[ ][ ] mg = R GM g = R GM = g R 2. Definition of G: From (1) we have, 2 NEWTON S LAW OF GRAVITATION:

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1 NEWTON S LAW OF GAVITATION: Statement: Every particle in the universe attracts every other particle with a force which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them and this force acts alon the line joinin the two particles and independent of the presence of other bodies. Gravitational force is a central force. m attracts m with a ravitational force F directed towards m, m attracts m with a force - F towards m. The force F and - F form action reaction pair, equal in manitude and opposite in direction. F m. m F r Gmm F () r Where, G is constant of proportionality known as universal ravitational constant its Value is 6.67 x 0 - Nm /k. In vector form it is written as Gm ^ m F r r Where r ^ is the unit vector from m to m. Dimensions of G: G F. r m. m [ ][ ] 0 0 M L T M L T G 0 0 [ M L T ] G [ M L T ] Pae of Definition of G: From () we have, F. r G m. m If r, m m, G F Hence, It is defined as the manitude of force of attraction between two bodies each of unit mass and separated by a unit distance. ELATION BETWEEN G AND (ACCELEATION DUE TO GAVITY): Let be the acceleration due to ravity on the earth s surface. Let M be the mass of the earth of radius and m be the mass of the body. On the surface the surface of earth: Weiht of the body ravitational force of attraction. G. Mm m VAIATIONS IN THE VALUE OF : The value of acceleration due to ravity () varies as we o above or below the surface of the earth. It also varies from place to place on the surface of the earth. Variation of with altitude:

2 Consider earth to be a sphere of radius and mass M. The acceleration due to ravity on the surface of earth (point Q in Fi.) is () Consider a point P at a heiht h above the surface of the earth. The acceleration due to ravity at point P is- h () ( + h) Dividin () by () h ( + h) h () ( + h) ( + h) h < h Thus, as we o above the earth s surface acceleration due to ravity oes on decreasin. h h + h h + h h + Q h < < < hiher powers of h/ can be nelected Usin Binomial Theorem, h h h h (4) Usin eq (4) value of acceleration due to ravity can be determined when h is small as compared to. Variation of with depth. Consider the earth to be a sphere of radius and mass M. The acceleration due to ravity at the surface of the earth is () If ρ is density of the earth, then, mass(m) Density (ρ) Volume(V) M ρv 4 But, V π 4 M π ρ Thus, substitutin in eq.() acceleration due to ravity in terms of density is iven by- G 4 π ρ 4 π ρ G () Consider a point P which is in side the earth below the earth s surface at depth d. Its distance from point O is (-d). Pae of

3 A body at point P will experience force only due to the portion of earth of radius (-d). The outer spherical shell, whose thickness is d, will not exert any force on the body at point P. Let M be the mass of the earth of portion of radius (-d) then ' d () ( d) 4 But, M π ( d ) ρ G 4 d π ( d ) ρ ( d) 4 d π ( d ) Gρ (4) Dividin equation (4) by () d d d d (5) Therefore, the value of acceleration due to ravity decreases with depth. The acceleration due to ravity at the centre of earth can be found by substitutin d in equation (5). d 0 centre Variation in the value of due to rotation of the earth : Earth is rotatin about its own axis at an anular velocityω ( one revolution per 4 hours). The line joinin the north and south poles is the axis of rotation. As a result of rotation, every point on the earth moves alon a circular path with the same anular velocity ω. A point at the equator moves in a circle of radius equal to the radius of earth and the centre of the circle is the same as the centre of the earth. For any other point on the earth, the circle of rotation is smaller than this. Consider earth to be a sphere of mass M and radius. Suppose a particle of mass m is situated at point P on the surface of the earth as shown in fiure. Let λ be the latitude of the point i.e. POE λ. Suppose is the acceleration due to ravity in the absence of rotational motion of the earth. In that case, the particle at P would have been attracted toward the centre O of the earth. Therefore true weiht m of the particle is directed toward O and is represented by the vector PA. Due to rotation of earth, let, anular speed of earth is ω the particle at P moves alon a circular path whose centre is C and radius r (CP). Since OPC λ, r cos λ. The centrifual force on the particle due To rotational motion of the earth acts alon the radius of the circular path in outward direction. The manitude of centrifual force is iven by; F C mrω mω cos λ The centrifual force is represented by the vector PB. The apparent The apparent weiht m of the particle is equal to the resultant of actual weiht (m) and the centrifual force F C ( mω cos λ). Complete the paralleloram PAC B. Then diaonal PC of the paralleloram represents the apparent weiht m of the particle. Now, 0 PC ' PA + PB + PA PB COS 80 λ ( ) ( m ') ( m ) + ( m ω cos λ ) + m m ω cos λ Or ( cos λ ) ' m ( ) m ( ) + m ( ω cosλ) ( ω cos λ) m ' 4 ( ) + ω cos λ ω cos λ 4 ' ω ωs λ ω cos λ ( ) + 4 ω cos ' + λ ω cos λ Pae of

4 ω Now, the numerical value of is 4 very small. Hence, ω / will be still smaller. Therefore, nelectin the factor 4 containin ω /, we et, ω cos λ ' ω cos λ + hiher terms Nelectin the terms containin hiher powers of ω cos λ /, we et, ω cos λ ' ' ω cos λ () From equation (), it is clear that acceleration due to ravity a) decreases on account of rotation of the earth b) increases with the increase in the latitude of the place (Q cos λ decreases as λ increases). This means that value of increases as we o from equator to the poles. At equator: At equator λ 0 0 so that cos λ cos 0 0 ' ω () Therefore, value of acceleration due to ravity is minimum at the equator. This is expected because the particle at the equator executes a circle of maximum radius. Therefore, the centrifual force is maximum. At poles: At poles, λ 90 0 so that cos λ cos '. Maximum Hence the value of is maximum at the poles. This is expected because the particle at the pole moves in a circle of zero radius. Therefore, no centrifual force acts on the particle. Definition of Escape velocity (Ve) : Maximum vertical velocity required to take the satellite just out side the earth s ravitational influence is called as escape velocity. If vertical velocity is less than escape velocity, body falls on the earth s surface but if vertical velocity is reater than or equal to escape velocity body oes out of earth s ravitational influence permanently. Definition of critical velocity / orbital velocity (Vc) : It is the horizontal velocity imparted to the satellite so that the satellite orbits around the earth in a stable circular orbit with constant manitude of this velocity. Satellite: A body which orbits in a closed orbit around another larer body, like planets, under the ravitational influence are called satellites. Moon is a natural satellite of earth, INSAT B is an artificial satellite of the earth. Q.Why minimum of two stae rocket is used for the projection of satellite? For a sinle stae rocket carryin satellite, if the velocity of projection is less than escape velocity, the satellite will come back to the earth, and if the velocity of projection is reater than or equal to the escape velocity, the satellite will escape earth s ravitational influence. Hence, is not possible to put a satellite into earth s orbit usin a sinle stae rocket. Hence minimum of two stae rocket is used. The first stae is used to carry the satellite to a certain heiht the second stae and the satellite is rotated throuh 90 0 and with the help of second stae it is projected in the horizontal direction to put the satellite in a stable circular orbit. Pae 4 of

5 POJECTION / LAUNCHING OF A SATELLITE: The satellite is held at the tip of rocket of minimum of two staes. The fuel of the first stae is inited. The up thrust takes the rocket vertically to a certain heiht and the first stae ets detached. By uidance and trackin system, the second stae and satellite is rotated throuh This second stae is used to ive the horizontal velocity to the satellite and finally ets detached. The satellite alone starts orbitin around the earth in a pre-determined orbit. DIFFEENT CASES OF SATELITE MOTION: The nature of the path of satellite depends on the horizontal velocity V h. from the earth s ravitational field after describin a hyperbola. Keplar s Laws FIST LAW: LAW OF OBIT Statement: Each planet revolves around the sun in an elliptical orbit with the sun as one of the foci. Planet (P) revolves in an elliptical orbit as shown in fi with sun (S) at the focus F. When the planet reaches at A it is nearest to sun and this minimum distance (AF ) is called periee. When it is at B, it is at farthest distance from the sun. The distance BF is known as apoee. AB and CD are major and minor axes of ellipse. Case I: If the horizontal velocity is less than a certain velocity called critical velocity, the satellite falls to the earth after followin a spiral path and falls on the earth. Case II: If horizontal velocity is slihtly less than critical velocity then it follows elliptical path with point of projection at apoee. Case III: If the horizontal velocity equals the critical velocity, the satellite orbits in a stable circular orbit. Case IV: If the horizontal velocity is reater than critical velocity but less than escape velocity, the satellite orbits around the earth in a elliptical orbit with point of projection at periee. Case V: If the horizontal velocity equals the escape, velocity the satellite will escape from the earth s ravitational field and be lost in space after describin a parabola. Case V: If the horizontal velocity is reater than escape velocity the satellite will escape Pae 5 of SECOND LAW: LAW OF AEA Statement: The position vector of planet from the sun sweeps out equal area in equal time of the area velocity of the planet around the sun always remains constant. Suppose that at certain time, planet is at B and after time dt it moves to B. Then position vector of planet sweeps out area BF B in time dt. Aain in time δ t the planet moves from A to A and sweeps out area AF A then accordin to second law, areas AF A and BF B must be same or equal. If da is area swept out in dt then areal da velocity is defined as and accordin to dt da second law is constant. dt THID LAW: LAWS OF PEIOD The square of the period of any planet

6 around the sun is directly proportional to the cube of the semi major axis of the elliptical orbit. If T is the time period of the planet and a is the semi major axis then T a If T and T are the periods of any two planets and a and aare their semi major axes then T a T a Q.OBTAIN AN EXPESSION FO CITICAL VELOCITY O OBITAL VELOCITY. Consider a satellite of mass m raised to a heiht h above the earth s surface. It is projected with critical velocity V c. If the satellite is orbitin close to the earth. h << then + h v c Substitutin in above equation v c v c () v c 7.9 km/s This is the expression for critical velocity of satellite orbitin close to earth. Substitutin h ( + h ) in eqn. () ( + h) h vc + h v ( h) () c h + This is an expression for critical velocity in terms of acceleration due to ravity at a heiht h. The satellite starts orbitin in a stable circular orbit of radius (+h) where is the radius of the earth. Let M be the mass of the earth. For a stable circular orbit, Centripetal force ravitational force of attraction. m vc G. Mm + h + h ( ) ( ) v c + h v c ( () h) As there is no, mass term, critical velocity is independent of the mass of the satellite and as is constant critical velocity is inversely proportional to the square root of the radius of the orbit. Q. Obtain an expression for time period a satellite. Show that the square of the period of satellite is directly proportional to the cube of the radius of the orbit. Consider a satellite of mass m raised to a heiht h above the earth s surface. It is projected with a critical velocity V c. The satellite starts orbitin in a stable circular orbit of radius ( + h), where is the radius of the earth. Let M be the mass of the earth. Time period is the time taken by the satellite to complete one rotation around the earth. Time distance / velocity circumfer enceof circular orbit Time period criticalvelocity Pae 6 of

7 T ( + h) π V c ( + h) π + h π ( + h) + h ( + h) T π () This is expression for time period for satellite. Squarin () 4 T π ( + h) () Let + h r 4π T.r 4π As, constant T Constant. r T r Thus, the square of the period o satellite is directly proportional to the cube of the radius of the orbit. This is Kepler s third law of planetary motion Substitutin in equation () ( + h) T π ( + h) T π () Substitutin h (+h) in equation () ( + h) T π (4) h EXTA FO COMPETETIVE EXAMS: For a satellite orbitin very close to earth + h Substitutin in equation () T π T π 84 min EXPESSION FO THE ADIUS OF THE OBIT: Let r be the radius of the circular orbit i.e. r + h We have T π. ( + h) r T π Squarin 4π T.r T r 4π Takin cube root, / T r 4 π Q. OBTAIN AN EXPESSION FO BINDING ENEGY OF THE SATELLITE OBITING AOUND THE EATH Definition: The minimum amount of enery required or work that must be done to move a body from a point in the earth s ravitational field to infinity, aainst the earth s ravitational force of attraction. Consider a satellite of mass m orbitin around the earth with critical velocity v c in a Pae 7 of

8 stable circular orbit of radius (+h). Let M be the mass of the earth. Expression for kinetic enery: K.E. mv c Substitutin v c + h E. m h + m K.E. ( + h) Expression for ravitational potential enery: Gravitational potential at a point is the work done in movin a unit mass (k) from infinity to that point. Gravitational potential due to mass M at a distance ( + h) from the earth s surface is, V - ( + h) Neative sin INDICATES ravitational force is the force of attraction towards the centre of mass M. Gravitational Potential Enery ravitational potential X mass -.m ( + h) Gravitational Potential Enery m - ( + h) Expression for total enery: T.E. K. E. + P. E. m m - ( + h) ( + h) m ( + h) m T.E ( + h) Neative sin for the total enery of the satellite shows that the satellite is bound to the earth by an attractive force and cannot m leave. If amount of enery is ( + h) supplied to the satellite, the total enery satellite will become zero. Hence, the satellite will move to infinity which from definition is bindin enery. m B. E () ( + h) K.E. B.E, P.E. - B.E. Substitutin in equation () m B.E () ( + h) and h (+h) in equation () h ( + h) m B.E. ( + h) h ( + h). m B.E () Q. OBTAIN AN EXPESSION FO BINDING ENEGY OF A BODY AT EST ON THE EATH S SUFACE AND ABOVE THE EATH S UFACE. The minimum amount of enery required or minimum amount of work that must be done to move a body from a point in earth s ravitational field to infinity aainst the earth s ravitational force of attraction. Consider a body of mass m situated at a heiht h above the earth s surface m be the mass of the earth of radius. As the body is at rest K. E. 0. Expression for ravitational potential enery: Gravitational potential at a point is the work done in movin a unit mass (k) from infinity to that point. Gravitational potential due to mass M at a distance ( + h) from the earth s centre is V - ( + h) Pae 8 of

9 Neative sin because, ravitational force is the force of attraction towards the centre of mass M Gravitational Potential Enery ravitational potential x mass V. m -. m ( + h) Expression for total enery: T. E. K. E. + P. E. m 0 ( + h) m T.E + h Neative sin for the total enery of the body shows that the body is bound to the earth by an attractive force and cannot leave it. m If amount of enery is + h supplied to the body, the total enery will become zero. The body will move the infinity which from definition is bindin enery. m B. E () ( + h) On the earth s surface, h 0 m B. E () Substitutin in equation (). m B.E. B.E. m () Definition of ravitational field intensity or ravitational field strenth: The ravitational intensity at a point in a ravitational field is the force actin on a unit mass placed at that point. F E m Pae 9 of m ( + h). m E ( + h) It is a vector quantity directed toward the centre of the earth. S. I. unit N/k On the earth s surface h0 and E E. Gravitational acceleration Gravitational field intensity. Q. Obtain an expression Escape velocity of a body at rest on the earth s surface O Prove that V e. Definition: It is the minimum velocity with which a body must be projected vertically upwards, so that it escapes, the earth s ravitational field. Such a body never returns to the earth s surface. EXPESSION: Let m be the mass of the body which is projected upwards with an escape velocity V e. In order to escape, if the kinetic enery iven to it is equal or reater bindin enery. Such a body will escape from the ravitational attraction of the earth. Bindin enery of a body at rest on the earth s surface m B. E. Where M is the mass of the earth is the

10 radius of earth. Kinetic Enery of projection m V e K.E. of projection m V e Kinetic Enery of projection Bindin Enery Kinetic Enery of projection B. E. m V e m V e V e () This is an expression for escape velocity of body a rest on the surface Substitutin, We et Ve V e () From the above expressions, escape velocity is independent of the mass of the body. Q. POVE THAT THE ESCAPE VELOCITY OF A SATELLITE IN OBITAL MOTION IS EQUAL TO CITICAL VELOCITY. (V c V e ) Consider a satellite orbitin around the earth with a critical velocity V c in a stable circular orbit of radius ( + h). In order to escape, the kinetic enery iven to the satellite is equal or reater than bindin enery of the satellite. The satellite will escape the ravitational attraction of the earth. Bindin enery of a satellite orbitin at a heiht h, m B. E. ( + h) Where M mass of the earth, the radius of the earth. m V e V e m + h ( + h) V e () h But, V c ( () h) From equation () and equation () V c V e Prove that the escape velocity of a body on the earth s surface is times velocity of the body when it moves in a circular orbit close to the earth. Or Ve Vc When a body is orbitin close to the earth s surface, radius of the orbit is equal to the radius of the earth + h Centripetal force ravitational force of attraction. c m.v m V c () Let Ve be the escape velocity of the earth surface. K.E. of projection B.E. m. V e V e V e m Pae 0 of

11 V e V c From equation () COMMUNICATION O SYNCHONOUS O GEOSTATIONAY SATELLITE: Those satellites orbit which once in 4 hours in the equatorial plane of the earth the same direction as that of the spin of the earth i.e. from west to east. Since, the period of rotations of earth about its own axis is also 4 hours, the relative velocity of the satellite w.r.t. earth is zero hence the satellite will always appear stationary above a iven place from the earth s surface. Therefore it is called eostationary satellite and the orbits are called eostationary orbit. The heiht of the eostationary satellite above the earth s surface is approx. 6,000 km. Condition for communication satellite: ) These satellites must be in equatorial plane of the earth. ) The direction of rotation must be same as the direction of rotation of earth in its own axis ) Period of the satellite must be equal to the period of the earth in its own axis. 4) The heiht of the satellite from the earth s surface should be 6000 km USES: ) Used to receive and transmit radio and television sinals. ) It is also used to study earth s atmosphere and forecast weather. Pae of ) It is used in telephone communication. WEIGHTLESSNESS: Feelin of weihtlessness in movin satellite. ) Gravitational attraction towards the centre of earth is by definition weiht of body. ) Weihtlessness is feelin in movin satellite. It is not due to weiht equal to zero. ) When an astronaut is on surface of earth, ravitational force acts on him. This ravitational force is weiht of astronaut. The earth s surface exerts an upward reaction on astronaut and due to this reaction astronaut feels his weiht. 4) In an orbitin satellite astronaut alon with centripetal acceleration directed towards centre of earth. Hence, astronaut can t produce any action on floor of satellite. So floor cannot produce any reaction on him. Due to absence of this reaction astronaut has feelin of weihtlessness. NUMEICALS. Show that Escape velocity of a body from the surface of the earth is Vc. Where Vc is critical velocity of body when it is orbitin very close to earth surface.. Prove that escape velocity of a body from the surface of the planet of mean radius Gd and density d is π. Where G is ravitational constant.. Show that critical velocity of satellite close to surface of the earth of radius and mean density (d) is iven by π G d 4. Show that h + h where h is acceleration due to ravity at a heiht h from the surface of the earth and is acceleration due to ravity on the surface of the earth.

12 5. An artificial satellite has to be set up to revolve around the planet in circular orbit close to its surface, if ρ is mean density and is radius of the planet show that its π period of revolution is ρg 6. Distance of the planet from the earth is.5 x 0 7 km and ravitational force between them is.8 x 0 8 N. Mass of planet and that of earth is equal, each bein 5.98 x 0 4. Calculate universal ravitation constant. 7. Two homoenous sphere one of mass 00 k and other of mass.75 k attracts each other with force of 9.6 x 0-7 N when kept with their centre 0. m apart. Estimate G. 8. Calculate the force of attraction between two metal spheres each of mass 90 k if the distance between their centre is 40 cm. Given G 6.67 x 0 - Nm /k. 9. Find the ravitational force of attraction if mass of moon is /8 times mass of earth. Given G 6.67 x 0 - Nm /k. adius of moon s orbit.85 x 0 5 km M 6 x 0 4 k 0. Calculate the acceleration due to ravity at the surface of the earth. Given 6.4 x 0 6 m G 6.67 x 0 - Nm /k Mean density ρ 5.5 x 0 k/m.. Mean radius of earth is 6400 km the acceleration due to ravity at its surface is 9.8 m/s. Estimate the mass of earth. Given G 6.67 x 0 - Nm /k..find the acceleration due to ravity on the surface of moon. Given that mass of moon is 80 times that of earth and diameter of moon is 4 times that of earth. Given : 9.8 m/s.. A body weiht 4.5 k on the surface of the earth. How much it will weih on the surface of planet whose mass is /9 th mass of earth and radius is half that of earth. 4. Two satellites X and Y are movin in circular orbit of radius r and r respectively around the same planet. What is the ratio of their critical velocity? 5. Satellite orbitin around the earth havin critical velocity in the ratio 4:5. Compare their orbital radii. 6. Compare the critical speeds of two satellites if the ratio of their periods is 8:. 7. adius of earth is 6400 km. Calculate the velocity in km/sec with which body should be projected so as to just escape from the earth s ravitational influence. 8. What would be the speed of satellite orbitin around the earth very close to its surface Given: 6400km 9.8 m/s. 9. Find the heiht of satellite from the surface of the earth whose critical velocity is 5 km/s. By considerin followin data Given: G 6.67 x 0 - Nm /k km M 5.98 x 0 4 k 0. Find the heiht of satellite from the surface of the earth whose critical velocity is 4 km/s. Given : G 6.67 x 0 - Nm /k.m 5.98 x 0 4 k 6.4 x 0 6 m. Find the heiht above the earth surface at which acceleration due to ravity is 50% or of the earth surface. Given: 6.4 x 0 6 m. Find the heiht above the earth surface at which acceleration due to ravity is 90% of the earth surface. Given: 6400 km. How far away from the centre of the earth does the acceleration due to ravity will be reduced by % of its value on the earth surface? Given: 6400 km 4. What would be the duration of a year if the distance between earth and sun ets doubled. Assumin present period o earth to be 65 days. 5. What would be the duration of a year if the distance between earth and sun ets half. Assumin present period o earth to be 65 days 6. Calculate period of revolution of planet Jupiter around the sun. Given that ratio of Pae of

13 radius of Jupiter s orbit to that of earth orbit 5.:. 7. Communication satellite appears stationary from the place of projection. Find the distance of satellite from the surface of the earth.(g 6.67 x 0 - Nm /k, 6400 km,m 6 x 0 4 k) 8. Escape velocity from the surface of the earth is. km/s. If the mass of Jupiter /8 times that of earth and its radius is. times that of earth. Find Escape velocity from the Jupiter surface. 9. A body is raised to a heiht of 600 km above the earth surface and projected with horizontal velocity of 6 km/s. Will it remove around the earth as satellite? (G 6.67 x 0 - Nm /k, 6400 km, M 6 x 0 4 k) 0. Satellite is taken to a heiht equal to radius of the earth and it is projected horizontal with speed 7 km/s. State the nature of its orbit. (G 6.67 x 0 - Nm /k, 6400 km M 5.98 x 0 4 k). A satellite is revolvin in circular orbit around a planet with velocity of 8 km/sec. at a heiht where value of acceleration due to ravity is 8 m/sec. How hih is the satellite from the plant s surface. adius of planet is 6000 km.. An artificial satellite is revolvin in circular orbit around the earth at a heiht of 000 km with speed 764 m/sec. Find the period of revolution. Given: 6400 km. Communication satellite is at a heiht 6000 km from the earth surface. What will be its new period when it is brouht down to a heiht of 000 km. Given: 6400 km 4. Calculate bindin enery, potential enery, kinetic enery and total enery of a artificial satellite of mass 000 k orbitin at a heiht of 600 km from the earth s surface.(m 6 x 0 4 k, 6.4 x 0 6 m, G 6.67 x 0 - Nm /k.) 5. Earth moves around the sun in circular orbit of radius.5 x 0 8 km. Calculate B.E. of the earth. (mass of sun x 0 0 k, adius of earth () 6400 km 9.8 m/sec.) 6. elation between Ve and Vc for orbitin satellite 7. Satellite is orbitin around the earth with orbital radius 7000 km. If Escape velocity of satellite is 7.5 x 0 m/s. Calculate its period. ******************************** Pae of