PHYSICS NOTES GRAVITATION

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1 GRAVITATION Newton s law of gavitation The law states that evey paticle of matte in the univese attacts evey othe paticle with a foce which is diectly popotional to the poduct of thei masses and invesely popotional to the squae of the distance between them. The diection of the foce is along the line joining the line of paticles Gavitational foce is always attactive. Conside two bodies of masses m1 and m2 with thei centes sepaated by a distance. The gavitational foce between them is Theefoe fom Newton s law of gavitation in vecto fom is F = Gm 1m 2 Hee F12 epesents foce on mass 1 due to mass 2 And F21 is foce of mass 2 due to mass 1 Note F12 and F21 ae equal and opposite. The gavitational foce foms between two paticles fom an action eaction pai. Solved Numeical Q ) How a mass M be divided so that gavitational foce is maximum between the pats Solution: Let be the distance between two pats m and M-m, the gavitational foce between them is Fo F to be maximum F = G m(m m) 2 = G 2 [Mm m2 ] df dm = 0 d dm ( G [Mm 2 m2 ]) = 0 O M -2m = 0 [ as G/ 2 0] O m/m = ½ The foce is maximum when two pats ae equal Q) Find the gavitational foce of attaction between a unifom sphee of mass M and a unifom od of length L and mass m oiented as shown in figue. (given in solution) Solution 2 1

2 Since the sphee is unifom its entie mass may be consideed at the cente. The foce on the elementay mass dm is +L F = GM x 2 m L dx F = GMm [ 1 L x ] 1 F = GMm L F = GMm L df = F = +L [ + L 1 ] L ( + L) +L GM dm x 2 GM dm x 2 but dm = m L dx F = GMm ( + L) Gavitational Field o Intensity (I) Pocess of action at a distance in which gavitational foce is exeted mutually on two bodies sepaated by some distance is explained though the field (i) Evey object poduces a gavitational field aound it, due to mass (ii) This field exets a foce on anothe body bought (o lying) in this field The gavitational foce exeted by the given body on a body of unit mass at a given point is called the intesity of gavitational field (I) at that point It is also known as the gavitational field o gavitational intensity The gavitational field intensity is a vecto quanity and its diection is the diection along which the unit mass ha a tendensy to move. The unit of gavitational field intesity is N/Kg and its dimensions ae [LT -2 ] Calculation of gavitational field (a)gavitational field intensity due to a point mass. Conside a point mass M at O and let us calculate gavitational intesity at A due to this point mass. Suppose a test mass is placed at A By Newton s law of gavitation, foce on test mass F = GMm 2 along AO 2

3 I = F m = GM 2 e eq(1) (b) Gavitational fild intensity due to a unifom cicula ing at a point on its axis Figue shows a ing of mass M and adius R. Let P is the point at a distance fom the cente of the ing. By symmety the field must be towads the cente that is along PO Let us assume that a paticle of mass dm on the ing say, at point A. Now the distance APis R Again the gavitational field at P due to dm is along PA and the magnitude is G dm di = Z 2 G dm decosθ = cosθ Sine components will be canceled when we conside magnetic field due to entie ing and only cos components will be added Net gavitational field I I = Gcosθ Z 2 dm But cosθ = /Z Cases (i) If >>R, 2 + R 2 = 2 I = GM 3 (ii) If <<R, 2 + R 2 = R 2 Z 2 I = GM Z 2 cosθ I = GM Z 2 Z = GM Z 3 GM I = (R ) 3 2 along PO = GM 2 [ negative sign indicates attaction] (iii) Fo maximum I I = 0 I = GM R 3 I 3

4 GM [( 2 + R 2 ) (2 + R 2 ) ] [ 2 + R 2 ] 3 = 0 [( 2 + R 2 ) (2 + R 2 ) ] = 0 [( 2 + R 2 ) ] = 0 [( 2 + R 2 ) 3 2 ] = 0 = ± R 2 (c) Gavitational field intensity due to a unifom disc at a point on its axis Let the mass of disc be M and its adius is R and P is the point on its axis whee gavitational field is to be calculated Let us daw a ing of adius x and thickness dx O is the cente of cicle. Aea of ing is 2πxdx The mass of ing dm = M 2Mxdx 2πxdx = πr2 R 2 Gavitational field at p due to ing is G ( 2Mxdx di = R 2 ) ( 2 + x 2 ) 3 2 di = 2GM R 2 0 R xdx ( 2 + x 2 ) 3 2 I = 2GM 1 R 2 [ 2 + x 2] 0 R In tems of θ I = 2GM R 2 [ x 2] I = 2GM R 2 [ 2 + x 2] 4

5 I = 2GM R 2 (1 cosθ) (d) Gavitational field due to a unifom solid sphee Case I Field at an extenal point Let the mass of the sphee be M and its adius be R. We have to calculate the gavitational field at P di = Gdm 2 di = G Gm dm = 2 2 Thus, a unifom sphee may be teated as a single paticle of equal mass placed at its cente fo calculating the gavitational field at an extenal point Case II Find at an intenal point Suppose the point P is inside the solid sphee, in this case <R the sphee may be divided into thin spheical shells all centeed at O. Suppose the mass of such a shell is dm. then gavitational field due to this spheical shell di = Gdm along PO 2 But dm = density volume dm = di = G 2 dm M 4 3 πr3 di = Gdm π3 = M3 R 3 I = GM R 3 Theefoe gavitational field due to a unifom sphee at an intenal point is popotional to the distance of the point fom the cente of the sphee. At the cente = 0 the field is zeo. At the suface of the sphee = R I = GM R 2 5

Note : (I)Gavitational field due to solid sphee is continuous but it is not diffeentiable function (ii) Gavitational field at point inside the sphee is only due to the mass enclosed by the suface

6 Note : (I)Gavitational field due to solid sphee is continuous but it is not diffeentiable function (ii) Gavitational field at point inside the sphee is only due to the mass enclosed by the suface passing though the point, volume enclosed is shown by shaded potion in diagam and field due to oute volume is zeo. (e) Field due to unifom thin spheical shell Case I When point lies inside the spheical shell suface passing though point do not enclose any mass thus I = 0 Case II Point P lies outside the spheical shell I = di = G GM dm = 2 2 Note : Gavitational field due to thin spheical shell is both discontinuous and nondiffeentiable function Solved Numeical Q) Two concentic shells of masses M1 and M2 ae situated as shown in figue. Find the foce on a paticle of mass m when the paticle is located at (a) point A (b) point B (c) point C. The distance is measued fom the cente of the shell 6

7 Solution: We know that attaction at an extenal point due to spheical shell of mass M is GM 2 While at an intenal point is zeo. So (a) At point A let = a, the extenal point fo both shells so field intensity I A = G(M 1 + M 2 ) a 2 F A = mi A = mg(m 1 + M 2 ) a 2 (b) Fo point B, let = b, the point is extenal to shell of mass M2 and intenal to the shell of mass M1, so I B = GM 2 b F B = mi B = GmM 2 b 2 (c) Fo point C, let = c, the point is intenal to both the shells; so I C = = 0 F C = mi C = 0 Gavitational potential Gavitational potential (V) at a point is defined as the amount of wok done in moving unit mass fom the point to infinity against the gavitational field. It is a scala quantity. Its unit is N m kg 1. O J kg -1 dimensional fomula M 0 L 2 T -2 Mathematically V = W/m By the definition of potential enegy U = W So V = U/m O U = mv Thus gavitational potential at a point epesents potential enegy of unit point mass at that point Wok is done against gavitational foce thus W = F gavitation d i.e dv = _Id V = W m = F gavitation m F gavitation But = I m V = I d o E = dv d d 7

8 Calculation of Gavitational potential (a) Gavitational potential at a point (P) due to a point mass(m) We have gavitational field due to a point mass I = GM 2 The negative sign is used as gavitational foce is attactive V = GM d 2 V = GM d 2 V = GM [ 1 ] = GM [ 1 1 ] = GM (b) Gavitational potential at a point due to a ing Let M be the mass and R be the adius of thin ing. Consideing a small element of the ing and teating it as a point mass, the potential at the point P is G dm G dm dv = = Z R Hence, the total potential at the point P is given by G dm V = R 2 + = GM 2 R At = 0 V = GM and dv R d = 0 Thus at cente of ing gavitational field is zeo but potential is not zeo Also dv d = d d ( GM R 2 + 2) dv GM 2 = d R 2 = 0 V is minimum at =

9 (c) Gavitational potential at a point due to a spheical shell ( hollow sphee) Conside a spheical shell of mass M and adius R. P is a point at a distance fom the cente O of the shell. Conside a ing at angle to OP. Let θ be the angula position of the ing fom the line OP. The adius of the ing = Rsinθ The width of the ing = Rdθ Suface aea of ing = (2πRsinθ)Rdθ Suface aea of ing = 2πR 2 sinθdθ The mass of the ing = (2πR 2 M sinθdθ) 4πR 2 = Msinθdθ 2 If x is the distance of the point P fom a point on the ing, then the potential at P due to the ing dv = GMsinθdθ eq(1) 2x Fom cosine popety of tiangle OAP x 2 = R Rcosθdθ Diffeentiating 2xdx = 2R sinθdθ x dx sinθdθ = R On substituting above value of sinθdθ dv = GM 2x x dx R dv = GM 2R dx Case I When point P lies outside the spheical shell V = GM +R 2R dx = R GM 2R [x] +R R V = GM 2R [( + R) ( R)] = GM This is the potential at P due to a point mass M at O 9

10 Fo an extenal point, a spheical shell behaves as a point mass supposed to be placed at its cente Case II When the point P lies inside the spheical shell V = GM R+ 2R dx = R GM 2R [x] R+ R V = GM R This expession is independent of. Thus, the potential at evey point inside the spheical shell is the same and is equal to the potential of the suface of the shell (d) Gavitational potential due to a homogeneous solid sphee Case (I) When the point P lies outside the sphee. Fo extenal point, a solid sphee behaves as if its entie mass is concentated at the cente. Case(II) When the point O lies inside the sphee Let us conside a concentic spheical suface though the point O. The potential at P aises out of the inne sphee and the oute thick spheical shell V = V1 + V2, whee V1 = potential due to the inne sphee and V2 = potential due to oute thick shell The mass of the inne sphee = 4 3 π3 ρ ρ = density of the sphee = M 3M = 4 4πR 3 3 πr3 The potential at P due to this sphee G [ 4π3 3 ] ρ V 1 = = 4πGρ 3 2 To find V2, conside a thin concentic shell of adius x and thickness dx The volume of the shell = 4πx 2 dx The mass of the shell = 4πx 2 dx ρ The potential at P due to this shell V 2 = 4πGρxdx R V 2 = 4πGρ [ x2 R 2 ] 10

11 At = 0 V 2 = 4πGρ [ R ] V 2 = 2πGρ[R 2 2 ] V = V 1 + V 2 = 4πGρ 3 2 = 2πGρ[R 2 2 ] V = 4πGρ 3 [2 + 3R ] V = 4πGρ 3 [3R ] V = 4πG 3 3M 4πR 3 [3R ] V = GM 2R 3 [3R2 2 ] V = 3GM 2R dv d = 0 Hence gavitational field is 0 at the cente of a solid sphee Gavitational potential enegy The gavitational potential enegy of a mass m at a distance fom anothe mass M is defined as the amount of wok done against gavitational foce in moving the mass m fom infinity to a distance U () = F d 11

12 Wok is done against gavitational foce so negative sign U () = GMm d Gavitational foce is attactive hence negative sign taken 2 Gavitational potential diffeence U () = GMm [ 1 ] U () = GMm If we take point at P at a distance p and othe point Q at a distance Q. Object of mass m is moved fom P to Q then, Wok done Q U = GMm 2 d P U = GMm [ 1 Q ] P O U = U P U Q = GMm [ 1 Q 1 P ] U = U Q U P = GMm [ 1 P 1 Q ] Solved Numeical Q) A paticle of mass m is placed on each vetex of a squae of side l. Calculate the gavitational potential enegy of this system of fou paticles. Also calculate the gavitational potential at the cente of the squae Solution: Hee we can wite enegy due to evey pai of paticles as U ij = Gm im j ij Whee mi and mj espectively ae the masses of the paticles I and j espectively and ij is the distance between them. mi = mj = m Theefoe potential enegy 12

13 U = Gm 2 [ 1 ] ij i<j U = Gm 2 [ ] U = Gm 2 [ 1 l + 1 2l + 1 l + 1 l + 1 2l + 1 l ] U = Gm 2 [ ] l Gavitational potential at the cente, due to each paticle is same The total gavitational potential at the cente of the squae is V= 4( potential due to evey paticle) Whee = 2l 2 V = 4 ( Gm ) V = 4 2Gm l Q) Two objects of masses 1kg and 2kg espectively ae eleased fom est when thei sepaation is 10m. Assuming that on it mutual gavitational foce act on them, find the velocity of each of them when sepaation becomes 5m ( Take G = Nm 2 /kg 2 ) Solution: Let v1 and v2 be the final velocity of masses, m1 = 1kg, m2 = 2 initial velocity is zeo Fom law of consevation of momentum m 1 v 1 + m 2 v 2 = 0 v 1 = m 2 v m 2 1 v 1 = 2 v 2 eq(1) Initial potential enegy Final potential enegy U i = Gm 1m 2 = ( )(1 2) i 10 U i = J U f = Gm 1m 2 f = ( )(1 2) 5 U f = J Change in Potential enegy = J Accoding to law of consevation of enegy 13

14 K = - U Fom equation (1) 1 2 m 1v m 1v 2 2 = v (2)v 2 2 = v 2 2 = v 2 = m/s v 1 = m/s Vaiation in acceleation due to gavity (a) With altitude At the suface of eath At height h above the suface of eath g g = g = g = GM e R e 2 GM e (R e + h) 2 2 R e (R e + h) 2 = 1 g = (1 + h ) R e g (1 + h 2 ) R e So, with incease in height, g deceases. If h <<R, then fo binomial theoem g = g [1 + h 2 ] = g [1 2h ] R e R e (b) With depth At the suface of the eath g = GM e R2 e Fo a point at the depth d below the suface Mass the eath enclosed by the suface passing though point P as shown in figue be m then Gm g = (R e d) 2 We know that gavitational at point P due to shaded potion is zeo thus 2 14

15 m = 4 3 π(r e d) 3 M e 4 3 πr e 3 m = M e R (R 3 e d) 3 e G M e g = (R e d) 2 R (R 3 e d) 3 e Thus g g = g = GM e R e 3 GM e R e 3 (R e d) GM e R e 2 (R e d) = R e d R e g = g [1 d R e ] So with incease in depth below the suface of the eath, g deceases and at the cente of the eath it becomes zeo It should be noted that value of g deceases, if we move above the suface o below the suface of the eath (C) Due to otation of the eath The eath is otating about its axis fom west to east. So, the eath is a non-inetial fame of efeence. Eveybody on its suface expeiences a centifugal foce. Conside a point P. Pependicula distance fom point with axis of otation is. Then centifugal foce at point is mω 2 cosα, going ouwad whee α is the latitude of the place. Hee α is the angle made by the line joining a given place on the Eath s suface to the cente of the Eath with the equatoial line is called latitude of the place. Hence fo equato latitude is and fo poles latitude is 90 O Gavitational foce mg is acting towads the cente of eath. Thus esultant foce Is mg = mg - mω 2 cosα ----eq(1) Fom the geomety of figue = Recosα theus fom equation (1) g = g - ω 2 Recos 2 α 15

Cases (i) At equato α = 0 cos α = 1 g = g - ω 2 Re Which shows minimum value of the effective gavitational acceleation (ii) At poles, α = 90 cos α = 0 g = g, which shows the maximum value of the

16 Cases (i) At equato α = 0 cos α = 1 g = g - ω 2 Re Which shows minimum value of the effective gavitational acceleation (ii) At poles, α = 90 cos α = 0 g = g, which shows the maximum value of the effective gavitational acceleation Solved Numeical Q) The density of the coe of planet is ρ1 and that of the oute shell is ρ2. The adii of the coe and that of the planet ae R and 2R espectively. Gavitational acceleation at the suface of the planet is same as at a depth R. Find the atio ρ1/ ρ2 Solution: Mass of inne sphee M1 M 1 = 4 3 πr3 ρ 1 Volume of oute shell 4 3 π(2r)3 4 3 πr3 = 4 3 π(7r3 ) Mass of oute shell M2 4 3 π(7r3 )ρ 2 Gavitational acceleation at suface of planet: a = G (2R) (M M 2 ) a = G (2R) 2 (4 3 πr3 ρ π7r3 ρ 2 ) a = GRπ 3 (ρ 1 + 7ρ 2 ) eq(1) Gavitational acceleation at depth R Given a = a thus a = G 4 3 πr3 ρ 1 R 2 = 4GπRρ 1 3 GRπ 3 (ρ 1 + 7ρ 2 ) = 4GπRρ 1 3 (ρ 1 + 7ρ 2 ) = 4ρ 1, eq(2) ρ 1 ρ 2 = 7 3 Satellite (a) Obital speed of satellite The velocity of a satellite in its obit is called obital velocity. Let vo be the obital velocity Gavitational foce povides necessay centipetal acceleation 16

17 GM em 2 = mv 2 o As = Re + h v 0 = GM e v 0 = GM e R e + h Notes Obital velocity is independent of the mass of the body and is always along the tangent to the obit Close to the suface of the eath, = R as h=0 v 0 = GM R = gr = = 8 km/s (b) Time peiod of a Satellite The time taken by a satellite to complete one evolution is called the time peiod (T) of the satellite It is given by T = 2π = 2π v o GM T 2 = ( 4π2 GM ) 3 T 2 3 (c) Angula momentum of a satellite (L) In case of satellite motion, angula momentum will be given by L = mv = m GM O L = (m 2 GM) 1 2 In the case of satellite motion, the net foce on the satellite is centipetal foce. The toque of this foce about the cente of the obit is zeo. Hence, angula momentum of the satellite is conseved. i.e L is constant (d) Enegy of satellite The P.E. of a satellite is U = GMm The kinetic enegy of the satellite is 17

18 K = 1 2 mv 0 2 But v 0 = GM K = GMm 2 Total mechanical enegy of the satellite E = GMm + GMm 2 = GMm 2 Note We have K = -E Also U = 2E Total enegy of a satellite in its obit is negative. Negative enegy means that the satellite is bounded to the cental body (eath) by an attactive foce and enegy must be supplied to emove it fom the obit to infinity. (e) Binding enegy of the satellite The enegy equied to emove the satellite fom its obit to infinity is called binding enegy of the satellite. i.e. Binding enegy = E = GMm 2 Solved Numeical Q) An atificial satellite is moving in a cicula obit aound the eath with a speed equal to half the magnitude of escape velocity fom the eath a) Detemine the height of satellite above the eath s suface b) If the satellite is suddenly stopped, find the speed with which the satellite will hit the eath s suface afte falling down Solution: Escape velocity = (2gR), whee g is the acceleation due to gavity on the suface of eath and R is the adius Obital velocity = 1 2 v e = 1 2 2gR = gr 2 eq(1) (1/2)Ve = (1/2) (2gR) = (gr/2) a) If h is the height of satellite above eath s suface, the gavitational foce povides the centipetal foce fo cicula motion mv 0 2 R + h = v 0 2 = GMm (R + h) 2 GM R + h = gr2 (R + h) 18

19 Fom equation (1) ( v 0) = gr2 R + h gr 2 = gr2 R + h R+h = 2R H = R b) If the satellite is stopped in obit, the kinetic enegy is zeo and its potential enegy is GMm 2R When it eaches the eath, let v be its velocity Hence kinetic enegy = (1/2) mv 2 Potential enegy = GMm R By the law of consevation of enegy 1 2 mv2 GMm R = GMm 2R v 2 = 2GM ( 1 R 1 2R ) = 2gR2 2R = gr Velocity with which the satellite will hit the eath s suface afte falling down is v = gr Q) Two satellites of same masses ae launched in the same obit aound the eath so as to otate opposite to each othe. They collide inelastically and stick togethe as weckage. Obtain the total enegy of the system befoe and afte collisions. Descibe the subsequent motion of weckage. Solution Potential enegy of satellite in obit If v is the velocity in obit, we have mv 2 GMn = GMm 2 Kinetic enegy Total enegy GMm 2 v 2 = GM 1 2 mv2 = GMm 2 GMm = GMm 2 19

20 Fo the two satellites, the total enegy befoe collision 2 ( GMm 2 ) = GMm Afte collision, let v be the velocity of the weckage. By the law of consevation of momentum, since they ae appoaching each othe mv mv = 2mv v = 0 The weckage has no kinetic enegy afte collision but has potential enegy P. E. = GM(2m) Total enegy afte collision = 2GMm Afte collision, the centipetal foce disappeas and the weckage falls down unde the action of gavity. Geostationay satellite If thee is a satellite otating in the diection of eath s otation. i.e. fom west to east, then fo an obseve on the eath the angula velocity of the satellite will be same as that of eath ωs = ωe Howeve, if ωs = ωe =0, satellite will appea stationay elative to the eath. Such a satellite is called Geostationay satellite and is used fo communication puposes The obit of geostationay satellite is called Paking Obit We know that T 2 = 4π2 GM 3 Fo geostationay satellite, T = 24 Hous Putting this value of T in the above equation, we get R = 42000Km O h= km Whee h is height of the satellite fom the suface of the eath Weightlessness in a satellite When the astonaut is in an obiting satellite, both the satellite and astonaut have the same acceleation towads the cente of the Eath. Hence, the astonaut does not exet any foce on the floo of the satellite. So, the floo of the satellite also does not exet any foce of eaction on the astonaut. As thee is no eaction, the astonaut has a feeling of weightlessness. The adial acceleation of the satellite is given by a = F m = GMm 2 1 m = GM 2 Fo an astonaut of mass ma inside the satellite, we have following foces 20

21 Downwad foce = GMm a 2 Upwad pseudo foce as motion of satellite is acceleated motion Upwad foce = GMm a 2 Thus esultant foce on Astonaut is zeo, o nomal foce is zeo Hence, the astonaut feels weightlessness 21

m1 m2 M 2 = M -1 L 3 T -2

m1 m2 M 2 = M -1 L 3 T -2 GAVITATION Newton s Univesal law of gavitation. Evey paticle of matte in this univese attacts evey othe paticle with a foce which vaies diectly as the poduct of thei masses and invesely as the squae of