9.18 For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. Therefore, fmax = 0.75 m.

9.6 REFRACTION THROUGH A PRISM Figure 9.21 shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and r1, while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation, d. In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°. FIGURE 9.21 A ray of light passing through a triangular glass prism. ÐA + ÐQNR = 180° From the triangle QNR, r1 + r2 + ÐQNR = 180° Comparing these two equations, we get r1 + r2 = A (9.34) The total deviation d is the sum of deviations at the two faces, d = (i – r1 ) + (e – r2 ) that is, d = i + e – A (9.35) Thus, the angle of deviation depends on the angle of incidence. A plot between the angle of deviation and angle of incidence is shown in Fig. 9.22. You can see that, in general, any given value of d, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in Eq. (9.35), i.e., d remains the same if i 239 Reprint 2025-26 Physics and e are interchanged. Physically, this is related to the fact that the path of ray in Fig. 9.21 can be traced back, resulting in the same angle of deviation. At the minimum deviation Dm, the refracted ray inside the prism becomes parallel to its base. We have d = Dm, i = e which implies r1 = r2. Equation (9.34) gives A 2r = A or r = (9.36) 2 In the same way, Eq. (9.35) gives Dm = 2i – A, or i = (A + Dm)/2 (9.37) The refractive index of the prism is FIGURE 9.22 Plot of angle of deviation (d) n 2 sin[( A + D m )/2] versus angle of incidence (i) for a n 21 = = (9.38) triangular prism. n1 sin[ A /2] The angles A and Dm can be measured experimentally. Equation (9.38) thus provides a method of determining refractive index of the material of the prism. For a small angle prism, i.e., a thin prism, Dm is also very small, and we get sin[( A + Dm )/2] A + Dm ) /2 ≃ ( n 21 = sin[ A /2] A /2 Dm = (n21–1)A It implies that, thin prisms do not deviate light much. 9.7 OPTICAL INSTRUMENTS A number of optical devices and instruments have been designed utilising reflecting and refracting properties of mirrors, lenses and prisms. Periscope, kaleidoscope, binoculars, telescopes, microscopes are some examples of optical devices and instruments that are in common use. Our eye is, of course, one of the most important optical device the nature has endowed us with. We have already studied about the human eye in Class X. We now go on to describe the principles of working of the microscope and the telescope. 9.7.1 The microscope A simple magnifier or microscope is a converging lens of small focal length (Fig. 9.23). In order to use such a lens as a microscope, the lens is held near the object, one focal length away or less, and the eye is positioned close to the lens on the other side. The idea is to get an erect, magnified and virtual image of the object at a distance so that it can be viewed comfortably, i.e., at 25 cm or more. If the object is at a distance f, the 240 image is at infinity. However, if the object is at a distance slightly less Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.23 A simple microscope; (a) the magnifying lens is located such that the image is at the near point, (b) the angle subtanded by the object, is the same as that at the near point, and (c) the object near the focal point of the lens; the image is far off but closer than infinity. than the focal length of the lens, the image is virtual and closer than infinity. Although the closest comfortable distance for viewing the image is when it is at the near point (distance D @ 25 cm), it causes some strain on the eye. Therefore, the image formed at infinity is often considered most suitable for viewing by the relaxed eye. We show both cases, the first in Fig. 9.23(a), and the second in Fig. 9.23(b) and (c). The linear magnification m, for the image formed at the near point D, by a simple microscope can be obtained by using the relation 241 Reprint 2025-26 Physics v  1 1   v  m = = v – 1 – u  v f =  f  Now according to our sign convention, v is negative, and is equal in magnitude to D. Thus, the magnification is  D  m = 1 + (9.39)  f  Since D is about 25 cm, to have a magnification of six, one needs a convex lens of focal length, f = 5 cm. Note that m = h¢/h where h is the size of the object and h¢ the size of the image. This is also the ratio of the angle subtended by the image to that subtended by the object, if placed at D for comfortable viewing. (Note that this is not the angle actually subtended by the object at the eye, which is h/u.) What a single-lens simple magnifier achieves is that it allows the object to be brought closer to the eye than D. We will now find the magnification when the image is at infinity. In this case we will have to obtained the angular magnification. Suppose the object has a height h. The maximum angle it can subtend, and be clearly visible (without a lens), is when it is at the near point, i.e., a distance D. The angle subtended is then given by  h  tan θo =  D » qo (9.40) We now find the angle subtended at the eye by the image when the object is at u. From the relations h ′ v = m = h u we have the angle subtended by the image h ′ h v h tan θi = = ⋅ = »q. The angle subtended by the object, when it −v −v u −u is at u = –f.  h  θi = (9.41)  f  as is clear from Fig. 9.23(c). The angular magnification is, therefore  θi  D m = (9.42)  θo = f This is one less than the magnification when the image is at the near point, Eq. (9.39), but the viewing is more comfortable and the difference in magnification is usually small. In subsequent discussions of optical instruments (microscope and telescope) we shall assume the image to be 242 at infinity. Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.24 Ray diagram for the formation of image by a compound microscope. The A simple microscope has a limited maximum magnification (£ 9) for realistic focal lengths. For much larger magnifications, one uses two lenses, one compounding the effect of the other. This is known as a compound world’s microscope. A schematic diagram of a compound microscope is shown in Fig. 9.24. The lens nearest the object, called the objective, forms a largestreal, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged optical and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly,the final image is inverted with respect to the original object. http://astro.nineplanets.org/bigeyes.html telescopes We now obtain the magnification due to a compound microscope. The ray diagram of Fig. 9.24 shows that the (linear) magnification due to the objective, namely h¢/h, equals h ′ L m O = = (9.43) h f o where we have used the result  h   h ′  tanβ =  f o =  L  Here h¢ is the size of the first image, the object size being h and fo being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope. 243 Reprint 2025-26 Physics As the first inverted image is near the focal point of the eyepiece, we use the result from the discussion above for the simple microscope to obtain the (angular) magnification me due to it [Eq. (9.39)], when the final image is formed at the near point, is  D  m e =  1 +  [9.44(a)]  f e  When the final image is formed at infinity, the angular magnification due to the eyepiece [Eq. (9.42)] is me = (D/fe ) [9.44(b)] Thus, the total magnification [(according to Eq. (9.33)], when the image is formed at infinity, is L   D  m = m om e =    (9.45)  f o   f e  Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths. In practice, it is difficult to make the focal length much smaller than 1 cm. Also large lenses are required to make L large. For example, with an objective with fo = 1.0 cm, and an eyepiece with focal length fe = 2.0 cm, and a tube length of 20 cm, the magnification is L   D  m = m o m e =     f o   f e  20 25 250 1 2 Various other factors such as illumination of the object, contribute to the quality and visibility of the image. In modern microscopes, multi- component lenses are used for both the objective and the eyepiece to improve image quality by minimising various optical aberrations (defects) in lenses. 9.7.2 Telescope The telescope is used to provide angular magnification of distant objects (Fig. 9.25). It also has an objective and an eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image. The magnifying power m is the ratio of the angle b subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye. Hence h f o f o m . (9.46) 244 f e h f e Reprint 2025-26 Ray Optics and Optical Instruments In this case, the length of the telescope tube is fo + fe. Terrestrial telescopes have, in addition, a pair of inverting lenses to make the final image erect. Refracting telescopes can be used both for terrestrial and astronomical observations. For example, consider a telescope whose objective has a focal length of 100 cm and the eyepiece a focal length of 1 cm. The magnifying power of this telescope is m = 100/1 = 100. Let us consider a pair of stars of actual separation 1¢ (one minute of arc). The stars appear as though they are separated by an angle of 100 × 1¢ = 100¢ =1.67°. FIGURE 9.25 A refracting telescope. The main considerations with an astronomical telescope are its light gathering power and its resolution or resolving power. The former clearly depends on the area of the objective. With larger diameters, fainter objects can be observed. The resolving power, or the ability to observe two objects distinctly, which are in very nearly the same direction, also depends on the diameter of the objective. So, the desirable aim in optical telescopes is to make them with objective of large diameter. The largest lens objective in use has a diameter of 40 inch (~1.02 m). It is at the Yerkes Observatory in Wisconsin, USA. Such big lenses tend to be very heavy and therefore, difficult to make and support by their edges. Further, it is rather difficult and expensive to make such large sized lenses which form images that are free from any kind of chromatic aberration and distortions. For these reasons, modern telescopes use a concave mirror rather than a lens for the objective. Telescopes with mirror objectives are called reflecting telescopes. There is no chromatic aberration in a mirror. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality, and can be supported over its entire back surface, not just over its rim. One obvious problem with a reflecting telescope is that the objective mirror focusses light inside 245 Reprint 2025-26 Physics FIGURE 9.26 Schematic diagram of a reflecting telescope (Cassegrain). the telescope tube. One must have an eyepiece and the observer right there, obstructing some light (depending on the size of the observer cage). This is what is done in the very large 200 inch (~5.08 m) diameters, Mt. Palomar telescope, California. The viewer sits near the focal point of the mirror, in a small cage. Another solution to the problem is to deflect the light being focussed by another mirror. One such arrangement using a convex secondary mirror to focus the incident light, which now passes through a hole in the objective primary mirror, is shown in Fig. 9.26. This is known as a Cassegrain telescope, after its inventor. It has the advantages of a large focal length in a short telescope. The largest telescope in India is in Kavalur, Tamil Nadu. It is a 2.34 m diameter reflecting telescope (Cassegrain). It was ground, polished, set up, and is being used by the Indian Institute of Astrophysics, Bangalore. The largest reflecting telescopes in the world are the pair of Keck telescopes in Hawaii, USA, with a reflector of 10 metre in diameter. SUMMARY 1. Reflection is governed by the equation Ði = Ðr¢ and refraction by the Snell’s law, sini/sinr = n, where the incident ray, reflected ray, refracted ray and normal lie in the same plane. Angles of incidence, reflection and refraction are i, r ¢ and r, respectively. 2. The critical angle of incidence ic for a ray incident from a denser to rarer medium, is that angle for which the angle of refraction is 90°. For i > ic, total internal reflection occurs. Multiple internal reflections in diamond (ic @ 24.4°), totally reflecting prisms and mirage, are some examples of total internal reflection. Optical fibres consist of glass fibres coated with a thin layer of material of lower refractive index. Light incident at an angle at one end comes out at the other, after multiple internal reflections, even if the fibre is bent. Reprint 2025-26 Ray Optics and Optical Instruments 3. Cartesian sign convention: Distances measured in the same direction as the incident light are positive; those measured in the opposite direction are negative. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis. The heights measured upwards above x-axis and normal to the principal axis of the mirror/ lens are taken as positive. The heights measured downwards are taken as negative. 4. Mirror equation: 1 1 1 + = v u f where u and v are object and image distances, respectively and f is the focal length of the mirror. f is (approximately) half the radius of curvature R. f is negative for concave mirror; f is positive for a convex mirror. 5. For a prism of the angle A, of refractive index n 2 placed in a medium of refractive index n1, n 2 sin ( A + D m ) / 2  n 21 = = n 1 sin ( A / 2 ) where Dm is the angle of minimum deviation. 6. For refraction through a spherical interface (from medium 1 to 2 of refractive index n1 and n 2, respectively) n 2 n 1 n 2 − n 1 − = v u R Thin lens formula 1 1 1 − = v u f Lens maker’s formula 1 ( n 2 − n1 )  1 1  = − f n1  R1 R 2  R1 and R2 are the radii of curvature of the lens surfaces. f is positive for a converging lens; f is negative for a diverging lens. The power of a lens P = 1/f. The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1. If several thin lenses of focal length f1, f2, f3,.. are in contact, the effective focal length of their combination, is given by 1 1 1 1 = + + + … f f 1 f 2 f 3 The total power of a combination of several lenses is P = P1 + P2 + P3 + … 7. Dispersion is the splitting of light into its constituent colour. 247 Reprint 2025-26 Physics 8. Magnifying power m of a simple microscope is given by m = 1 + (D/f), where D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens. If the image is at infinity, m = D/f. For a compound microscope, the magnifying power is given by m = me × m0 where me = 1 + (D/fe), is the magnification due to the eyepiece and mo is the magnification produced by the objective. Approximately, L D m = × f o f e where fo and fe are the focal lengths of the objective and eyepiece, respectively, and L is the distance between their focal points. 9. Magnifying power m of a telescope is the ratio of the angle b subtended at the eye by the image to the angle a subtended at the eye by the object. β f o m = = α f e where f0 and fe are the focal lengths of the objective and eyepiece, respectively. POINTS TO PONDER 1. The laws of reflection and refraction are true for all surfaces and pairs of media at the point of the incidence. 2. The real image of an object placed between f and 2f from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? This question puzzles many, because it is difficult to reconcile ourselves with an image suspended in air without a screen. But the image does exist. Rays from a given point on the object are converging to an image point in space and diverging away. The screen simply diffuses these rays, some of which reach our eye and we see the image. This can be seen by the images formed in air during a laser show. 3. Image formation needs regular reflection/refraction. In principle, all rays from a given point should reach the same image point. This is why you do not see your image by an irregular reflecting object, say the page of a book. 4. Thick lenses give coloured images due to dispersion. The variety in colour of objects we see around us is due to the constituent colours of the light incident on them. A monochromatic light may produce an entirely different perception about the colours on an object as seen in white light. 5. For a simple microscope, the angular size of the object equals the angular size of the image. Yet it offers magnification because we can keep the small object much closer to the eye than 25 cm and hence have it subtend a large angle. The image is at 25 cm which we can see. Without the microscope, you would need to keep the small object at 25 cm which would subtend a very small angle. Reprint 2025-26 Ray Optics and Optical Instruments EXERCISES

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