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Gravitation

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Subject : PHYSICS CHAPTER 5 : GRAVITATION MAHARASHTRA STATE BOARD
CAN YOU RECALL? β€’ When released from certain height why do objects tend to fall vertically downwards? When released from certain height it falls vertically downward because of Earth's gravitational force. β€’ What is the shape of the orbits of planets? All orbits are elliptical, which means they are an ellipse, similar to an oval. β€’ What are Kepler’s laws? First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.
Introduction β€’ All material objects have a natural tendency to get attracted towards the Earth. β€’ In many natural phenomena like coconut falling from trees, raindrops falling from the clouds, etc., the same tendency is observed. β€’ All bodies are attracted towards the Earth with constant acceleration. This fact was recognized by Italian physicist Galileo. β€’ Every massive object in the universe experiences gravitational force. It is the force of mutual attraction between any two objects by virtue of their masses. It is always an attractive force with infinite range. It does not depend upon intervening medium. It is much weaker than other fundamental forces.
Kepler’s Laws β€’ Kepler’s laws of planetary motion describe the orbits of the planets around the Sun. 1. Law of orbit β€’ All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse. β€’ Here, S and S’ are the foci of the ellipse the Sun being at S. β€’ P is the closest point along the orbit from S and is, called β€˜Perihelion’. β€’ A is the farthest point from S and is, called β€˜Aphelion’. β€’ PA is the major axis = 2a. β€’ PO and AO are the semimajor axes = a. β€’ MN is the minor axis =2b. β€’ MO and ON are the semiminor axes = b Fig.: An ellipse traced by a planet with the Sun at the focus.
Kepler’s Laws 2. Law of areas β€’ The line that joins a planet and the Sun sweeps equal areas in equal intervals of time. β€’ Kepler observed that planets do not move around the Sun with uniform speed. They move faster when they are nearer to the Sun while they move slower when they are farther from the Sun. This is explained by this law. Fig.: The orbit of a planet P moving around the Sun. The area swept by the planet of mass m in given interval βˆ†π‘‘ is βˆ†π΄ = 1 2 π‘Ÿ Γ— 𝑣 βˆ†π‘‘ βˆ†π΄ βˆ†π‘‘ = 1 2 (π‘Ÿ Γ— 𝑣 ) 𝑝 = π‘š 𝑣 βˆ†π΄ βˆ†π‘‘ = 1 2 π‘Ÿ Γ— 𝑝 π‘š ∴ 𝐿 = π‘Ÿ Γ— 𝑝 βˆ†π΄ βˆ†π‘‘ = 𝐿 2π‘š = πΆπ‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
Kepler’s Laws 3. Law of periods β€’ The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet. β€’ If r is length of semimajor axis then, this law states that 𝑇2 ∝ π‘Ÿ3 Or 𝑇2 π‘Ÿ3 = πΆπ‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ β€’ Kepler’s laws were based on regular observations of the motion of planets. β€’ Kepler did not know why the planets obey these laws,. i.e. he had not derived these laws.
Example 5.1: What would be the average duration of year if the distance between the Sun and the Earth becomes (A) thrice the present distance. (B) twice the present distance. Solution: (A) Let, π‘Ÿ1 = present distance between the earth and sun T = 365 days If π‘Ÿ2 = 3π‘Ÿ1, 𝑇2 = ? According to kepler’s law of period 𝑇1 2 ∝ π‘Ÿ1 3 and 𝑇2 2 ∝ π‘Ÿ2 3 ∴ 𝑇2 2 𝑇1 2 = π‘Ÿ2 3 π‘Ÿ1 3 = 3 π‘Ÿ1 π‘Ÿ1 3 ∴ 𝑇2 𝑇1 = 27 ∴ 𝑇2 = 𝑇1 Γ— 27 ∴ 𝑇2 = 365 Γ— 27 = 1897 days (B) If π‘Ÿ2 = 2π‘Ÿ1, 𝑇2 = ? According to kepler’s law of period 𝑇1 2 ∝ π‘Ÿ1 3 and 𝑇2 2 ∝ π‘Ÿ2 3 ∴ 𝑇2 2 𝑇1 2 = π‘Ÿ2 3 π‘Ÿ1 3 = 2 π‘Ÿ1 π‘Ÿ1 3 ∴ 𝑇2 𝑇1 = 8 ∴ 𝑇2 = 𝑇1 Γ— 8 ∴ 𝑇2 = 365 Γ— 8 = 1032 days
5.3 Universal law of Gravitation 𝐹 = π‘š π‘Ÿ πœ”2 Also, F = ma m a = π‘š π‘Ÿ πœ”2 π‘Ž = π‘Ÿ πœ”2 We know, πœ” = 2 πœ‹ 𝑇 We get, π‘Ž = π‘Ÿ 2πœ‹ 𝑇 2 π‘Ž = 3.85 Γ— 105 Γ— 103 Γ—4 Γ— πœ‹2 (27.3 Γ—24 Γ—60 Γ—60)2 π‘Ž = 0.0027 π‘š/𝑠2 We have, π‘Žπ‘œπ‘π‘—π‘’π‘π‘‘ π‘Žπ‘šπ‘œπ‘œπ‘› = 9.8 π‘š/𝑠2 0.0027 π‘š/𝑠2 β‰ˆ 3600 π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘œπ‘› π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘’π‘Žπ‘Ÿπ‘‘β„Žβ€²π‘  π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘œπ‘π‘—π‘’π‘π‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘’π‘Žπ‘Ÿπ‘‘β„Žβ€²π‘  π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ = 3.85 Γ— 105 π‘˜π‘š 6378 π‘˜π‘š β‰ˆ 60 ∴ π‘Žπ‘œπ‘π‘—π‘’π‘π‘‘ π‘Žπ‘šπ‘œπ‘œπ‘› = π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘œπ‘› π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘œπ‘π‘—π‘’π‘π‘‘ 2
5.3 Universal law of Gravitation ∴ π‘Ž ∝ 1 π‘Ÿ2 As, F = ma 𝐹 ∝ π‘š π‘Ÿ2 𝐹𝐸 ∝ 𝑀 π‘Ÿ2 𝐹 ∝ π‘€π‘š π‘Ÿ2 𝐹 ∝ π‘š1π‘š2 π‘Ÿ2 𝐹 = 𝐺 π‘š1π‘š2 π‘Ÿ2 𝐺 = 6.67 Γ— 10βˆ’11 π‘π‘š2 /π‘˜π‘”2 𝐺 = [𝐿3 π‘€βˆ’1 π‘‡βˆ’2 ]
Example: The gravitational force between two bodies is 1 N. If distance between them is doubled, what will be the gravitational force between them? Solution: Let the masses of the two bodies be π‘š1 and π‘š2 and the distance between them be r. The force between them, 𝐹1 = 𝐺 π‘š1π‘š2 π‘Ÿ2 When the distance between them is doubled the force becomes, 𝐹2 = 𝐺 π‘š1π‘š2 (2π‘Ÿ)2 = 𝐺 π‘š1π‘š2 4 π‘Ÿ2 ∴ 𝐹1 𝐹2 = πΊπ‘š1π‘š2 π‘Ÿ2 Γ— 4 π‘Ÿ2 𝐺 π‘š1π‘š2 ∴ 𝐹1 𝐹2 = 4 ∴ 𝐹2 = 1 4 𝑁 (∡ 𝐹1 = 1 𝑁) ∴ 𝐹2 = 0.25𝑁 ∴ πΉπ‘œπ‘Ÿπ‘π‘’ π‘π‘’π‘π‘œπ‘šπ‘’ π‘œπ‘›π‘’ π‘“π‘œπ‘Ÿπ‘‘β„Ž (0.25 𝑁)
5.3 Universal law of Gravitation Fig.: Gravitational force between masses π’ŽπŸ and π’ŽπŸ . Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Figure shows two point masses π‘š1 and π‘š2 with position vectors π‘Ÿ1 & π‘Ÿ2 and respectively from origin O. The position vector of π‘š2 with respect to π‘š2 is then given by, π‘Ÿ21 = π‘Ÿ2 βˆ’ π‘Ÿ1 Similarly, π‘Ÿ12 = π‘Ÿ1 βˆ’ π‘Ÿ2 = βˆ’π‘Ÿ21 If, π‘Ÿ = I π‘Ÿ12 I = I π‘Ÿ21 I, the formula for force on π‘š2 due to π‘š1 can be expressed in vector form as, 𝐹21 = 𝐺 π‘š1π‘š2 π‘Ÿ2 (βˆ’π‘Ÿ21) 𝐹12 = 𝐺 π‘š1π‘š2 π‘Ÿ2 (βˆ’π‘Ÿ12) 𝐹12 = βˆ’πΉ21
5.3 Universal law of Gravitation Fig.: gravitational force due to a collection of masses . β€’ For a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by all the other point masses. For n particles, Force on π‘–π‘‘β„Ž mass 𝐹𝑖 = 𝑗=𝑖 𝑗 β‰ 1 𝑛 𝐹𝑖𝑗 β€’ The gravitational force between an extended object like the Earth and a point mass A can be obtained by obtaining the vector sum of forces on the point mass A due to each of the point mass which make up the extended object.
5.3 Universal law of Gravitation We can consider the two special cases, for which we can get a simple result. (1) The gravitational force of attraction due to a hollow, thin spherical shell of uniform density, on a point mass situated inside it is zero. (2) The gravitational force of attraction between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside is just as if the entire mass of the shell or sphere is concentrated at the centre of the shell or sphere.
Example: Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = 𝑙. A fourth particle D is kept on the perpendicular bisecter of AC at a distance l from B. Determine the gravitational force on D. Solution: 𝐢𝐷 = 𝐴𝐷 = 𝐴𝐡2 + 𝐡𝐷2 = 2 𝑙 Gravitational force on D = Vector sum of gravitational forces due to A, B & C. Force due to 𝐴 = πΊπ‘šπ‘š (𝐴𝐷)2 = πΊπ‘š2 2𝑙2 . This will be along 𝐷𝐴 Force due to C = πΊπ‘šπ‘š (𝐢𝐷)2 = πΊπ‘š2 2𝑙2 . This will be along 𝐷𝐢 Force due to B = πΊπ‘šπ‘š (𝐡𝐷)2 = πΊπ‘š2 𝑙2 . This will be along 𝐷𝐡 We can resolve the forces along horizontal and vertical directions. Let the unit vector along horizontal direction 𝐴𝐢 be 𝑖 and along the vertical direction 𝐡𝐷 be 𝑗 Net horizontal force on D = πΊπ‘š2 2𝑙2 cos 450 βˆ’π‘– + 𝐺 π‘š2 𝑙2 cos 900 𝑖 + 𝐺 π‘š2 2 𝑙2 cos 450 𝑖 Net horizontal force on D = βˆ’πΊπ‘š2 2 2 𝑙2 + 𝐺 π‘š2 2 2 𝑙2 = 0 Net vertical force on D = πΊπ‘š2 2𝑙2 cos 450 βˆ’π‘— + 𝐺 π‘š2 𝑙2 βˆ’π‘— + 𝐺 π‘š2 2 𝑙2 cos 450 βˆ’π‘— Net vertical force on D = βˆ’πΊπ‘š2 𝑙2 1 2 + 1 𝑗 = βˆ’πΊπ‘š2 𝑙2 1 2 + 1 βˆ’π‘— βˆ’π‘— shows that the net force is directed along DB.
5.5 Acceleration due to Gravity β€’ The magnitude of the gravitational force on a point object of mass m due to another point object of mass M at a distance r from it is, 𝐹 = 𝐺 π‘šπ‘€ π‘Ÿ2 β€’ This formula can be used to calculate the gravitational force on an object due to the Earth. We know that the Earth is an extended object. In many practical applications Earth can be assumed to be a uniform sphere. β€’ Thus if the mass of the Earth is M and that of the point object is m and the distance of the point object from the centre of the Earth is r then the force of attraction between them is, 𝐹 = 𝐺 π‘€π‘š π‘Ÿ2 β€’ If the point object is not acted upon by any other force, it will be accelerated towards the centre of the Earth under the action of this force. Its acceleration can be calculated by using Newton’s second law F = ma. β€’ Acceleration due to the gravity of the Earth = 𝐺 π‘€π‘š π‘Ÿ2 Γ— 1 π‘š = 𝐺𝑀 π‘Ÿ2 β€’ If the object is close to the surface of the Earth, r β‰… R, the radius of the Earth then, π‘”πΈπ‘Žπ‘Ÿπ‘‘β„Žβ€²π‘  π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ = 𝐺𝑀 π‘Ÿ2
Example: Calculate mass of the Earth from given data, Acceleration due to gravity g = 9.81 π‘š/𝑠2 Radius of the Earth 𝑅𝐸 = 6.37 Γ— 106 m G = 6.67 Γ— 10βˆ’11 N π‘š2 /π‘˜π‘”2 Solution: 𝑔 = 𝐺 𝑀𝐸 𝑅𝐸 2 ∴ 𝑀𝐸 = 𝑔 𝑅𝐸 2 𝐺 ∴ 𝑀𝐸 = 9.81 Γ— (6.37 Γ— 106)2 6.67 Γ— 10βˆ’11 ∴ 𝑀𝐸 = 5.97 Γ— 1024 π‘˜π‘” β€’ The value of g depends only on the properties of the Earth and does not depend on the mass of the object. This is exactly what Galileo had found from his experiments of dropping objects with different masses from the same height.
Do you know ? An object of mass m (much smaller than the mass of the Earth) is attracted towards the Earth and falls on it. The Earth is also attracted by the same force (magnitude) toward the mass m. However, its acceleration towards m will be π‘Žπ‘’π‘Žπ‘Ÿπ‘‘β„Ž = 𝐺 π‘€π‘š π‘Ÿ2 𝑀 = πΊπ‘š π‘Ÿ2 ∴ π‘Žπ‘’π‘Žπ‘Ÿπ‘‘β„Ž 𝑔 = π‘š 𝑀 π‘Žπ‘  𝑔 = 𝐺𝑀 π‘Ÿ2 As m << M, π‘ŽπΈπ‘Žπ‘Ÿπ‘‘β„Ž<< g and is nearly zero. Thus, practically only the mass m moves towards the Earth.
Example: Calculate the acceleration due to gravity on the surface of moon if mass of the moon is 1/80 times that of the Earth and diameter of the moon is 1/4 times that of the Earth (g = 9.8 π‘š/𝑠2) Solution: π‘€π‘š = Mass of the moon = M/80, Where M is mass of the Earth. π‘…π‘š = Radius of the moon = R/4, Where R is Radius of the Earth. Acceleration due to gravity on the surface of the Earth, 𝑔 = 𝐺𝑀 π‘Ÿ2 -(1) Acceleration due to gravity on the surface of the moon, π‘”π‘š = πΊπ‘€π‘š π‘…π‘š 2 -(2) From equation (1) and (2) π‘”π‘š 𝑔 = π‘€π‘š 𝑀 Γ— 𝑅 π‘…π‘š 2 π‘”π‘š 𝑔 = 1 80 Γ— 4 1 2 π‘”π‘š 𝑔 = 1 5 ∴ π‘”π‘š = 𝑔 5 = 9.8 5 ∴ π‘”π‘š = 1.96 π‘š/𝑠2
Example: Find the acceleration due to gravity on a planet that is 10 times as massive as the Earth and with radius 20 times of the radius of the Earth (g = 9.8 π‘š/𝑠2). Solution: Let mass of the planet be 𝑀𝑃, radius of the Earth and that of the planet be 𝑅𝐸 and 𝑅𝑃 respectively. Let mass of the Earth be 𝑀𝐸 and 𝑔𝑃 be acceleration due to gravity on the planet. 𝑀𝑃 = 10𝑀𝐸, 𝑅𝑃 = 20 𝑅𝐸, g = 9.8 π‘š/𝑠2 , 𝑔𝑃 = ? g = 𝐺𝑀𝐸 𝑅𝐸 2 , 𝑔𝑃 = 𝐺𝑀P 𝑅𝑃 2 ∴ 𝑔𝑃 = 𝐺(10 𝑀𝐸) (20 𝑅𝐸)2 = 10 𝐺 𝑀𝐸 400 𝑅𝐸 2 ∴ 𝑔𝑃 = 1 40 𝑔 ∴ 𝑔𝑃 = 0.245 π‘š/𝑠2
5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (A) Variation in g with Altitude: Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is, 𝑔 = 𝐺𝑀 π‘Ÿ2 Fig. Acceleration due to gravity at height h above the Earth’s surface. For small altitude h, i.e., for β„Ž 𝑅 β‰ͺ 1, ∴ 𝑔′ = 𝑔 1 βˆ’ 2β„Ž 𝑅 (By neglecting higher power terms of h/R as h/R << 1) β€’ This expression can be used to calculate the value of g at height h above the surface of the Earth as long as h << R
Example: At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface? (Radius of Earth = 6400 km). Assume the distance above the surface to be small compared to the radius of the Earth. Solution: g’ = 90% of g (g decreases by 10% hence it becomes 90%) or, 𝑔′ 𝑔 = 90 100 = 0.9 ∴ 𝑔′ = 𝑔 1 βˆ’ 2β„Ž 𝑅 ∴ 𝑔′ 𝑔 = 1 βˆ’ 2β„Ž 𝑅 ∴ 0.9 = 1 βˆ’ 2β„Ž 𝑅 ∴ β„Ž = 𝑅 20 ∴ β„Ž = 6400 20 ∴ β„Ž = 320 π‘˜π‘š
5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (B) Variation in g with Depth: Fig. Acceleration due to gravity at depth d below the surface of the Earth
5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (B) Variation in g with Depth: Special Case: At the centre of the Earth, where d = R, 𝑔𝑑 = 0 The value of acceleration due to gravity is maximum at the surface of the Earth. The value goes on decreasing with 1) increase in depth below the Earth’s surface. (varies linearly with (R-d) = r) 2) increase in height above the Earth’s surface. (varies inversely with (𝑅 + β„Ž)2= π‘Ÿ2) For r < R, 𝑔′ = 𝑔 1 βˆ’ 𝑑 𝑅 Writing R - d = r, the distance from the centre of the Earth, g(r) = g 𝑔 𝑅 which is the equation of a straight line with slope g/R and passing through the origin. Fig. - Variation of g due to depth and altitude from the Earth’s surface β€’ For r > R Writing R + h = r g(r) = 𝑔 𝑅2 π‘Ÿ2 which is plotted.
5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (C) Variation in g with Latitude and Rotation of the Earth: Fig. Variation of g with latitude. Latitude is an angle made by radius vector of any point from centre of the Earth with the equatorial plane PO’ = r ∠ 𝐸𝑂𝑃 = πœƒ ∠ 𝑂𝑃𝑂′ = πœƒ In βˆ† 𝑂𝑃𝑂′, cos πœƒ = 𝑃𝑂′ 𝑃𝑂 = π‘Ÿ 𝑅 ∴ π‘Ÿ = 𝑅 cos πœƒ π‘Ž = π‘Ÿ πœ”2 π‘Ž = 𝑅 πœ”2 cos πœƒ The component of this centripetal acceleration along PO, i.e., towards the centre of the Earth is π‘Žπ‘Ÿ = π‘Ž cos πœƒ π‘Žπ‘Ÿ = 𝑅 πœ”2 π‘π‘œπ‘ 2πœƒ π‘šπ‘”β€² = π‘šπ‘” βˆ’ π‘šπ‘… πœ”2π‘π‘œπ‘ 2πœƒ 𝑔′ = 𝑔 βˆ’ 𝑅 πœ”2π‘π‘œπ‘ 2πœƒ
5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (C) Variation in g with Latitude and Rotation of the Earth: special case I: At equator πœƒ = 0 cos πœƒ = 1 𝑔′ = 𝑔 βˆ’ π‘…πœ”2 Case II: At poles πœƒ = 900 cos πœƒ = 0 ∴ 𝑔′ = 𝑔 βˆ’ π‘Ÿπœ”2 cos πœƒ = 𝑔 βˆ’ 0 = 𝑔 Effect of the shape of the Earth: Weight of an object is the force with which the Earth attracts that object. Thus, weight w = mg where m is the mass of the object. As the value of g changes with altitude, depth and latitude, the weight also changes. Weight of an object is minimum at the equator. Similarly, the weight of an object reduces with increasing height above the Earth’s surface and with increasing depth below the its surface.
5.7 Gravitational Potential and Potential Energy There is a universal principle that states, Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy. Examples: (I) A spring in its natural state, possesses minimum potential energy. (II) When an object is lying on the Earth, the system of that object and the Earth has minimum potential energy. This is the gravitational potential energy of the system as these two are bound by the gravitational force. Work done against a conservative force acting on an object = Increase in the potential energy of the system. ∴ 𝐹. 𝑑π‘₯ = π‘‘π‘ˆ
5.7 Gravitational Potential and Potential Energy 5.7.1 Expression for Gravitational Potential Energy: ∴ π‘‘π‘ˆ = βˆ’πΉ 𝑔. π‘‘π‘Ÿ ∴ βˆ†π‘ˆ = π‘Ÿπ‘– π‘Ÿπ‘“ (π‘‘π‘ˆ) = π‘Ÿπ‘– π‘Ÿπ‘“ ( βˆ’ πΉβ„Ž. π‘‘π‘Ÿ) 𝐹 𝑔 = βˆ’ πΊπ‘€π‘š π‘Ÿ2 π‘Ÿ ∴ βˆ†π‘ˆ = π‘Ÿπ‘– π‘Ÿπ‘“ (π‘‘π‘ˆ) = π‘Ÿπ‘– π‘Ÿπ‘“ βˆ’ βˆ’ πΊπ‘€π‘š π‘Ÿ2 π‘Ÿ π‘‘π‘Ÿ ∴ βˆ†π‘ˆ = πΊπ‘€π‘š π‘Ÿπ‘– π‘Ÿπ‘“ π‘‘π‘Ÿ π‘Ÿ2 as π‘‘π‘Ÿ 𝑖𝑠 π‘Žπ‘™π‘œπ‘›π‘” π‘Ÿ ∴ βˆ†π‘ˆ = πΊπ‘€π‘š βˆ’ 1 π‘Ÿ π‘Ÿπ‘– π‘Ÿπ‘“ = GMm 1 π‘Ÿπ‘– βˆ’ 1 π‘Ÿπ‘“ Change in potential energy corresponds to the work done against conservative forces. For gravitational force, such point is taken at π‘Ÿ = ∞. ∴ π‘ˆ π‘Ÿπ‘– = 0 π‘Žπ‘‘ π‘Ÿπ‘– = ∞ ∴ π‘Ÿπ‘“ = π‘Ÿ ∴ π‘ˆ(π‘Ÿ)π‘”π‘Ÿπ‘Žπ‘£.= πΊπ‘€π‘š 1 π‘Ÿπ‘– βˆ’ 1 π‘Ÿπ‘“ ∴ π‘ˆ(π‘Ÿ)π‘”π‘Ÿπ‘Žπ‘£.= πΊπ‘€π‘š 1 ∞ βˆ’ 1 π‘Ÿ = βˆ’ πΊπ‘€π‘š π‘Ÿ
Example: What will be the change in potential energy of a body of mass m when it is raised from height 𝑅𝐸 above the Earth’s surface to 5/2 𝑅𝐸 above the Earth’s surface? 𝑅𝐸 and 𝑀𝐸 are the radius and mass of the Earth respectively. Solution: βˆ†π‘ˆ = πΊπ‘š 𝑀𝐸 1 π‘Ÿπ‘– βˆ’ 1 π‘Ÿπ‘“ βˆ†π‘ˆ = πΊπ‘š 𝑀𝐸 1 2 𝑅𝐸 βˆ’ 1 3.5 𝑅𝐸 βˆ†π‘ˆ = πΊπ‘šπ‘€πΈ 𝑅𝐸 Γ— 1.5 2Γ—3.5 βˆ†π‘ˆ = 2.14πΊπ‘šπ‘€πΈ 𝑅𝐸
5.7 Gravitational Potential and Potential Energy 5.7.2 Connection of potential energy formula with mgh: If the object is on the surface of Earth, r = R ∴ π‘ˆ1 = βˆ’ πΊπ‘€π‘š 𝑅 If the object is lifted to height h above the surface of Earth, the potential energy becomes ∴ π‘ˆ1 = βˆ’ πΊπ‘€π‘š 𝑅+β„Ž βˆ†π‘ˆ = π‘ˆ2 βˆ’ π‘ˆ1 = πΊπ‘€π‘š βˆ’ 1 𝑅+β„Ž βˆ’ βˆ’ 1 𝑅 βˆ†π‘ˆ = 𝐺𝑀 π‘š β„Ž 𝑅 𝑅+β„Ž = πΊπ‘€π‘šβ„Ž 𝑅(𝑅+β„Ž) If g is acceleration due to the Earth on the surface of Earth, 𝐺𝑀 = 𝑔𝑅2 βˆ†π‘ˆ = π‘šπ‘”β„Ž β„Ž 𝑅+β„Ž 𝐼𝑓 β„Ž β‰ͺ 𝑅, 𝑅 + β„Ž β‰… 𝑅. βˆ†π‘ˆ = π‘šπ‘”β„Ž Thus, mgh is increase in the gravitational potential energy of the Earth -mass system if an object of mass m is lifted to a height h, provided h is negligible compared to radius of the Earth (up to a few kilometers).
5.7 Gravitational Potential and Potential Energy 5.7.3 Concept of Potential: The gravitational potential energy of the system of Earth and any mass m at a distance r from the centre of the Earth is, π‘ˆ = βˆ’ πΊπ‘€π‘š π‘Ÿ = βˆ’ 𝐺𝑀 π‘Ÿ π‘š = 𝑉𝐸 π‘Ÿ π‘š The factor βˆ’ 𝐺𝑀 π‘Ÿ = 𝑉𝐸 π‘Ÿ , depends only upon mass of Earth and the location. Gravitational potential energy, U = Gravitational potential 𝑉 π‘Ÿ Γ— mass m or Gravitational potential is Gravitational potential energy per unit mass, i.e., 𝑉 π‘Ÿ = π‘ˆ π‘š . Gravitational potential difference between any two points in gravitational field, 𝑉2 βˆ’ 𝑉1 = π‘ˆ2βˆ’π‘ˆ1 π‘š = π‘‘π‘Š π‘š = Work done (or change in potential energy) per unit mass Gravitational potential energy, π‘ˆ = βˆ’ πΊπ‘š1π‘š2 π‘Ÿ = 𝑉1 π‘š2 = 𝑉2 π‘š1
5.7 Gravitational Potential and Potential Energy 5.7.4 Escape Velocity: β€’ When any object is thrown vertically up, it falls back to the Earth after reaching a certain height. Higher the speed with which the object is thrown up, greater will be the height. If we keep on increasing the velocity, a stage will come when the object will reach heights so large that it will escape the gravitational field of the Earth and will not fall back on the Earth. This initial velocity is called the escape velocity. 𝐾. 𝐸. = 1 2 π‘š 𝑣𝑒 2 , 𝑃. 𝐸. = βˆ’ πΊπ‘€π‘š 𝑅 Total energy = P.E. + K.E. Total energy = 1 2 π‘š 𝑣𝑒 2 βˆ’ πΊπ‘€π‘š 𝑅 K.E. = 0 Also, P.E. = βˆ’ πΊπ‘€π‘š ∞ = 0 ∴ π‘‡π‘œπ‘‘π‘Žπ‘™ πΈπ‘›π‘’π‘Ÿπ‘” = 𝑃. 𝐸. +𝐾. 𝐸 = 0 As energy is conserved 1 2 π‘š 𝑣𝑒 2 βˆ’ πΊπ‘€π‘š 𝑅 = 0 Or 𝑣𝑒 = 2𝐺𝑀 𝑅
5.8 Earth Satellites The objects which revolve around the Earth are called Earth satellites. moon is the only natural satellite of the Earth. It revolves in almost a circular orbit around the Earth with period of revolution of nearly 27.3 days. Communication Satellites: These are geostationary satellites. They revolve around the Earth in equatorial plane. They have same sense of rotation as that of the Earth and the same period of rotation as that of the Earth, i. e., one day or 24 hours. Due to this, they appear stationary from the Earth’s surface. Hence they are called geostationary satellites or geosynchronous satellites. Polar Satellites: Polar satellites are used for weather forecasting and meteorological purpose. They are also used for astronomical observations and study of Solar radiations. They go around the poles of the Earth in a north-south direction while the Earth rotates in an east-west direction about its own axis.
5.8 Earth Satellites 5.8.1 Projection of Satellite: The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbit round the Earth is called the critical velocity or orbital velocity (𝑣𝑐). A satellite follows different paths depending upon the horizontal velocity provided to it. Different possible cases are: Fig. Various possible orbits depending on the value of 𝒗𝒉 . Case (I) π‘£β„Ž < 𝑣𝑐 It experiences a non-conservative force of air resistance. Case (II) π‘£β„Ž = 𝑣𝑐 The satellite moves in a stable circular orbit round the Earth. Case (III) 𝑣𝑐 < π‘£β„Ž < 𝑣𝑒 The satellite again moves in an elliptical orbit round the Earth with the point of projection as perigee. Case (IV) π‘£β„Ž = 𝑣𝑒 Its speed will be zero at infinity. Case (V) π‘£β„Ž > 𝑣𝑒
5.8 Earth Satellites Expression for critical speed If the satellite is moving in a circular orbit of radius (R+h) = r, its speed must be the magnitude of critical velocity 𝑣𝑐. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth. ∴ Centripetal force = Gravitational force π‘šπ‘£π‘ 2 π‘Ÿ = πΊπ‘€π‘š π‘Ÿ2 ∴ 𝑣𝑐 2 = 𝐺𝑀 π‘Ÿ ∴ 𝑣𝑐 = 𝐺𝑀 π‘Ÿ ∴ 𝑣𝑐 = 𝐺𝑀 (𝑅+β„Ž) = π‘”β„Ž(𝑅 + β„Ž) This is the expression for critical speed at the orbit of radius (R + h) Fig. Various possible orbits depending on the value of 𝒗𝒉 .
5.8 Earth Satellites β€’ Special case When the satellite is revolving close to the surface of the Earth, the height is very small as compared to the radius of the Earth. Hence the height can be neglected and radius of the orbit is nearly equal to R (i.e R >> h, R+h β‰ˆ R) ∴ πΆπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ 𝑆𝑝𝑒𝑒𝑑, 𝑣𝑐 = 𝐺𝑀 𝑅 As G is related to acceleration due to gravity by the relation, 𝑔 = 𝐺𝑀 𝑅2 ∴ 𝐺𝑀 = 𝑔𝑅2 ∴ Critical speed in terms of acceleration due to gravity, 𝑣𝑐 = 𝑔𝑅2 𝑅 = 𝑔𝑅 = 7.92 π‘˜π‘š 𝑠 Obviously, this is the maximum possible critical speed. This is at least 25 times the speed of the fastest passenger aeroplanes.
Example: Show that the critical velocity of a body revolving in a circular orbit very close to the surface of a planet of radius R and mean density 𝜌 is 2𝑅 πΊπœ‹πœŒ 3 Solution: Since the body is revolving very close to the planet, h = 0 Density, 𝜌 = 𝑀 𝑉 = 𝑀 4 3 πœ‹π‘…3 = 4 3 πœ‹π‘…3𝜌 Critical Velocity, 𝑣𝑐 = 𝐺𝑀 𝑅 = 𝐺 4 3 πœ‹π‘…3𝜌 𝑅 ∴ 𝑣𝑐 = 2𝑅 πΊπœ‹πœŒ 3 β€’ When a satellite revolves very close to the surface of the Earth, motion of satellite gets affected by the friction produced due to resistance of air. In deriving the above expression the resistance of air is not considered.
5.8 Earth Satellites 5.8.2 Weightlessness in a Satellite: (i) Gravitational force mg directed vertically downwards (towards centre of the earth) & (ii) Normal reaction force N directed vertically upwards, exerted by the floor of the lift. As these forces are oppositely directed, the net force in the downward direction will be F = ma - N . Though the weight of a body is the gravitational force acting upon it, we feel our weight only due to the normal reaction force N exerted by the floor. According to Newton’s second law of motion, F = ma Let us consider the example of a lift or elevator from an inertial frame of reference. Whether the lift is at rest or in motion, a passenger in it experiences only two forces:
5.8 Earth Satellites 5.8.2 Weightlessness in a Satellite: Case I: Lift having zero acceleration The net force F = 0 = mg - N ∴ mg =N Hence in this case we feel our normal weight mg . Case II:Lift having net upward acceleration 𝒂𝒖 ∴ 𝐹 = π‘š π‘Žπ‘’ = 𝑁 βˆ’ π‘šπ‘” ∴ 𝑁 = π‘šπ‘” + π‘šπ‘Žπ‘’, i.e., N > mg (we feel heavier) Case III (a): Lift having net downward acceleration 𝒂𝒅 ∴ 𝐹 = π‘š π‘Žπ‘‘ = π‘šπ‘” βˆ’ 𝑁 ∴ 𝑁 = π‘šπ‘” βˆ’ π‘šπ‘Žπ‘‘, i.e., N < mg (we feel lighter) Case III (b): State of free fall The downward acceleration π‘Žπ‘‘ = 𝑔. ∴ 𝑁 = π‘šπ‘” βˆ’ π‘šπ‘Žπ‘‘ = 0 β€’ The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.
5.8 Earth Satellites ∴ 𝐺𝑀 = π‘”β„Ž(𝑅 + β„Ž)2 ∴ 𝑇 = 2πœ‹ (𝑅+β„Ž)3 π‘”β„Ž(𝑅+β„Ž)2 ∴ 𝑇 = 2πœ‹ 𝑅+β„Ž π‘”β„Ž = 2πœ‹ π‘Ÿ π‘”β„Ž Special case: When satellite revolves close to the surface of the Earth, R + h β‰ˆ R and π‘”β„Ž β‰ˆ 𝑔. (𝑇)π‘šπ‘–π‘›= 2πœ‹ 𝑅 𝑔 5.8.3 Time Period of a Satellite: The time taken by a satellite to complete one revolution round the Earth is its time period. The satellite revolves in a circular orbit of radius (R+h) = r. ∴ πΆπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ 𝑠𝑝𝑒𝑒𝑑 = πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘π‘–π‘‘ π‘‡π‘–π‘šπ‘’ π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑣𝑐 = 2πœ‹π‘Ÿ 𝑇 But we have, 𝑣𝑐 = πΊπ‘š π‘Ÿ ∴ πΊπ‘š π‘Ÿ = 2πœ‹π‘Ÿ 𝑇 Or, 𝐺𝑀 π‘Ÿ = 4πœ‹2π‘Ÿ2 𝑇2 ∴ 𝑇2 = 4πœ‹2π‘Ÿ2 𝐺𝑀 𝑇 = 2πœ‹ π‘Ÿ3 𝐺𝑀 = 2πœ‹ (𝑅+β„Ž)3 𝐺𝑀
Example: Calculate the period of revolution of a polar satellite orbiting close to the surface of the Earth. Given R = 6400 km, g = 9.8 π‘š/𝑠2 Solution: h is negligible as satellite is close to the Earth surface. ∴ 𝑅 + β„Ž β‰ˆ 𝑅 π‘”β„Ž β‰ˆ 𝑔 𝑅 = 6400 π‘˜π‘š = 6.4 Γ— 106 π‘š 𝑇 = 2πœ‹ 𝑅 𝑔 𝑇 = 2 Γ— 3.14 6.4Γ—106 9.8 𝑇 = 5.705 Γ— 103 π‘ π‘’π‘π‘œπ‘›π‘‘ 𝑇 = 85 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘  π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦
Example: An artificial satellite revolves around a planet in circular orbit close to its surface. Obtain the formula for period of the satellite in terms of density 𝜌 and radius R of planet. Solution: Period of satellite is, ∴ 𝑇 = 2πœ‹ (𝑅+β„Ž)3 𝐺𝑀 …..(1) Here, the satellite revolves close to the surface of planet, hence h is negligible, hence 𝑅 + β„Ž β‰ˆ 𝑅 𝐷𝑒𝑛𝑠𝑖𝑑𝑦 𝜌 = π‘šπ‘Žπ‘ π‘  (𝑀) π‘£π‘œπ‘™π‘’π‘šπ‘’ (𝑉) ∴ 𝑀 = 𝜌 𝑉 …..(2) As planet is spherical in shape, volume of planet is given as, 𝑉 = 4 3 πœ‹π‘…3 ∴ 𝑀 = 4 3 πœ‹π‘…3 𝜌 …….(3) Substituting the values from equation (2) and (3) in equation (1), we get 𝑇 = 2πœ‹ 𝑅3 𝐺× 4 3 πœ‹π‘…3𝜌 = 3 πœ‹ 𝐺 𝜌
5.8 Earth Satellites 5.8.4 Binding Energy of an orbiting satellite: The minimum energy required by a satellite to escape from Earth’s gravitational influence is the binding energy of the satellite. Let M be the mass of the Earth, R be the Radius of the Earth, 𝑣𝑐 be critical velocity of satellite, r = (R+h) be the radius of the orbit. 𝐾𝑖𝑛𝑒𝑑𝑖𝑐 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘“ π‘ π‘Žπ‘‘π‘’π‘™π‘™π‘–π‘‘π‘’ = 1 2 π‘š 𝑣𝑐 2 𝐾𝑖𝑛𝑒𝑑𝑖𝑐 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘“ π‘ π‘Žπ‘‘π‘’π‘™π‘™π‘–π‘‘π‘’ = 1 2 πΊπ‘€π‘š π‘Ÿ The gravitational potential at a distance r from the centre of the Earth is, βˆ’ 𝐺𝑀 π‘Ÿ ∴ Potential energy of satellite = Gravitational potential Γ— mass of satellite ∴ Potential energy of satellite = βˆ’ πΊπ‘€π‘š π‘Ÿ The total energy of satellite is given as, T.E. = K.E. + P.E. π‘‡π‘œπ‘‘π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ = 1 2 πΊπ‘€π‘š π‘Ÿ βˆ’ πΊπ‘€π‘š π‘Ÿ = βˆ’ 1 2 πΊπ‘€π‘š π‘Ÿ The minimum energy to be supplied to unbind the satellite is + 1 2 πΊπ‘€π‘š π‘Ÿ

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chapter5-gravitationppt-copy-211229151431 (2).pdf

  • 1. Subject : PHYSICS CHAPTER 5 : GRAVITATION MAHARASHTRA STATE BOARD
  • 2. CAN YOU RECALL? β€’ When released from certain height why do objects tend to fall vertically downwards? When released from certain height it falls vertically downward because of Earth's gravitational force. β€’ What is the shape of the orbits of planets? All orbits are elliptical, which means they are an ellipse, similar to an oval. β€’ What are Kepler’s laws? First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.
  • 3. Introduction β€’ All material objects have a natural tendency to get attracted towards the Earth. β€’ In many natural phenomena like coconut falling from trees, raindrops falling from the clouds, etc., the same tendency is observed. β€’ All bodies are attracted towards the Earth with constant acceleration. This fact was recognized by Italian physicist Galileo. β€’ Every massive object in the universe experiences gravitational force. It is the force of mutual attraction between any two objects by virtue of their masses. It is always an attractive force with infinite range. It does not depend upon intervening medium. It is much weaker than other fundamental forces.
  • 4. Kepler’s Laws β€’ Kepler’s laws of planetary motion describe the orbits of the planets around the Sun. 1. Law of orbit β€’ All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse. β€’ Here, S and S’ are the foci of the ellipse the Sun being at S. β€’ P is the closest point along the orbit from S and is, called β€˜Perihelion’. β€’ A is the farthest point from S and is, called β€˜Aphelion’. β€’ PA is the major axis = 2a. β€’ PO and AO are the semimajor axes = a. β€’ MN is the minor axis =2b. β€’ MO and ON are the semiminor axes = b Fig.: An ellipse traced by a planet with the Sun at the focus.
  • 5. Kepler’s Laws 2. Law of areas β€’ The line that joins a planet and the Sun sweeps equal areas in equal intervals of time. β€’ Kepler observed that planets do not move around the Sun with uniform speed. They move faster when they are nearer to the Sun while they move slower when they are farther from the Sun. This is explained by this law. Fig.: The orbit of a planet P moving around the Sun. The area swept by the planet of mass m in given interval βˆ†π‘‘ is βˆ†π΄ = 1 2 π‘Ÿ Γ— 𝑣 βˆ†π‘‘ βˆ†π΄ βˆ†π‘‘ = 1 2 (π‘Ÿ Γ— 𝑣 ) 𝑝 = π‘š 𝑣 βˆ†π΄ βˆ†π‘‘ = 1 2 π‘Ÿ Γ— 𝑝 π‘š ∴ 𝐿 = π‘Ÿ Γ— 𝑝 βˆ†π΄ βˆ†π‘‘ = 𝐿 2π‘š = πΆπ‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
  • 6. Kepler’s Laws 3. Law of periods β€’ The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet. β€’ If r is length of semimajor axis then, this law states that 𝑇2 ∝ π‘Ÿ3 Or 𝑇2 π‘Ÿ3 = πΆπ‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ β€’ Kepler’s laws were based on regular observations of the motion of planets. β€’ Kepler did not know why the planets obey these laws,. i.e. he had not derived these laws.
  • 7. Example 5.1: What would be the average duration of year if the distance between the Sun and the Earth becomes (A) thrice the present distance. (B) twice the present distance. Solution: (A) Let, π‘Ÿ1 = present distance between the earth and sun T = 365 days If π‘Ÿ2 = 3π‘Ÿ1, 𝑇2 = ? According to kepler’s law of period 𝑇1 2 ∝ π‘Ÿ1 3 and 𝑇2 2 ∝ π‘Ÿ2 3 ∴ 𝑇2 2 𝑇1 2 = π‘Ÿ2 3 π‘Ÿ1 3 = 3 π‘Ÿ1 π‘Ÿ1 3 ∴ 𝑇2 𝑇1 = 27 ∴ 𝑇2 = 𝑇1 Γ— 27 ∴ 𝑇2 = 365 Γ— 27 = 1897 days (B) If π‘Ÿ2 = 2π‘Ÿ1, 𝑇2 = ? According to kepler’s law of period 𝑇1 2 ∝ π‘Ÿ1 3 and 𝑇2 2 ∝ π‘Ÿ2 3 ∴ 𝑇2 2 𝑇1 2 = π‘Ÿ2 3 π‘Ÿ1 3 = 2 π‘Ÿ1 π‘Ÿ1 3 ∴ 𝑇2 𝑇1 = 8 ∴ 𝑇2 = 𝑇1 Γ— 8 ∴ 𝑇2 = 365 Γ— 8 = 1032 days
  • 8. 5.3 Universal law of Gravitation 𝐹 = π‘š π‘Ÿ πœ”2 Also, F = ma m a = π‘š π‘Ÿ πœ”2 π‘Ž = π‘Ÿ πœ”2 We know, πœ” = 2 πœ‹ 𝑇 We get, π‘Ž = π‘Ÿ 2πœ‹ 𝑇 2 π‘Ž = 3.85 Γ— 105 Γ— 103 Γ—4 Γ— πœ‹2 (27.3 Γ—24 Γ—60 Γ—60)2 π‘Ž = 0.0027 π‘š/𝑠2 We have, π‘Žπ‘œπ‘π‘—π‘’π‘π‘‘ π‘Žπ‘šπ‘œπ‘œπ‘› = 9.8 π‘š/𝑠2 0.0027 π‘š/𝑠2 β‰ˆ 3600 π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘œπ‘› π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘’π‘Žπ‘Ÿπ‘‘β„Žβ€²π‘  π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘œπ‘π‘—π‘’π‘π‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘’π‘Žπ‘Ÿπ‘‘β„Žβ€²π‘  π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ = 3.85 Γ— 105 π‘˜π‘š 6378 π‘˜π‘š β‰ˆ 60 ∴ π‘Žπ‘œπ‘π‘—π‘’π‘π‘‘ π‘Žπ‘šπ‘œπ‘œπ‘› = π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘œπ‘› π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ π‘œπ‘π‘—π‘’π‘π‘‘ 2
  • 9. 5.3 Universal law of Gravitation ∴ π‘Ž ∝ 1 π‘Ÿ2 As, F = ma 𝐹 ∝ π‘š π‘Ÿ2 𝐹𝐸 ∝ 𝑀 π‘Ÿ2 𝐹 ∝ π‘€π‘š π‘Ÿ2 𝐹 ∝ π‘š1π‘š2 π‘Ÿ2 𝐹 = 𝐺 π‘š1π‘š2 π‘Ÿ2 𝐺 = 6.67 Γ— 10βˆ’11 π‘π‘š2 /π‘˜π‘”2 𝐺 = [𝐿3 π‘€βˆ’1 π‘‡βˆ’2 ]
  • 10. Example: The gravitational force between two bodies is 1 N. If distance between them is doubled, what will be the gravitational force between them? Solution: Let the masses of the two bodies be π‘š1 and π‘š2 and the distance between them be r. The force between them, 𝐹1 = 𝐺 π‘š1π‘š2 π‘Ÿ2 When the distance between them is doubled the force becomes, 𝐹2 = 𝐺 π‘š1π‘š2 (2π‘Ÿ)2 = 𝐺 π‘š1π‘š2 4 π‘Ÿ2 ∴ 𝐹1 𝐹2 = πΊπ‘š1π‘š2 π‘Ÿ2 Γ— 4 π‘Ÿ2 𝐺 π‘š1π‘š2 ∴ 𝐹1 𝐹2 = 4 ∴ 𝐹2 = 1 4 𝑁 (∡ 𝐹1 = 1 𝑁) ∴ 𝐹2 = 0.25𝑁 ∴ πΉπ‘œπ‘Ÿπ‘π‘’ π‘π‘’π‘π‘œπ‘šπ‘’ π‘œπ‘›π‘’ π‘“π‘œπ‘Ÿπ‘‘β„Ž (0.25 𝑁)
  • 11. 5.3 Universal law of Gravitation Fig.: Gravitational force between masses π’ŽπŸ and π’ŽπŸ . Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Figure shows two point masses π‘š1 and π‘š2 with position vectors π‘Ÿ1 & π‘Ÿ2 and respectively from origin O. The position vector of π‘š2 with respect to π‘š2 is then given by, π‘Ÿ21 = π‘Ÿ2 βˆ’ π‘Ÿ1 Similarly, π‘Ÿ12 = π‘Ÿ1 βˆ’ π‘Ÿ2 = βˆ’π‘Ÿ21 If, π‘Ÿ = I π‘Ÿ12 I = I π‘Ÿ21 I, the formula for force on π‘š2 due to π‘š1 can be expressed in vector form as, 𝐹21 = 𝐺 π‘š1π‘š2 π‘Ÿ2 (βˆ’π‘Ÿ21) 𝐹12 = 𝐺 π‘š1π‘š2 π‘Ÿ2 (βˆ’π‘Ÿ12) 𝐹12 = βˆ’πΉ21
  • 12. 5.3 Universal law of Gravitation Fig.: gravitational force due to a collection of masses . β€’ For a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by all the other point masses. For n particles, Force on π‘–π‘‘β„Ž mass 𝐹𝑖 = 𝑗=𝑖 𝑗 β‰ 1 𝑛 𝐹𝑖𝑗 β€’ The gravitational force between an extended object like the Earth and a point mass A can be obtained by obtaining the vector sum of forces on the point mass A due to each of the point mass which make up the extended object.
  • 13. 5.3 Universal law of Gravitation We can consider the two special cases, for which we can get a simple result. (1) The gravitational force of attraction due to a hollow, thin spherical shell of uniform density, on a point mass situated inside it is zero. (2) The gravitational force of attraction between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside is just as if the entire mass of the shell or sphere is concentrated at the centre of the shell or sphere.
  • 14. Example: Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = 𝑙. A fourth particle D is kept on the perpendicular bisecter of AC at a distance l from B. Determine the gravitational force on D. Solution: 𝐢𝐷 = 𝐴𝐷 = 𝐴𝐡2 + 𝐡𝐷2 = 2 𝑙 Gravitational force on D = Vector sum of gravitational forces due to A, B & C. Force due to 𝐴 = πΊπ‘šπ‘š (𝐴𝐷)2 = πΊπ‘š2 2𝑙2 . This will be along 𝐷𝐴 Force due to C = πΊπ‘šπ‘š (𝐢𝐷)2 = πΊπ‘š2 2𝑙2 . This will be along 𝐷𝐢 Force due to B = πΊπ‘šπ‘š (𝐡𝐷)2 = πΊπ‘š2 𝑙2 . This will be along 𝐷𝐡 We can resolve the forces along horizontal and vertical directions. Let the unit vector along horizontal direction 𝐴𝐢 be 𝑖 and along the vertical direction 𝐡𝐷 be 𝑗 Net horizontal force on D = πΊπ‘š2 2𝑙2 cos 450 βˆ’π‘– + 𝐺 π‘š2 𝑙2 cos 900 𝑖 + 𝐺 π‘š2 2 𝑙2 cos 450 𝑖 Net horizontal force on D = βˆ’πΊπ‘š2 2 2 𝑙2 + 𝐺 π‘š2 2 2 𝑙2 = 0 Net vertical force on D = πΊπ‘š2 2𝑙2 cos 450 βˆ’π‘— + 𝐺 π‘š2 𝑙2 βˆ’π‘— + 𝐺 π‘š2 2 𝑙2 cos 450 βˆ’π‘— Net vertical force on D = βˆ’πΊπ‘š2 𝑙2 1 2 + 1 𝑗 = βˆ’πΊπ‘š2 𝑙2 1 2 + 1 βˆ’π‘— βˆ’π‘— shows that the net force is directed along DB.
  • 15. 5.5 Acceleration due to Gravity β€’ The magnitude of the gravitational force on a point object of mass m due to another point object of mass M at a distance r from it is, 𝐹 = 𝐺 π‘šπ‘€ π‘Ÿ2 β€’ This formula can be used to calculate the gravitational force on an object due to the Earth. We know that the Earth is an extended object. In many practical applications Earth can be assumed to be a uniform sphere. β€’ Thus if the mass of the Earth is M and that of the point object is m and the distance of the point object from the centre of the Earth is r then the force of attraction between them is, 𝐹 = 𝐺 π‘€π‘š π‘Ÿ2 β€’ If the point object is not acted upon by any other force, it will be accelerated towards the centre of the Earth under the action of this force. Its acceleration can be calculated by using Newton’s second law F = ma. β€’ Acceleration due to the gravity of the Earth = 𝐺 π‘€π‘š π‘Ÿ2 Γ— 1 π‘š = 𝐺𝑀 π‘Ÿ2 β€’ If the object is close to the surface of the Earth, r β‰… R, the radius of the Earth then, π‘”πΈπ‘Žπ‘Ÿπ‘‘β„Žβ€²π‘  π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ = 𝐺𝑀 π‘Ÿ2
  • 16. Example: Calculate mass of the Earth from given data, Acceleration due to gravity g = 9.81 π‘š/𝑠2 Radius of the Earth 𝑅𝐸 = 6.37 Γ— 106 m G = 6.67 Γ— 10βˆ’11 N π‘š2 /π‘˜π‘”2 Solution: 𝑔 = 𝐺 𝑀𝐸 𝑅𝐸 2 ∴ 𝑀𝐸 = 𝑔 𝑅𝐸 2 𝐺 ∴ 𝑀𝐸 = 9.81 Γ— (6.37 Γ— 106)2 6.67 Γ— 10βˆ’11 ∴ 𝑀𝐸 = 5.97 Γ— 1024 π‘˜π‘” β€’ The value of g depends only on the properties of the Earth and does not depend on the mass of the object. This is exactly what Galileo had found from his experiments of dropping objects with different masses from the same height.
  • 17. Do you know ? An object of mass m (much smaller than the mass of the Earth) is attracted towards the Earth and falls on it. The Earth is also attracted by the same force (magnitude) toward the mass m. However, its acceleration towards m will be π‘Žπ‘’π‘Žπ‘Ÿπ‘‘β„Ž = 𝐺 π‘€π‘š π‘Ÿ2 𝑀 = πΊπ‘š π‘Ÿ2 ∴ π‘Žπ‘’π‘Žπ‘Ÿπ‘‘β„Ž 𝑔 = π‘š 𝑀 π‘Žπ‘  𝑔 = 𝐺𝑀 π‘Ÿ2 As m << M, π‘ŽπΈπ‘Žπ‘Ÿπ‘‘β„Ž<< g and is nearly zero. Thus, practically only the mass m moves towards the Earth.
  • 18. Example: Calculate the acceleration due to gravity on the surface of moon if mass of the moon is 1/80 times that of the Earth and diameter of the moon is 1/4 times that of the Earth (g = 9.8 π‘š/𝑠2) Solution: π‘€π‘š = Mass of the moon = M/80, Where M is mass of the Earth. π‘…π‘š = Radius of the moon = R/4, Where R is Radius of the Earth. Acceleration due to gravity on the surface of the Earth, 𝑔 = 𝐺𝑀 π‘Ÿ2 -(1) Acceleration due to gravity on the surface of the moon, π‘”π‘š = πΊπ‘€π‘š π‘…π‘š 2 -(2) From equation (1) and (2) π‘”π‘š 𝑔 = π‘€π‘š 𝑀 Γ— 𝑅 π‘…π‘š 2 π‘”π‘š 𝑔 = 1 80 Γ— 4 1 2 π‘”π‘š 𝑔 = 1 5 ∴ π‘”π‘š = 𝑔 5 = 9.8 5 ∴ π‘”π‘š = 1.96 π‘š/𝑠2
  • 19. Example: Find the acceleration due to gravity on a planet that is 10 times as massive as the Earth and with radius 20 times of the radius of the Earth (g = 9.8 π‘š/𝑠2). Solution: Let mass of the planet be 𝑀𝑃, radius of the Earth and that of the planet be 𝑅𝐸 and 𝑅𝑃 respectively. Let mass of the Earth be 𝑀𝐸 and 𝑔𝑃 be acceleration due to gravity on the planet. 𝑀𝑃 = 10𝑀𝐸, 𝑅𝑃 = 20 𝑅𝐸, g = 9.8 π‘š/𝑠2 , 𝑔𝑃 = ? g = 𝐺𝑀𝐸 𝑅𝐸 2 , 𝑔𝑃 = 𝐺𝑀P 𝑅𝑃 2 ∴ 𝑔𝑃 = 𝐺(10 𝑀𝐸) (20 𝑅𝐸)2 = 10 𝐺 𝑀𝐸 400 𝑅𝐸 2 ∴ 𝑔𝑃 = 1 40 𝑔 ∴ 𝑔𝑃 = 0.245 π‘š/𝑠2
  • 20. 5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (A) Variation in g with Altitude: Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is, 𝑔 = 𝐺𝑀 π‘Ÿ2 Fig. Acceleration due to gravity at height h above the Earth’s surface. For small altitude h, i.e., for β„Ž 𝑅 β‰ͺ 1, ∴ 𝑔′ = 𝑔 1 βˆ’ 2β„Ž 𝑅 (By neglecting higher power terms of h/R as h/R << 1) β€’ This expression can be used to calculate the value of g at height h above the surface of the Earth as long as h << R
  • 21. Example: At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface? (Radius of Earth = 6400 km). Assume the distance above the surface to be small compared to the radius of the Earth. Solution: g’ = 90% of g (g decreases by 10% hence it becomes 90%) or, 𝑔′ 𝑔 = 90 100 = 0.9 ∴ 𝑔′ = 𝑔 1 βˆ’ 2β„Ž 𝑅 ∴ 𝑔′ 𝑔 = 1 βˆ’ 2β„Ž 𝑅 ∴ 0.9 = 1 βˆ’ 2β„Ž 𝑅 ∴ β„Ž = 𝑅 20 ∴ β„Ž = 6400 20 ∴ β„Ž = 320 π‘˜π‘š
  • 22. 5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (B) Variation in g with Depth: Fig. Acceleration due to gravity at depth d below the surface of the Earth
  • 23. 5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (B) Variation in g with Depth: Special Case: At the centre of the Earth, where d = R, 𝑔𝑑 = 0 The value of acceleration due to gravity is maximum at the surface of the Earth. The value goes on decreasing with 1) increase in depth below the Earth’s surface. (varies linearly with (R-d) = r) 2) increase in height above the Earth’s surface. (varies inversely with (𝑅 + β„Ž)2= π‘Ÿ2) For r < R, 𝑔′ = 𝑔 1 βˆ’ 𝑑 𝑅 Writing R - d = r, the distance from the centre of the Earth, g(r) = g 𝑔 𝑅 which is the equation of a straight line with slope g/R and passing through the origin. Fig. - Variation of g due to depth and altitude from the Earth’s surface β€’ For r > R Writing R + h = r g(r) = 𝑔 𝑅2 π‘Ÿ2 which is plotted.
  • 24. 5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (C) Variation in g with Latitude and Rotation of the Earth: Fig. Variation of g with latitude. Latitude is an angle made by radius vector of any point from centre of the Earth with the equatorial plane PO’ = r ∠ 𝐸𝑂𝑃 = πœƒ ∠ 𝑂𝑃𝑂′ = πœƒ In βˆ† 𝑂𝑃𝑂′, cos πœƒ = 𝑃𝑂′ 𝑃𝑂 = π‘Ÿ 𝑅 ∴ π‘Ÿ = 𝑅 cos πœƒ π‘Ž = π‘Ÿ πœ”2 π‘Ž = 𝑅 πœ”2 cos πœƒ The component of this centripetal acceleration along PO, i.e., towards the centre of the Earth is π‘Žπ‘Ÿ = π‘Ž cos πœƒ π‘Žπ‘Ÿ = 𝑅 πœ”2 π‘π‘œπ‘ 2πœƒ π‘šπ‘”β€² = π‘šπ‘” βˆ’ π‘šπ‘… πœ”2π‘π‘œπ‘ 2πœƒ 𝑔′ = 𝑔 βˆ’ 𝑅 πœ”2π‘π‘œπ‘ 2πœƒ
  • 25. 5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape: (C) Variation in g with Latitude and Rotation of the Earth: special case I: At equator πœƒ = 0 cos πœƒ = 1 𝑔′ = 𝑔 βˆ’ π‘…πœ”2 Case II: At poles πœƒ = 900 cos πœƒ = 0 ∴ 𝑔′ = 𝑔 βˆ’ π‘Ÿπœ”2 cos πœƒ = 𝑔 βˆ’ 0 = 𝑔 Effect of the shape of the Earth: Weight of an object is the force with which the Earth attracts that object. Thus, weight w = mg where m is the mass of the object. As the value of g changes with altitude, depth and latitude, the weight also changes. Weight of an object is minimum at the equator. Similarly, the weight of an object reduces with increasing height above the Earth’s surface and with increasing depth below the its surface.
  • 26. 5.7 Gravitational Potential and Potential Energy There is a universal principle that states, Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy. Examples: (I) A spring in its natural state, possesses minimum potential energy. (II) When an object is lying on the Earth, the system of that object and the Earth has minimum potential energy. This is the gravitational potential energy of the system as these two are bound by the gravitational force. Work done against a conservative force acting on an object = Increase in the potential energy of the system. ∴ 𝐹. 𝑑π‘₯ = π‘‘π‘ˆ
  • 27. 5.7 Gravitational Potential and Potential Energy 5.7.1 Expression for Gravitational Potential Energy: ∴ π‘‘π‘ˆ = βˆ’πΉ 𝑔. π‘‘π‘Ÿ ∴ βˆ†π‘ˆ = π‘Ÿπ‘– π‘Ÿπ‘“ (π‘‘π‘ˆ) = π‘Ÿπ‘– π‘Ÿπ‘“ ( βˆ’ πΉβ„Ž. π‘‘π‘Ÿ) 𝐹 𝑔 = βˆ’ πΊπ‘€π‘š π‘Ÿ2 π‘Ÿ ∴ βˆ†π‘ˆ = π‘Ÿπ‘– π‘Ÿπ‘“ (π‘‘π‘ˆ) = π‘Ÿπ‘– π‘Ÿπ‘“ βˆ’ βˆ’ πΊπ‘€π‘š π‘Ÿ2 π‘Ÿ π‘‘π‘Ÿ ∴ βˆ†π‘ˆ = πΊπ‘€π‘š π‘Ÿπ‘– π‘Ÿπ‘“ π‘‘π‘Ÿ π‘Ÿ2 as π‘‘π‘Ÿ 𝑖𝑠 π‘Žπ‘™π‘œπ‘›π‘” π‘Ÿ ∴ βˆ†π‘ˆ = πΊπ‘€π‘š βˆ’ 1 π‘Ÿ π‘Ÿπ‘– π‘Ÿπ‘“ = GMm 1 π‘Ÿπ‘– βˆ’ 1 π‘Ÿπ‘“ Change in potential energy corresponds to the work done against conservative forces. For gravitational force, such point is taken at π‘Ÿ = ∞. ∴ π‘ˆ π‘Ÿπ‘– = 0 π‘Žπ‘‘ π‘Ÿπ‘– = ∞ ∴ π‘Ÿπ‘“ = π‘Ÿ ∴ π‘ˆ(π‘Ÿ)π‘”π‘Ÿπ‘Žπ‘£.= πΊπ‘€π‘š 1 π‘Ÿπ‘– βˆ’ 1 π‘Ÿπ‘“ ∴ π‘ˆ(π‘Ÿ)π‘”π‘Ÿπ‘Žπ‘£.= πΊπ‘€π‘š 1 ∞ βˆ’ 1 π‘Ÿ = βˆ’ πΊπ‘€π‘š π‘Ÿ
  • 28. Example: What will be the change in potential energy of a body of mass m when it is raised from height 𝑅𝐸 above the Earth’s surface to 5/2 𝑅𝐸 above the Earth’s surface? 𝑅𝐸 and 𝑀𝐸 are the radius and mass of the Earth respectively. Solution: βˆ†π‘ˆ = πΊπ‘š 𝑀𝐸 1 π‘Ÿπ‘– βˆ’ 1 π‘Ÿπ‘“ βˆ†π‘ˆ = πΊπ‘š 𝑀𝐸 1 2 𝑅𝐸 βˆ’ 1 3.5 𝑅𝐸 βˆ†π‘ˆ = πΊπ‘šπ‘€πΈ 𝑅𝐸 Γ— 1.5 2Γ—3.5 βˆ†π‘ˆ = 2.14πΊπ‘šπ‘€πΈ 𝑅𝐸
  • 29. 5.7 Gravitational Potential and Potential Energy 5.7.2 Connection of potential energy formula with mgh: If the object is on the surface of Earth, r = R ∴ π‘ˆ1 = βˆ’ πΊπ‘€π‘š 𝑅 If the object is lifted to height h above the surface of Earth, the potential energy becomes ∴ π‘ˆ1 = βˆ’ πΊπ‘€π‘š 𝑅+β„Ž βˆ†π‘ˆ = π‘ˆ2 βˆ’ π‘ˆ1 = πΊπ‘€π‘š βˆ’ 1 𝑅+β„Ž βˆ’ βˆ’ 1 𝑅 βˆ†π‘ˆ = 𝐺𝑀 π‘š β„Ž 𝑅 𝑅+β„Ž = πΊπ‘€π‘šβ„Ž 𝑅(𝑅+β„Ž) If g is acceleration due to the Earth on the surface of Earth, 𝐺𝑀 = 𝑔𝑅2 βˆ†π‘ˆ = π‘šπ‘”β„Ž β„Ž 𝑅+β„Ž 𝐼𝑓 β„Ž β‰ͺ 𝑅, 𝑅 + β„Ž β‰… 𝑅. βˆ†π‘ˆ = π‘šπ‘”β„Ž Thus, mgh is increase in the gravitational potential energy of the Earth -mass system if an object of mass m is lifted to a height h, provided h is negligible compared to radius of the Earth (up to a few kilometers).
  • 30. 5.7 Gravitational Potential and Potential Energy 5.7.3 Concept of Potential: The gravitational potential energy of the system of Earth and any mass m at a distance r from the centre of the Earth is, π‘ˆ = βˆ’ πΊπ‘€π‘š π‘Ÿ = βˆ’ 𝐺𝑀 π‘Ÿ π‘š = 𝑉𝐸 π‘Ÿ π‘š The factor βˆ’ 𝐺𝑀 π‘Ÿ = 𝑉𝐸 π‘Ÿ , depends only upon mass of Earth and the location. Gravitational potential energy, U = Gravitational potential 𝑉 π‘Ÿ Γ— mass m or Gravitational potential is Gravitational potential energy per unit mass, i.e., 𝑉 π‘Ÿ = π‘ˆ π‘š . Gravitational potential difference between any two points in gravitational field, 𝑉2 βˆ’ 𝑉1 = π‘ˆ2βˆ’π‘ˆ1 π‘š = π‘‘π‘Š π‘š = Work done (or change in potential energy) per unit mass Gravitational potential energy, π‘ˆ = βˆ’ πΊπ‘š1π‘š2 π‘Ÿ = 𝑉1 π‘š2 = 𝑉2 π‘š1
  • 31. 5.7 Gravitational Potential and Potential Energy 5.7.4 Escape Velocity: β€’ When any object is thrown vertically up, it falls back to the Earth after reaching a certain height. Higher the speed with which the object is thrown up, greater will be the height. If we keep on increasing the velocity, a stage will come when the object will reach heights so large that it will escape the gravitational field of the Earth and will not fall back on the Earth. This initial velocity is called the escape velocity. 𝐾. 𝐸. = 1 2 π‘š 𝑣𝑒 2 , 𝑃. 𝐸. = βˆ’ πΊπ‘€π‘š 𝑅 Total energy = P.E. + K.E. Total energy = 1 2 π‘š 𝑣𝑒 2 βˆ’ πΊπ‘€π‘š 𝑅 K.E. = 0 Also, P.E. = βˆ’ πΊπ‘€π‘š ∞ = 0 ∴ π‘‡π‘œπ‘‘π‘Žπ‘™ πΈπ‘›π‘’π‘Ÿπ‘” = 𝑃. 𝐸. +𝐾. 𝐸 = 0 As energy is conserved 1 2 π‘š 𝑣𝑒 2 βˆ’ πΊπ‘€π‘š 𝑅 = 0 Or 𝑣𝑒 = 2𝐺𝑀 𝑅
  • 32. 5.8 Earth Satellites The objects which revolve around the Earth are called Earth satellites. moon is the only natural satellite of the Earth. It revolves in almost a circular orbit around the Earth with period of revolution of nearly 27.3 days. Communication Satellites: These are geostationary satellites. They revolve around the Earth in equatorial plane. They have same sense of rotation as that of the Earth and the same period of rotation as that of the Earth, i. e., one day or 24 hours. Due to this, they appear stationary from the Earth’s surface. Hence they are called geostationary satellites or geosynchronous satellites. Polar Satellites: Polar satellites are used for weather forecasting and meteorological purpose. They are also used for astronomical observations and study of Solar radiations. They go around the poles of the Earth in a north-south direction while the Earth rotates in an east-west direction about its own axis.
  • 33. 5.8 Earth Satellites 5.8.1 Projection of Satellite: The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbit round the Earth is called the critical velocity or orbital velocity (𝑣𝑐). A satellite follows different paths depending upon the horizontal velocity provided to it. Different possible cases are: Fig. Various possible orbits depending on the value of 𝒗𝒉 . Case (I) π‘£β„Ž < 𝑣𝑐 It experiences a non-conservative force of air resistance. Case (II) π‘£β„Ž = 𝑣𝑐 The satellite moves in a stable circular orbit round the Earth. Case (III) 𝑣𝑐 < π‘£β„Ž < 𝑣𝑒 The satellite again moves in an elliptical orbit round the Earth with the point of projection as perigee. Case (IV) π‘£β„Ž = 𝑣𝑒 Its speed will be zero at infinity. Case (V) π‘£β„Ž > 𝑣𝑒
  • 34. 5.8 Earth Satellites Expression for critical speed If the satellite is moving in a circular orbit of radius (R+h) = r, its speed must be the magnitude of critical velocity 𝑣𝑐. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth. ∴ Centripetal force = Gravitational force π‘šπ‘£π‘ 2 π‘Ÿ = πΊπ‘€π‘š π‘Ÿ2 ∴ 𝑣𝑐 2 = 𝐺𝑀 π‘Ÿ ∴ 𝑣𝑐 = 𝐺𝑀 π‘Ÿ ∴ 𝑣𝑐 = 𝐺𝑀 (𝑅+β„Ž) = π‘”β„Ž(𝑅 + β„Ž) This is the expression for critical speed at the orbit of radius (R + h) Fig. Various possible orbits depending on the value of 𝒗𝒉 .
  • 35. 5.8 Earth Satellites β€’ Special case When the satellite is revolving close to the surface of the Earth, the height is very small as compared to the radius of the Earth. Hence the height can be neglected and radius of the orbit is nearly equal to R (i.e R >> h, R+h β‰ˆ R) ∴ πΆπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ 𝑆𝑝𝑒𝑒𝑑, 𝑣𝑐 = 𝐺𝑀 𝑅 As G is related to acceleration due to gravity by the relation, 𝑔 = 𝐺𝑀 𝑅2 ∴ 𝐺𝑀 = 𝑔𝑅2 ∴ Critical speed in terms of acceleration due to gravity, 𝑣𝑐 = 𝑔𝑅2 𝑅 = 𝑔𝑅 = 7.92 π‘˜π‘š 𝑠 Obviously, this is the maximum possible critical speed. This is at least 25 times the speed of the fastest passenger aeroplanes.
  • 36. Example: Show that the critical velocity of a body revolving in a circular orbit very close to the surface of a planet of radius R and mean density 𝜌 is 2𝑅 πΊπœ‹πœŒ 3 Solution: Since the body is revolving very close to the planet, h = 0 Density, 𝜌 = 𝑀 𝑉 = 𝑀 4 3 πœ‹π‘…3 = 4 3 πœ‹π‘…3𝜌 Critical Velocity, 𝑣𝑐 = 𝐺𝑀 𝑅 = 𝐺 4 3 πœ‹π‘…3𝜌 𝑅 ∴ 𝑣𝑐 = 2𝑅 πΊπœ‹πœŒ 3 β€’ When a satellite revolves very close to the surface of the Earth, motion of satellite gets affected by the friction produced due to resistance of air. In deriving the above expression the resistance of air is not considered.
  • 37. 5.8 Earth Satellites 5.8.2 Weightlessness in a Satellite: (i) Gravitational force mg directed vertically downwards (towards centre of the earth) & (ii) Normal reaction force N directed vertically upwards, exerted by the floor of the lift. As these forces are oppositely directed, the net force in the downward direction will be F = ma - N . Though the weight of a body is the gravitational force acting upon it, we feel our weight only due to the normal reaction force N exerted by the floor. According to Newton’s second law of motion, F = ma Let us consider the example of a lift or elevator from an inertial frame of reference. Whether the lift is at rest or in motion, a passenger in it experiences only two forces:
  • 38. 5.8 Earth Satellites 5.8.2 Weightlessness in a Satellite: Case I: Lift having zero acceleration The net force F = 0 = mg - N ∴ mg =N Hence in this case we feel our normal weight mg . Case II:Lift having net upward acceleration 𝒂𝒖 ∴ 𝐹 = π‘š π‘Žπ‘’ = 𝑁 βˆ’ π‘šπ‘” ∴ 𝑁 = π‘šπ‘” + π‘šπ‘Žπ‘’, i.e., N > mg (we feel heavier) Case III (a): Lift having net downward acceleration 𝒂𝒅 ∴ 𝐹 = π‘š π‘Žπ‘‘ = π‘šπ‘” βˆ’ 𝑁 ∴ 𝑁 = π‘šπ‘” βˆ’ π‘šπ‘Žπ‘‘, i.e., N < mg (we feel lighter) Case III (b): State of free fall The downward acceleration π‘Žπ‘‘ = 𝑔. ∴ 𝑁 = π‘šπ‘” βˆ’ π‘šπ‘Žπ‘‘ = 0 β€’ The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.
  • 39. 5.8 Earth Satellites ∴ 𝐺𝑀 = π‘”β„Ž(𝑅 + β„Ž)2 ∴ 𝑇 = 2πœ‹ (𝑅+β„Ž)3 π‘”β„Ž(𝑅+β„Ž)2 ∴ 𝑇 = 2πœ‹ 𝑅+β„Ž π‘”β„Ž = 2πœ‹ π‘Ÿ π‘”β„Ž Special case: When satellite revolves close to the surface of the Earth, R + h β‰ˆ R and π‘”β„Ž β‰ˆ 𝑔. (𝑇)π‘šπ‘–π‘›= 2πœ‹ 𝑅 𝑔 5.8.3 Time Period of a Satellite: The time taken by a satellite to complete one revolution round the Earth is its time period. The satellite revolves in a circular orbit of radius (R+h) = r. ∴ πΆπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ 𝑠𝑝𝑒𝑒𝑑 = πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘π‘–π‘‘ π‘‡π‘–π‘šπ‘’ π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑣𝑐 = 2πœ‹π‘Ÿ 𝑇 But we have, 𝑣𝑐 = πΊπ‘š π‘Ÿ ∴ πΊπ‘š π‘Ÿ = 2πœ‹π‘Ÿ 𝑇 Or, 𝐺𝑀 π‘Ÿ = 4πœ‹2π‘Ÿ2 𝑇2 ∴ 𝑇2 = 4πœ‹2π‘Ÿ2 𝐺𝑀 𝑇 = 2πœ‹ π‘Ÿ3 𝐺𝑀 = 2πœ‹ (𝑅+β„Ž)3 𝐺𝑀
  • 40. Example: Calculate the period of revolution of a polar satellite orbiting close to the surface of the Earth. Given R = 6400 km, g = 9.8 π‘š/𝑠2 Solution: h is negligible as satellite is close to the Earth surface. ∴ 𝑅 + β„Ž β‰ˆ 𝑅 π‘”β„Ž β‰ˆ 𝑔 𝑅 = 6400 π‘˜π‘š = 6.4 Γ— 106 π‘š 𝑇 = 2πœ‹ 𝑅 𝑔 𝑇 = 2 Γ— 3.14 6.4Γ—106 9.8 𝑇 = 5.705 Γ— 103 π‘ π‘’π‘π‘œπ‘›π‘‘ 𝑇 = 85 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘  π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦
  • 41. Example: An artificial satellite revolves around a planet in circular orbit close to its surface. Obtain the formula for period of the satellite in terms of density 𝜌 and radius R of planet. Solution: Period of satellite is, ∴ 𝑇 = 2πœ‹ (𝑅+β„Ž)3 𝐺𝑀 …..(1) Here, the satellite revolves close to the surface of planet, hence h is negligible, hence 𝑅 + β„Ž β‰ˆ 𝑅 𝐷𝑒𝑛𝑠𝑖𝑑𝑦 𝜌 = π‘šπ‘Žπ‘ π‘  (𝑀) π‘£π‘œπ‘™π‘’π‘šπ‘’ (𝑉) ∴ 𝑀 = 𝜌 𝑉 …..(2) As planet is spherical in shape, volume of planet is given as, 𝑉 = 4 3 πœ‹π‘…3 ∴ 𝑀 = 4 3 πœ‹π‘…3 𝜌 …….(3) Substituting the values from equation (2) and (3) in equation (1), we get 𝑇 = 2πœ‹ 𝑅3 𝐺× 4 3 πœ‹π‘…3𝜌 = 3 πœ‹ 𝐺 𝜌
  • 42. 5.8 Earth Satellites 5.8.4 Binding Energy of an orbiting satellite: The minimum energy required by a satellite to escape from Earth’s gravitational influence is the binding energy of the satellite. Let M be the mass of the Earth, R be the Radius of the Earth, 𝑣𝑐 be critical velocity of satellite, r = (R+h) be the radius of the orbit. 𝐾𝑖𝑛𝑒𝑑𝑖𝑐 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘“ π‘ π‘Žπ‘‘π‘’π‘™π‘™π‘–π‘‘π‘’ = 1 2 π‘š 𝑣𝑐 2 𝐾𝑖𝑛𝑒𝑑𝑖𝑐 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘“ π‘ π‘Žπ‘‘π‘’π‘™π‘™π‘–π‘‘π‘’ = 1 2 πΊπ‘€π‘š π‘Ÿ The gravitational potential at a distance r from the centre of the Earth is, βˆ’ 𝐺𝑀 π‘Ÿ ∴ Potential energy of satellite = Gravitational potential Γ— mass of satellite ∴ Potential energy of satellite = βˆ’ πΊπ‘€π‘š π‘Ÿ The total energy of satellite is given as, T.E. = K.E. + P.E. π‘‡π‘œπ‘‘π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ = 1 2 πΊπ‘€π‘š π‘Ÿ βˆ’ πΊπ‘€π‘š π‘Ÿ = βˆ’ 1 2 πΊπ‘€π‘š π‘Ÿ The minimum energy to be supplied to unbind the satellite is + 1 2 πΊπ‘€π‘š π‘Ÿ