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

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

1.6 FORCES BETWEEN MULTIPLE CHARGES The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges q1, q2, q3, ..., qn in vacuum. What is the force on q1 due to q2, q3, ..., qn? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition. To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in Fig. 1.5(a). The force on one charge, say q1, due to two other charges q2, q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12, F12 is given by Eq. (1.3) even though other charges are present. FIGURE 1.5 A system of 1 q1q 2 (a) three charges Thus, F12 = 2 ˆr12 (b) multiple charges. 11 4 πε0 r12 Reprint 2025-26 Physics In the same way, the force on q1 due to q3, denoted by F13, is given by 1 q1q 3 F13 = 2 rˆ13 4 πε0 r13 which again is the Coulomb force on q1 due to q3, even though other charge q2 is present. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as 1 q1q 2 1 q1q 3 F1 = F12 + F13 = rˆ12 + rˆ13 (1.4) 4 π ε0 r122 4 πε0 r132 The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.5(b). The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n: i.e., 1  q1q 2 q1q 3 q1q n  F1 = F12 + F13 + ...+ F1n =  2 rˆ12 + 2 rˆ13 + ... + 2 1rˆ n  4 πε0  r12 r13 r1n  q1 n q i = ˆr1i 2 (1.5) 4πε0 =∑i 2 r1 i The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle. Example 1.5 Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.6? FIGURE 1.6 1.5 Solution In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O EXAMPLE 12 from A is (2/3) AD = (1/ 3 ) l. By symmatry AO = BO = CO. Reprint 2025-26 Electric Charges and Fields Thus, 3 Qq Force F1 on Q due to charge q at A = 2 along AO 4 πε0 l 3 Qq 2 along BOForce F2 on Q due to charge q at B = 4 πε0 l 3 Qq 2 along COForce F3 on Q due to charge q at C = 4 πε0 l 3 Qq 2 along OA, by theThe resultant of forces F2 and F3 is 4 πε0 l 3 Qq parallelogram law. Therefore, the total force on Q = 2 ( rˆ − rˆ ) 4 πε0 l = 0, where ˆr is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. EXAMPLE Consider what would happen if the system was rotated through 60° about O. 1.5 Example 1.6 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge? FIGURE 1.7 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.7. By the parallelogram law, the total force F1 on the charge q at A is given by F1 = F 1ˆr where 1ˆr is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the q 2 same magnitude F = 4 π ε0 l 2 EXAMPLE The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a unit vector along AC. 1.6 13 Reprint 2025-26 Physics Similarly the total force on charge –q at C is F3 = 3 F ˆn , where ˆn is the unit vector along the direction bisecting the ÐBCA. It is interesting to see that the sum of the forces on the three charges 1.6 is zero, i.e., F1 + F2 + F3 = 0 The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof EXAMPLE is left to you as an exercise. 1.7 ELECTRIC FIELD Let us consider a point charge Q placed in vacuum, at the origin O. If we place another point charge q at a point P, where OP = r, then the charge Q will exert a force on q as per Coulomb’s law. We may ask the question: If charge q is removed, then what is left in the surrounding? Is there nothing? If there is nothing at the point P, then how does a force act when we place the charge q at P. In order to answer such questions, the early scientists introduced the concept of field. According to this, we say that the charge Q produces an electric field everywhere in the surrounding. When another charge q is brought at some point P, the field there acts on it and produces a force. The electric field produced by the charge Q at a point r is given as 1 Q 1 Q rˆ rˆ = E ( r ) = (1.6) 4 πε0 r 2 4 πε0 r 2 where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6) specifies the value of the electric field for each value of the position vector r. The word “field” signifies how some distributed quantity (which could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. We obtain the force F exerted by a charge Q on a charge q, as 1 Qq F = 2 rˆ (1.7) 4 πε0 r Note that the charge q also exerts an equal and opposite force on the charge Q. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. If we denote the position of charge q by the vector r, it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. Thus, F(r) = q E(r) (1.8) Equation (1.8) defines the SI unit of electric field as N/C*. Some important remarks may be made here: (i) From Eq. (1.8), we can infer that if q is unity, the electric field due to FIGURE 1.8 Electric a charge Q is numerically equal to the force exerted by it. Thus, the field (a) due to a electric field due to a charge Q at a point in space may be defined charge Q, (b) due to a as the force that a unit positive charge would experience if placed charge –Q. 14 * An alternate unit V/m will be introduced in the next chapter. Reprint 2025-26 Electric Charges and Fields at that point. The charge Q, which is producing the electric field, is called a source charge and the charge q, which tests the effect of a source charge, is called a test charge. Note that the source charge Q must remain at its original location. However, if a charge q is brought at any point around Q, Q itself is bound to experience an electrical force due to q and will tend to move. A way out of this difficulty is to make q negligibly small. The force F is then negligibly small but the ratio F/q is finite and defines the electric field:  F  E = lim (1.9) q → 0  q  A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice. When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1.14), the charges on the sheet are held to their locations by the forces due to the unspecified charged constituents inside the sheet. (ii) Note that the electric field E due to Q, though defined operationally in terms of some test charge q, is independent of q. This is because F is proportional to q, so the ratio F/q does not depend on q. The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. Thus, the electric field E due to Q is also dependent on the space coordinate r. For different positions of the charge q all over the space, we get different values of electric field E. The field exists at every point in three-dimensional space. (iii) For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is negative, the electric field vector, at each point, points radially inwards. (iv) Since the magnitude of the force F on charge q due to charge Q depends only on the distance r of the charge q from charge Q, the magnitude of the electric field E will also depend only on the distance r. Thus at equal distances from the charge Q, the magnitude of its electric field E is same. The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge at its centre; in other words, it has a spherical symmetry. 1.7.1 Electric field due to a system of charges Consider a system of charges q1, q2, ..., qn with position vectors r1, r2, ..., rn relative to some origin O. Like the electric field at a point in space due to a single charge, electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point, without disturbing the original positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by 15position vector r. Reprint 2025-26 Physics Electric field E1 at r due to q1 at r1 is given by 1 q1 E1 = 2 ˆr1P 4 πε0 r1 P where ˆr1P is a unit vector in the direction from q1 to P, and r1P is the distance between q1 and P. In the same manner, electric field E2 at r due to q2 at r2 is 1 q 2 E2 = 2 ˆr2P 4 πε0 r2 P where ˆr2P is a unit vector in the direction from q2 to P FIGURE 1.9 Electric field at a point and r2P is the distance between q2 and P. Similar due to a system of charges is the expressions hold good for fields E3, E4, ..., En due to vector sum of the electric fields at charges q3, q4, ..., qn. the point due to individual charges. By the superposition principle, the electric field E at r due to the system of charges is (as shown in Fig. 1.9) E(r) = E1 (r) + E2 (r) + … + En(r) 1 q1 1 q 2 1 q n = 2 rˆ1P + 2 rˆ2 P + ... + 2 rˆn P 4 πε0 r1P 4 πε0 r2 P 4 πε0 rn P 1 n q i E(r) = 2 ˆri P (1.10) 4 πε0 =∑i 1 ri P E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges. 1.7.2 Physical significance of electric field You may wonder why the notion of electric field has been introduced here at all. After all, for any system of charges, the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Eq. (1.5)]. Why then introduce this intermediate quantity called the electric field? For electrostatics, the concept of electric field is convenient, but not really necessary. Electric field is an elegant way of characterising the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. Electric field is a vector field, since force is a vector quantity. The true physical significance of the concept of electric field, however, emerges only when we go beyond electrostatics and deal with time- dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. Now the greatest speed with which a signal or information can go from one point to another 16 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot Reprint 2025-26 Electric Charges and Fields arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus, a source of time- dependent electromagnetic fields, turned on for a short interval of time and then switched off, leaves behind propagating electromagnetic fields transporting energy. The concept of field was first introduced by Faraday and is now among the central concepts in physics. Example 1.7 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’. FIGURE 1.10 Solution In Fig. 1.10(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae = eE/me where me is the mass of the electron. Starting from rest, the time required by the electron to fall through a 2 h 2 h m e distance h is given by t e = = a e e E For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg, E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m, te = 2.9 × 10–9s In Fig. 1.10 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is ap = eE/mp EXAMPLE where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of fall for the proton is 1.7 17 Reprint 2025-26 Physics 2 h 2 h m p –7 t p = = = 1. 3 × 10 s a p e E Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field: e E a p = m p (1 . 6 × 10 − 19 C) × (2. 0 × 10 4 N C −1 ) = −27 1 .67 × 10 kg 12 –2 1.7 = 1 .9 × 10 m s which is enormous compared to the value of g (9.8 m s–2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored EXAMPLE in this example. Example 1.8 Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11. FIGURE 1.11 Solution The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude (9 × 10 9 Nm 2 C -2 ) × (10 −8 C) E1A = 2 = 3.6 × 104 N C–1 (0.05m) 1.8 The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7.2 × 104 N C–1 EXAMPLE EA is directed toward the right.18 Reprint 2025-26 Electric Charges and Fields The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E1B = 2 = 3.6 × 104 N C–1 (0.05 m) The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude (9 × 10 9 Nm 2 C –2 ) × (10 −8 C) E 2B = 2 = 4 × 103 N C–1 (0.15 m) The magnitude of the total electric field at B is EB = E1B – E2B = 3.2 × 104 N C–1 EB is directed towards the left. The magnitude of each electric field vector at point C, due to charge q1 and q2 is (9 × 10 9 Nm 2 C –2 ) × (10 − 8 C) E1C = E 2C = 2 = 9 × 103 N C–1 (0.10 m) The directions in which these two vectors point are indicated in Fig. 1.11. The resultant of these two vectors is π π EXAMPLE E C = E1c cos + E 2 c cos = 9 × 103 N C–1 3 3 1.8 EC points towards the right.