Gravitation PDF

CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything 7.1 Introd...

CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything 7.1 Introduction thrown up falls down towards the earth, going uphill is lot 7.2 Kepler’s laws more tiring than going downhill, raindrops from the clouds 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 7.4 The gravitational constant (1564-1642) who recognised the fact that all bodies, 7.5 Acceleration due to irrespective of their masses, are accelerated towards the earth gravity of the earth with a constant acceleration. It is said that he made a public 7.6 Acceleration due to demonstration of this fact. To find the truth, he certainly 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) 2024-25 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 laws. These laws were known to Newton and the sum of the distances from F1 and F2 is a enabled him to make a great scientific leap in constant. F1, F 2 are called the focii. Join the proposing his universal law of gravitation. points F 1 and F 2 and extend the line to 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 slower when they are farther from the sun P S S' A 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. 2024-25 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 out by the planet. ⊳ Example 7.1 Let the speed of 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 ≡ 10 Semi-major axis in units of 10 m. momentum at P is Lp = mp rp vp, since inspection T ≡ Time period of revolution of the planet tells us that r p and v p are mutually in years(y). perpendicular. Similarly, LA = mp rA vA. From Q ≡ The quotient ( T2/a3 ) in units of angular momentum conservation 10 -34 y2 m-3.) mp rp vp = mp rA vA Planet a T Q vp rA or = ⊳ Mercury 5.79 0.24 2.95 vA rp Venus 10.8 0.615 3.00 Earth 15.0 1 2.96 Since rA > rp, vp > vA. 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 The law of areas can be understood as a Legend has it that observing an apple falling from consequence of conservation of angular a tree, Newton was inspired to arrive at an momentum whch is valid for any central universal law of gravitation that led to an force. A central force is such that the force explanation of terrestrial gravitation as well as on the planet is along the vector joining the of Kepler’s laws. Newton’s reasoning was that Sun and the planet. Let the Sun be at the the moon revolving in an orbit of radius Rm was origin and let the position and momentum subject to a centripetal acceleration due to of the planet be denoted by r and p earth’s gravity of magnitude respectively. Then the area swept out by the V2 4π 2 Rm planet of mass m in time interval ∆ t is (Fig. am = = (7.3) Rm T2 7.2) ∆ A given by ∆A = ½ (r × v∆t) (7.1) where V is the speed of the moon related to the Hence time period T by the relation V = 2π Rm / T. The ∆A /∆t =½ (r × p)/m, (since v = p/m) = L / (2 m) (7.2) time period T is about 27.3 days and Rm was where v is the velocity, L is the angular already known then to be about 3.84 × 108m. If momentum equal to ( r × p). For a central we substitute these numbers in Eq. (7.3), we get force, which is directed along r, L is a constant a value of am much smaller than the value of acceleration due to gravity g on the surface of as 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. 2024-25 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 1 square 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 Rm −2 ; g α R− E 2 and we get as F12 = – F21. Before we can apply Eq. (7.5) to objects under g R2 = m2 3600 (7.4) consideration, we have to be careful since the am RE 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 m1 m 2 |F | = G (7.5) r2 Equation (7.5) can be expressed in vector form as F= G m1 m 2 r 2 ( ) – rɵ = – G m1 m 2 ɵ r2 r Fig. 7.4 Gravitational force on point mass m1 is the 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 m1 ɵ Gm 3 m1 ɵ Gm 4 m1 ɵ F1 = r 21 + r 31 + r 41 rɵ is the unit vector from m1 to m2 and r = r2 – r1 2 r21 2 r31 2 r41 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 2024-25 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 ) ˆ spherical shell of uniform density, on a FGB = 1 ( −i cos 30ο − ˆj sin 30ο ) point mass situated inside it is zero. Qualitatively, we can again understand this Gm (2m ) ˆ FGC = 1 ( + i cos 30ο − ˆj sin 30ο ) result. Various regions of the spherical shell 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. 2024-25 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 θ is 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 balls is the same as if their masses are Mr 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 (7.6) d2 concentric shells as before and a point mass m If L is the length of the bar AB , then the situated at a distance r from the centre. The torque arising out of F is F multiplied by L. At point P lies outside the sphere of radius r. For equilibrium, this is equal to the restoring torque the shells of radius greater than r, the point P and hence lies inside. Hence according to result stated in the last section, they exert no gravitational force Mm on mass m kept at P. The shells with radius ≤ r G L =τ θ (7.7) d2 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 = (7.9) r2 We assume that the entire earth is of uniform 7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH 4π 3 density and hence its mass is M E = RE ρ 3 The earth can be imagined to be a sphere made where ME is the mass of the earth RE is its radius of a large number of concentric spherical shells and ρ is the density. On the other hand the with the smallest one at the centre and the largest one at its surface. A point outside the 4π mass of the sphere Mr of radius r is ρ r 3 and earth is obviously outside all the shells. Thus, 3 2024-25 GRAVITATION 133 hence its distance from the centre of the earth is (R E + h ). If F (h) denoted the magnitude of the force on the point mass m , we get from G m ME Eq. (7.5) : = r (7.10) RE 3 If the mass m is situated on the surface of GM E m earth, then r = RE and the gravitational force on F (h ) = (7.13) (R E + h )2 it is, from Eq. (7.10) M Em The acceleration experienced by the point F =G (7.11) mass is F (h )/ m ≡ g(h ) and we get R E2 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 F GM E surface of earth : g = R 2. For h m, the total energy of the system is GMm 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 GMm K= 2a GMm V =− a 8. The escape speed from the surface of the earth is 2 G ME ve = = 2 gRE RE 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. 2024-25 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 m1 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 m1 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 ? 2024-25 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/r 2 – 1/r 1 ) 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 ? 2024-25 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 ? 2024-25