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

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

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