7.3 Universal law of above fall towards the earth and there are many other such gravitation phenomena. Historically it was the Italian Physicist Galileo

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

7.4 The gravitational (1564-1642) who recognised the fact that all bodies, constant irrespective of their masses, are accelerated towards the earth7.5 Acceleration due to gravity of the earth with a constant acceleration. It is said that he made a public demonstration of this fact. To find the truth, he certainly7.6 Acceleration due to gravity below and above did experiments with bodies rolling down inclined planes and the surface of earth arrived at a value of the acceleration due to gravity which is 7.7 Gravitational potential close to the more accurate value obtained later. energy A seemingly unrelated phenomenon, observation of stars, 7.8 Escape speed planets and their motion has been the subject of attention 7.9 Earth satellites in many countries since the earliest of times. Observations 7.10 Energy of an orbiting since early times recognised stars which appeared in the satellite sky with positions unchanged year after year. The more Summary interesting objects are the planets which seem to have regular Points to ponder motions against the background of stars. The earliest Exercises recorded model for planetary motions proposed by Ptolemy about 2000 years ago was a ‘geocentric’ model in which all celestial objects, stars, the sun and the planets, all revolved around the earth. The only motion that was thought to be possible for celestial objects was motion in a circle. Complicated schemes of motion were put forward by Ptolemy in order to describe the observed motion of the planets. The planets were described as moving in circles with the centre of the circles themselves moving in larger circles. Similar theories were also advanced by Indian astronomers some 400 years later. However a more elegant model in which the Sun was the centre around which the planets revolved – the ‘heliocentric’ model – was already mentioned by Aryabhatta (5th century A.D.) in his treatise. A thousand years later, a Polish monk named Nicolas Copernicus (1473-1543) Reprint 2025-26 128 PHYSICS proposed a definitive model in which the planets of the ellipse (Fig. 7.1a). This law was a moved in circles around a fixed central sun. His deviation from the Copernican model which theory was discredited by the church, but allowed only circular orbits. The ellipse, of notable amongst its supporters was Galileo who which the circle is a special case, is a closed had to face prosecution from the state for his curve which can be drawn very simply as beliefs. follows. It was around the same time as Galileo, a Select two points F1 and F2. Take a length nobleman called Tycho Brahe (1546-1601) of a string and fix its ends at F1 and F2 by hailing from Denmark, spent his entire lifetime pins. With the tip of a pencil stretch the string recording observations of the planets with the taut and then draw a curve by moving the naked eye. His compiled data were analysed pencil keeping the string taut throughout.(Fig. later by his assistant Johannes Kepler (1571- 7.1(b)) The closed curve you get is called an 1640). He could extract from the data three ellipse. Clearly for any point T on the ellipse, elegant laws that now go by the name of Kepler’s the sum of the distances from F1 and F2 is alaws. These laws were known to Newton and constant. F1, F2 are called the focii. Join theenabled him to make a great scientific leap in points F1 and F2 and extend the line toproposing his universal law of gravitation. intersect the ellipse at points P and A as shown 7.2 KEPLER’S LAWS in Fig. 7.1(b). The midpoint of the line PA is The three laws of Kepler can be stated as the centre of the ellipse O and the length PO = follows: AO is called the semi-major axis of the ellipse. 1. Law of orbits : All planets move in elliptical For a circle, the two focii merge into one and orbits with the Sun situated at one of the foci the semi-major axis becomes the radius of the circle. B 2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time (Fig. 7.2). This law comes from the observations that planets appear to move 2b P S S' A slower when they are farther from the sun than when they are nearer. C 2a Fig. 7.1(a) An ellipse traced out by a planet around the sun. The closest point is P and the farthest point is A, P is called the perihelion and A the aphelion. The semimajor axis is half the distance AP. Fig. 7.2 The planet P moves around the sun in an elliptical orbit. The shaded area is the area Fig. 7.1(b) Drawing an ellipse. A string has its ends ∆A swept out in a small interval of time ∆t. fixed at F1 and F2. The tip of a pencil holds the string taut and is moved around. Reprint 2025-26 GRAVITATION 129 3. Law of periods : The square of the time period the law of areas. Gravitation is a central force of revolution of a planet is proportional to the and hence the law of areas follows. cube of the semi-major axis of the ellipse traced ⊳ Example 7.1 Let the speed of the planetout by the planet. at the perihelion P in Fig. 7.1(a) be vP and Table 7.1 gives the approximate time periods the Sun-planet distance SP be rP. Relate of revolution of eight* planets around the sun {rP, vP} to the corresponding quantities at along with values of their semi-major axes. the aphelion {rA, vA}. Will the planet take Table 7.1 Data from measurement of equal times to traverse BAC and CPB ? planetary motions given below confirm Kepler’s Law of Periods Answer The magnitude of the angular (a ≡ Semi-major axis in units of 1010 m. momentum at P is Lp = mp rp vp, since inspection T ≡ Time period of revolution of the planet tells us that rp and v p are mutually in years(y). perpendicular. Similarly, LA = mp rA vA. From Q ≡ The quotient ( T2/a3 ) in units of -34 angular momentum conservation 10 y2 m-3.) mp rp vp = mp rA vA Planet a T Q v p r A = Mercury 5.79 0.24 2.95 or v A r p ⊳ Venus 10.8 0.615 3.00 Since rA > rp, vp > vA . Earth 15.0 1 2.96 Mars 22.8 1.88 2.98 The area SBAC bounded by the ellipse and Jupiter 77.8 11.9 3.01 the radius vectors SB and SC is larger than SBPC Saturn 143 29.5 2.98 in Fig. 7.1. From Kepler’s second law, equal areas Uranus 287 84 2.98 are swept in equal times. Hence the planet will Neptune 450 165 2.99 take a longer time to traverse BAC than CPB. 7.3 UNIVERSAL LAW OF GRAVITATION Legend has it that observing an apple falling from The law of areas can be understood as a a tree, Newton was inspired to arrive at anconsequence of conservation of angular universal law of gravitation that led to anmomentum whch is valid for any central explanation of terrestrial gravitation as well asforce . A central force is such that the force of Kepler’s laws. Newton’s reasoning was thaton the planet is along the vector joining the the moon revolving in an orbit of radius Rm wasSun and the planet. Let the Sun be at the subject to a centripetal acceleration due toorigin and let the position and momentum earth’s gravity of magnitudeof the planet be denoted by r and p respectively. Then the area swept out by the 2 2 V 4π R m =planet of mass m in time interval ∆t is (Fig. a m = T 2 (7.3) R m 7.2) ∆A given by ∆A = ½ (r × v∆t) (7.1) where V is the speed of the moon related to the Hence m / T . The ∆A /∆t =½ (r × p)/m, (since v = p/m) time period T by the relation V = 2πR time period T is about 27.3 days and Rm was = L / (2 m) (7.2) already known then to be about 3.84 × 108m. If where v is the velocity, L is the angular momentum equal to ( r × p). For a central we substitute these numbers in Eq. (7.3), we get a value of am much smaller than the value offorce, which is directed along r, L is a constant acceleration due to gravity g on the surface ofas the planet goes around. Hence, ∆A /∆t is a the earth, arising also due to earth’s gravitational constant according to the last equation. This is attraction. Reprint 2025-26 130 PHYSICS This clearly shows that the force due to The gravitational force is attractive, i.e., the earth’s gravity decreases with distance. If one force F is along – r. The force on point mass m1 assumes that the gravitational force due to the due to m2 is of course – F by Newton’s third law. earth decreases in proportion to the inverse Thus, the gravitational force F12 on the body 1square of the distance from the centre of the due to 2 and F21 on the body 2 due to 1 are related earth, we will have am α R−;m2 g α R−E 2 and we get as F12 = – F21. 2 Before we can apply Eq. (7.5) to objects under g R m = 2 3600 (7.4) consideration, we have to be careful since the a m R E law refers to point masses whereas we deal with in agreement with a value of g 9.8 m s-2 and extended objects which have finite size. If we have the value of am from Eq. (7.3). These observations a collection of point masses, the force on any led Newton to propose the following Universal Law one of them is the vector sum of the gravitational of Gravitation : forces exerted by the other point masses as Every body in the universe attracts every other shown in Fig 7.4. body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short). Stated Mathematically, Newton’s gravitation law reads : The force F on a point mass m2 due to another point mass m1 has the magnitude m 1 m 2 | F | = G 2 (7.5) r Equation (7.5) can be expressed in vector form as m 1 m 2 ɵ m 1 m 2 ɵ – r r F = G = – G ) Fig. 7.4 Gravitational force on point mass m1 is the r 2 ( r 2 vector sum of the gravitational forces exerted m1 m 2 ɵ by m2, m3 and m4. = – G 3 r r The total force on m1 is where G is the universal gravitational constant, Gm 2 m 1 ɵ Gm 3 m 1 ɵ Gm 4 m 1 ɵ r 41 r 21 + r 31 + 1 = ɵr is the unit vector from m1 to m2 and r = r2 – r1 F r212 r412 r312 as shown in Fig. 7.3. ⊳ Example 7.2 Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass 2m placed at the centroid G of the triangle? (b) What is the force if the mass at the O vertex A is doubled ? Take AG = BG = CG = 1 m (see Fig. 7.5) Answer (a) The angle between GC and the positive x-axis is 30° and so is the angle between Fig. 7.3 Gravitational force on m1 due to m2 is along GB and the negative x-axis. The individual forces r where the vector r is (r2– r1). in vector notation are Reprint 2025-26 GRAVITATION 131 cases, a simple law results when you do that : (1) The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell. Qualitatively this can be understood as follows: Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components Fig. 7.5 Three equal masses are placed at the three prependicular to this line cancel out when vertices of the ∆ ABC. A mass 2m is placed summing over all regions of the shell leaving at the centroid G. only a resultant force along the line joining the point to the centre. The magnitude of Gm ( 2m ) ˆ FGA = j this force works out to be as stated above. 1 (2) The force of attraction due to a hollow Gm ( 2m ) FGB = − ˆi cos 30ο − ˆj sin 30ο ( ) sphericalpoint massshellsituatedof uniforminside density,it is zero.on a 1 Qualitatively, we can again understand this Gm 2m ( ) FGC = + ˆi cos 30ο − ˆj sin 30ο result. Various regions of the spherical shell 1 ( ) attract the point mass inside it in various From the principle of superposition and the law directions. These forces cancel each other of vector addition, the resultant gravitational completely. force FR on (2m) is FR = FGA + FGB + FGC 7.4 THE GRAVITATIONAL CONSTANT FR = 2Gm 2 ˆj + 2Gm 2 (− ˆi cos 30ο− ˆj sin 30ο ) The value of the gravitational constant G entering the Universal law of gravitation can be + 2Gm 2 ( ˆi cos 30ο − ˆj sin 30ο ) = 0 determined experimentally and this was first done Alternatively, one expects on the basis of by English scientist Henry Cavendish in 1798. symmetry that the resultant force ought to be The apparatus used by him is schematically zero. shown in Fig.7.6 (b) Now if the mass at vertex A is doubled then ⊳ For the gravitational force between an extended Fig. 7.6 Schematic drawing of Cavendish’s object (like the earth) and a point mass, Eq. (7.5) is not experiment. S1 and S2 are large spheres directly applicable. Each point mass in the extended which are kept on either side (shown object will exert a force on the given point mass and shades) of the masses at A and B. When the big spheres are taken to the other side these force will not all be in the same direction. We of the masses (shown by dotted circles), have to add up these forces vectorially for all the point the bar AB rotates a little since the torque masses in the extended object to get the total force. reverses direction. The angle of rotation can This is easily done using calculus. For two special be measured experimentally. Reprint 2025-26 132 PHYSICS The bar AB has two small lead spheres all the shells exert a gravitational force at the attached at its ends. The bar is suspended from point outside just as if their masses are a rigid support by a fine wire. Two large lead concentrated at their common centre according spheres are brought close to the small ones but to the result stated in section 7.3. The total mass on opposite sides as shown. The big spheres of all the shells combined is just the mass of the attract the nearby small ones by equal and earth. Hence, at a point outside the earth, the opposite force as shown. There is no net force gravitational force is just as if its entire mass of on the bar but only a torque which is clearly the earth is concentrated at its centre. equal to F times the length of the bar,where F is For a point inside the earth, the situation the force of attraction between a big sphere and is different. This is illustrated in Fig. 7.7. its neighbouring small sphere. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque . If θis the angle of twist of the suspended wire, the restoring torque is proportional to θ, equal to τθ. Where τ is the restoring couple per unit angle of twist. τ can be measured independently e.g. by applying a known torque and measuring the angle of twist. The gravitational force between the spherical Mrballs is the same as if their masses are concentrated at their centres. Thus if d is the separation between the centres of the big and Fig. 7.7 The mass m is in a mine located at a depth its neighbouring small ball, M and m their d below the surface of the Earth of mass masses, the gravitational force between the big ME and radius RE. We treat the Earth to be sphere and its neighouring small ball is. spherically symmetric. Mm Again consider the earth to be made up of F = G 2 (7.6) d concentric shells as before and a point mass m situated at a distance r from the centre. The If L is the length of the bar AB , then the torque arising out of F is F multiplied by L. At point P lies outside the sphere of radius r. For the shells of radius greater than r, the point Pequilibrium, this is equal to the restoring torque lies inside. Hence according to result stated inand hence the last section, they exert no gravitational force Mm G 2 L = τθ (7.7) on mass m kept at P. The shells with radius ≤r d make up a sphere of radius r for which the point Observation of θ thus enables one to P lies on the surface. This smaller sphere calculate G from this equation. therefore exerts a force on a mass m at P as if Since Cavendish’s experiment, the its mass Mr is concentrated at the centre. Thus measurement of G has been refined and the the force on the mass m at P has a magnitude currently accepted value is Gm ( M r ) G = 6.67×10-11 N m2/kg2 (7.8) F = 2 (7.9) r We assume that the entire earth is of uniform