CBSE Class 11 Physics Notes - Chapter 8 - Gravitation PDF

Revision Notes Class – 11 Physics Chapter 8 – Gravitation 1. Introduction The constituents of the universe are galaxy, stars, planets, comets, asteroids, meteoroids. The force which keeps them bound together is called gravitational force. G...

Revision Notes Class – 11 Physics Chapter 8 – Gravitation 1. Introduction The constituents of the universe are galaxy, stars, planets, comets, asteroids, meteoroids. The force which keeps them bound together is called gravitational force. Gravitation is a nature phenomenon by which material objects attract towards one another. In 1687 A.D. English Physicist, Sir Isaac Newton published principia Mathematica, which explains the inverse-square law of gravitation. 2. Newtons Law of Gravitation 2.1 Definition Every particle of matter attracts every other particle of matter with a force that is proportional to the product of their masses and inversely proportional to the square of their separation. 2.2 Mathematical Form If m1 and m2 are the masses of the particles and r is the distance between them, the m1m2 force of attraction F between the particles is given by: F  r2 Class XI Physics www.vedantu.com 1  m1m2 ∴ F G. (G is the universal constant of gravitation.) r 2.3 Vector Form In vector form, Newton’s law of gravitation is represented in the following manner. The force  F21  exerted on the particle m2 by the particle m1 is given by: m1m2 F 21  G (r̂12 ) …(1) r2 Where ( r̂12 ) is a unit vector drawn from particle m1 to particle m2. Similarly, the force  F21  exerted on particle m1 by particle m2 is given by x …(2) B Where ( r̂12 ) is a unit vector drawn from particle m1 to particle m2. From (1) and (2): F12  F21 3. Universal Constant for Gravitation Fr 2 Universal gravitation constant is given as, G  m1m 2 Suppose that, m1  m2  1 , and r  1 then G  F. The force of attraction between two unit masses put at a unit distance apart is numerically equal to the universal gravitation constant. 3.1 Unit newton ( metre )2 Nm2 SI unit:  ( kilogram )2 kg 2 CGS Unit: dyne cm2 / gm2. Class XI Physics www.vedantu.com 2 3.2 Value of G The value of G; G  6.67  1011 Nm2 / kg2 Dimensions of G:  M1L1T 2   M0 L2T 0  [F]  r 2  [G]  = =  M1L3T 2   m1m 2   M L T  2 0 0 Notes: 1. The gravitational force is independent of the intervening medium. 2. The gravitational force is a conservative force. 3. The first particle exerts a force on the second that is exactly equal to and opposite to the second particle's force on the first. 4. The gravitational force between two particles acts along the line that connects them, and they are part of an action-reaction pair. 4. VARIATION IN ‘g’ 4.1 The Acceleration due to Gravity at a height h above the Earth’s surface Let M and R be the earth's mass and radius, respectively, and g denote the acceleration due to gravity at the surface. Assume that a mass of m is placed on the earth's surface. The weight ‘mg’ of the body is equal to the gravitational force acting on it. GMm GM mg  2 => g  2 …(1) R R Now suppose that the body is raised to a height h , above the earth’s surface, the GMm weight of the body is now mg and the gravitational force acting on it is. (R  h) 2 Class XI Physics www.vedantu.com 3  GMm ∴ mg h  (R  h) 2 GM gh  …(2) ( R  h) 2 gh R2 Dividing eq (2) by eq (1), we get,  g (R  h) 2  R2   gh   2 g.  (R  h)  4.2 Acceleration due to gravity at a very small height 2  Rh   2  h 2h h 2 gh  g   => gh  g 1   => g h  g 1   2 .   R   R  R R  If h ve  2 r R 11.2 Expression for ‘ Ve ’ in term’s of ‘ g ’ 2GM The escape velocity for any object on the earth’s surface is given by: ve  R If m is the mass of the object, its weight mg is equal to the gravitational force acting on it. Class XI Physics www.vedantu.com 16  GMm mg  => G.M  gR 2 R2 Substituting this value in the expression for Ve we get, ve  2gR 11.3 Expression for the escape velocity of a body from Earth in terms of mean density of the planet 2GM 1. Derive expression for ve  R 4 2. Let  be the mean density of the planet. Then, M   R 3  3 2G 4 2 G ve    R 3  => ve  2R R 3 3 11.4 The escape velocity of a body from the surface of the earth is 2 times its critical velocity when it revolves close to the earth’s surface Let M and R be the earth's mass and radius, respectively, and m be the body's mass. The radius of the orbit is nearly equal to R while orbiting close to the earth's surface. If vc is the body's critical velocity, then the orbit is circular. Centripetal force = Gravitational force GMm GM mv c2  => vc  …(1) R2 R If ve is the escape velocity from the earth’s surface, K.E. of projection = Binding energy mG 1 GMm mve2  2 R2 2GM ∴ ve  …(2) R Class XI Physics www.vedantu.com 17 From Eq (1) and Eq. (2), we get, ve  2vc 12. COMMUNICATION SATELLITE An artificial satellite that revolves in a circular orbit around the earth in the same sense as the earth's rotation and has the same period of revolution as the earth's rotation( i.e.1 day  24 hours  86400 seconds ) is called as geo-stationary or communication satellite. As relative velocity of the satellite with respective to the earth is zero it appears stationary from the earth’s surface. Therefore it is know as geo-stationary satellite or geosynchronous satellite. 1. The height of the communication satellite above the earth’s surface is about 36000 km and its period of revolution is 24 hours or 24  60  60 seconds. 2. The satellite appears to be at rest, because its speed relative to the earth is zero, hence it is called as geostationary or geosynchronous satellite. 12.1 Uses of the communication satellite 1. For sending TV signals over large distances on the earth’s surface 2. Telecommunication. 3. Weather forecasting. 4. For taking photographs of astronomical objects. 5. For studying of solar and cosmic radations. 13. WEIGHTLESSNESS 1. The weight of a body refers to the gravitational force that pulls it towards the earth's centre a feeling of weightlessness is like a moving satellite. It isn't because the weight is zero. Class XI Physics www.vedantu.com 18 2. When an astronaut is on the surface of earth, gravitational force acts on him. This gravitational force is the weight of astronaut and astronaut exerts this force on the surface of earth. The surface of earth exerts an equal and opposite reaction and due to this reaction he feels his weight on the earth. 3. for an astronaut in an orbiting satellite, the satellite and astronaut both have same acceleration towards the centre of earth and this acceleration is equal to the acceleration due to gravity of earth. 4. As a result, astronaunt does not have any effect on the satellite's floor. Naturally, the astronaut is unaffected by the floor's reaction force. The astronaut experiences a sense of weightlessness since there is no reaction. (i.e. he has no idea how heavy he is). NOTE: 1. The sensation of weightlessness experienced by an astronaut is not the result of there being zero gravitational acceleration, but of there being zero difference between the acceleration of the spacecraft and the acceleration of the astronaut. 2. The most common problem experienced by astronauts in the initial hours of weightlessness is known as space adaptation syndrome (space sickness). 14. KEPLER’S LAWS 14.1 Law of Orbit Each Planet move surround the sun in an elliptical orbit with the sun at one of the foci as shown in figure. The eccentricity of an ellipse is defined as the ratio of the SO distance SO and AO i.e. e  AO SO e => SO  ea a Class XI Physics www.vedantu.com 19 The distance of closest approach with 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  14.2 Law of Area 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. 1 r(rd ) Area swept 2 1 d Area velocity    r2  constant time dt 2 dt 1 2 Hence, r   constant 2 14.3 Law of Periods Class XI Physics www.vedantu.com 20 The square of the time for the planet to complete a revolution about the sun is proportional to the cube of semimajor axis of the elliptical orbit. i.e. Centripetal force = Gravitational force mv 2 GMm GM  2   v2 R R R circumference of the circular orbit 2 R Now, velocity of the planet is, v   Time Period T GM 4 2 R 2 4 2 R 3 Substituting Value in above equation  or T 2  R T2 GM  4 2  Since,   is constant.  GM  T2  T 2  R 3 or  constant R3 14.4 Gravity The force of attraction exerted by the earth towards its centre on a body lying on or near the surface of the earth is known as gravity. Gravity is a type of gravity that is also known as the earth's gravitational pull. Weight of a body is defined as the force of attraction exerted by the earth on the body towards its centre. The units and dimensions of gravity pull or weight are the same as those of force. Body Sun Earth Moon Mean radius, (m) 6.95 108 6.37  106 1.74  106 Class XI Physics www.vedantu.com 21  Mass, (kg) 1.97 1030 5.96  10 24 7.30 10 22 Mean density ( 1.41 5.52 3.30 3 10 kg / m3 ) Period of rotation 25.4 1.00 27.3 about axis, (days) LAUNCHING OF AN ARTIFICIAL SATELLITE AROUND THE EARTH 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 (figure) producing a horizontal velocity v. The total energy of the satellite at A is thus, 1 2 GMm E mv  2 Rh The orbit will be an ellipse (closed path), a parabola, or an hyperbola depending on whether E is negative, zero, or positive. In all cases the centre of the earth is at one focus of the path. If the energy is too low, the elliptical orbit will intersect the earth and the satellite will fall back. Otherwise it will keep moving in a closed orbit, or will escape from the earth, depending on the values of v and R. Hence a satellite carried to a height h   R  and given a horizontal velocity of 8 km / sec will be placed almost in a circular orbit around the earth (figure). If launched at less than 8 km / sec , it would get closer and closer to earth until it hits the ground. Thus 8 km / sec is the critical (minimum) velocity. 14.5 Inertial mass Class XI Physics www.vedantu.com 22 Newton's second law of motion defines a body's inertial mass, which is related to its inertia in linear motion. Let a body of mass mi move with acceleration a under the action of an external force F. According to Newton’s second law of motion, F  mi a or mi  F / a As a result, a body's inertial mass is equal to the magnitude of external force necessary to produce unit acceleration. 14.6 Gravitational mass Gravitational mass of a body is related to gravitational pull on the body, and is defined by Newton’s law of gravitational. GMm G F F F or mG  . R 2  GM / R 2 I The mass mG of the body in this sence is the gravitational mass of the body. The intertia of the body has no effect on the gravitational mass of the body. m G  F Thus, Gravitational mass of a body is defined as the magnitude of gravitational pull experienced by the body in a gravitational field of unit intensity. 14.7 Centre of Gravity Centre of gravity of a body placed in the gravitational field is that point where the net gravitational force of the field acts. Class XI Physics www.vedantu.com 23