Gravitation Study Guide

Gravitation INTRODUCTION The motion of celestial bodies such as the sun, the moon, the earth and the planets etc. has been a subject of fascination since time immemorial. Indian astronomers of the ancient times have done brilliant work in this field, the most notable among them being Arya Bhatt the first person to assert that all planets including the earth revolve round the sun. A millennium later the Danish astronomer Tycobrahe (1546-1601) conducted a detailed study of planetary motion which was interpreted by his pupil Johnaase Kepler (1571-1630), ironically after the master himself had passed away. Kepler formulated his important findings in three laws of planetary motion where G = 6.67 × 10–11 Nm2 kg–2 is the universal gravitational constant. This law holds good irrespective of the nature of two objects (size, shape, mass etc.) at all places and all times. That is why it is known as universal law of gravitation. Dimensional formula of G : Fr 2 F= m m = 1 2 [MLT 2 ] [L2 ] [M2 ] = [M–1 L3 T–2 ] Newton's Law of gravitation in vector form : Gm1m 2  r̂12 & F12 = r2 Gm1m2  r̂21 F2 1 = r2 UNIVERSAL LAW OF GRAVITATION According to this law "Each particle attracts every other particle. The force of attraction between them is directly proportional to the product of their masses and inversely proportional to square of the distance between them". F m1 m 2 r2 or F = G m1 2m 2 r  Where F12 is the force on mass m1 exerted by mass m2and vice-versa. Now r̂12   r̂21 ,  Thus F21   G m21 m2 r̂12 . r   Comparing above, we get F12   F21 Important characteristics of gravitational force (i) Gravitational force between two bodies form an action and reaction pair i.e. the forces are equal in magnitude but opposite in direction. (ii) Gravitational force is a central force i.e. it acts along the line joining the centres of the two interacting bodies. (iii) Gravitational force between two bodies is independent of the nature of the medium, in which they lie. (iv) Gravitational force between two bodies does not depend upon the presence of other bodies. (v) Gravitational force is negligible in case of light bodies but becomes appreciable in case of massive bodies like stars and planets. (vi) Gravitational force is long range-force i.e., gravitational force between two bodies is effective even if their separation is very large. For example, gravitational force between the sun and the earth is of the order of 1027 N although distance between them is 1.5 × 107 km Ex.1 The centres of two identical spheres are at a distance 1.0 m apart. If the gravitational force between them is 1.0 N, then find the mass of each sphere. (G = 6.67 × 10–11 m3 kg–1 sec–1) Sol. Gravitational force F = Gm.m r Gravitation Sol. The gravitational force of attraction on m1 due to m2 at a separation r is G m1m 2 F1 = r2 Therefore, the acceleration of m1 is F Similarly, the acceleration of m2 due to m1 is G m1 a2 = – r2 the negative sign being put as a2 is directed opposite to a1. The relative acceleration of approach is a = a1 – a2 = G = 6.67 × 10–11 m3 kg–1 sec–1 we get m = 1.225 × 105 kg G (m1  m 2 ) r2 .... (1) If v is the relative velocity, then dv dt a= = dv dr dr dt . dr But – dt = v (negative sign shows that r decreases with increaing t ). a=– dv dr v.. .... (2) From (1) and (2), we have v dv = – G (m1  m 2 ) r2 2 on substituting F = 1.0 N , r = 1.0 m and Gm 1 2 a1 = m  r 2 1 dr Integrating, we get G (m1  m 2 ) v2 = +C r 2 At r = , v = 0 (given), and so C = 0. Ex-2 Two particles of masses m1 and m2, initially at rest at infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant their relative velocity of approach is 2 G (m1  m 2 ) / R that instant. , where R is their separation at  v2 = 2 G (m1  m 2 ) r Let v = vR when r = R. Then  2G (m1  m 2 )   R   vR =  Ex-3 Three identical bodies of mass M are located at the vertices of an equilateral triangle with side L. At what speed must they move if they all revolve under the influence of one another's gravity in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ? Sol. Let A, B and C be the three masses and O the centre of the circumscribing circle. The radius of this circle is L L 2 R= 2 sec 30° = 2  = 3 L 3 . Gravitation Ex.- 5 A solid sphere of lead has mass M and radius R.A spherical hollow is dug out from it (see figure). Its boundary passing through the centre and also touching the boundary of the solid sphere. Deduce the gravitational force on a mass m placed at P, which is distant r from O along the line of centres. Sol. Let O be the centre of the sphere and O' that of the hollow (figure). For an external point the sphere behaves as if its entire mass is concentrated at its centre. Therefore, the gravitatinal force on a mass `m` at P due to the original sphere (of mass M) is Let v be the speed of each mass M along the circle. Let us consider the motion of the mass at A. The force of gravitational attraction on it due to the masses at B and C are GM2 L2 along AB GM2 and L2 along AC The resultant force is therefore 2 GM2 cos30° = 2 L 3 GM2 L2 along AD. F=G Mm r2 , along PO. The diameter of the smaller sphere (which would be cut off) is R, so that its radius OO' is R/2. The force on m at P due to this sphere of mass M' (say) would be M m F = G (r  R )2 along PO. 2 [ distance PO = r – This, for preserving the triangle, must be equal to the necessary centripetal force. That is , 3 GM 2 L2 Mv 2  R (2L3 / 2 ) 3GM M . 8 F = G Mm 8 (r  R ) 2 along PO. As both F and F point along the same direction, the force due to the hollowed sphere is GM L Ex.4 Find out the time period of circular motion in above example Ans. M = 2 3 Mv 2 L [ R = L/ 3 ] or v = ] As the radius of this sphere is half of that of the original sphere, we have  = R 2 F – F = = G Mm r 2 – GMm R 8r 2 (1  )2 2r   G Mm  1  1   2 2 . R r  8 (1  2 r )    Gravitation GRAVITATIONAL POTENTIAL GRAVITATIONAL FIELD The gravitational potential at a point in the The space surrounding the body within which its gravitational field of a body is defined as the amount gravitational force of attraction is experienced by other bodies is called gravitational field. Gravitational of work done by an external agent in bringing a body field is very similar to electric field in electrostatics of unit mass from infinity to that point, slowly (no where charge 'q' is replaced by mass 'm' and electric change in kinetic energy). Gravitational potential is constant 'K' is replaced by gravitational constant 'G'. very similar to electric potential in electrostatics. The intensity of gravitational field at a points is defined Let the unit mass be displaced through a distance dr as the force experienced by a unit mass placed at towards mass M, then work done is given by that point. E= F m = GM r2 dW = F dr = GM r 2 dr .   dW   r GM  2 The unit of the intensity of gravitational field is N kg– 1 . In vector form  GM E   2 r̂ r Thus gravitational potential, V   r GM r dr  GM . r . The unit of gravitational potential is J kg –1. Dimensional Formula of gravitational potential Dimensional formula of intensity of gravitational field = Work [ ML2 T 2 ]  = [M°L2 T–2]. mass [M] F [ MLT 2 ] 0 2 = m  [ M ]  [ M LT ] RELATION BETWEEN GRAVITATIONAL Ex-6 Find the relation between the gravitational field on the surface of two planets A & B of masses mA, mB FIELD AND POTENTIAL & radius RA & RB respectively if The work done by an external agent to move unit mass from a point to another point in the direction of (i) they have equal mass the field E, slowly through an infinitesimal distance dr (ii) they have equal (uniform) density = Force by external agent × distance moved = – Edr. Let EA & EB be the gravitational field intensities on the surface of planets A & B. dV Thus dV = – Edr  E=– . then, EA = Gm A R 2A 4 G R 3A A 3  R 2A dr = 4G   RA 3 A Therefore, gravitational field at any point is equal to the negative gradient at that point. Gm 4G B   R Similarly, EB  R 2 = B 3 B B R B2 (i) For mA = mB EA EB (ii) For & A = B EA RA  EB R B = R2 A  Ex.7 The gravitational field in a region is given by E = – (20N/kg) ( î  ĵ ) . Find the gravitational potential at the origin (0, 0) – (in J/kg) (A*) zero (B) 20 2 (C) – 20 2 (D) can not be defined Sol. V = –  E. dr =  Ex. dx   Ey. dy Gravitation II. A linear mass of finite length on its axis : = 20x + 20y at origin V = 0 Ex.8 In above problem, find the gravitational potential at a point whose co-ordinates are (5, 4) – (in J/kg) (A) – 180 (C) – 90 (B*) 180 (D) zero Sol. V = 20 × 5 + 20 × 4 = 180 J/kg Ex.9 In the above problem, find the work done in shifting a particle of mass 1 kg from origin (0, 0) to a point (5, 4) – (In J) (A) – 180 (C) – 90 (a) Potential : V = – =– GM n (sec 0 + tan 0 ) L GM L  L  L2  d2  d  ln (B*) 180 (D) zero Sol. W = m (V – Vi) = 1 (180 – 0) = 180 J (b) Field intensity : GM Ld E=–    sin 0 GM = GRAVITATIONAL POTENTIAL & FIELD FOR DIFFERENT OBJECTS I. d L2  d2 III. An infinite uniform linear mass distribution of  GM Ring. V = x or (a 2  r 2 )1/ 2  GM r & E = (a 2  r 2 )3 / 2 r̂ or E = – GM cos  x 2 Gravitational field is maximum at a distance, r = ± a 2 and it is – 2GM 3 3 a 2  . 2 linear mass density , Here 0 = M And noting that  = 2 L in case of a finite rod we get, for field intensity E= 2G d Potential for a mass-distribution extending to infinity is not defined. However even for such mass distributions potential-difference is defined. Here potential difference between points P 1 and P2 respectively at distances d1 and d2 from the infinite d2 rod, v12 = 2G n d 1 IV. Uniform Solid Sphere (a) Point P inside the shell. r < a, then V=  GM 2a 3 (3a 2  r 2 ) & E = – and at the centre V = – 3GM 2a GM r a3 , and E = 0 Gravitation (b) Point P outside the shell. r > a, then V = & E=– GM  r GM r2 (c) Point between the two surface GM 2r V=– Ex.10 Calculate the gravitational field intensity at the center of the base of a hollow hemisphere of mass M and radius R. (Assume the base of hemisphere to be open) Sol. We consider the shaded elemental ring of mass, dm V. Uniform Thin Spherical Shell (a) Point P Inside the shell.  GM a & E=0 M (b) Point P outside shell. r > a, then V =  GM r & E=– ; 3 3 GM r  R1 3 3 r 2 R 2  R1 E=– r < a , then V =  3rR 22  r 3  2R13      R 32  R13   GM r2 = (2R 2 ) 2  R sin (Rd) Field due to this ring at 0, dE = Gdm R cos  R3 (see formulae for field due to a ring) or, dE = GM R2 sin  cos  d /2 /2 Hence, E =  dE = 0 0 VI. Uniform Thick Spherical Shell (a) Point outside the shell V=–G M r ; E=–G M r2 (b) Point inside the Shell V=– E=0 3 GM 2   R  R1  2 2  2  R 2  R1R 2  R1  or,  GM E = 2R 2 GM R2 sin  cos  d Gravitation Ex.11 Calculate the gravitational field intensity and potential at the centre of the base of a solid hemisphere of mass m, radius R. Sol. We consider the shaded elemental disc of radius R sin and thickness Rd M Its mass, dM = 2  R3 3  3GM sin 3 (cos ec   cot ) d (R sin ) or, dV = /2  3GM V=– R  (R sin )2 (Rd sin ) or dM = sin3  d 0 /2 2 3GM  cos   cos     2  R  0 =– 3M 2  (sin   cos  sin ) d or, v = – 3GM 2R Aliter : Consider a hemispherical shell of radius r and thickness dr Field due to this plate at O, dE = 2Gd M (1  cos ) (R sin ) Its mass,dm = 2 2 3 (see field due to a uniform disc) or dE = R2  dE /2 0 =  0 3GM sin (1  cos ) R (2r 2 dr ) R3 or,, dm = 3 Mr 2 d r R3 Since all points of this hemispherical shell are at the same distance r from O. Hence potential at O due to it is , 3GM sin  (1  cos ) d /2  E= M 2 Gdm r dV = = 3GMrdr R3 R /2 d = cos 2   3GM   cos     2  R 2  0 3GM  V =  dv = 2R 0 GRAVITATIONAL POTENTIAL ENERGY Gravitational potential energy of two mass system is equal to the work done by an external agent in Now potential due to the element under consideration assembling them, while their initial separation was at the centre of the base of the hemisphere, infinity. Consider a body of mass m placed at a 2GdM distance x from another body of mass M. The dV = (cosec  – cot ) r gravitational force of attraction between them is (see potential due to a circular plate) or E = 3GM 2R 2 given by, F = GM m r2 . Now, Let the body of mass m is displaced from point. C to B through a distance 'dr' towards the mass M, then work done by internal conservative force (gravitational) is given by, Gravitation (iv) When the body of mass m is moved from the surface of earth (i.e., r1 = Re) to a height h (i.e., r2 = Re + h), then change in P.E. of the earth-mass system s given by  1 1  dU = – GMe m  R  h  R  e  e = 1  GMem   GMem   1 h        1 1 1   R e   1  h / R e  R e   R e     Using binomial expansion  dU = dW = F dr = GMm r 2 r dr   dW   Gravitational potential energy,  GM m  U r2 dr GMm r Special Cases: (i) From above equation, it is clear that gravitational potential energy of two mass system increases with increase in separation (r) (i.e. it becomes less negative). (ii) Gravitational P.E. becomes maximum (or zero) at r = . (iii) If the body of mass m moves from a distance r1 to r2 (r1 > r2), then work done or change in gravitational P.E. is given by r2 dU =  r1 r2 GMem r2 GMem   h   GMemh   1  1  R e   R e   R e2 GMe Since g = R 2 then dU = mgh e Gravitational potential difference is defined as the work done by an external agent to move a unit mass from one point to the other point in the gravitational field. According to the definition, E is the force experienced by a unit mass at A. The direction of this force is towards the body of mass M. Now the work done to move the unit mass from A to B is given by   dW = F . d x = Edx cos 180º = – Edx dr =GMe m 1 1  r 2dr  GMem  r2  r1  r1 Since r1 > r2, so change in gravitational P.E. of the body is negative. It means, when the body is brought near to the earth, P.E. of the earth-mass system decrease. This work done is equal to the gravitational potential difference (dV). Where dV dx is called potential gradient. Gravitation Ex.13 Distance between centres of two stars is 10 a. The masses of these stars are M and 16 M and their Ex.12 Calculate the velocity with which a body must be radii are a & 2a resp. A body is fired straight from thrown vertically upward from the surface of the earth the surface of the larger star towards the smaller so that it may reach a height of 10 R, where R is the star. What should be its minimum initial speed to radius of the earth and is equal to 6.4 × 10 8 m. reach the surface of the smaller star? (Earth's mass = 6 × 1024 kg, Gravitational constant Sol. Let P be the point on the line joining the centres of G = 6.7 × 10–11 nt-m2/kg2) the two planets s.t. the net field at it is zero Sol. The gravitational potential energy of a body of mass m on earth's surface is U (R) = – G Mm R where M is the mass of the earth (supposed to be concentrated at its centre) and R is the radius of the earth (distance of the particle from the centre of the earth). The gravitational energy of the same body at a height 10 R from earth's surface, i.e. at a distance 11R from earth's centre is U (11 R) = – GM G.16M Then, r 2  (10a  r )2 = 0  (10 a–r)2 = 16 r2  10a – r = 4r r = 2a Potential at point P, vP = G Mm R GM  change in potential energy U(11R) – U(R) =– G Mm 11 R  GMm  This difference must come from the initial kinetic energy given to the body in sending it to that height. Now, supose the body is thrown up with a vertical Then 1 mv2 2 = 10 GMm 11 R or v = 1 2 mv2.  20 GMm     11 R  . Putting the given values : v=  20  (6.7  10 11nt  m 2 / kg2 )  (6  10 24 kg)      11(6.4  10 6 m)   = 1.07 × 104 m/s. 2GM = 2a  a  10 GMm –  R  =   11 R speed v, so that its initial kinetic energy is GM G.16M  r (10 a  r ) 5 GM 2a . Now if the particle projected from the larger planet has enough energy to cross this point, it will reach the smaller planet. For this, the K.E. imparted to the body must be just enough to raise its total mechanical energy to a value which is equal to P.E. at point P. 1 2 2 i.e. mv  or, G(16M)m GMm 2a 8a = mvP v 2  8GM  GM  5GMm  2 a 8a 2a v2 = 45GM 4a or,, vmin = or,, 3 5GM 2 a Gravitation GRAVITATIONAL SELF-ENERGY The gravitational self-energy of a body (or a system of particles) is defined as the workdone by an external agent in assembling the body (or system of particles) from infinitesimal elements (or particles) that are initially an infinite distance apart. Gravitational self energy of a system of n particles =– 1 3 G (4)2 r4 dr,, R Ustar = – 1 G 3 (4) =– 1 G 3 2 =– 3 5 2 r 4 dr 0 R (4) r 5     5  0 2  4 3  1 G  3 R  . R Potential energy of n particles at an average distance 3 GM2 'r' due to their mutual gravitational attraction is equal  Ustar = – 5 R to the sum of the potential energy of all pairs of particle, i.e., ACCELERATION DUE TO GRAVITY mim j It is the acceleration, a freely falling body near the Us = – G  earth’s surface acquires due to the earth’s all pairs rij gravitational pull. The property by virtue of which a ji body experiences or exerts a gravitational pull on This expression can be written as another body is called gravitational mass mG, and the property by virtue of which a body opposes any j n m m i j 1 in G Us = – change in its state of rest or uniform motion is called   2 i  1 j  1 rij  j i its inertial mass m thus if E is the gravitational field  intensity due to the earth at a point P, and g is If consider a system of 'n' particles, each of same acceleration due to gravity at the same point, then mass 'm' and seperated from each other by the same   m g = m G E . average distance 'r', then self energy Now the value of inertial & gravitational mass happen n n to be exactly same to a great degree of accuracy for  m2  1   or Us = – G     g=E r  all bodies. Hence, 2 i 1 j 1  ij ji The gravitational field intensity on the surface of earth is therefore numerically equal to the acceleration due Thus on the right handside 'i' comes 'n' times while 'j' to gravity (g), there. Thus we get, comes (n – 1) times. Thus Us = – 1 m2 Gn (n – 1) 2 r g Gravitational Self energy of a Uniform Sphere (star) Ushell = – G 4 3  2  r  4r dr  3  r M where = 4     R  3 3 GMe Re 2 where , Me = Mass of earth Re = Radius of earth Note : Here the distribution of mass in the earth is taken to be spherical symmetrical so that its entire mass can be assumed to be concentrated at its center for the purpose of calculation of g. Gravitation VARIATION OF ACCELERATION DUE TO GRAVITY (a) Effect of Altitude Acceleration due to gravity on the surface of the IMPORTANT POINTS (i) A t the center of the earth, d = R e, so gcentre = g  Re 1   R e      = 0. Thus weight (mg) of the body at the centre of the earth is zero. GMe earth is given by, g = R 2 e Now, consider the body at a height 'h' above the surface of the earth, then the acceleration due to gravity at height 'h' given by GMe  h   e 2  gh = R  h 2 = g 1  e R  ~g  2h  1    R  e    when h << R. The decrease in the value of 'g' with height h = g – gh 2gh = R . Then percentage decrease in the value of e 'g' = g  gh 2h  100   100% g Re (ii) Percentage decrease in the value of 'g' with the depth  g  gd  d   100 =  100 . g   Re = (b) Effect of depth The gravitational pull on the surface is equal to its (c) Effect of the surface of Earth weight i.e. mg = G  mg = GMe m The equatorial radius is about 21 km longer than its polar radius. R 2e 4 R 3e  m 3 R 2e or 4 g =  G Re ....(1) 3 We know, g = GMe R 2e Hence gpole > gequator. The weight of the body increase as the body taken from the equator to the pole. When the body is taken to a depth d, the mass of the sphere of radius (Re – d) will only be effective for the gravitational pull and the outward shall will have no resultant effect on the mass. If the acceleration due to gravity on the surface of the solid sphere is gd, then 4 gd =  G (Re – d)  3 ...............(2) By dividing equation (2) by equation (1)  d  gd = g 1    Re  (d) Effect of rotation of the Earth The earth rotates around its axis with angular velocity . Consider a particle of mass m at latitude . The angular velocity of the particle is also . Gravitation ESCAPE SPEED The minimum speed required to project a body from the surface of the earth so that it never returns to the surface of the earth is called escape speed. A body thrown with escape speed goes out of the gravitational pull of the earth. Work done to displace the body from the surface of the earth (r = Re) to infinity (r = ) is given by   dW =  GMem Re r2  dr or W = GMe m 1  r 2 dr Re  1 1   1 = – GMe m  r  = – GMe m    R    Re e  According to parallelogram law of vector addition, the resultant force acting on mass m along PQ is W= F = [(mg)2 + (m2 Re cos)2 + {2mg × m2 Re cos} cos (180 – )]1/2 = [(mg)2 + (m2 Re cos)2 – (2m2 g2 Re cos) cos]1/2 2 = mg    2  2 1   R e   cos 2   2 R e  cos 2    g   g   Let ve be the escape speed of the body of mass m, then kinetic energy of the body is given by GMem 1 mv 2e  2 Re 1/ 2 If the body is taken from pole to the equator, then  2   R e g = g 1  g  . Hence % change in weight =  R 2  mg mg1 e   g   mg  ve = 2 g R e = 11.2 km s–1. Important Points 1. Escape speed depends on the mass and size of the planet. That is why escape velocity on the Jupiter is more than on the earth. 2. Escape speed is independent of the mass of the body. 3. Any body thrown upward with escape speed start moving around the sun. At pole  = 90°  gpole = g , At equator  = 0   R e 2  gequator = g 1  g  . Hence gpole > gequator   GMem Re MOTION OF SATELLITES AND KEPLER LAWS  100 mRe 2 R 2  100 e 100 mg g A heavenly body revolving around a planet in an orbit is called natural satellite. For example, moon revolves around the planet the earth, so moon is the satellite of the earth. Their motions can be sttudied with the help of kepler's laws, as stated : Gravitation III. Law of periods : The square of the time for the I . L aw of or bit : Each Planet moves arround the sun planet to complete a revolution about the sun is in an elliptical orbit with the sun at one of the foci as proportional to the cube of semimajor axis of the shown in figure. The eccentricity of an ellipse is elliptical orbit. defined as the ratio of the distance SO and AO i.e. e i.e. Centripetal force = Gravitational force SO = AO m v 2 GMm GM   v2  R R R2 Now, speed of the planet is v= Circumference of the circular orbit 2  R  Time period T Substituting value in above equation   e= GM 42 R 2 4 2 R 3 2  or T  R GM T2 SO a SO = ea The distance of closest approach with the sun at F1 is AS. This distance is called perigee. The greatest distance (BS) of the planet from the sun is called apogee. Perigee (AS) = AO – OS = a – ea = a (1 – e) Apogee (BS) = OB + OS = a + ea = a (1 + e)  4 2  Since  GM  is constant,    T2  R3 or T2 R3 = constant Ex.14 A satellite is launched into a circular orbit 1600 km above the surface of the earth. Find the period of revolution if the radius of the earth is R = 6400 km and the acceleration due to gravity is 9.8 m/sec2. At what height from the ground should it be launched so that it may appear stationary over a point on the earth's equator ? Sol. The orbiting period of a satellite at a height h from 2r 3 / 2 II . Law of Areas : The line joining the sun and a planet sweeps out equal areas in equal intervals of time. A planet takes the same time to travel from A to B as from C to D as shown in figure. (The shaded areas are equal). Naturally the planet has to move faster from C to D. The law of areas is identical with the law of conservation of angular momentum. Areal velocity = = 1 2 r2 d dt area swept time 1 r (rd) = 2 dt = constant Hence 1 2 r2  = constant. earth's surface is T = gR 2 where r = R + h then, T = 2 (R  h) R R h    g  Here, R = 6400 km, h = 1600 km = R/4. Then 2 R  R T= 4 R R  R   4  g    R = 2(1 – 2J)3/2 g Putting the given values : T = 2 × 3.14 ×  6.4  10 6 m    3/2  9.8 m / s 2  (1.25) = 7092 sec = 1.97 hours   Gravitation Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution round the earth is equal to the period of revolution of the earth round its own axis which is 24 hours. Let us find the height h of such a satellite above the earth's surface in terms of the earth's radius. Let it be nR. then T=  R  nR     g  2 (R  n R) R      R = 2  g  (1 + n)3./2 6  6.4  10 meter / sec = 2 × 3.14  9.8 meter / sec 2  (1 + n)3/2   = (5075 sec) (1 + n)3/2 = (1.41hours) (1 + n)3/2 The gravitational force on either star is G (2m)m d2 . If we consider the rotation of the smaller star, the   2d   2 centripetal force (m r 2) is m 3   and for      2md2   i.e.same  3  bigger star   G (2m) m d 2  2d  2 = m 3     3 Gm   3  d  or  =  Therefore, the period of revolution is given by T = For T = 24 hours, we have (24 hours) = (1.41) hours) (1 + n)3/2 or (1 + n)3/2 = 24 1.41 = 17 2   3  d = 2   3 Gm    The ratio of the angular momenta is  d (2m)   3 or 1 + n = (17)2/3 = 6.61 or n = 5.61 The height of the geo-stationary satellite above the earth's surface is nR = 5.61 × 6400 km = 3.59 × 104 km. Ex.15 In a double star, two stars (one of mass m and the other of 2m) distant d apart rotate about their common centre of mass. Deduce an expression ofr the period of revolution. Show that the ratio of their angular momenta about the centre of mass is the same as the ratio of their kinetic energies. ( )big ( )small   big  small =  2d  m   3  2 2 = 1 , 2 since  is same for both. The ratio of their kinetic ( 21 2 )big  big 1  energies is ( 1 2 ) = ,  small 2 small 2 which is the same as the ratio of their angular momenta. Sol. The centre of mass C will be at distances d/3 and SATELLITE SPEED 2d/3 from the masses 2m and m respectively. Both (OR ORBITAL SPEED) The speed required to put the satellite into its orbit the stars rotate round C in their respective orbits around the earth is called orbital speed. with the same angular velocity . The gravitational The gravitational attraction between satellite and the force acting on each star due to the other supplies earth provides the necessary centripetal force. the necessary centripetal force. Gravitation Energy of a Satellite P.E. of a satellite of mass m revolving around the earth in a circular orbit of the earth is given by U= GMe m Re  h m v 02 Re  h  2 1 2  GM and K.E. = GM GM em 2r Hence K.E. = 1 2 e e or, v0 =    =  R  R  h  e h   e  Total Energy E = U + K.E. When h << Re then v0 = g R e =  v0 = 9.8  6.4  10 6 = 7.92 × 103 GMem r + GMem 2r GMem 2r or E = – Since total energy is negative, so it implies that satellite is bound to the earth. If satellite is close to the surface of the earth then total energy  ms–1 = 7.92 km s–1 Time period of Satellite Time period, T = 1 mv 02 2 mv 02 GMem GMem .  or mv 02  2 r r r 2 e or v 0  R  h e  gR 2 GMem r E=– Circumference of the orbit 2 R e  h  orbital speed v0 GM e m 2R e GEO-STATIONARYSATELLITES OR GEO-SYNCHRONOUS SATELLITES  But v0 =  1 2 g R 2e  T=   R e  h  2 R e  h gR 2e (i) The time period of the satellite around the earth must be equal to the rotational period of the earth (i.e. 24 hours.) 1 2     R e  h  (ii) The direction of motion of the satellite must be same as that of the earth. i.e. from west to east. 1 3 2  R e  h  = R e  g 2   Height of the satellite above the earth's surface Time period of satellite is given by, The height of the geio-stationary satellite from the surface of the earth can be calculated from the 1 equation h = T= 1 3 2 2  R e  h   R e  g  Now  T 2 R 2e g  2    Re  42    T = 24 hours = 24 × 3600 s, Re = 6.4 × 106 m, g = 9.8 ms–2 2 T = 42 R e  h g R 2e 3 or (Re + h)3 = T 2 R 2e g 1 4 2  h (Re + h) = 1  T 2 R 2e g  3   2    4  h= 1  T 2 R 2e g  3   2    4  – Re =  24  3600    2  6.4  10 4 2 6 2 2  9.8    6.4  10 6   or h = 35930 × 103 m = 35930 km. (i) (ii) (iii) (iv) (v) Uses of Artificial Satellites Some important uses of artificial satellites are : They are used as communication satellites to send messages to distant places. They are used as weather satellites to forecast weather. They are used to explore the upper region of the atmosphere. They are used to telecast T.V. programs to distant places. They are used to know the exact shape of the earth. Gravitation intersect the earth and the satellite will fall back. Otherwise it will keep on moving in a closed orbit, or will escape from the Earth, depending on the values of v and R. LAUNCHING OFAN ARTIFICIAL SATELLITE AROUND THE EARTH (a) The satellite is placed upon the rocket which is launched from the earth. After the rocket reaches its maximum vertical height h, a spherical mechanism gives a thrust to the satellite at point A (fig.) producing a horizontal speed v. The total energy of the satellite at A is thus. E= 1 mv 2 2 – GMm Rh (b) Hence a satellite carried to a height h (<< R) and given a horizontal speed of 8 km/sec will be placed almost in a circular orbit around the earth (fig.) If launched at less than 8 km/sec, it would get closer and closer to the earth until it hits the ground. Thus 8 km/sec is the critical (minimum) speed. The orbit will be an ellipse (closed path), a parabola, or an hyperbola depending on whether (a) Orbits and Speed: For a body on the earth’s surface, projected E is negative, zero, or positive. In all cases the horizontally with a speed v, the trajectory depends centre of the earth is at one focus of the path. If on the value of its speed v. the energy is too low, the elliptical orbit will Velocity (v) (i) Less than the orbital speed v < gR e Trajectory (i) Body returns to the earth (ii) Equal to orbital speed v = gR e (iii) Between orbital and escape speed gR e < v < 2gR e (ii) Body acquires a near the earth circular orbit (iii) Body acquires an eiliptical orbit with the earth as the near focus (iv) Equal to escape speed v = 2gR e (iv) Body just escapes the earth’s gravity along in a parabolic path. (v) Body escape’s the earth’s gravity in ah hyperbolic path. (v) Greater then escape speedv = 2gR e Gravitation Ex- 17 For a particle projected in a transverse direction from a height h above Earth’s surface, find the minimum initial velocity so that it just grazes the 2 hg 2 2 h surface of earth path of this particle would be an = , v0  v 1 R ellipse with center of earth as the farther focus, point where R is the radius of the earth and g is acceleration of projection as the apojee and a diametrically due to gravity at earth's surface. Hence deduce an opposite point on earth’s surface as perigee. expression for maximum height reached by a rocket Sol. Suppose velocity of projection at point A is vA & at fired with speed 0.9 times the escape velocity. point B, the velocity of the particle is vB. Sol. The gravitational potential energy of a mass m on earth's surface and that a height h is given by Ex.16 A rocket starts vertically upward with speed v0. Shown that its speed v at height h is given by U (R) = – GMm R and U (R + h) = –  1 GMm Rh 1 then applying Newton’s 2nd law at point A & B, we  U(R + h) – U(R) = – GMm  R  h  R    mv 2A m hg GMmh [  GM = gR2] = (R  h) R = 1  h R This increase in potential energy occurs at the cost of kinetic energy which correspondingly decreases. If v is the velocity of the rocket at height h, then the 2 2 decrease in kinetic energy is 21 mv 0  21 mv . Thus, 1 mv 2  1 mv 2 0 2 2 = 1 h R mv 2 GMe m = (R  n)2 &  B = GMem R2 Where A & B are radius of curvature of the orbit at points A & B of the ellipse, but A = B = (say). Now applying conservation of energy at points A & B GM em 1 GMem 1 2  mv 2A   mv B Rh 2 R 2 m hg , 1 1   GMem  R  (R  h)  2 gh  v 02  v 2 = 1  h or get,  A  R Let hmax be the maximum height reached by the rocket, at which its velocity has been reduced to zero. Thus, substituting v = 0 and h = hmax in the last expression, we have v 20  2 g hmax 1 hmax R  h  v 02 1  max  R   or or v02 = hmax 2    2g  v 0   R   = 2 ghmax v 02 2 or hmax = 2g  v 0 R Now, it is given that v0 = 0.9 × escape velocity = 0.9 × (2 g R )  hmax = (09  0.9 ) 2 g R (09  0.9) 2 g R 2g  R 1.62 gR 1.62 R = 2g  1.62R = 0.38 = 4.26 R = 1 (mvB2 – mvA2) = 2 or,  = 2R(R  h) 2R  h GMe = 1   1   GMem  1    R 2 (R  h)2   2    2Rr Rr R  VA2 = (R  h)2 = 2GMe r (r  R) where r = distance of point of projection from earth’s centre = R + h.