r × rdθDividing by dt on both sides2dA1d θ = ×r ×dt2dt2dA1 = rωdt2 where ω isthe angular velocity.The angular momentum is given by L = mr2ω∴ ωr 2 = LmHence, dAdt = 1L2mSince the line of action of gravitational force passes through theaxis, the external torque is zero. Hence, the angular momentum isconserved.dAdt ∴ = constant.(i.e) the area swept by the radius vector in unit time is the same.(iii)The law of periodsThe square of the period of revolution of a planet around the SunPSunHigher SpeedLower SpeedS 1S 2A 1A 2Fig. 4.17 Law of areasdrA BOFig. 4.18 Proof for the lawof areas

176is directly proportional to the cube of the mean distance between theplanet and the Sun.(i.e)T 2α r 323 Tr = constantProof for the law of periodsLet us consider a planet of mass moving with the velocity Pmvaround the Sun of mass M in a circular orbit of radius .rThe gravitational force of attraction of the Sun on the planet is,F = 2GMmrThe centripetal force is, = F2mvrEquating the two forces22mvGMm = rr2GMv = r.....(1)If be the period of revolution of theTplanet around the Sun, thenv = 2 r πT.....(2)Substituting (2) in (1) 2224 πrGM= rT322rGM= T4 πGM is a constant for any planet∴ T 2αr34.10.3 Distance of a heavenly body in the Solar systemThe distance of a planet can be accurately measured by the radarecho method. In this method, the radio signals are sent towards theplanet from a radar. These signals are reflected back from the surfaceof a planet. The reflected signals or pulses are received and detected onrSunvPlanetFig. 4.19 Proof for thelaw of periods

177Earth. The time t taken by the signal in going to the planet and comingback to Earth is noted. The signal travels with the velocity of the light .cThe distance of the planet from the Earth is given by = ss2 ct4.10.4 Size of a planetIt is possible to determine the size of any planetonce we know the distance of the planet. The image ofSevery heavenly body is a disc when viewed through aoptical telescope. The angle between two extreme pointsθA and B on the disc with respect to a certain point on theEarth is determined with the help of a telescope. Theangle is called the angular diameter of the planet. Theθlinear diameter of the planet is then given bydd = distance × angular diameterd = × sθ4.10.5 Surface temperatures of the planetsThe planets do not emit light of their own. They reflect the Sun’slight that falls on them. Only a fraction of the solar radiation is absorbedand it heats up the surface of the planet. Then it radiates energy. Wecan determine the surface temperature of the planet using Stefan’sTlaw of radiation = Eσ T 4 where is the Stefan’s constant and is theσEradiant energy emitted by unit area in unit time.In general, the temperature of the planets decreases as we goaway from the Sun, since the planets receive less and less solar energyaccording to inverse square law. Hence, the planets farther away fromthe Sun will be colder than those closer to it. Day temperature ofMercury is maximum (340 C) since it is a planet closest to the Sun andothat of Pluto is minimum ( 240 C). However Venus is an exception as−oit has very thick atmosphere of carbon di oxide. This acts as a blanket− −and keeps its surface hot. Thus the temperature of Venus is comparitivelylarge of the order of 480 C.o4.10.6 Mass of the planets and the SunIn the universe one heavenly body revolves around another massiveheavenly body. (The Earth revolves around the Sun and the moon revolvesAB dPSEarthFig. 4.20 Sizeof a planet

178around the Earth). The centripetal force required by the lighter body torevolve around the heavier body is provided by the gravitational force ofattraction between the two. For an orbit of given radius, the mass of theheavier body determines the speed with which the lighter body mustrevolve around it. Thus, if the period of revolution of the lighter bodyis known, the mass of the heavier body can be determined. For example,in the case of Sun planet system, the mass of the Sun −M can becalculated if the distance of the Sun from the Earth , the period ofrrevolution of the Earth around the Sun and the gravitational constantTG are known using the relation M = 2324 πrGT4.10.7 AtmosphereThe ratio of the amount of solar energy reflected by the planet tothat incident on it is known as albedo. From the knowledge of albedo,we get information about the existence of atmosphere in the planets.The albedo of Venus is 0.85. It reflects 85% of the incident light, thehighest among the nine planets. It is supposed to be covered with thicklayer of atmosphere. The planets Earth, Jupiter, Saturn, Uranus andNeptune have high albedoes, which indicate that they possessatmosphere. The planet Mercury and the moon reflect only 6% of thesunlight. It indicates that they have no atmosphere, which is alsoconfirmed by recent space probes.There are two factors which determine whether the planets haveatmosphere or not. They are (i) acceleration due to gravity on its surfaceand (ii) the surface temperature of the planet.The value of for moon is very small (¼th of the Earth).gConsequently the escape speed for moon is very small. As the averagevelocity of the atmospheric air molecules at the surface temperature ofthe moon is greater than the escape speed, the air molecules escape.Mercury has a larger value of than moon. Yet there is nogatmosphere on it. It is because, Mercury is very close to the Sun andhence its temperature is high. So the mean velocity of the gas moleculesis very high. Hence the molecules overcome the gravitational attractionand escape.

1794.10.8 Conditions for life on any planetThe following conditions must hold for plant life and animal life toexist on any planet.(i) The planet must have a suitable living temperature range.(ii) The planet must have a sufficient and right kind of atmosphere.(iii) The planet must have considerable amount of water on itssurface.4.10.9 Other objects in the Solar system(i) AsteroidsAsteroids are small heavenly bodies which orbit round the Sunbetween the orbits of Mars and Jupiter. They are the pieces of muchlarger planet which broke up due to the gravitational effect of Jupiter.About 1600 asteroids are revolving around the Sun. The largest amongthem has a diameter of about 700 km is called Ceres. It circles the Sunonce in every 4½ years.(ii) CometsA comet consists of a small mass of rock like material surrounded−by large masses of substances such as water, ammonia and methane.These substances are easily vapourised. Comets move round the Sun inhighly elliptical orbits and most of the time they keep far away from theSun. As the comet approaches the Sun, it is heated by the Sun’s radiantenergy and vapourises and forms a head of about 10000 km in diameter.The comet also develops a tail pointing away from the Sun. Some cometsare seen at a fixed regular intervals of time. Halley’s comet is a periodiccomet which made its appearance in 1910 and in 1986. It would appearagain in 2062.(iii) Meteors and MeteoritesThe comets break into pieces as they approach very close to theSun. When Earth’s orbit cross the orbit of comet, these broken piecesfall on the Earth. Most of the pieces are burnt up by the heat generateddue to friction in the Earth’s atmosphere. They are called meteors(shooting stars). We can see these meteors in the sky on a clear moonlessnight.

180Some bigger size meteors may survive the heat produced by frictionand may not be completely burnt. These blazing objects which manageto reach the Earth are called meteorites.The formation of craters on the surface of the moon, Mercury andMars is due to the fact that they have been bombarded by large numberof meteorites.4.10.10 StarsA star is a huge, more or less spherical mass of glowing gasemitting large amount of radiant energy. Billions of stars form a galaxy.There are three types of stars. They are (i) double and multiple stars (ii)intrinsically variable stars and (iii) Novae and super novae.In a galaxy, there are only a few single stars like the Sun. Majorityof the stars are either double stars (binaries) or multiple stars. Thebinary stars are pairs of stars moving round their common centre ofgravity in stable equilibrium. An intrinsically variable star shows variationin its apparent brightness. Some stars suddenly attain extremely largebrightness, that they may be seen even during daytime and then theyslowly fade away. Such stars are called novae. Supernovae is a largenovae.The night stars in the sky have been given names such as Sirius(Vyadha), Canopas (Agasti), Spica (Chitra), Arcturus (Swathi), Polaris(Dhruva) ... etc. After the Sun, the star Alpha Centauri is nearest toEarth.SunThe Sun is extremely hot and self luminous body. It is made of−hydrogeneous matter. It is the star nearest to the Earth. Its mass isabout 1.989 × 10 kg. Its radius is about 6.95 × 10 m. Its distance308 from the Earth is 1.496 × 1011 m. This is known as astronomicalunit (AU). Light of the sun takes 8 minutes 20 seconds to reach theEarth. The gravitational force of attraction on the surface of the Sun isabout 28 times that on the surface of the Earth.Sun rotates about its axis from East to West. The period ofrevolution is 34 days at the pole and 25 days at the equator. The densityof material is one fourth that of the Earth. The inner part of the Sun

181is a bright disc of temperature 14 × 10 K known as photosphere. The6 outer most layer of the Sun of temperature 6000 K is called chromosphere.4.10.11 ConstellationsMost of the stars appear to be grouped together forming interestingpatterns in the sky. The configurations or groups of star formed in thepatterns of animals and human beings are called constellations. Thereare 88 constellations into which the whole sky has been divided.If we look towards the northern sky on a clear moonless nightduring the months of July and August, a group of seven bright starsresembling a bear, the four stars forming a quadrangle form the body,the remaining three stars make the tail and some other faint stars formthe paws and head of the bear. This constellation is called Ursa Majoror Saptarishi or Great Bear. The constellation Orion resembles the figureof a hunter and Taurus (Vrishabha) resembles the shape of a bull.4.10.12 GalaxyA large band of stars, gas and dust particles held together bygravitational forces is called a galaxy. Galaxies are really complex innature consisting of billions of stars. Some galaxies emit a comparativelysmall amount of radio radiations compared to the total radiations emitted.They are called normal galaxies. Our galaxy Milky Way is a normalgalaxy spiral in shape.The nearest galaxy to us known as Andromeda galaxy, is also anormal galaxy. It is at a distance of 2 × 10 light years. (The distance6travelled by the light in one year [9.467 × 10 km] is called light year).12Some galaxies are found to emit millions of times more radio wavescompared to normal galaxies. They are called radio galaxies.4.10.13 Milky Way galaxyMilky Way looks like a stream of milk across the sky. Some of theimportant features are given below.(i) Shape and sizeMilky Way is thick at the centre and thin at the edges. Thediameter of the disc is 10 light years. The thickness of the Milky Way5varies from 5000 light years at the centre to 1000 light years at the

182position of the Sun and to500 light years at the edges.The Sun is at a distance ofabout 27000 light yearsfrom the galactic centre.(ii) Interstellar matterThe interstellar spacein the Milky Way is filledwith dust and gases calledinter stellar matter. It is found that about 90% of the matter is in theform of hydrogen.(iii) ClustersGroups of stars held by mutual gravitational force in the galaxyare called star clusters. A star cluster moves as a whole in the galaxy.A group of 100 to 1000 stars is called galactic cluster. A group of about10000 stars is called globular cluster.(iv) RotationThe galaxy is rotating about an axis passing through its centre. Allthe stars in the Milky Way revolve around the centre and complete onerevolution in about 300 million years. The Sun, one of the many starsrevolves around the centre with a velocity of 250 km/s and its periodof revolution is about 220 million years.(v) MassThe mass of the Milky Way is estimated to be 3 × 1041kg.4.10.14 Origin of the UniverseThe following three theories have been proposed to explain theorigin of the Universe.(i) Big Bang theoryAccording to the big bang theory all matter in the universe wasconcentrated as a single extremely dense and hot fire ball. An explosionoccured about 20 billion years ago and the matter was broken intopieces, thrown off in all directions in the form of galaxies. Due to<>10 Light years5SunGalactic centre<>27,000Light yearsFig. 4.21 Milky Way galaxy

183continuous movement more and more galaxies will go beyond theboundary and will be lost. Consequently, the number of galaxies perunit volume will go on decreasing and ultimately we will have an emptyuniverse.(ii) Pulsating theorySome astronomers believe that if the total mass of the universe ismore than a certain value, the expansion of the galaxies would bestopped by the gravitational pull. Then the universe may again contract.After it has contracted to a certain critical size, an explosion againoccurs. The expansion and contraction repeat after every eight billionyears. Thus we may have alternate expansion and contraction givingrise to a pulsating universe.(iii) Steady state theoryAccording to this theory, new galaxies are continuously createdout of empty space to fill up the gap caused by the galaxies whichescape from the observable part of the universe. This theory, thereforesuggests that the universe has always appeared as it does today and therate of expansion has been the same in the past and will remain thesame in future. So a steady state has been achieved so that the totalnumber of galaxies in the universe remains constant.

184Solved Problems4.1Calculate the force of attraction between two bodies, each of mass200 kg and 2 m apart on the surface of the Earth. Will the forceof attraction be different, if the same bodies are placed on themoon, keeping the separation same?Data :m = m = 200 kg ; r = 2 m ; G = 6.67 × 1012-11 Nm kg ;2 -2F = ?Solution : F =-1112 22G m m6.67 ×10× 200× 200= r(2)Force of attraction, F = 6.67 × 10 N-7The force of attraction on the moon will remain same, since G is theuniversal constant and the masses do not change.4.2The acceleration due to gravity at the moon’s surface is 1.67 m s . If–2the radius of the moon is 1.74 × 10 m, calculate the mass of the6 moon.Data :g = 1.67 m s–2 ; R = 1.74 × 10 m 6;G = 6.67 × 10–11 N m kg ; M = ?2 -2 Solution :M = 26 2111.67 (1.74 10 )= 6.67 10gRG−×× ×M = 7.58 × 10 kg224.3Calculate the height above the Earth’s surface at which the valueof acceleration due to gravity reduces to half its value on theEarth’s surface. Assume the Earth to be a sphere of radius6400 km.Data :h = ?; gh = g2; R = 6400 x 10 m3Solution :h gg= 222RR =R+ h(R + h)⎛⎞ ⎜⎟ ⎝⎠ 22 gRgRh⎛⎞ =⎜⎟+⎝⎠ R1 =R+ h2

185h = ( 2-1) R = (1.414 - 1) 6400 × 10√3h = 2649.6 × 10 m3 At a height of 2649.6 km from the Earth’s surface, the accelerationdue to gravity will be half of its value at the Earth’s surface.4.4Determine the escape speed of a body on the moon. Given : radiusof the moon is 1.74 × 10 m and mass of the moon is67.36 × 10 kg.22Data :G = 6.67 × 10-11 N m kg ; R = 1.74 × 10 m ;2 -26 M = 7.36 × 10 kg; v = ?22eSolution : v = e-1 12 262G M2 × 6 .6 7 × 1 0× 7 .3 6 × 1 0=R1.7 4 × 1 0v = 2.375 km se–14.5The mass of the Earth is 81 times that of the moon and thedistance from the centre of the Earth to that of the moon is about4 × 10 km. Calculate the distance from the centre of the Earth5where the resultant gravitational force becomes zero when aspacecraft is launched from the Earth to the moon.Solution :Let the mass of the spacecraft be m. The gravitational force on thespacecraft at S due to the Earth is opposite in direction to that of themoon. Suppose the spacecraft S is at a distance x from the centreof the Earth and at a distance of (4 × 10 - x) from the moon.5∴mE252GM mGM m=x(4×10-x)Em MM= 81 = 252x(4 × 10 - x)∴ x = 3.6 × 10 km.5The resultant gravitational force is zero at a distance of3.6 × 10 km from the centre of the Earth. The resultant force on S5due to the Earth acts towards the Earth until 3.6 × 10 km is reached.5Then it acts towards the moon.SF mM mM EF Ex

1864.6A stone of mass 12 kg falls on the Earth’s surface. If the mass ofthe Earth is about 6 × 10 kg and acceleration due to gravity is249.8 m s , calculate the acceleration produced on the Earth by the-2stone.Data :m = 12 kg; M = 6 × 10 kg;24g = a = 9.8 m s ; a = ?s-2ESolution : Let F be the gravitational force between the stone andthe Earth.The acceleration of the stone (g) a = F/mSThe acceleration of the Earth, a = F/MEE24Sam12===aM6×10 2 × 10–24a = 2 × 10E–24 × 9.8a = 19.6 × 10E–24 m s–24.7The maximum height upto which astronaut can jump on the Earthis 0.75 m. With the same effort, to what height can he jump onthe moon? The mean density of the moon is (2/3) that of theEarth and the radius of the moon is (1/4) that of the Earth.Data : ρm = 23ρE; R = m14R ;Eh = 0.75 m ; h = ?EmSolution : The astronaut of mass m jumps a height h on the EarthEand a height h on the moon. If he gives himself the same kineticmenergy on the Earth and on the moon, the potential energy gainedat h and h will be the same.Em∴mgh = constantmg h = mg hmmEEmEEmhg= hg... (1)For the Earth, g = EE2EGM4=3Rπ G RE Eρ

187For the moon, g = mm2m GM4=3Rπ G Rm mρ∴ρρEEEmmmgR=.gR... (2)Equating (1) and (2)h = mρρ EEmmRR × hEh = mρ12ρ43EEEERR×× 0.75h = 4.5 mm4.8Three point masses, each of mass , are placed at the vertices ofman equilateral triangle of side . What is the gravitational field andapotential due to the three masses at the centroid of the triangle.Solution :The distance of each mass from the centroid0 is OA = OB = OCFrom the ∆ ODC, cos 30 = oa/2OC∴ OC =o a/2a =3cos 30Similarly, OB = a3and OA= a3(i)The gravitational field E =2GMr∴ Field at O due to A is, E = A23GMa(towards A)Field at O due to B is, E = B23GMa(towards B)Field at O due to C is, E = C23GMa(towards C)ABCE AE BE CD E RO 30 oA a / 2

188The resultant field due to E and E isBCE = E + E + 2E E cos 120R√B 2C 2B CoE = E + E - E = ER√B 2B 2B 2B[ E = E ]∵BCThe resultant field E = R23GMa acts along OD.Since E along OA and E along OD are equal and opposite, the netARgravitational field is zero at the centroid.(ii)The gravitational potential is, v = – GMrNet potential at ‘O’ isv = GMGMGM---a/ 3 a/ 3 a/ 3 = GMGMGM-3aaa⎛⎞ ++⎜⎟ ⎝⎠ = –3 3 √GMa4.9A geo-stationary satellite is orbiting the Earth at a height of 6Rabove the surface of the Earth. Here R is the radius of the Earth.What is the time period of another satellite at a height of 2.5Rfrom the surface of the Earth?Data : The height of the geo-stationary satellite from the Earth’ssurface, h = 6RThe height of another satellite from the Earth’s surface,h = 2.5RSolution : The time period of a satellite is T = 2 π3(R+h)GM∴T (R+h)α3For geo-stationary satellite,T (R + 6R)1α √3T (7R)1α √3... (1)For another satellite,T 2α √(R + 2.5R)3T 2α √(3.5R)3...(2)Dividing (2) by (1)3231T(3.5R)1==T(7R)22T = 21 T24 = 2222T =28 hours 29 minutes [ T = 24 hours)∵1

189Self evaluation(The questions and problems given in this self evaluation are only samples.In the same way any question and problem could be framed from the textmatter. Students must be prepared to answer any question and problemfrom the text matter, not only from the self evaluation.)4.1If the distance between two masses is doubled, the gravitationalattraction between them(a) is reduced to half(b) is reduced to a quarter(c) is doubled(d) becomes four times4.2The acceleration due to gravity at a height (1/20)th the radius ofthe Earth above the Earth’s surface is 9 m s . Its value at a point at-2an equal distance below the surface of the Earth is(a) 0(b) 9 m s-2(c) 9.8 m s-2(d) 9.5 m s-24.3The weight of a body at Earth’s surface is W. At a depth half wayto the centre of the Earth, it will be(a) W(b) W/2(c) W/4(d) W/84.4Force due to gravity is least at a latitude of(a) 0o(b) 45o(c) 60o(d) 90o4.5If the Earth stops rotating, the value of g at the equator will(a) increase(b) decrease(c) remain same(d) become zero4.6The escape speed on Earth is 11.2 km s . Its value for a planet–1having double the radius and eight times the mass of the Earth is(a) 11.2 km s–1(b) 5.6 km s–1(c) 22.4 km s–1(d) 44.8 km s–14.7If r represents the radius of orbit of satellite of mass m movingaround a planet of mass M. The velocity of the satellite is given by(a) v =2GMr(b) v =GMr(c) v = 2GMmr(d) v = Gmr

1904.8If the Earth is at one fourth of its present distance from the Sun, theduration of the year will be(a) one fourth of the present year(b) half the present year(c) one - eighth the present year(d) one - sixth the present year4.9Which of the following objects do not belong to the solar system?(a) Comets(b) Nebulae(c) Asteroids(d) Planets4.10According to Kepler’s law, the radius vector sweeps out equal areasin equal intervals of time. The law is a consequence of theconservation of(a) angular momentum(b) linear momentum(c) energy(d) all the above4.11Why is the gravitational force of attraction between the two bodiesof ordinary masses not noticeable in everyday life?4.12State the universal law of gravitation.4.13Define gravitational constant. Give its value, unit and dimensionalformula.4.14The acceleration due to gravity varies with (i) altitude and (ii) depth.Prove.4.15Discuss the variation of g with latitude due to the rotation of theEarth.4.16The acceleration due to gravity is minimum at equator and maximumat poles. Give the reason.4.17What are the factors affecting the ‘g’ value?4.18Why a man can jump higher on the moon than on the Earth?4.19Define gravitational field intensity.4.20Define gravitational potential.4.21Define gravitational potential energy. Deduce an expression for itfor a mass in the gravitational field of the Earth.4.22Obtain an expression for the gravitational potential at a point.4.23Differentiate between inertial mass and gravitational mass.

1914.24The moon has no atmosphere. Why?4.25What is escape speed? Obtain an expression for it.4.26What is orbital velocity? Obtain an expression for it.4.27What will happen to the orbiting satellite, if its velocity varies?4.28What are the called geo-stationary satellites?4.29Show that the orbital radius of a geo-stationary satellite is36000 km.4.30Why do the astronauts feel weightlessness inside the orbitingspacecraft?4.31Deduce the law of periods from the law of gravitation.4.32State and prove the law of areas based on conservation of angularmomentum.4.33State Helio-Centric theory.4.34State Geo-centric theory.4.35What is solar system?4.36State Kepler’s laws of planetary motion.4.37What is albedo?4.38What are asteroids?4.39What are constellations?4.40Write a note on Milky Way.Problems4.41Two spheres of masses 10 kg and 20 kg are 5 m apart. Calculatethe force of attraction between the masses.4.42What will be the acceleration due to gravity on the surface of themoon, if its radius is 14th the radius of the Earth and its mass is180th the mass of the Earth? (Take g as 9.8 m s )-24.43The acceleration due to gravity at the surface of the moon is1.67 m s . The mass of the Earth is about 81 times more massive-2than the moon. What is the ratio of the radius of the Earth to that ofthe moon?4.44If the diameter of the Earth becomes two times its present valueand its mass remains unchanged, then how would the weight of anobject on the surface of the Earth be affected?

1924.45Assuming the Earth to be a sphere of uniform density, how muchwould a body weigh one fourth down to the centre of the Earth, if itweighed 250 N on the surface?4.46What is the value of acceleration due to gravity at an altitude of500 km? The radius of the Earth is 6400 km.4.47What is the acceleration due to gravity at a distance from the centreof the Earth equal to the diameter of the Earth?4.48What should be the angular velocity of the Earth, so that bodieslying on equator may appear weightless? How many times thisangular velocity is faster than the present angular velocity?(Given ; g = 9.8 m s ; R = 6400 km)-24.49Calculate the speed with which a body has to be projected verticallyfrom the Earth’s surface, so that it escapes the Earth’s gravitationalinfluence. (R = 6.4 × 10 km ; g = 9.8 m s )3–24.50Jupiter has a mass 318 times that of the Earth and its radius is11.2 times the radius of the Earth. Calculate the escape speed of abody from Jupiter’s surface. (Given : escape speed on Earth is 11.2km/s)4.51A satellite is revolving in circular orbit at a height of 1000 km fromthe surface of the Earth. Calculate the orbital velocity and time ofrevolution. The radius of the Earth is 6400 km and the mass of theEarth is 6 × 1024 kg.4.52An artificial satellite revolves around the Earth at a distance of3400 km. Calculate its orbital velocity and period of revolution.Radius of the Earth = 6400 km ; g = 9.8 m s .-24.53A satellite of 600 kg orbits the Earth at a height of 500 km from itssurface. Calculate its (i) kinetic energy (ii) potential energy and(iii) total energy ( M = 6 × 10 kg ; R = 6.4 × 10 m)2464.54A satellite revolves in an orbit close to the surface of a planet ofdensity 6300 kg m . Calculate the time period of the satellite. Take-3the radius of the planet as 6400 km.4.55A spaceship is launched into a circular orbit close to the Earth’ssurface. What additional velocity has to be imparted to the spaceshipin the orbit to overcome the gravitational pull.(R = 6400 km, g = 9.8 m s ).–2

193Answers4.1(b)4.2(d)4.3(b)4.4(a)4.5(a)4.6(c)4.7(a)4.8(c)4.9(b)4.10(a)4.4153.36 × 10-11 N4.421.96 m s-24.433.714.44W/44.45187.5 N4.468.27 m s-24.472.45 m s-24.481.25 × 10 rad s ; 17-3–14.4911.2 km s–14.5059.67 km s ;–14.517.35 km s ; 1 hour 45 minutes 19 seconds–14.526.4 km s ; 9614 seconds–14.531.74 × 10 J; -3.48 × 10 J; -1.74 × 10 J1010104.544734 seconds4.553.28 km s–1

207Matter is a substance, which has certain mass and occupiessome volume. Matter exists in three states namely solid, liquid andgas. A fourth state of matter consisting of ionised matter of bare nucleiis called plasma. However in our forth coming discussions, we restrictourselves to the first three states of matter. Each state of matter hassome distinct properties. For example a solid has both volume andshape. It has elastic properties. A gas has the volume of the closedcontainer in which it is kept. A liquid has a fixed volume at a giventemperature, but no shape. These distinct properties are due to twofactors: (i) interatomic or intermolecular forces (ii) the agitation orrandom motion of molecules due to temperature.In solids, the atoms and molecules are free to vibrate about theirmean positions. If this vibration increases sufficiently, molecules willshake apart and start vibrating in random directions. At this stage, theshape of the material is no longer fixed, but takes the shape of itscontainer. This is liquid state. Due to increase in their energy, if themolecules vibrate at even greater rates, they may break away from oneanother and assume gaseous state. Water is the best example for thischanging of states. Ice is the solid form of water. With increase intemperature, ice melts into water due to increase in molecular vibration.If water is heated, a stage is reached where continued molecularvibration results in a separation among the water molecules andtherefore steam is produced. Further continued heating causes themolecules to break into atoms.5.1Intermolecular or interatomicforcesConsider two isolated hydrogenatoms moving towards each other asshown in Fig. 5.1.As they approach each other,the following interactions areobserved.5. Mechanics of Solids and FluidsAAFig. 5.1 Electrical origin ofinteratomic forces

208Fig. 5.2. Variation of potential energy with interatomic distanceGasesLiquidsInteratomic distance between hydrogen atomsPR r0SolidsO(i) Attractive force A between the nucleus of one atom and electronof the other. This attractive force tends to decrease the potential energyof the atomic system.(ii) Repulsive force R between the nucleus of one atom and thenucleus of the other atom and electron of one atom with the electronof the other atom. These repulsive forces always tend to increase theenergy of the atomic system.There is a universal tendency of all systems to acquire a state ofminimum potential energy. This stage of minimum potential energycorresponds to maximum stability.If the net effect of the forces of attraction and repulsion leads todecrease in the energy of the system, the two atoms come closer toeach other and form a covalent bond by sharing of electrons. On theother hand, if the repulsive forces are more and there is increase in theenergy of the system, the atoms will repel each other and do not forma bond.The variation of potential energy with interatomic distance betweenthe atoms is shown in Fig. 5.2.

209It is evident from the graph that as the atoms come closer i.e.when the interatomic distance between them decreases, a stage isreached when the potential energy of the system decreases. When thetwo hydrogen atoms are sufficiently closer, sharing of electrons takesplace between them and the potential energy is minimum. This resultsin the formation of covalent bond and the interatomic distance is .roIn solids the interatomic distance is and in the case of liquidsroit is greater than . For gases, it is much greater than .roroThe forces acting between the atoms due to electrostatic interactionbetween the charges of the atoms are called interatomic forces. Thus,interatomic forces are electrical in nature. The interatomic forces areactive if the distance between the two atoms is of the order of atomicsize ≈10-10 m. In the case of molecules, the range of the force is of theorder of 10 m–9.5.2 ElasticityWhen an external force is applied on a body, which is not freeto move, there will be a relative displacement of the particles. Due tothe property of elasticity, the particles tend to regain their originalposition. The external forces may produce change in length, volumeand shape of the body. This external force which produces these changesin the body is called deforming force. A body which experiences sucha force is called deformed body. When the deforming force is removed,the body regains its original state due to the force developed within thebody. This force is called restoring force. The property of a material toregain its original state when the deforming force is removed is calledelasticity. The bodies which possess this property are called elasticbodies. Bodies which do not exhibit the property of elasticity are calledplastic. The study of mechanical properties helps us to select thematerial for specific purposes. For example, springs are made of steelbecause steel is highly elastic.Stress and strainIn a deformed body, restoring force is set up within the bodywhich tends to bring the body back to the normal position. Themagnitude of these restoring force depends upon the deformationcaused. This restoring force per unit area of a deformed body is knownas stress.

210∴ Stress = restoring forceareaN m–2Its dimensional formula is ML T .–1 –2Due to the application of deforming force, length, volume orshape of a body changes. Or in other words, the body is said to bestrained. Thus, strain produced in a body is defined as the ratio ofchange in dimension of a body to the original dimension.∴ Strain = change in dimensionoriginal dimensionStrain is the ratio of two similar quantities. Therefore it has nounit.Elastic limitIf an elastic material is stretched or compressed beyond a certainlimit, it will not regain its original state and will remain deformed. Thelimit beyond which permanent deformation occurs is called the elasticlimit.Hooke’s lawEnglish Physicist Robert Hooke (1635 - 1703) in the year 1676put forward the relation between the extension produced in a wire andthe restoring force developed in it. The law formulated on the basis ofthis study is known as Hooke’s law. According to Hooke’s law, withinthe elastic limit, strain produced in a body is directly proportional to thestress that produces it.(i.e) stress α strainStressStrain = a constant, known as modulus ofelasticity.Its unit is N m and its dimensional formula-2is ML T .-1 -25.2.1 Experimental verification of Hooke’s lawA spring is suspended from a rigid supportas shown in the Fig. 5.3. A weight hanger and alight pointer is attached at its lower end suchSpringSlottedWeights123456Fig. 5.3 Experimentalsetup to verifyHooke’s law

211that the pointer can slide over a scale graduated in millimeters. Theinitial reading on the scale is noted. A slotted weight of kg is addedmto the weight hanger and the pointer position is noted. The sameprocedure is repeated with every additional m kg weight. It will beobserved that the extension of the spring is proportional to the weight.This verifies Hooke’s law.5.2.2 Study of stress - strain relationshipLet a wire be suspended from a rigid support. At the free end, aweight hanger is provided on which weights could be added to studythe behaviour of the wire under different load conditions. The extensionof the wire is suitablymeasured and a stress - straingraph is plotted as in Fig. 5.4.(i) In the figure the regionOP is linear. Within a normalstress, strain is proportional tothe applied stress. This isHooke’s law. Upto P, when theload is removed the wireregains its original length alongPO. The point P represents theelastic limit, PO represents theelastic range of the materialand OB is the elastic strength.(ii) Beyond P, the graph is not linear. In the region PQ the materialis partly elastic and partly plastic. From Q, if we start decreasing theload, the graph does not come to O via P, but traces a straight line QA.Thus a permanent strain OA is caused in the wire. This is calledpermanent set.(iii) Beyond Q addition of even a very small load causes enormousstrain. This point Q is called the yield point. The region QR is theplastic range.(iv) Beyond R, the wire loses its shape and becomes thinner andthinner in diameter and ultimately breaks, say at S. Therefore S is thebreaking point. The stress corresponding to S is called breaking stress.OAStrainStressBPQRSFig. 5.4 Stress - Strain relationship

2125.2.3 Three moduli of elasticityDepending upon the type of strain in the body there are threedifferent types of modulus of elasticity. They are(i) Young’s modulus(ii) Bulk modulus(iii) Rigidity modulus(i) Young’s modulus of elasticityConsider a wire of length and cross sectional area lAstretched by a force F acting along its length. Let d be thelextension produced.∴ Longitudinal stress = = ForceFAreaALongitudinal strain = change in lengthoriginal length = dllYoung’s modulus of the material of the wire is definedas the ratio of longitudinal stress to longitudinal strain. It isdenoted by q.Young’s modulus = longitudinal stresslongitudinal strain(i.e)q = // FAdl l or q = FlA dl(ii) Bulk modulus of elasticitySuppose euqal forces actperpendicular to the six faces of a cubeof volume as shown in Fig. 5.6. DueVto the action of these forces, let thedecrease in volume be dV .Now, Bulk stress = ForceFAreaA =Bulk Strain =change in volumeoriginal volume = −dVV(The negative sign indicates thatvolume decreases.)Fig. 5.5Young’smodulus ofelasticityFig. 5.6 Bulk modulusof elasticityabcdefgFFFh

213Bulk modulus of the material of the object is defined as the ratiobulk stress to bulk strain.It is denoted by .k∴ Bulk modulus = Bulk stressBulk strain(i.e)k = −/FAdVV= −PdVVF PA⎡⎤=⎢⎥ ⎣⎦ ∵ or k = -P VdV(iii)Rigidity modulus or shear modulusLet us apply a force Ftangential to the top surface ofa block whose bottom AB isfixed, as shown in Fig. 5.7.Under the action of thistangential force, the body suffersa slight change in shape, itsvolume remaining unchanged.The side AD of the block issheared through an angle toθthe position AD .’If the area of the top surface is , then shear stress = AF/A.Shear modulus or rigidity modulus of the material of the object isdefined as the ratio of shear stress to shear strain. It is denoted by n.Rigidity modulus = shear stressshear strain(i.e) n = θ /FA= θ FATable 5.1 lists thevalues of the threemoduli of elasticity forsome commonly usedmaterials.ABCC /DD /FFFig. 5.7 Rigidity modulusTable 5.1 Values for themoduli of elasticityMaterialModulus of elasticity (× 10 Pa)11qknAluminium0.700.700.30Copper1.11.40.42Iron1.91.00.70Steel2.01.60.84Tungsten3.62.01.5

2145.2.4 Relation between the threemoduli of elasticitySuppose three stresses , andP QR act perpendicular to the three facesABCD, ADHE and ABFE of a cube ofunit volume (Fig. 5.8). Each of thesestresses will produce an extension inits own direction and a compressionalong the other two perpendiculardirections. If is the extension per unitλstress, then the elongation along thedirection of will be Pλ P. If is theµcontraction per unit stress, then the contraction along the direction ofP due to the other two stresses will be µ Q and µ R .∴ The net change in dimension along the direction of due to allPthe stresses is e = λ P - µ Q - µ R .Similarly the net change in dimension along the direction of isQf = λ Q - µ P - µ R and the net change in dimension along the directionof is = Rgλ R - µ P - µ Q .Case (i)If only acts and = = 0 then it is a case of longitudinal stress.PQR∴ Linear strain = = eλ P∴ Young’s modulus = qlinear stresslinear strain = Pλ P(i.e) = q1λorλ = 1q...(1)Case (ii)If R = O and P = – Q, then the change in dimension along P ise = P - (-P)λµ(i.e)e = (λ + µ) PAngle of shear θ = 2e* = 2 (λ + µ) P∴Rigidity modulusn = θ P = P2( + )Pλµ (or) 2 (λ + µ) = 1n.....(2)Fig. 5.8 Relation between thethree moduli of elasticityABCDEFGRPQH* The proof for this is not given here

215Case (iii)If = = , the increase in volume is = PQRe + f + g= 3e = 3 (λ − 2µ) P (since e = f = g)∴ Bulk strain = 3(λ−2µ) PBulk modulus =kP3( -2 )Pλµor(λ − 2µ) =13 k...(3)From (2), 2(λ + µ) = 1n2λ + 2µ = 1n...(4)From (3), (λ − 2µ) = 13 k...(5)Adding (4) and (5),3 λ= 113 nk +λ= 1139 nk +∴ From (1), 1q = 1139 nk +or 931qnk=+This is the relation between the threemoduli of elasticity.5.2.5 Determination of Young’s modulusby Searle’s methodThe Searle’s apparatus consists of tworectangular steel frames A and B as shownin Fig. 5.9. The two frames are hingedtogether by means of a frame F. A spiritlevel L is provided such that one of its endsis pivoted to one of the frame B whereas theother end rests on top of a screw workingthrough a nut in the other frame. The bottomABFCVPQSRLWFig. 5.9 Searle’sapparatus

216of the screw has a circular scale C which can move along a verticalscale V graduated in mm. This vertical scale and circular scalearrangement act as pitch scale and head scale respectively of amicrometer screw.The frames A and B are suspended from a fixed support by meansof two wires PQ and RS respectively. The wire PQ attached to the frameA is the experimental wire. To keep the reference wire RS taut, a constantweight W is attached to the frame B. To the frame A, a weight hangeris attached in which slotted weights can be added.To begin with, the experimental wire PQ is brought to the elasticmood by loading and unloading the weights in the hanger in the frameA four or five times, in steps of 0.5 kg. Then with the dead load, themicrometer screw is adjusted to ensure that both the frames are at thesame level. This is done with the help of the spirit level. The readingof the micrometer is noted by taking the readings of the pitch scaleand head scale. Weights are added to the weight hanger in steps of 0.5kg upto 4 kg and in each case the micrometer reading is noted byadjusting the spirit level. The readings are again noted during unloadingand are tabulated in Table 5.2. The mean extension for M kg of loaddlis found out.Table 5.2 Extension for M kg weightLoad in weightMicrometer readingExtensionhangerkgLoadingUnloadingMeanfor M kg weightWW + 0.5W + 1.0W + 1.5W + 2.0W + 2.5W + 3.0W + 3.5W + 4.0