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https://docs.google.com/document/d/1722ia95qw8P6lVhoyVOBrcez12FEd_0N/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true NEWTON'S LAW OF UNIVERSAL GRAVITATION Between any two particles of masses m1 and m2 at separation r from each other there exist attractive forces and directed from one body to the other and equal in magnitude which is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between the two. In vector form . Where G is universal gravitational constant. G = 6.67 ×10 -11 Nm2 / kg2 Dimensional formula of G: Gravitation force due to a rod A uniform rod of mass M of length . We will calculate gravitational force on this rod due to a point particle of mass m placed at a distance r from one of its ends as shown. To find this we consider a small element of width dx on the rod as shown in figure at a distance x from the point mass in. the mass of this element can be given as The force on mass m due to dm is, To find the net force on rod we integrate the above expression for the whole rod i.e. from a distance r to r+l, 1Illustration 1: A mass M is split into two parts m and (M-m), which are then separated by a certain distance. What ratio (m/M) maximises the gravitational force between the parts. Solution: If r is the distance between m and (M-m), the gravitational force will be F= For F to be maximum as M and r are constants i.e. i.e. M – 2m = 0 ⇒ GRAVITATIONAL FIELD intensity (I or E) The intensity of gravitational field at point is the gravitational force exerted on a unit mass kept at that point. Dimensional formula of gravitational field intensity 2Illustration 4: The distance between earth and moon is about km. At what point or points will the gravitational field strength of earth-moon system be zero? Given mass of earth is 81 times the moon’s mass. Solution: In this problem earth and moon will be treated as point masses as they are far apart and as in case of a point mass , the field due to both the masses will be zero at infinite. In addition to it, the field will also be zero at a point in between earth and moon where the pull of earth balances the pull of moon. If this point is at a distance x from the earth, or [as ME = 81 MM or or GRAVITATIONAL POTENTIAL ENERGY The amount of work done in bringing a body from infinity to a point within the gravitational field is called the gravitational potential energy of the body at that point. GRAVITATIONAL SELF ENERGY It is the energy possessed by a body due to the interaction forces inside the body. This can be defined as the work done in assembling all the particles of a body in a definite shape and size or it is the work done in creating a body. INTERACTION ENERGY OF A SYSTEM OF PARTICLES In a system of N particles with masses m1, m2 ... mn separated from each other by a distance r12, r13 where r12 is the separation between m1 and m2 and so on. The total interaction energy of system is, The factor 1/2 is taken because the interaction energy for each possible pair of system is taken twice during summation as for masses m1 and m3 (i ≠ j) Example:- Gravitational Self energy of a Uniform Sphere (star) GRAVITATIONAL POTENTIAL Gravitational potential Dimensional Formula of gravitational potential 4Illustration 7: At a point above the surface of earth, the gravitational potential is – 5.12 x 107 J/kg and the acceleration due to gravity is 6.4 m/s2. Assuming the mean radius of the earth to be 6400 km, calculate the height of this point above the earth’s surface. Solution: Let r be the distance of the given point from the centre of the earth. Then Gravitational potential = = – 5.12 x x107 ….(1) and acceleration due to gravity, g = . . . (2) Dividing (1) by (2), we get r = height of the point from earth’s surface = r – R = 8000 – 6400 = 1600 km GRAVITATIONAL POTENTIAL & FIELD DUE TO VARIOUS OBJECTS Causing Shape Gravitational Potential (V) Gravitational Field (I or E) Graph V vs R POINT MASS AT A POINT ON THE AXIS OF RING 0 ≤ r ≤ ∞ ROD 1. AT AN AXIAL POINT V = – 2. AT AN EQUATORIAL POINT V CIRCULAR ARC HOLLOW SPHERE SOLID SPHERE LONG THREAD V = ∞ RELATION BETWEEN GRAVITATIONAL POTENTIAL (V) AND INTENSITY (I OR E) gradient V = – grad V OR RELATION BETWEEN G AND g On the surface of earth, Substituting the values of G, M, R we get g = 9.81 m/s2. VARIATION IN ACCELERATION DUE TO GRAVITY (a) Variation of g with altitude g′ = for h < Ve The satellite escapes the field of earth along a hyperbolic trajectory. GEO STATIONARY SATELLITES (i) The time period of the satellite around the earth must be equal to the rotational period of the earth (i.e. 24 hours). (ii) The direction of motion of the satellite must be same as that of the earth. 8Illustration 18: A body is orbiting around the earth at mean radius 9 times as great as the orbit of a geostationary satellite. In how many days it will complete one revolution around the earth and what is its angular velocity? Solution: ⇒ ⇒ For the geostationary satellite, T1 = 1 day (24 hrs.) ∴ T2 = (9R3 / R3) ½ = 27days So it will complete one revolution in 27 days ω = 2π/T2 = rad/s = 2.693 × 10−6 rad/sec KEPLER’S LAWS 1. First law (Law of orbit) : Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse. This is called law of orbits. 2. Second law (Law of area): A line joining any planet to the sun (i.e. radius vector of the planet form sun) sweeps equal areas in equal interval of time Third law (Law of period) : The square of the time periods of the planets are proportional to the cube of semi-major axis of ellipse. 9Illustration 21: A planet of mass m moves along an ellipse around the sun so that its maximum and minimum distances from the sun are equal to R and r respectively. Find the angular momentum of this planet relative to the centre of the sun. Solution: According to Kepler’s Second Law the angular momentum of the planet is constant, we have mv1R = mv2r, v1R = v2r If the mass of the Sun is M conserving total mechanical energy of the system at two given positions we have, – Or Or or Now angular momentum = mv1R = m AT A GLANCE Newton's Law ; ; where and are point masses. Variation of acceleration due to gravity (a) inside the earth (b) outside the earth Where is the acceleration due to gravity on the surface of earth , R = radius of earth. Variation of acceleration due to gravity close to the earth's surface (inside the earth) (outside the earth) h is the small distance measured from the earth's surface. Variation of gravity with latitude α. (due to rotation of earth) ; T = 24 hours. Where g is the gravity at poles. Gravitational energy of a system of point masses Eg. For a system of four identical point masses located at corners of a square of side length 'a'. For a satellite of mass 'm' moving in a circular orbit of radius 'r' about a central body of mass M. Potential energy Total energy Escape velocity of a body is the velocity at which total energy of the body becomes zero. at a distance 'r' (>Re) from centre of earth ASSIGNMENT 1. A planet of mass m is revolving round the sun (of mass Ms) in an elliptical orbit. If is the velocity of the planet when its position vector from the sun is , then areal velocity of the planet is. (A) × (B) × (C) (D) MCQ - PG- 159 - Q-1 Ans. (D) Solution: Areal velocity 2. A system consists of n identical particles each of mass m. The total number of interactions possible is. (A) n(n + 1) (B) (C) n(n –1) (D) MCQ - PG- 159 - Q-4 Ans. (D) Solution: Total number of interactions 3. The ratio of the time period of a simple pendulum of length with a pendulum of infinite length is. (A) zero (B) (C) (D) (Where, R is the radius of earth) MCQ - PG- 160 - Q-5 Solution: (B) and 4. The rotation of the earth having R radius about its axis speeds up to a value such that a man at latitude angle 60 feels weightlessness. The duration of the day in such a case is. (A) (B) (C) (D) MCQ - PG- 160 - Q-6 Solution: (B) 5. At what height the gravitational field reduces by 75% the gravitational field at the surface of earth? (A) R (B) 2R (C) 3R (D) 4R MCQ - PG- 160 - Q-7 Solution: (A) 6. In a certain region of space gravitational field is given by . Taking the reference point to be at r = ro with v = vo, the potential (V) is given by (A) V = K (B) V = K (C) V = K (D) V = K Solution: (C) We know that ⇒ ⇒ ⇒ 7. Two different planets have same density but different radii. The acceleration due to gravity (g) on the surface of the planets is dependent on its radius (R) as. (A) (B) (C) (D) MCQ - PG- 160 - Q- 9. Solution: (D) 8. A particle of mass m lies at a distance r from the centre of earth. The force of attraction the particle and earth is F(r). (A) for r < R (B) for r R (C) for r < R (D) for r < R MCQ - PG- 161 - Q-10 Solution: (B, C) where ρ is density of earth. 9. A shell of mass M and radius R has another point mass m placed at a distance r from its centre (r > R). The force of attraction between the shell and point mass is. (A) (B) (C) F = zero (D) None of above MCQ - PG- 161 - Q-13 Solution: (B) 10. If gh and gd be the accelerations due to gravity at a height h and at depth d, above and below the surface of earth respectively. Assuming h << R and d < v0 but < ve, the body follows elliptical path around the earth. (C) v < v0, the body follows elliptical path and returns to surface of earth. (D) v > ve, the body follows hyperbolic path and escapes the gravitational pull of the earth. (A) A, B (B) B, C (C) A, B, C (D) A, B, C, D MCQ - PG- 169 - Q-35 Solution: (D) Velocity of Satellite Nature of Path 1. v = v0 Circular path around the earth. 2. v < v0 Elliptical path and body returns to earth. 3. v > v0 but < ve Elliptical path around the earth and will not escape. 4. v = ve Parabolic path and it escapes from the earth. 5. v > ve Hyperbolic path and escapes from earth. 9. A particle is launched from the surface of earth with speed v. For the particle to move as a satellite, which statement is correct? (A) < v < ve (B) < v < ve (C) ve < v < (D) < v < MCQ - PG- 169 - Q-55 Solution: (B) From SOLUTION to PROBLEM 8, we have 10. Two bodies of masses m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. (A) (B) (C) (D) MCQ - PG- 169 - Q-56 Solution: (D) Let gravitational field be zero at a point lying at distance x from M. Then, ...(1) ...(2) Since, ...(3) Substituting (1) and (2) in (3), we get 11. The orbital period of revolution of a planet round the sun is T0. Suppose we make a model of Solar system scaled down in the ratio η but of materials of the same mean density as the actual material of planet and the sun has. The new orbital period is. (A) ηT0 (B) η2T0 (C) η3T0 (D) T0 MCQ - PG- 170 - Q-58 Solution: (D) 12. P is a point at a distance r from the centre of a solid sphere of radius R0. The gravitational potential at P is V. If V is plotted as a function of r, then the curve representing the plot correctly is (A) (B) (C) (D) MCQ - PG- 170 - Q-59 Solution: (D) 13. A point P is lying at a distance r (< a) from the centre of shell of radius a. If E and V be the gravitational field and potential at the point P then E = ? (A) E = 0 (B) (C) v = 0 (D) MCQ - PG- 170 - Q-60 Solution: (A) The field inside the shell is zero. 14. A point P is lying at a distance r (< a) from the centre of shell of radius a. If E and V be the gravitational field and potential at the point P then, V = ? (A) E = 0 (B) (C) v = 0 (D) MCQ - PG- 170 - Q-60 Solution: (D) The field inside the shell is zero and so potential inside the shell is constant equal to the value that exists at the surface i.e. 15. A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then the time period of a spy satellite orbiting a few hundred kilometre above the earth's surface (Rearth = 6400 km) will approximately be (A) h (B) 1 h (C) 2 h (D) 4 h MCQ - PG- 170 - Q-61 Solution: (C) is the time period of satellite revolving very close to earth's surface. So time period of spy satellite must be slightly greater and so C is the appropriate answer. 16. Consider a thin uniform spherical layer of mass M and radius R. The potential energy of gravitational interaction of matter forming this shell is (A) (B) (C) (D) MCQ - PG- 171 - Q-63 Solution: (B) Let us consider the shell when a mass m is already piled on it by the agency. If V is the potential on the shell, then To add a mass dm further we have dW = Vdm = Potential Energy of Interaction. 17. Consider a thin uniform spherical layer of mass M and radius R, if we consider a solid sphere of mass M and radius R, then the potential energy of gravitational interaction of matter forming this solid sphere is (A) (B) (C) (D) MCQ - PG- 171 - Q-64 Solution: (C) 18. What should be the period of rotation of earth so as to make any object on the equator weigh half of its present value? (A) 2 hrs (B) 24 hrs (C) 8 hrs (D) 12 hrs Solution: (A) ge = g0 − Rω2 ⇒ go/2 = go − Rω2 ⇒ ω2R = ⇒ ω = ⇒ T = 2π , putting R = 6.4 × 106 m and g0 = 9.8 m/sec2, we obtain, T = 1.99 hrs ≈ 2 hrs 19. An artificial satellite is describing an equatorial orbit at 3600 km above the earth’s surface. Calculate its period of revolution ? (A) 8.71 hrs (B) 9.71 hrs (C) 10.71 hrs (D) 11.71 hrs Solution: (A) The time period of satellite is given by T2 = (R + h)3 = T = = [as R+h = (6400 + 3600) × 103 m = 107m] =31360.78 sec = 8.71 hrs 20. An artificial satellite is describing an equatorial orbit at 3600 km above the earth’s surface. Calculate its orbital speed. (A) 6.335 km/sec (B) 7.335 km/sec (C) 8.335 km/sec (D) 9.335 km/sec Solution: (A) The time period of satellite is given by T2 = (R + h)3 = T = = [as R+h = (6400 + 3600) × 103 m = 107m] =31360.78 sec = 8.71 hrs Orbital speed is Vo = = = 6335 m/s = 6.335 km/sec 21. Three particles each having a mass of 100 gm are placed on the vertices of an equilateral triangle of side 20 cm. The work done in increasing the side of the triangle to 40 cm is (A) 5.0 × 10-12J (B) 2.25 × 10-10J (C) 4.0 × 10-11J (D) 6.0 × 10-15J Solution: (A) W = = Putting , m and and m = 0.1 kg We get J 22. If the time of revolution of a satellite is T, then Kinetic energy is proportional to (A) 1/T (B) 1/T2 (C) 1/T3 (D) T-2/3 Solution: Orbital velocity Hence, KE ∝ 23. The magnitude of gravitational potential energy of the earth-satellite system is U with zero potential energy at infinite separation. The kinetic energy of satellite is K. Mass of satellite << mass of the earth. Then (A) K = 2U (B) (C) K = U (D) K = 4U Solution: (A) Kinetic energy of the satellite while magnitude of the potential energy Hence 24. A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.5 x 108 km away from the sun? (A) 1.43 × 109 km (B) 2.43 × 109 km (C) 3.43 × 109 km (D) 4.43 × 109 km Solution: (A) It is given that Ts = 29.5 Te; Re = 1.5 x 1011 m Now, according to Kepler’s third law ⇒ Rs = = 1.5 × 1011 = 1.43 × 1012 m = 1.43 × 109 km 25. A spherical planet for out in space has a mass Mo and diameter Do. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to (A) (B) (C) (D) Solution: (C) Let go to the acceleration thin 26. A body falls freely towards the earth from a height 2R, above the surface of the earth, where initially it was at rest. If R is the radius of the earth then its velocity on reaching the surface of the earth is (A) (B) (C) (D) 2gR Solution: (A) Initial energy of the body = − Final energy of the body = − where m is the mass of the body and M is the mass of the earth. Applying conservation of mechanical energy − = − ⇒ mv2 = ⇒ v2 = ⇒ v = = 27. Two satellite (I) and (II) are moving round a planet in circular orbit having radii R and 3R respectively, if the speed of satellite (I) is v the speed of satellite B will be (A) v /3 (B) (C) 3v (D) data insufficient Solution: (C) T = 2πr/v T1 = 2πR/v1 and T2 = 2π3R/v2 But by Kepler’s third law ⇒ v2 = 28. The radius of a planet is R. A satellite revolves around it in a circle of radius r with angular speed ω. The acceleration due to gravity on planet’s surface will be: (A) (B) (C) (D) Solution: (C) Let M be the mass of the planet and m the mass of satellite. Then ⇒ GM = Now 29. The gravitational field in a region is given by N/kg. Work done by this field is zero when the particle is moved along the line (A) y+4x = 2 (B) 4y+x = 6 (C) x+y = 5 (D) all of the above Solution: (A) Work done will be zero when displacement is perpendicular to the filed. The field makes an angle with positive x-axis While the line y + 4x = 2 makes an angle with positive x–axis i.e. the line y + 4x = 2 is perpendicular to 30. A particle of mass m is placed inside a spherical shell, away from its centre. The mass of the shell is M. (A) The particle will move towards the centre. (B) The particle will move away from the centre, towards the nearest wall. (C) The particle will move towards the centre if m < M and away from the centre if m > M. (D) The particle will remain stationary. Solution: (D) As the gravitational field inside the shell is zero, the particle will remain stationary 57: Three solid spheres each of mass m and radius R are released from the position shown in figure. The speed of any one sphere at the time of collision would be (A) (B) (C) (D) Solution: From conservation of mechanical energy Hence, (D) is correct. CMP 1. A satellite of mass 5000kg is projected in space with an initial speed of 4000m/s making an angle of 30o with the radial direction from a distance 3.6×107m away from the centre of the earth. 58: The angular momentum of satellite (A) 3.6×107 joule ×sec (B) 4.9×107 joule ×sec (C) 9.2×107 joule ×sec (D) 3.6×1014 joule ×sec Solution: (D) Given, r = 3.6 × 107 m, m = 5000 kg, v = 4000 m s–1, φ = 30° Angular momentum of satellite L = mvr sinφ = 5000 × 4000 × 3.6 × 107 × sin 30° = 3.6 × 1014 joule × second. 59: The semi-major axis of the orbit of satellite (A) 6.6×107 m (B) 14.9×107 m (C) 19.2×1017 m (D) 1.6×104 m Solution: (A) Now semi–major axis a = = 6.6 × 107m. 60: Semi-minor axis of the orbit of satellite (A) 16.6×107 m (B) 3.92×107 m (C) 10.2×1017 m (D) 2.6×104 m Solution: (B) Eccentricity of the orbit = 0.804 Hence, semi–minor axis b = a 61: The minimum distance of satellite from earth (A) 66.6×107 m (B) 14.9×107 m (C) 1.29×107 m (D) 1.6×104 m Solution: (C) Minimum distance rmin = a(1 –e) = 6.6 × 107 × (1 – 0.804) = 1.29 × 107m. 62: The maximum distance of satellite from earth. (A) 6.6×107 m (B) 24.9×107 m (C) 11.9×107 m (D) 1.6×104 m Solution: (C) Maximum distance rmax = a (1 + e) = 6.6 × 107 × (1 + 0.804) = 11.9 × 107m 63: The energy of satellite (A) 1.6×107 joule (B) 4.9×107 joule (C) 0.2×107 joule (D) -1.5×1010 joule Solution: (D) Energy of the satellite E = = = –1.5 × 1010 joule.