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

5.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of in Chapter 3, 50.0 m s-1. (a) What is the work done by the v2 − u2 = 2 as (5.2) gravitational force ? What is the work done where u and v are the initial and final speeds by the unknown resistive force? and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the sides by m/2, we have drop is 1 2 1 2 1 2 mv − mu = mas = Fs (5.2a) ∆ K = m v − 0 2 2 2 where the last step follows from Newton’s Second 1 -3 = × 10 × 50 × 50 Law. We can generalise Eq. (5.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement Assuming that g is a constant with a value vectors of the object respectively. 10 m/s2, the work done by the gravitational force Once again multiplying both sides by m/2 , we obtain is, 1 1 mv 2 − mu 2 = m a.d = F.d (5.2b) Wg = mgh 2 2 = 10-3 ×10 ×103 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the quantity ‘half the mass times the square of the ∆ K = W g + W r speed’ from its initial value to its final value. We where Wr is the work done by the resistive force call each of these quantities the ‘kinetic energy’, on the raindrop. Thus denoted by K. The right side is a product of the Wr = ∆K − Wg displacement and the component of the force = 1.25 −10 along the displacement. This quantity is called = − 8.75 J ‘work’ and is denoted by W. Eq. (5.2b) is then is negative. ⊳ Kf − Ki = W (5.3) 5.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 5.2. Equation (5.2) is also a special case of the work-energy (WE) theorem : The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section. ⊳ Example 5.2 It is well known that a raindrop falls under the influence of the Fig. 5.2 An object undergoes a displacement d downward gravitational force and the under the influence of the force F. opposing resistive force. The latter is known Reprint 2025-26 74 PHYSICS The work done by the force is defined to be Table 5.1 Alternative Units of Work/Energy in J the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ)d = F.d (5.4) We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, ⊳ the force you exert on the wall does no work. Yet Example 5.3 A cyclist comes to a skidding your muscles are alternatively contracting and stop in 10 m. During this process, the force relaxing and internal energy is being used up on the cycle due to the road is 200 N and and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How in physics is different from its usage in everyday much work does the road do on the cycle ? language. (b) How much work does the cycle do on the road ? No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force does no work on the load during this time. on the cycle due to the road. (ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make horizontal table is not acted upon by a an angle of 180o (π rad) with each other. horizontal force (since there is no friction), but Thus, work done by the road, may undergo a large displacement. Wr = Fd cosθ (iii) the force and displacement are mutually = 200 × 10 × cos π perpendicular. This is so since, for θ= π/2 rad = – 2000 J (= 90o), cos (π/2) = 0. For the block moving on It is this negative work that brings the cycle a smooth horizontal table, the gravitational to a halt in accordance with WE theorem. force mg does no work since it acts at right (b) From Newton’s Third Law an equal and angles to the displacement. If we assume that opposite force acts on the road due to the the moon’s orbits around the earth is cycle. Its magnitude is 200 N. However, the perfectly circular then the earth’s road undergoes no displacement. Thus, gravitational force does no work. The moon’s work done by cycle on the road is zero. ⊳ instantaneous displacement is tangential while the earth’s force is radially inwards and The lesson of Example 5.3 is that though the θ = π/2. force on a body A exerted by the body B is always Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s between 0o and 90o, cos θ in Eq. (5.4) is positive. Third Law); the work done on A by B is not If θ is between 90o and 180o, cos θ is negative. necessarily equal and opposite to the work done In many examples the frictional force opposes on B by A. displacement and θ = 180o. Then the work done 5.4 KINETIC ENERGY by friction is negative (cos 180o = –1). As noted earlier, if an object of mass m has From Eq. (5.4) it is clear that work and energy velocity v, its kinetic energy K ishave the same dimensions, [ML2T–2]. The SI unit of these is joule (J), named after the famous British 1 1 2physicist James Prescott Joule (1811-1869). Since K = m v. v = mv (5.5) 2 2work and energy are so widely used as physical concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic these are listed in Table 5.1. energy of an object is a measure of the work an Reprint 2025-26 WORK, ENERGY AND POWER 75 Table 5.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 5.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 5.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing x f ships employ the kinetic energy of the wind. Table W ≅ F (x )∆x (5.6) ∑

5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳