GRAVITATION - Lecture Notes

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 CHAPTER - 00 GRAVITATION Universal law of Gravitation /Newton’s law of Gravitation Every body in the universe attract each other with a force which is directly proportional to product of masses and inversely proportional to the square of distance between them Newton’s law possible for point masses or spherical symmetry F m1m2 F 1 G  Universa l g ravitational cons tan t r2  6.67 1011 Nm2 / Kg2  6.67 108 dyne cm2 / g2 F m1m2  M1L3T2  r2 F  Gm1m2 r2 Fr 2 G  m1m2 , if r = 1, m1 = 1, m2 = 1, then G = F It is the gravitational force, between 2 unit masses, separated by a unit distance. Characteristic Properties of Gravitational Force * Always attractive * Weakest force in the universe * Long-range force * Central force  force acts along the line joining the centres of interacting bodies * Conservative force  work done by gravitational force is independent of the path or dependent only on initial and final positions 1

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 * forms action, reaction pairs * Independent of the intervening medium (sheilding  gravitations not possible) * Independent of the presence of other masses Vector Notation F12  Force on m1 due to m2 r21  From m2 to m1   Gm1m2 rˆ21 F12 r2 –ve sign indicate attraction F21 = Force on m2 due to m r12 = from 1 to 2   Gm1m2 rˆ12 F21 r2 r12   r21 F12  F21 Principle of Superposition 2

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Principle of superposition says that force exerted by a particle or other particle in same.    Fnet  F01  F02  F03.......  Fon , if there are n masses Force on a point mass due to a shell, of uniform density and uniform thickness Case I : The point mass is outside the shell. F  GMm r2 Case II : Point mass is inside the shell In this case also, net force = 0 The particle is situated close to the top side, so the magnitude of force experienced will be more, but it is compensated, as more particles are situated in the bottom portion.  net force = 0 Force on a point mass due to a sphere, of uniform density GMm r  R(outside) F  r2 r  R(surface) F  GMm R2 3

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 r  R(inside) F  GMm  r R3 r = 0 (centre) F = 0 Note : If equal masses are distributed symmetrically on the vertices of any regular polygon, the net force (gravitational force / field) acting at the geometrical centre will be zero. Acceleration due to gravity Gravity on a Body Force with which the planets attracts the body Wt. of a body The force with which the earth attracts the body Weight  Gravity mg  GMm R2 g  GM R2 “Independent of mass of the object” Variation of Acceleration due to Gravity * Variation due to earth’s shape * Variation due to height * Variation due to depth * Variation due to earth’s rotation Variation due to earth’s shape Earth is not a perfect sphere. Its a big flattered at the poles and bulged at the equator. This shape is known as geoid. 4

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 g  GM or g  1 R2 R2 gp  ge Variation due to height gh  Gm ...........(2) R  h2 g  GM ...........(1) R2 2  gh  R2 1 g R  h2 gh  g R 2  R  h   gh  g  1 2   h  R    R      gh  g  1    2   1  h   R     = g 1 h  2 R  If h < < < R, the higher power of h/R can be neglected 5

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 gh  g 1 2h  R  Fractional Change in g gh  g 2h g R 2h  g  g  gh R 2h  g  gh  Fractional change in g OR Fractional decrease OR R g 2h 100  g  gh 100% Rg 3) Variation due to depth Assume, the earth to be a perfect sphere of radius R and uniform density  Acceleration due to gravity on the surface of the earth = GM  G   4 R3 R2 R2 3 Acceleration due to gravity at a depth d from the surface of earth gd  GM1  G  4 R  d 3 .............(2) R  d2 R  d2 3 gd  R  d as R  d  r  r  gd gR Rg gd 1 d g r  gd gR R 6

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 gd  g 1 d  R  At the centre, r = 0 gd = 0 * Relation between gd & gh If gd = gh g 1 d   g 1 2h  here, we have taken h < < < R R  R  d  2h Variation due to rotation / latitude Consider a mass ‘m’ situated on the surface of earth. Let  be the latitude at the place. Due to earth’s rotation, the masses executing circular motion of radius ‘r’ 7

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Centrifugal force acting on the mass is radially outward. The apparent wt. of the mass, mg '  mg  mR2 cos2  g '  g  R2 cos2  cos   r R r  cos R At equator,   0  Effect of rotation is max imum g '  g  R2  At the poles,   90o g'  g At the poles, there is no effect of rotation If speed  , g '  g  R2  value will be less so, wt. will be less Gravitational field It is a space around a mass upto which its gravitational force can be experienced # Strength of the field varies from place to place # To express the field strength at a point, we use a quantity called gravitational field intensity  Gravitational Field Intensity  I      N kg  LT2  I F m It is the force experienced by unit mass Gravitational field intensity at a point is the force experienced by a unit mass at a point A point mass ‘m’ is placed at a point ‘p’ which is at a distance ‘r’ from the centre of mass ‘M’. Force exerted by M as ‘m’ F GMm towards M r2 The gravitational field intensity at the point, P I  F  GM m r2 I=g 8

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Special cases I gravitational field intensity due to a shell. a) r  R I  GM rI r2 b) r  R Finside  0 I  0 c) r = R GM max R  I  I  R2 III Gravitational field due to solid sphere rR rR rR a) I  GM b) I  GM  max c) I GM r2 R2 R3 r from c) I  r At the centre r = 0, I = 0 (Min) 9

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 GRAVITATIONAL POTENTIAL Potential = V  W  T  L2T2  m kg It is the work done to bring a unit mass from infinity to a point. This work done is the gravitational potential of that point Consider a mass ‘M’ the potential of it at a distance ‘r’ from its centre The work done in moving a unit mass through a distance dx dW  F dx (unit mass = 1) dW  GM 1 dx r2 Then, the Total work done in bringing it from infinity to P r 1  GM r GM  GM x2  x r GM     dx  = W  GM  V r Gravitational Potential due to a shell 10

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 A) r  R, V  Gm B) r  R, V  GM C) r  R, V  GM (But f is zero) R RR Gravitational Potential due to a solid sphere A) r  R, V  Gm B) r  R, V  GM r  R, V  GM r R 2R3  C) At the centre r = 0 3R 2  r2  V  3GM 2R 11

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 GRAVITATIONAL POTENTIAL ENERGY GPE of a mass at a point is the work done to bring the mass from infinity to that point. This work done is stored as gravitation potential energy of the mass at that point. Work done to bring unit mass = V Work done to bring ‘m’ mass = U  GM m r I. Gravitational potential energy of an isolated system of masses U  U12  U23  U31    Gm1m2  Gm2m3  Gm1m3   r12 r23 r13    Gravitation potential energy  isolated system U12  Gm1m2 r12 12

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Gravitational Potential Energy on Earth 1) Consider a body of mass m is taken from the surface of the earth to a height ‘h’ The change in potential energy u  w u  GMm  GMm Rh R u  GMm  1  R 1 h   R   or u  w  mgh 1 h R 2) Consider a body of mass ‘m’ is vertically projected from the surface of the earth to a height ‘h’. the required velocity v  2gh 1 hR Satellites Satellite is an object which continuously revolve around a much larger object due to the gravity 13

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Geostationary / Geosynchronous Satellite It appears to be at rest for an observer on earth surface i.e., there is not relative motion between earth and satellite. 1) Direction of revelution from west to east 2) Time period = 24 hrs 3) Height h = 36000 km 4) Orbit of a geostationary satellite must be concentric or coplanar with the equatorial plane of earth Polar Satellite It revolves around the polar of earth. During each revolution, polar satellite scan the information on a slip of area of the earth, in the interval of 1 day, it scans the entire area of earth. Orbital velocity / Orbital speed Consider a satellite of mass ‘m’ revolving in an orbit which is at a height ‘h’ from the earth’s surface. Let V0 be the orbital velocity of the satellite. The necessary centripetal force for the satellite is provided by earth’s gravity. FC  FG  mv2  GMm r r2 v0  GM ; V0 1 r r 14

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Orbital velocity is orbit specific, i.e., for a particular orbital, value of orbital velocity is a constant. Condition for min. orbit / closest orbital speed Time period of satellite dis tan ce 2r  Time period = speed v0 r r3 R  h3  2r  2 GM GM or T  2 GM T  r3/2 or T2 r3 T  R  h 3/2 or T2 R  h 3 Consider for min. orbit, h < < < R, R  h  R , r = R T  2 R3 instead of GM = gR2 gR 2  2 R  1.4 hrs g Angular Momentum  L rp L  r  mv0  rm GM r = m GMr 15

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Energy of an orbiting Satellite Kinetic energy Potential energy U  GMm r Total energy TE  K  U  GMm  GMm  GMm 2r r 2r TE  GMm 2r TE = –k, TE = u/2; K  U 2 Height of a Satellite T  2 r3 r  R h GM T2  42r3 GM T2  42 R  h3 GM  T2GM  13  42  R  h    T2gR 2 1 3  42  R  h    16

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Escape Velocity / Escape Speed It is the minimum speed with which a body should be projected, so, that it can cross the gravitational field / boundary of earth Velocity from earth’s surface TE  TE GMm  1 mve2  0 R2 1 ve2  GM 2R ve  2GM R Ve  2GM  2gR 2 R R Ve  2gR V0  gR , Ve  2gR Ve  2V0 If the body is projected at a height h above the surface Ve  2GM Rh Escape velocity does not depends on mass of the body and angle of projection 17

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 Kepler’s Law of Planetary Motion Kepler’s first law  law of elliptical orbit Planets revolve around the sun in elliptical orbits, with the sun at one of it focer a  semi major axis b  semi minor axis ra  apogee rp  perigee P  perihelion A Aphelion Second law of Kepler  Law of Areas The line joining the planet and sun sweeps equal areas in equal interval of time This law is a consequence of law of conservation of angular momentum Areal velocity  A  cons tan t T means equal areas covered in equal interval of time. A  1 bh  1 r  vt 22 A  1 r  v  t t 2 t A  1 rv t 2 A  r  m t 2 m 18

Brilliant STUDY CENTRE LT-23M SPECIAL PHYSICS (ONLINE) -2023 = mr  L A  L 2m 2m t 2m L  constant as there is no external torque acting Hence proved areal velocity is a constant Angular momentum of the planet is conserved. Lp = La m  vp  rp  m  va  ra rp  a 1 e vprp  vara ra  a 1 e We know ra > rp Vp  ra Va rp vp  va L = mvr rp  1 e ra 1 e vp  1 e va 1 e vprp  vara GM 1 e VP  a 1 e vp  a 1 e  va  a 1 e VA  GM 1 e vp  1 e a 1e va 1 e Kepler’s 3rd law  Harmonic Law The square of time-period of revolution of a planet is proportional to the cube of semi major axis of its elliptical path T2 a3 a  r  circle 19