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GRAVITATION KEPLERโS LAWS OF PLANETARY MOTION 1. Law of orbits: All planets move in elliptical orbits with the sun situated at one of its foci of the ellipse. 2. Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time. Or areal velocity (areal velocity = area swept by the radius vector in unit time) of planets around the sun is a constant, ๐๐ด ๐๐ก = ๐๐๐๐ ๐ก๐๐๐ก. (figure 8.2) Note: Planets appear to move slower when they are farther from the sun and faster when they are closer than sun. 3. Law of periods: The square of time period(T) of revolution of a planet is directly proportional to the cube of the semi major axis(r) of the ellipse traced out by the planet. ๐ 2 ๐ 3 2 2 ๐ 2 โ ๐ 3 Note : ๐12 = ๐13 Q1. Page 201, question 8.3, NCRT TEXT Q2. Page 202, question 8.14, NCERT TEXT UNIVERSAL LAW OF GRAVITATION Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them |๐ | = G ๐1 ๐2 ๐2 ๐ ๐ In vector form, โโโโโโโ F12 = G ๐1 2 2 (โ๐ฬ), where โ๐ฬ indicate the direction of force which is always attractive Properties of gravitational force 1. Gravitational force between two masses is always attractive. 2. Gravitational force is a central force( a force which act along the line joining the centers of two bodies) 3. Gravitational force is a conservative force 4. Gravitational force between two masses is independent of the nature of the medium between the masses. 5. It obeys Newtonโs third law of motion. 6. It obeys inverse square law. 1/10 Intensity of gravitational field (gravitational field strength) at a point is defined as the gravitational force per unit mass. Intensity of gravitational field of a mass โMโ at a distance ๐น โrโ from it = ๐ = ๐บ๐ ๐ 2 ๐บ๐๐ ๐ ๐2 = ๐บ๐ ๐2 . At the surface of earth(r=R) intensity of gravitational field = = ๐, ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐ข๐ ๐ก๐ ๐๐๐๐ฃ๐๐ก๐ฆ Note: 1.The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is same as the entire mass of the shell is concentrated at the center of the shell. 2. The force of attraction due to a hollow spherical shell of uniform density on a point mass situated inside is zero. 3. Weight is the gravitational force with which earth attract an object towards its center W = mg. Even for same mass weight changes with change in value of g. Mass is the amount of matter in a body. It is always constant for a body. Expression for acceleration due to gravity. ME is the mass of earth. m is the mass of an object placed at a distance r from center of the earth.RE is the radius of earth. According to Newtonโs law of gravitation force on m due to ME , F = G According to second law of motion F =ma or F = mg From (1) and (2) mg = G ๐๐ธ ๐ ๐2 =G or g ๐๐ธ ๐2 Near the surface of the earth, r โRE , we have mass = volume x density ME = ๐ ๐๐ . ๐ ๐2 . (1) (2) RE + h = r, g=๐ ๐๐ ๐๐ธ ๐ ๐ด๐ฌ ๐น๐๐ฌ ๐น๐๐ฌ ๐, ๐ g = ๐ ๐น๐ ( ๐ ๐น๐๐ฌ ๐) = ๐ ๐ ๐ฎ๐น๐ฌ ๐ ๐ฌ Note 1. Acceleration due to gravity is independent of mass of falling object. Q1. Two shot put balls A and B of masses 5kg and 10kg are releases from same heights simultaneously. Which will reach the ground first? Why? Binomial theorem (1 + ๐ฅ)๐ = 1 + ๐๐ฅ + ๐(๐โ1) 2 ๐ฅ (1)(2) + โฆ .. + ๐ฅ๐ . For small values of x, (1 + ๐ฅ)๐ = 1 + ๐๐ฅ Variation g with altitude ( height from the surface of the ground) (add diagram) At the surface of the earth, mgs = G ๐๐ธ ๐ 2 ๐ ๐ธ or 2/10 gs = ๐บ ๐๐ธ 2 ๐ ๐ธ Similarly at a height โhโ above the ground ๐ ๐ก ๐ ๐ฌ ๐โ ๐๐ = = ๐๐๐ (๐ ๐ + ๐ก)๐ (1 + ๐บ ๐๐ธ (๐ ๐ธ + โ)2 (This formula is applicable to all heights) 2 ๐ ๐ธ 2 ๐ ๐ธ gh = ๐โ โ 2 ) ๐ ๐ธ ๐๐ = 1 (1 + ๐โ โ ๐ ๐ธ )2 ๐๐ = (1 + โ ๐ ๐ธ )โ2 Applying binomial theorem and neglecting higher power of ๐ฅ, ๐โ ๐๐ . gh = gs (1 โ 2โ ๐ ๐ธ 2โ = (1 โ ๐ ๐ธ 2โ ) or mgh = mgs (1 โ ๐ ๐ธ ) , for small heights (h<<RE) 2โ ) or Wh = Ws (1 โ ๐ ๐ธ ) . This is applicable to small heights only. Variation g with depth (distance from the surface of the earth to down ) (add diagram) At the surface of the earth, mgs = G ๐๐ธ ๐ or 2 ๐ ๐ธ ๐บ๐๐ธ gs = 2 ๐ ๐ธ Similarly at a depthโdโ below the ground gd gs = 4๐๐บ๐๐ธ (๐ ๐ธ โ๐) 3 gd = gs (1 โ ๐ ๐ ๐ธ x 3 ๐ ๐ ๐ธ 4๐๐บ๐๐ธ , ๐๐ = ๐ ) or mgd = mgs (1 โ gs = 4 x 3 ๐๐ ๐ธ3 = 4๐๐บ๐๐ธ gd = 3 (๐ ๐ธ โ๐) ๐๐ ๐ ๐ธ ๐๐ ๐ ๐ ๐ธ ๐บ ๐๐ธ 2 ๐ ๐ธ 4๐๐บ๐๐ธ 3 = ๐บ ๐๐ธ ๐๐ธ 2 ๐ ๐ธ 4 ๐๐ธ = 3 ๐๐ ๐ธ3 ๐ ๐ธ (๐ ๐ธ โ ๐) = (1 โ ) or Wd = Ws (1 โ ๐ ๐ ๐ธ ๐ ๐ ๐ธ ) ) Q: What is the value of g at the center of the earth? Q: A person has mass of 100 kg on the surface of the earth. Calculate the mass and weight of person (i) at the center of the earth (ii) at a height equal to half the radius of the earth (iii) at depth equal to half the radius of earth. Acceleration due to gravity is not a constant. Its value decreases with height (as we go up from the surface of earth) and depth (as we go deep in to earth from the surface of earth.) Acceleration due to gravity is maximum at the surface of earth (at the poles) Q: Draw a graph showing variation of acceleration due to gravity with distance from the surface of earth. g RE distance from center of earth 3/10 Gravitational potential energy of a particle of mass โmโ at a distance โrโ from the center of another mass โMโ is defined as the work done in bringing โmโ from infinity to โrโ. Consider a mass โmโ at a distance โxโ from center of another mass โMโ Gravitational force on โmโ due to โMโ, F = ๐บ๐๐ ๐ฅ2 work done to move โmโ through a small distance โdxโ , dw = F dx = ๐ ๐บ๐๐ Total work done in moving โmโ from infinity to โrโ, w = โซโ ๐ฅ2 ๐บ๐๐ ๐ฅ2 dx dx ๐ 1 W = ๐บ๐๐ โซโ ๐ฅ 2 dx ๐ W= ๐บ๐๐ โซโ ๐ฅ โ2 dx dx W= ๐บ๐๐[โ๐ฅ โ1 ] โ๐ 1 r 1 W = โ๐บ๐๐[๐ โ โ] m M 1 W = โ๐บ๐๐[๐ โ 0] W= โ๐ฎ๐ด๐ ๐ Q: Derive expression for work done to move an object from surface of earth to a height โhโ above it. What will be the expression if h <<R Gravitational Potential energy of a system of particles is the work done to assemble all the articles of the system from infinity to their respective locations against the gravitational force. Potential energy of two particle system in the absence of external field. U12 = W12 = โ๐บ ๐1 ๐2 ๐12 Potential energy of three particle system in the absence of external field. U123 = W12 + W13 + W23 = โ ๐บ ๐1 ๐2 ๐12 + โ ๐บ ๐1 ๐ 3 ๐13 + โ๐บ ๐2 ๐3 ๐23 Note: work done to dissociate a system= -( work done to assemble the system) Q: Write expressions for potential energy of four particle system and five particle system Gravitational potential of a mass โMโ at a distance r from its center is defined as the total work done to bring unit mass (m =1kg) from infinity to that point. It is the gravitational potential 4/10 energy per unit mass. Gravitational potential = ๐๐๐๐ฃ๐๐ก๐๐ก๐๐๐๐๐ ๐๐๐ก๐๐๐ก๐๐๐ ๐๐๐๐๐๐ฆ ๐๐๐ ๐ Gravitational potential at a distance r from a mass M, V(r) = โ๐ฎ๐ด ๐ Q: Draw a graph to show variation of gravitational potential with distance from center of a mass Q: Four identical masses โmโ each are placed at the four corners of a square of side โaโ each. Calculate the (i) gravitational potential energy of the system (ii) gravitational potential at the center of the system (iii) work done to dissociate the system. (iv) total gravitational force on any one mass of the system (v) total gravitational force on unit mass placed at the center of the system. Escape speed (Ve) from a planet is defined as the minimum speed with which an object must be launched from the planet such that it just escape from the gravitational field of the planet. Let M be the mass of the planet of radius R. let m be the mass of the object to be launched so that it just escape the gravitational field. Let Ve be the escape speed. 1 Total energy of the mass at the surface of earth = K.E. + P.E. = 2 ๐ ๐ฃ๐2 + โ๐ฎ๐ด๐ ๐น As the object move up speed and kinetic energy decreases and just becomes zero at infinity. Potential energy increases and becomes zero at infinity. Total energy at infinity becomes zero according to law of conservation of energy, Total energy at the surface of earth = total energy at infinity 1 2 ๐ ๐ฃ๐2 โ 1 2 ๐ฎ๐ด๐ ๐ ๐ฃ๐2 = ๐ฃ๐2 = ๐ฎ๐ด๐ ๐น ๐๐ฎ๐ด ๐น ๐ฃ๐ = โ At the surface of earth ๐ = ๐บ๐ ๐ 2 =0 ๐น ๐๐ฎ๐ด ๐น or gR2 = GM 5/10 ๐ฃ๐ = โ ๐๐๐น๐ ๐น ๐ฃ๐ = โ2gR From the surface of earth Ve = โ2x9.8x 6.4x106 = 11.2 Km/s. The escape speed from a planet is independent of the mass of the object projected. Escape speed is same for an ant and elephant from same planet. The force needed to impart this speed for ant and elephant is different. Q: Calculate the escape speed from the surface of moon. Radius of the moon is 1600km and acceleration due to gravity on the moon = 1/6 of that on earth. Hence explain why moon has very rarer atmosphere (even no atmosphere) The escape speed from the surface of moon is 2.3Km/s, which is about five times smaller compared to that from the surface of earth. Thermal speed of gas molecules on the surface of moon is greater than the escape speed from the surface of moon and the gas molecules escape out of the gravitational pull of the moon. Note : if initial velocity (Vi) is greater than escape speed (Ve) K.E. at infinity โ 0 1 1 ๐ ๐ฃ๐2 = 2 ๐ ๐ฃ๐2 + 2 If initial velocity is less than escape speed, = โ๐ฎ๐ด๐ ๐น โ๐ฎ๐ด๐ (๐น+๐) 1 = 2 ๐ ๐ฃ๐2 + โ๐ฎ๐ด๐ ๐น 1. Escape speed from the surface of earth is 11.2km/s. A rocket is projected with thrice this speed from the surface of earth. Calculate its speed far away from earth, neglecting the presence of sun and other planets. 2. A rocket is projected with a speed of 5km/s from the surface of the earth. how far from the surface of the earth does the rocket go before returning to earth. (M earth = , Rearth = and G = ) Satellites are objects which revolve around a planet. Orbital speed of a satellite (Vo) is the speed with which a satellite in a particular orbit around the planet. Let m be the mass of satellite revolving around a planet of mass M at a distance r from the center of the planet. Vo = orbital speed, R + h = r , R = radius of the planet and h = height of the satellite from the surface of the earth. Assuming the orbit to be circular, centripetal force for circular motion is provided by the gravitational force between the planet and the satellite. 6/10 ๐ ๐๐2 ๐ = ๐๐2 = ๐บ๐๐ ๐2 ๐บ๐ ๐ ๐ฎ๐ด Vo = โ ๐ . The orbital speed of satellite is independent of the mass of the satellite. If the orbit is very close to surface of earth, rโ ๐ (h<<R) ๐บ๐ Vo = โ ๐ Near to the surface of earth have ๐ = ๐๐ 2 Vo = โ ๐ ๐บ๐ ๐ 2 or gR2 = GM Vo = โ๐๐ Vo = โ9.8 x6.4 x 106 = 8.3 Km/s 1. Calculate orbital speed of a satellite revolving at a height equal to half the radius of the earth. Relation between orbital speed and escape speed near the surface of the earth (h<<R) Orbital speed ,Vo = โ๐๐ and escape speed ๐ฃ๐ = โ2gR ๐ฃ๐ = โ2 Vo Time period of satellite (T) is the time taken by the satellite to complete one revolution around the planet. Let m be the mass of satellite revolving around a planet of mass M at a distance r from the center of the planet. Vo = orbital speed, R + h = r , R = radius of the planet and h = height of the satellite from the surface of the earth. Assuming the orbit to be circular, time period, T = ๐= ๐= 2๐๐ ๐๐ 2๐๐ ๐ฎ๐ด ๐ โ ๐ T = 2๐๐ โ๐ฎ๐ด 7/10 ๐๐๐๐๐ข๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐ก ๐๐๐๐๐ก๐๐ ๐ ๐๐๐๐ ๐๐ T = 2๐ โ๐ฎ๐ด . The time period of satellite is independent of the mass of the satellite. ๐น๐ If the orbit is very close to the orbit rโ ๐ (h<<R), T = 2๐โ๐ฎ๐ด ๐บ๐ Near to the surface of earth have ๐ = ๐ 2 or gR2 = GM ๐น๐ T = 2๐โ๐๐น๐ ๐.๐ x 106 ๐น T = 2๐ โ๐ T = 2๐ โ ๐.๐ = 1.5 hours 1. Calculate time period of a satellite revolving at a height equal to half the radius of the earth. Total energy of a satellite Total energy of a satellite is the sum of potential and kinetic energy of the satellite Let m be the mass of satellite revolving around a planet of mass M at a distance r from the center of the planet. Vo = orbital speed, R + h = r , R = radius of the planet and h = height of the satellite from the surface of the earth. 1 ๐ฎ๐ด K.E. = 2 ๐๐๐2 using the value of orbital sped, Vo = โ K.E. = P.E. = ๐ฎ๐ด๐ ๐๐ โ๐ฎ๐ด๐ ๐ Total energy of the satellite, E= K.E. + P.E. E= E= E= ๐บ๐๐ 2๐ + โ๐บ๐๐ ๐ ๐บ๐๐ 1 ๐ ๐บ๐๐ ๐ ( โ 1) 2 1 (โ ) 2 8/10 ๐ E= โ๐ฎ๐ด๐ ๐๐ Total energy of a satellite is negative. It means that satellite is bound to the nucleus. To free the satellite from the gravitational field of planet, energy equal to total energy must be supplied to the satellite in the positive sense. 1. (a)A satellite of 400kg mass is revolving at a height equal to 1600km from the surface of earth. Calculate the (i) total energy, (ii) kinetic energy and (iii) potential energy of the satellite. (b) How much energy is needed to double the radius of the orbit? (c) how much energy is needed to rocket the satellite out of gravitational field of earth ( neglect the presence of sun and other planets) Note: 1. Energy needed to change the orbit of a satellite, โE = Efinal โ Einitial = โGMm โGMm โ 2r 2rf i ๐๐ โ๐ฌ = ๐ฎ๐ด๐ ๐ ๐ ( โ ) ๐ ๐๐ ๐๐ 2. Energy needed to rocket a satellite out of gravitational field of earth (planet) Gravitational field is zero when ๐๐ = โ 1 ๐๐ = 0 , โ๐ฌ = ๐ฎ๐ด๐ ๐ ๐ ๐๐ = โ(๐ญ๐จ๐ญ๐๐ฅ ๐๐ง๐๐ซ๐ ๐ฒ ๐จ๐ ๐ฌ๐๐ญ๐๐ฅ๐ฅ๐ข๐ญ๐) Relation between kinetic energy, potential energy and total energy K.E. = ๐บ๐๐ 2๐ P.E. = โ๐บ๐๐ ๐ and E = โ๐บ๐๐ 2๐ 1. Kinetic energy = โ(total energy) 2. Potential energy = 2 x total energy and 3. Potential energy = โ2(kinetic energy) Satellites are classified in to natural and artificial satellites. Natural satellites are natural bodies revolving around a planet. Moon is the natural satellite of earth. Artificial satellites are manmade satellites revolving around the planet. Aryabhatta, INSAT series of satellites are examples for artificial satellites. Uses of artificial satellites 1. Television broad cast 2. Communication purposes 3. Weather fore casting 9/10 4. Remote sensing 5. Meteorology and 6. Environmental studies Polar satellites and geostationary satellites Polar satellites go around the poles of earth in north south direction. These satellites orbit the earth at low altitudes (nearly 500 to 800Km from the earth surface.) .Time period of revolution these satellites are nearly 100 minutes. The whole earth can be scanned strip by strip in one day. Polar satellites are used for remote sensing, meteorology, and environmental studies Satellites which revolve in circular orbits around the earth in the equatorial plane along the direction of rotation of earth (from west to east) with time period 24hours are called as geostationary satellites. The height of a geostationary satellite from the surface of earth is nearly 35800Km. Q: Calculate the height of a geostationary satellite from the surface of earth Weightlessness: Weight of an object is the force with which earth attracts it towards the center of earth. Direction of gravity gives the sense of vertical downward direction for us . We experience (feel) our own weight when we stand on a surface, since the surface exerts a force opposite to our weight to keep us at rest. When an object is in free fall there is no force against gravity to keep us at rest, and we will not experience our weight. This state is known as weightlessness. In a satellite around the earth everything has acceleration towards the center of the earth which is exactly equal to acceleration due to gravity at that height. This situation is same as free fall. Thus in an artificial satellite people inside experience no gravity and hence there is no sense of vertical or horizontal direction for them. 10/10