1.4 DIMENSIONS OF PHYSICAL QUANTITIES formula of the given physical quantity. For example, the dimensional formula of the volume The nature of a physical quantity is described is [M° L3 T°], and that of speed or velocity is by its dimensions. All the physical quantities [M° L T-1]. Similarly, [M° L T–2] is the dimensional represented by derived units can be expressed formula of acceleration and [M L–3 T°] that of in terms of some combination of seven mass density. fundamental or base quantities. We shall call An equation obtained by equating a physical these base quantities as the seven dimensions quantity with its dimensional formula is called of the physical world, which are denoted with the dimensional equation of the physical square brackets [ ]. Thus, length has the quantity. Thus, the dimensional equations are dimension [L], mass [M], time [T], electric current the equations, which represent the dimensions [A], thermodynamic temperature [K], luminous of a physical quantity in terms of the base intensity [cd], and amount of substance [mol]. quantities. For example, the dimensional The dimensions of a physical quantity are the equations of volume [V], speed [v], force [F ] and powers (or exponents) to which the base mass density [ρ] may be expressed as quantities are raised to represent that [V] = [M0 L3 T0] quantity. Note that using the square brackets [v] = [M0 L T–1] [ ] round a quantity means that we are dealing [F] = [M L T–2] with ‘the dimensions of’ the quantity. [ρ] = [M L–3 T0] In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] The dimensional equation can be obtained and [T]. For example, the volume occupied by from the equation representing the relations between the physical quantities. Thean object is expressed as the product of length, dimensional formulae of a large number andbreadth and height, or three lengths. Hence the wide variety of physical quantities, derived fromdimensions of volume are [L] × [L] × [L] = [L]3 = [L3]. the equations representing the relationships As the volume is independent of mass and time, among other physical quantities and expressed it is said to possess zero dimension in mass [M°], in terms of base quantities are given in zero dimension in time [T°] and three Appendix 9 for your guidance and ready dimensions in length. reference. Similarly, force, as the product of mass and acceleration, can be expressed as 1.6 DIMENSIONAL ANALYSIS AND ITS Force = mass × acceleration APPLICATIONS = mass × (length)/(time)2 The recognition of concepts of dimensions, which The dimensions of force are [M] [L]/[T]2 = guide the description of physical behaviour is [M L T–2]. Thus, the force has one dimension in of basic importance as only those physical Reprint 2025-26 8 PHYSICS quantities can be added or subtracted which such as angle as the ratio (length/length), have the same dimensions. A thorough refractive index as the ratio (speed of light in understanding of dimensional analysis helps us vacuum/speed of light in medium) etc., has no in deducing certain relations among different dimensions. physical quantities and checking the derivation, Now we can test the dimensional consistency accuracy and dimensional consistency or or homogeneity of the equation homogeneity of various mathematical 2expressions. When magnitudes of two or more x = x 0 + v 0 t + (1/2 ) a t physical quantities are multiplied, their units for the distance x travelled by a particle or body should be treated in the same manner as in time t which starts from the position x0 with ordinary algebraic symbols. We can cancel an initial velocity v0 at time t = 0 and has uniformidentical units in the numerator and acceleration a along the direction of motion. denominator. The same is true for dimensions The dimensions of each term may be written as of a physical quantity. Similarly, physical [x] = [L] quantities represented by symbols on both sides of a mathematical equation must have the same [x0 ] = [L] dimensions. [v0 t] = [L T–1] [T] = [L] [(1/2) a t2] = [L T–2] [T2]1.6.1 Checking the Dimensional Consistency of Equations = [L] As each term on the right hand side of this The magnitudes of physical quantities may be equation has the same dimension, namely that added together or subtracted from one another of length, which is same as the dimension of only if they have the same dimensions. In other left hand side of the equation, hence this words, we can add or subtract similar physical equation is a dimensionally correct equation. quantities. Thus, velocity cannot be added to It may be noted that a test of consistency of force, or an electric current cannot be subtracted dimensions tells us no more and no less than a from the thermodynamic temperature. This test of consistency of units, but has the simple principle called the principle of advantage that we need not commit ourselves homogeneity of dimensions in an equation is to a particular choice of units, and we need not extremely useful in checking the correctness of worry about conversions among multiples and an equation. If the dimensions of all the terms sub-multiples of the units. It may be borne in are not same, the equation is wrong. Hence, if mind that if an equation fails this consistency we derive an expression for the length (or test, it is proved wrong, but if it passes, it is distance) of an object, regardless of the symbols not proved right. Thus, a dimensionally correct appearing in the original mathematical relation, equation need not be actually an exact when all the individual dimensions are (correct) equation, but a dimensionally wrong simplified, the remaining dimension must be (incorrect) or inconsistent equation must be that of length. Similarly, if we derive an equation wrong. of speed, the dimensions on both the sides of ⊳equation, when simplified, must be of length/ Example 1.3 Let us consider an equation time, or [L T–1]. Dimensions are customarily used as a 1 2 m v = m g h preliminary test of the consistency of an 2 equation, when there is some doubt about the where m is the mass of the body, v its correctness of the equation. However, the velocity, g is the acceleration due to dimensional consistency does not guarantee gravity and h is the height. Check correct equations. It is uof dimensionless quantities or functions. The correct. arguments of special functions, such as the trigonometric, logarithmic and exponential Answer The dimensions of LHS are functions must be dimensionless. A pure [M] [L T–1 ]2 = [M] [ L2 T–2] number, ratio of similar physical quantities, = [M L2 T–2] Reprint 2025-26 UNITS AND MEASUREMENT 9 The dimensions of RHS are string, that oscillates under the action of [M][L T–2] [L] = [M][L2 T–2] the force of gravity. Suppose that the period = [M L2 T–2] of oscillation of the simple pendulum depends on its length (l), mass of the bobThe dimensions of LHS and RHS are the same and hence the equation is dimensionally correct. ⊳ (m) and acceleration due to gravity (g). Derive the expression for its time period ⊳ using method of dimensions. Example 1.4 The SI unit of energy is J = kg m2 s–2; that of speed v is m s–1 and Answer The dependence of time period T on of acceleration a is m s–2. Which of the the quantities l, g and m as a product may be formulae for kinetic energy (K) given below written as : can you rule out on the basis of T = k lx gy mz dimensional arguments (m stands for the mass of the body) : where k is dimensionless constant and x, y (a) K = m2 v3 and z are the exponents. (b) K = (1/2)mv2 By considering dimensions on both sides, we (c) K = ma have (d) K = (3/16)mv2 o o 1 1 x 1 –2 y 1 z [L M T ]=[L ] [L T ] [M ] (e) K = (1/2)mv2 + ma = Lx+y T–2y Mz Answer Every correct formula or equation must On equating the dimensions on both sides, have the same dimensions on both sides of the we have equation. Also, only quantities with the same x + y = 0; –2y = 1; and z = 0 physical dimensions can be added or 1 1 subtracted. The dimensions of the quantity on So that x = , y = – , z = 0 2 2the right side are [M2 L3 T–3] for (a); [M L2 T–2] for (b) and (d); [M L T–2] for (c). The quantity on the Then, T = k l½ g–½ right side of (e) has no proper dimensions since two quantities of different dimensions have been l or, T = kadded. Since the kinetic energy K has the g dimensions of [M L2 T–2], formulas (a), (c) and (e) Note that value of constant k can not be obtainedare ruled out. Note that dimensional arguments by the method of dimensions. Here it does notcannot tell which of the two, (b) or (d), is the matter if some number multiplies the right sidecorrect formula. For this, one must turn to the of this formula, because that does not affect itsactual definition of kinetic energy (see dimensions.Chapter 5). The correct formula for kinetic energy is given by (b). ⊳ l Actually, k = 2π so that T = 2π ⊳ 1.6.2 Deducing Relation among the g Physical Quantities The method of dimensions can sometimes be Dimensional analysis is very useful in deducing used to deduce relation among the physical relations among the interdependent physical quantities. For this we should know the quantities. However, dimensionless constants dependence of the physical quantity on other cannot be obtained by this method. The method quantities (upto three physical quantities or of dimensions can only test the dimensional linearly independent variables) and consider it validity, but not the exact relationship between as a product type of the dependence. Let us take physical quantities in any equation. It does not an example. distinguish between the physical quantities having same dimensions.⊳ Example 1.5 Consider a simple A number of exercises at the end of this pendulum, having a bob attached to a chapter will help you develop skill in dimensional analysis. Reprint 2025-26 10 PHYSICS SUMMARY 1. Physics is a quantitative science, based on measurement of physical quantities. Certain physical quantities have been chosen as fundamental or base quantities (such as length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity). 2. Each base quantity is defined in terms of a certain basic, arbitrarily chosen but properly standardised reference standard called unit (such as metre, kilogram, second, ampere, kelvin, mole and candela). The units for the fundamental or base quantities are called fundamental or base units. 3. Other physical quantities, derived from the base quantities, can be expressed as a combination of the base units and are called derived units. A complete set of units, both fundamental and derived, is called a system of units. 4. The International System of Units (SI) based on seven base units is at present internationally accepted unit system and is widely used throughout the world. 5. The SI units are used in all physical measurements, for both the base quantities and the derived quantities obtained from them. Certain derived units are expressed by means of SI units with special names (such as joule, newton, watt, etc). 6. The SI units have well defined and internationally accepted unit symbols (such as m for metre, kg for kilogram, s for second, A for ampere, N for newton etc.). 7. Physical measurements are usually expressed for small and large quantities in scientific notation, with powers of 10. Scientific notation and the prefixes are used to simplify measurement notation and numerical computation, giving indication to the precision of the numbers. 8. Certain general rules and guidelines must be followed for using notations for physical quantities and standard symbols for SI units, some other units and SI prefixes for expressing properly the physical quantities and measurements. 9. In computing any physical quantity, the units for derived quantities involved in the relationship(s) are treated as though they were algebraic quantities till the desired units are obtained. 10. In measured and computed quantities proper significant figures only should be retained. Rules for determining the number of significant figures, carrying out arithmetic operations with them, and ‘rounding off ‘ the u 11. The dimensions of base quantities and combination of these dimensions describe the nature of physical quantities. Dimensional analysis can be used to check the dimensional consistency of equations, deducing relations among the physical quantities, etc. A dimensionally consistent equation need not be actually an exact (correct) equation, but a dimensionally wrong or inconsistent equation must be wrong. EXERCISES Note : In stating numerical answers, take care of significant figures. 1.1 Fill in the blanks (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3. 1.2 Fill in the blanks by suitable conversion of units (a) 1 kg m2 s–2 = ....g cm2 s–2 (b) 1 m = ..... ly (c) 3.0 m s–2 = .... km h–2 (d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1. Reprint 2025-26 UNITS AND MEASUREMENT 11 1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γ 2 in terms of the new units. 1.4 Explain this statement clearly : “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary : (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light. 1.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ? 1.6 Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ? 1.7 A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ? 1.8 Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ? 1.9 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement. 1.10 State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2 1.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. 1.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ? 1.13 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m 0 m = 2 1/2 . 1 − v ( ) Guess where to put the missing c. Reprint 2025-26 12 PHYSICS 1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ? 1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ? 1.16 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you). 1.17 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m. Reprint 2025-26 CHAPTER TWO MOTION IN A STRAIGHT LINE 2.1 INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves 2.1 Introduction into and out of our lungs and blood flows in arteries and 2.2 Instantaneous velocity and veins. We see leaves falling from trees and water flowing speed down a dam. Automobiles and planes carry people from one 2.3 Acceleration place to the other. The earth rotates once every twenty-four 2.4 Kinematic equations for hours and revolves round the sun once in a year. The sun uniformly accelerated motion itself is in motion in the Milky Way, which is again moving

27.3 days which is also roughly equal to the Which is approximately 85 minutes. rotational period of the moon about its own axis. ⊳ Example 7.5 The planet Mars has twoSince, 1957, advances in technology have enabled moons, phobos and delmos. (i) phobos hasmany countries including India to launch artificial a period 7 hours, 39 minutes and an orbitalearth satellites for practical use in fields like radius of 9.4 ×103 km. Calculate the masstelecommunication, geophysics and meteorology. of mars. (ii) Assume that earth and mars We will consider a satellite in a circular orbit move in circular orbits around the sun,of a distance (RE + h) from the centre of the earth, with the martian orbit being 1.52 timeswhere RE = radius of the earth. If m is the mass the orbital radius of the earth. What isof the satellite and V its speed, the centripetal the length of the martian year in days ?force required for this orbit is mV 2 Answer (i) We employ Eq. (7.38) with the sun’s F(centripetal) = (7.33) ( R E + h ) mass replaced by the martian mass Mm directed towards the centre. This centripetal force 2 4 π 2 3 T = Ris provided by the gravitational force, which is GM m G m M E 4 π 2 R 3 F(gravitation) = 2 (7.34) Mm = 2 ( R E + h ) G T where ME is the mass of the earth. 2 3 18 Equating R.H.S of Eqs. (7.33) and (7.34) and 4 × ( 3.14 ) × ( 9.4 ) × 10 = -11 2cancelling out m, we get 6.67 × 10 × ( 459 × 60 ) 2 G M E 2 3 18 V = (7.35) 4 × ( 3.14 ) × ( 9.4 ) × 10 ( R E + h ) M m = 2 -5 6.67 × ( 4.59 × 6 ) × 10 Thus V decreases as h increases. From = 6.48 × 1023 kg. equation (7.35),the speed V for h = 0 is (ii) Once again Kepler’s third law comes to our V 2 (h = 0) = GM / R E = gR E (7.36) aid, where we have used the relation T M2 R MS3 2 2 = 3 g = GM / R E . In every orbit, the satellite T E R ES Reprint 2025-26 138 PHYSICS where RMS is the mars -sun distance and RES is − 13  1 2   1    d the earth-sun distance. = 10    ( 24 × 60 × 60 ) 2  ( 1 / 1000 ) 3 km 3  ∴ TM = (1.52)3/2 × 365 = 1.33 ×10–14 d2 km–3 = 684 days Using Eq. (7.38) and the given value of k, We note that the orbits of all planets except the time period of the moon is Mercury and Mars are very close to being 2 T = (1.33 × 10-14)(3.84 × 105)3 circular. For example, the ratio of the semi- T = 27.3 d ⊳ minor to semi-major axis for our Earth is, Note that Eq. (7.38) also holds for elliptical b/a = 0.99986. ⊳ orbits if we replace (RE+h) by the semi-major axis ⊳ of the ellipse. The earth will then be at one of Example 7.6 Weighing the Earth : You the foci of this ellipse. are given the following data: g = 9.81 ms–2, RE = 6.37×106 m, the distance to the moon R 7.10 ENERGY OF AN ORBITING SATELLITE = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the Using Eq. (7.35), the kinetic energy of the satellite mass of the Earth ME in two different ways. in a circular orbit with speed v is 1 m v 2Answer From Eq. (7.12) we have K i E = 2 g R E2 M E = Gm M E G = , (7.40) 2( R E + h ) 6 2 Considering gravitational potential energy at 9.81 × ( 6.37 × 10 ) = -11 infinity to be zero, the potential energy at distance 6.67 × 10 (Re+h) from the centre of the earth is = 5.97× 1024 kg. The moon is a satellite of the Earth. From G m M E P .E = − (7.41)the derivation of Kepler’s third law [see Eq. ( R E + h ) (7.38)] The K.E is positive whereas the P.E is 2 4 π2R 3 negative. However, in magnitude the K.E. is half T = G M E the P.E, so that the total E is E 4 π2R 3 E = K .E + P .E = − G m M ME = G T 2 2( R E + h ) (7.42) 4 × 3.14 × 3.14 × ( 3.84 ) 3 × 10 24 The total energy of an circularly orbiting = -11 2 satellite is thus negative, with the potential 6.67 × 10 × ( 27.3 × 24 × 60 × 60 ) energy being negative but twice is magnitude of = 6.02 × 1024 kg the positive kinetic energy. Both methods yield almost the same answer, When the orbit of a satellite becomes the difference between them being less than 1%. elliptic, both the K.E. and P.E. vary from point ⊳ to point. The total energy which remains constant is negative as in the circular orbit case. ⊳ Example 7.7 Express the constant k of Eq. This is what we expect, since as we have (7.38) in days and kilometres. Given discussed before if the total energy is positive or k = 10–13 s2 m–3. The moon is at a distance zero, the object escapes to infinity. Satellites of 3.84 × 105 km from the earth. Obtain its are always at finite distance from the earth and time-period of revolution in days. hence their energies cannot be positive or zero. Answer Given k = 10–13 s2 m–3 Reprint 2025-26 GRAVITATION 139 The change in the total energy is⊳ Example 7.8 A 400 kg satellite is in a circular ∆E = Ef – Ei orbit of radius 2RE about the Earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in  E E = G M E m =  the kinetic and potential energies ?  G M2  m R 8 R E  R E  8 Answer Initially, g m R E = 9.81 × 400 × 6. 37 × 106 = 3.13 × 10 9 J ∆ E = G M E m 8 8 E i = − 4 R E The kinetic energy is reduced and it mimics While finally ∆E, namely, ∆K = Kf – Ki = – 3.13 × 109 J. The change in potential energy is twice the G M E m E f = − change in the total energy, namely 8 R E ∆V = Vf – Vi = – 6.25 × 109 J ⊳ SUMMARY 1. Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude m 1m 2 F = G 2 r where G is the universal gravitational constant, which has the value 6.672 ×10–11 N m2 kg–2. 2. If we have to find the resultant gravitational force acting on the particle m due to a number of masses M1, M2, ….Mn etc. we use the principle of superposition. Let F1, F2, ….Fn be the individual forces due to M1, M2, ….Mn, each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force FR is then found by vector addition n FR = F1 + F2 + ……+ Fn = ∑ Fi i = 1 where the symbol ‘Σ’ stands for summation. 3. Kepler’s laws of planetary motion state that (a) All planets move in elliptical orbits with the Sun at one of the focal points (b) The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved. (c) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet The period T and radius R of the circular orbit of a planet about the Sun are related by 2  4 π 2  3 T =   R  G M s  where Ms is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if R is replaced by the semi-major axis, a. 4. The acceleration due to gravity. (a) at a height h above the earth’s surface G M E g ( h ) = 2 ( R E + h ) G M E  2 h  ≈ 2 1 − for h << RE R E  R E  Reprint 2025-26 140 PHYSICS  2 h  G M E g (h ) = g ( 0 ) 1 − where g ( 0 ) = 2  R E  R E (b) at depth d below the earth’s surface is d   1 − g (d ) = G M2 E 1 − d  = g ( 0 ) R E  R E   R E  5. The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance r is given by G m1 m 2 V = − r where V is taken to be zero at r → ∞. The total potential energy for a system of particles is the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of superposition. 6. If an isolated system consists of a particle of mass m moving with a speed v in the vicinity of a massive body of mass M, the total mechanical energy of the particle is given by 1 G M m E = m v 2− 2 r That is, the total mechanical energy is the sum of the kinetic and potential energies. The total energy is a constant of motion. 7. If m moves in a circular orbit of radius a about M, where M >> m, the total energy of the system is G M m E = − 2a with the choice of the arbitrary constant in the potential energy given in the point 5., above. The total energy is negative for any bound system, that is, one in which the orbit is closed, such as an elliptical orbit. The kinetic and potential energies are G M m K = 2a G M m V = − a 8. The escape speed from the surface of the earth is 2 G M E ve = = 2 gR E R E and has a value of 11.2 km s–1. 9. If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric internal mass distribution, the sphere attracts the particle as though the mass of the sphere or shell were concentrated at the centre of the sphere. 10. If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle. Reprint 2025-26 GRAVITATION 141 POINTS TO PONDER 1. In considering motion of an object under the gravitational influence of another object the following quantities are conserved: (a) Angular momentum (b) Total mechanical energy Linear momentum is not conserved 2. Angular momentum conservation leads to Kepler’s second law. However, it is not special to the inverse square law of gravitation. It holds for any central force. 3. In Kepler’s third law (see Eq. (7.1) and T2 = KS R3. The constant KS is the same for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (7.38)]. 4. An astronaut experiences weightlessness in a space satellite. This is not because the gravitational force is small at that location in space. It is because both the astronaut and the satellite are in “free fall” towards the Earth. 5. The gravitational potential energy associated with two particles separated by a distance r is given by G m 1 m 2 V = – + constant r The constant can be given any value. The simplest choice is to take it to be zero. With this choice G m 1 m 2 V = – r This choice implies that V → 0 as r → ∞. Choosing location of zero of the gravitational energy is the same as choosing the arbitrary constant in the potential energy. Note that the gravitational force is not altered by the choice of this constant. 6. The total mechanical energy of an object is the sum of its kinetic energy (which is always positive) and the potential energy. Relative to infinity (i.e. if we presume that the potential energy of the object at infinity is zero), the gravitational potential energy of an object is negative. The total energy of a satellite is negative. 7. The commonly encountered expression m g h for the potential energy is actually an approximation to the difference in the gravitational potential energy discussed in the point 6, above. 8. Although the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centre of mass. For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central. 9. The gravitational force on a particle inside a spherical shell is zero. However, (unlike a metallic shell which shields electrical forces) the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible. EXERCISES 7.1 Answer the following : (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ? Reprint 2025-26 142 PHYSICS 7.2 Choose the correct alternative : (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula –G Mm(1/r2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth. 7.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ? 7.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun. 7.5 Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly. 7.6 Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence. 7.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched? 7.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun. 7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem. 7.10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0. Fig. 7.11 7.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. 7.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m). 7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km. 7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ? 7.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? Reprint 2025-26 GRAVITATION 143 7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ? 7.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.18 The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. 7.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.20 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G). 7.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ? Reprint 2025-26

7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH 4π 3 density and hence its mass is M E = R E ρ 3 The earth can be imagined to be a sphere made where ME is the mass of the earth RE is its radiusof a large number of concentric spherical shells and ρ is the density. On the other hand thewith the smallest one at the centre and the largest one at its surface. A point outside the 4π 3 ρr and mass of the sphere Mr of radius r isearth is obviously outside all the shells. Thus, 3 Reprint 2025-26 GRAVITATION 133 hence its distance from the centre of the earth is (RE + h ). If F (h) denoted the magnitude of the force on the point mass m , we get from G m M E Eq. (7.5) : = 3 r (7.10) R E If the mass m is situated on the surface of GM E m F (h ) = earth, then r = RE and the gravitational force on ( R E + h )2 (7.13) it is, from Eq. (7.10) The acceleration experienced by the point M E m F = G 2 (7.11) mass is F (h )/ m ≡ g (h ) and we get R E The acceleration experienced by the mass F (h ) GM E . g (h ) = = (7.14)m, which is usually denoted by the symbol g is m ( R E + h )2 related to F by Newton’s 2nd law by relation This is clearly less than the value of g on the F = mg. Thus GM E . g = surface of earth : GM F For h << R E , we can E R E2 g = = 2 (7.12) m R E expand the RHS of Eq. (7.14) : E Acceleration g is readily measurable. RE is a g (h ) = 2 GM 2 = g (1 + h / R E )−2known quantity. The measurement of G by R E (1 + h / R E ) Cavendish’s experiment (or otherwise), combined h << 1 , using binomial expression,with knowledge of g and RE enables one to For R E estimate ME from Eq. (7.12). This is the reason  2h why there is a popular statement regarding g (h ) ≅ g 1 − . (7.15)Cavendish : “Cavendish weighed the earth”.  RE  7.6 ACCELERATION DUE TO GRAVITY BELOW Equation (7.15) thus tells us that for small AND ABOVE THE SURFACE OF EARTH heights h above the value of g decreases by a factor (1 − 2h / R E ).Consider a point mass m at a height h above the Now, consider a point mass m at a depthsurface of the earth as shown in Fig. 7.8(a). The d below the surface of the earth (Fig. 7.8(b)),radius of the earth is denoted by RE . Since this so that its distance from the centre of thepoint is outside the earth, earth is ( R E − d ) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If Ms is the mass of the smaller sphere, then, Ms/ME = ( RE – d)3 / RE3 ( 7.16) Since mass of a sphere is proportional to be Fig. 7.8 (a) g at a height h above the surface of the cube of its radius. earth. Reprint 2025-26 134 PHYSICS close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a Ms ME point at a height h1 from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by W12 Fig. 7.8 (b) g at a depth d. In this case only the smaller W12 = Force × displacement = mg (h2 – h1) (7.20) sphere of radius (RE–d) contributes to g. Thus the force on the point mass is If we associate a potential energy W(h) at a point at a height h above the surface such that F (d) = G Ms m / (RE – d ) 2 (7.17) W(h) = mgh + Wo (7.21) Substituting for Ms from above , we get (where Wo = constant) ; F (d) = G ME m ( RE – d ) / RE 3 (7.18) then it is clear that and hence the acceleration due to gravity at W12 = W(h2) – W(h1) (7.22) a depth d, The work done in moving the particle is just the difference of potential energy between its F ( d ) final and initial positions.Observe that the g(d) = is m constant Wo cancels out in Eq. (7.22). Setting h = 0 in the last equation, we get W ( h = 0 ) = Wo. F (d ) GM E g ( d ) = = 3 ( R E − d ) . h = 0 means points on the surface of the earth. m R E Thus, Wo is the potential energy on the surface of the earth. R E − d = g = g (1 − d / R E ) (7.19) If we consider points at arbitrary distance R E from the surface of the earth, the result just Thus, as we go down below earth’s surface, derived is not valid since the assumption that the acceleration due gravity decreases by a factor the gravitational force mg is a constant is no (1 − d / R E ). The remarkable thing about longer valid. However, from our discussion we know that a point outside the earth, the force of acceleration due to earth’s gravity is that it is gravitation on a particle directed towards the maximum on its surface decreasing whether you centre of the earth is go up or down. G ME m F = 2 (7.23)7.7 GRAVITATIONAL POTENTIAL ENERGY r where ME = mass of earth, m = mass of theWe had discussed earlier the notion of potential particle and r its distance from the centre of theenergy as being the energy stored in the body at earth. If we now calculate the work done inits given position. If the position of the particle lifting a particle from r = r1 to r = r2 (r2 > r1) alongchanges on account of forces acting on it, then a vertical path, we get instead of Eq. (7.20) the change in its potential energy is just the amount of work done on the body by the force. r2 G M m W12 2 d rAs we had discussed earlier, forces for which the =∫r1 r work done is independent of the path are the conservative forces.  1 1  = − G M E m − (7.24) The force of gravity is a conservative force  r2 r1  and we can calculate the potential energy of a In place of Eq. (7.21), we can thus associate body arising out of this force, called the a potential energy W(r) at a distance r, such that gravitational potential energy. Consider points Reprint 2025-26 GRAVITATION 135 G M E m W (r ) =− + W1 , (7.25) r valid for r > R , so that once again W12 = W(r2) – W(r1). Setting r = infinity in the last equation, we get W ( r = infinity ) = W1 . Thus, W1 is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (7.22) and (7.24). One conventionally sets W1 equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point. We have calculated the potential energy at a point of a particle due to gravitational forces Fig. 7.9 on it due to the earth and it is proportional to the mass of the particle. The gravitational The gravitational potential at the centre of potential due to the gravitational force of the the square r = 2 l/2 is ( )earth is defined as the potential energy of a particle of unit mass at that point. From the G m U (r ) = − 4 2 . ⊳earlier discussion, we learn that the gravitational l potential energy associated with two particles of masses m1 and m2 separated by distance by a 7.8 ESCAPE SPEED distance r is given by If a stone is thrown by hand, we see it falls back Gm 1m 2 to the earth. Of course using machines we can V = – (if we choose V = 0 as r →∞) r shoot an object with much greater speeds and It should be noted that an isolated system of with greater and greater initial speed, the object particles will have the total potential energy that scales higher and higher heights. A natural equals the sum of energies (given by the above query that arises in our mind is the following: equation) for all possible pairs of its constituent ‘can we throw an object with such high initial particles. This is an example of the application speeds that it does not fall back to the earth?’ of the superposition principle. The principle of conservation of energy helps us to answer this question. Suppose the object ⊳ Example 7.3 Find the potential energy of did reach infinity and that its speed there was a system of four particles placed at the Vf. The energy of an object is the sum of potential vertices of a square of side l. Also obtain and kinetic energy. As before W1 denotes that the potential at the centre of the square. gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is Answer Consider four masses each of mass m at the corners of a square of side l; See Fig. 7.9. 2 mV fWe have four mass pairs at distance l and two E ( ∞=) W1 + (7.26) 2 diagonal pairs at distance 2 l If the object was thrown initially with a speed Hence, Vi from a point at a distance (h+RE) from the G m 2 G m 2 centre of the earth (RE = radius of the earth), its W (r ) = − 4 − 2 l 2 l energy initially was 2 2 1 2 GmM E 2 G m  1  G m E (h + R E ) = mVi – + W1 (7.27) 5.41 = − 2 (h + R E ) 2 +  = − l  l  2  Reprint 2025-26 136 PHYSICS By the principle of energy conservation ⊳ Eqs. (7.26) and (7.27) must be equal. Hence Example 7.4 Two uniform solid spheres 2 2 of equal radii R, but mass M and 4 M have mV f mVi GmM E a centre to centre separation 6 R, as shown – = (7.28) 2 (h + R E ) 2 in Fig. 7.10. The two spheres are held fixed. The R.H.S. is a positive quantity with a A projectile of mass m is projected from the minimum value zero hence so must be the L.H.S. surface of the sphere of mass M directly Thus, an object can reach infinity as long as Vi towards the centre of the second sphere. is such that Obtain an expression for the minimum speed v of the projectile so that it reaches mVi 2 GmM E the surface of the second sphere. – ≥ 0 (7.29) 2 (h + R E ) The minimum value of Vi corresponds to the case when the L.H.S. of Eq. (7.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) Fig. 7.10corresponds to 1 2 GmM E Answer The projectile is acted upon by two m = (7.30) mutually opposing gravitational forces of the two ( Vi )min 2 h + R E spheres. The neutral point N (see Fig. 7.10) is defined as the position where the two forces If the object is thrown from the surface of cancel each other exactly. If ON = r, we have the earth, h = 0, and we get G M m 4 G M m = 2GM E r 2 (6 R −r )2 (Vi )min = (7.31) (6R – r)2 = 4r2 R E 6R – r = ±2r r = 2R or – 6R. 2 The neutral point r = – 6R does not concern Using the relation g = GM E / R E , we get us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed (Vi )min = 2 gR E (7.32) which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface Using the value of g and RE, numerically of M is (Vi)min≈11.2 km/s. This is called the escape 1 2 G M m 4 G M mspeed, sometimes loosely called the escape E i = m v − − . velocity. 2 R 5 R Equation (7.32) applies equally well to an At the neutral point N, the speed approaches object thrown from the surface of the moon with zero. The mechanical energy at N is purely g replaced by the acceleration due to Moon’s potential. gravity on its surface and rE replaced by the G M m 4 G M m − EN = − .radius of the moon. Both are smaller than their 2 R 4 R values on earth and the escape speed for the From the principle of conservation of moon turns out to be 2.3 km/s, about five times mechanical energy smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the 1 2 GM 4GM GM GMsurface of the moon having velocities larger than v − − = − − this will escape the gravitational pull of the 2 R 5 R 2R R moon. or Reprint 2025-26 GRAVITATION 137 + h) with speed V. Its 2 2 G M  4 1  traverses a distance 2π(RE v = − time period T therefore is R  5 2  2π( R E + h ) 2π( R E + h )3 / 2 T = = (7.37)  3 G M 1/2 V ⊳ G M E v =  5 R  on substitution of value of V from Eq. (7.35). A point to note is that the speed of the projectile Squaring both sides of Eq. (7.37), we get is zero at N, but is nonzero when it strikes the 2 T = k ( RE + h)3 (where k = 4 π2 / GME) (7.38)heavier sphere 4 M. The calculation of this speed is left as an exercise to the students. which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a 7.9 EARTH SATELLITES satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (7.38). Earth satellites are objects which revolve around Hence, for such satellites, T is To, where the earth. Their motion is very similar to the motion of planets around the Sun and hence T 0 = 2π R E / g (7.39) Kepler’s laws of planetary motion are equally If we substitute the numerical values applicable to them. In particular, their orbits g ≃ 9.8 m s-2 and RE = 6400 km., we get around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near 6.4 × 10 6 T 0 = 2π s circular orbit with a time period of approximately 9.8