gavitation.pdf

CHAPTER EIGHT GRAVITATION 8.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything 8.1 I...

CHAPTER EIGHT GRAVITATION 8.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything 8.1 Introduction thrown up falls down towards the earth, going uphill is lot 8.2 Kepler’s laws more tiring than going downhill, raindrops from the clouds 8.3 Universal law of above fall towards the earth and there are many other such gravitation phenomena. Historically it was the Italian Physicist Galileo 8.4 The gravitational constant (1564-1642) who recognised the fact that all bodies, irrespective of their masses, are accelerated towards the earth 8.5 Acceleration due to gravity of the earth with a constant acceleration. It is said that he made a public 8.6 Acceleration due to demonstration of this fact. To find the truth, he certainly did gravity below and above experiments with bodies rolling down inclined planes and the surface of earth arrived at a value of the acceleration due to gravity which is 8.7 Gravitational potential close to the more accurate value obtained later. energy A seemingly unrelated phenomenon, observation of stars, 8.8 Escape speed planets and their motion has been the subject of attention in 8.9 Earth satellites many countries since the earliest of times. Observations since 8.10 Energy of an orbiting early times recognised stars which appeared in the sky with satellite positions unchanged year after year. The more interesting 8.11 Geostationary and polar objects are the planets which seem to have regular motions satellites against the background of stars. The earliest recorded model 8.12 Weightlessness for planetary motions proposed by Ptolemy about 2000 years Summary ago was a ‘geocentric’ model in which all celestial objects, Points to ponder stars, the sun and the planets, all revolved around the earth. Exercises The only motion that was thought to be possible for celestial Additional exercises 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 184 PHYSICS Copernicus (1473-1543) proposed a definitive of the ellipse (Fig. 8.1a). This law was a deviation model in which the planets moved in circles from the Copernican model which allowed only around a fixed central sun. His theory was circular orbits. The ellipse, of which the circle is discredited by the church, but notable amongst a special case, is a closed curve which can be its supporters was Galileo who had to face drawn very simply as follows. prosecution from the state for his beliefs. Select two points F1 and F2. Take a length It was around the same time as Galileo, a of a string and fix its ends at F1 and F2 by pins. nobleman called Tycho Brahe (1546-1601) With the tip of a pencil stretch the string taut hailing from Denmark, spent his entire lifetime and then draw a curve by moving the pencil recording observations of the planets with the keeping the string taut throughout.(Fig. 8.1(b)) naked eye. His compiled data were analysed The closed curve you get is called an ellipse. later by his assistant Johannes Kepler (1571- Clearly for any point T on the ellipse, the sum of 1640). He could extract from the data three the distances from F1 and F2 is a constant. F1, elegant laws that now go by the name of Kepler’s laws. These laws were known to Newton and F2 are called the focii. Join the points F1 and F2 enabled him to make a great scientific leap in and extend the line to intersect the ellipse at proposing his universal law of gravitation. points P and A as shown in Fig. 8.1(b). The midpoint of the line PA is the centre of the ellipse 8.2 KEPLER’S LAWS O and the length PO = AO is called the semi- The three laws of Kepler can be stated as follows: major axis of the ellipse. For a circle, the two 1. Law of orbits : All planets move in elliptical focii merge into one and the semi-major axis orbits with the Sun situated at one of the foci becomes the radius of the circle. 2. Law of areas : The line that joins any planet B to the sun sweeps equal areas in equal intervals of time (Fig. 8.2). This law comes from the observations that planets appear to move slower when they are farther from the sun than when 2b they are nearer. P S S' A C 2a Fig. 8.1(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. 8.2 The planet P moves around the sun in an elliptical orbit. The shaded area is the area ∆A swept out in a small interval of time ∆t. 3. Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet. Fig. 8.1(b) Drawing an ellipse. A string has its ends Table 8.1 gives the approximate time periods fixed at F1 and F2. The tip of a pencil holds of revolution of eight* planets around the sun the string taut and is moved around. along with values of their semi-major axes. * Refer to information given in the Box on Page 182 GRAVITATION 185 Table 8.1 Data from measurement of as the planet goes around. Hence, ∆ A /∆t is a planetary motions given below constant according to the last equation. This is confirm Kepler’s Law of Periods the law of areas. Gravitation is a central force (a ≡ Semi-major axis in units of 1010 m. T ≡ Time period of revolution of the planet and hence the law of areas follows. in years(y). Q ≡ The quotient ( T2/a3 ) in units of t Example 8.1 Let the speed of the planet 10 -34 y2 m-3.) at the perihelion P in Fig. 8.1(a) be vP and the Sun-planet distance SP be rP. Relate Planet a T Q {rP, vP} to the corresponding quantities at the aphelion {rA, vA}. Will the planet take Mercury 5.79 0.24 2.95 equal times to traverse BAC and CPB ? Venus 10.8 0.615 3.00 Earth 15.0 1 2.96 Answer The magnitude of the angular Mars 22.8 1.88 2.98 momentum at P is Lp = mp rp vp, since inspection Jupiter 77.8 11.9 3.01 tells us that r p and v p are mutually Saturn 143 29.5 2.98 perpendicular. Similarly, LA = mp r A vA. From Uranus 287 84 2.98 Neptune 450 165 2.99 angular momentum conservation Pluto* 590 248 2.99 mp rp vp = mp rA vA The law of areas can be understood as a vp rA consequence of conservation of angular = or vA rp t momentum whch is valid for any central force. A central force is such that the force on the Since rA > rp, vp > vA. planet is along the vector joining the sun and The area SBAC bounded by the ellipse and the planet. Let the sun be at the origin and let the radius vectors SB and SC is larger than SBPC the position and momentum of the planet be in Fig. 8.1. From Kepler’s second law, equal areas denoted by r and p respectively. Then the area are swept in equal times. Hence the planet will swept out by the planet of mass m in time take a longer time to traverse BAC than CPB. interval ∆t is (Fig. 8.2) ∆A given by 8.3 UNIVERSAL LAW OF GRAVITATION ∆A = ½ (r × v∆t) (8.1) Legend has it that observing an apple falling Hence from a tree, Newton was inspired to arrive at an ∆A /∆t =½ (r × p)/m, (since v = p/m) universal law of gravitation that led to an = L / (2 m) (8.2) explanation of terrestrial gravitation as well as where v is the velocity, L is the angular of Kepler’s laws. Newton’s reasoning was that momentum equal to ( r × p ). For a central the moon revolving in an orbit of radius Rm was force, which is directed along r, L is a constant subject to a centripetal acceleration due to earth’s gravity of magnitude Johannes Kepler (1571–1630) was a V2 4π 2 Rm scientist of Ger man am = = (8.3) origin. He formulated Rm T2 the three laws of where V is the speed of the moon related to the planetary motion based on the painstaking time period T by the relation V = 2π Rm / T. The observations of Tycho time period T is about 27.3 days and Rm was Brahe and coworkers. Kepler himself was an already known then to be about 3.84 × 108m. If assistant to Brahe and it took him sixteen long we substitute these numbers in Eq. (8.3), we years to arrive at the three planetary laws. He get a value of am much smaller than the value of is also known as the founder of geometrical acceleration due to gravity g on the surface of optics, being the first to describe what happens to light after it enters a telescope. the earth, arising also due to earth’s gravitational attraction. * Refer to information given in the Box on Page 182 186 PHYSICS Central Forces We know the time rate of change of the angular momentum of a single particle about the origin is dl =r ×F dt The angular momentum of the particle is conserved, if the torque τ = r × F due to the force F on it vanishes. This happens either when F is zero or when F is along r. We are interested in forces which satisfy the latter condition. Central forces satisfy this condition. A ‘central’ force is always directed towards or away from a fixed point, i.e., along the position vector of the point of application of the force with respect to the fixed point. (See Figure below.) Further, the magnitude of a central force F depends on r, the distance of the point of application of the force from the fixed point; F = F(r). In the motion under a central force the angular momentum is always conserved. Two important results follow from this: (1) The motion of a particle under the central force is always confined to a plane. (2) The position vector of the particle with respect to the centre of the force (i.e. the fixed point) has a constant areal velocity. In other words the position vector sweeps out equal areas in equal times as the particle moves under the influence of the central force. Try to prove both these results. You may need to know that the areal velocity is given by : dA/dt = ½ r v sin α. An immediate application of the above discussion can be made to the motion of a planet under the gravitational force of the sun. For convenience the sun may be taken to be so heavy that it is at rest. The gravitational force of the sun on the planet is directed towards the sun. This force also satisfies the requirement F = F(r), since F = G m1m2/r2 where m1 and m2 are respectively the masses of the planet and the sun and G is the universal constant of gravitation. The two results (1) and (2) described above, therefore, apply to the motion of the planet. In fact, the result (2) is the well-known second law of Kepler. Tr is the trejectory of the particle under the central force. At a position P, the force is directed along OP, O is the centre of the force taken as the origin. In time ∆t, the particle moves from P to P′, arc PP′ = ∆s = v ∆t. The tangent PQ at P to the trajectory gives the direction of the velocity at P. The area swept in ∆t is the area of sector POP′ ≈ (r sin α ) PP′/2 = (r v sin a) ∆t/2.) GRAVITATION 187 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 earth, we will have a m α R m−2 ; g α R E–2 and we get related as F12 = – F21. Before we can apply Eq. (8.5) to objects under g R2 consideration, we have to be careful since the = m2
3600 (8.4) law refers to point masses whereas we deal with am RE extended objects which have finite size.. If we in agreement with a value of g
9.8 m s-2 and have a collection of point masses,the force on the value of a m from Eq. (8.3). These any one of them is the vector sum of the observations led Newton to propose the following gravitational forces exerted by the other point Universal Law of Gravitation : masses as shown in Fig 8.4. Every body in the universe attracts every other 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 (8.5) r2 Equation (8.5) can be expressed in vector form as m1 m 2 m1 m 2  F= G r 2 ( ) – r = – G r2 r m1 m 2  = –G 3 r r Fig. 8.4 Gravitational force on point mass m1 is the where G is the universal gravitational constant, vector sum of the gravitational forces exerted r is the unit vector from m1 to m2 and r = r2 – r1 by m2, m3 and m4. as shown in Fig. 8.3. The total force on m1 is Gm 2 m 1  Gm 3 m 1  Gm 4 m1  F1 = 2 r 21 + 2 r 31 + 2 r 41 r21 r31 r41 t Example 8.2 Three equal masses of m kg each are fixed at the vertices of an O 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 vertex A is doubled ? Fig. 8.3 Gravitational force on m1 due to m2 is along Take AG = BG = CG = 1m (see Fig. 8.5) r where the vector r is (r2– r1). 188 PHYSICS ( ) + 2Gm 2 ˆi cos 30ο − ˆj sin 30ο = 0 Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero. (b) By symmetry the x-component of the force cancels out. The y-component survives. FR = 4Gm 2ˆj − 2Gm 2ˆj = 2Gm 2ˆj t For the gravitational force between an extended object (like the earth) and a point mass, Eq. (8.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in Fig. 8.5 Three equal masses are placed at the three vertices of the ∆ ABC. A mass 2m is placed the same direction. We have to add up these forces at the centroid G. vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple Answer (a) The angle between GC and the law results when you do that : positive x-axis is 30° and so is the angle between (1) The force of attraction between a hollow GB and the negative x-axis. The individual forces spherical shell of uniform density and a in vector notation are point mass situated outside is just as if Gm (2m ) ˆ the entire mass of the shell is FGA = j concentrated at the centre of the shell. 1 Qualitatively this can be understood as Gm (2m ) ˆ follows: Gravitational forces caused by the FGB = 1 ( − i cos 30ο − ˆj sin 30ο ) various regions of the shell have components along the line joining the point Gm (2m ) ˆ mass to the centre as well as along a FGC = 1 ( + i cos 30ο − ˆj sin 30ο ) direction prependicular to this line. The From the principle of superposition and the law components prependicular to this line of vector addition, the resultant gravitational cancel out when summing over all regions force FR on (2m) is of the shell leaving only a resultant force FR = FGA + FGB + FGC along the line joining the point to the centre. The magnitude of this force works out to ( FR = 2Gm2 ˆj + 2Gm 2 −ˆi cos 30ο −ˆj sin 30ο ) be as stated above. Newton’s Principia Kepler had formulated his third law by 1619. The announcement of the underlying universal law of gravitation came about seventy years later with the publication in 1687 of Newton’s masterpiece Philosophiae Naturalis Principia Mathematica, often simply called the Principia. Around 1685, Edmund Halley (after whom the famous Halley’s comet is named), came to visit Newton at Cambridge and asked him about the nature of the trajectory of a body moving under the influence of an inverse square law. Without hesitation Newton replied that it had to be an ellipse, and further that he had worked it out long ago around 1665 when he was forced to retire to his farm house from Cambridge on account of a plague outbreak. Unfortunately, Newton had lost his papers. Halley prevailed upon Newton to produce his work in book form and agreed to bear the cost of publication. Newton accomplished this feat in eighteen months of superhuman effort. The Principia is a singular scientific masterpiece and in the words of Lagrange it is “the greatest production of the human mind.” The Indian born astrophysicist and Nobel laureate S. Chandrasekhar spent ten years writing a treatise on the Principia. His book, Newton’s Principia for the Common Reader brings into sharp focus the beauty, clarity and breath taking economy of Newton’s methods. GRAVITATION 189 (2) The force of attraction due to a hollow its neighbouring small sphere. Due to this spherical shell of uniform density, on a torque, the suspended wire gets twisted till such point mass situated inside it is zero. time as the restoring torque of the wire equals Qualitatively, we can again understand this the gravitational torque. If θ is the angle of result. Various regions of the spherical shell twist of the suspended wire, the restoring torque attract the point mass inside it in various is proportional to θ, equal to τθ. Where τ is the directions. These forces cancel each other restoring couple per unit angle of twist. τ can be completely. measured independently e.g. by applying a known torque and measuring the angle of twist. 8.4 THE GRAVITATIONAL CONSTANT The gravitational force between the spherical balls is the same as if their masses are The value of the gravitational constant G concentrated at their centres. Thus if d is the entering the Universal law of gravitation can be separation between the centres of the big and determined experimentally and this was first its neighbouring small ball, M and m their done by English scientist Henry Cavendish in masses, the gravitational force between the big 1798. The apparatus used by him is sphere and its neighouring small ball is. schematically shown in figure.8.6 Mm F =G (8.6) d2 If L is the length of the bar AB , then the torque arising out of F is F multiplied by L. At equilibrium, this is equal to the restoring torque and hence Mm G L =τ θ (8.7) d2 Observation of θ thus enables one to calculate G from this equation. Since Cavendish’s experiment, the measurement of G has been refined and the currently accepted value is G = 6.67×10-11 N m2/kg2 (8.8) Fig. 8.6 Schematic drawing of Cavendish’s experiment. S1 and S2 are large spheres 8.5 ACCELERATION DUE TO GRAVITY OF which are kept on either side (shown shades) of the masses at A and B. When THE EARTH the big spheres are taken to the other side The earth can be imagined to be a sphere made of the masses (shown by dotted circles), the bar AB rotates a little since the torque of a large number of concentric spherical shells reverses direction. The angle of rotation can with the smallest one at the centre and the be measured experimentally. largest one at its surface. A point outside the earth is obviously outside all the shells. Thus, 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 8.3. 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. 8.7. 190 PHYSICS The acceleration experienced by the mass m, which is usually denoted by the symbol g is related to F by Newton’s 2nd law by relation F = mg. Thus F GM E g= = (8.12) m R E2 Acceleration g is readily measurable. RE is a known quantity. The measurement of G by Cavendish’s experiment (or otherwise), combined Mr with knowledge of g and RE enables one to estimate ME from Eq. (8.12). This is the reason why there is a popular statement regarding Fig. 8.7 The mass m is in a mine located at a depth d below the surface of the Earth of mass Cavendish : “Cavendish weighed the earth”. ME and radius RE. We treat the Earth to be spherically symmetric. 8.6 ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF Again consider the earth to be made up of EARTH concentric shells as before and a point mass m situated at a distance r from the centre. The Consider a point mass m at a height h above the point P lies outside the sphere of radius r. For surface of the earth as shown in Fig. 8.8(a). The the shells of radius greater than r, the point P radius of the earth is denoted by RE. Since this lies inside. Hence according to result stated in point is outside the earth, the last section, they exert no gravitational force on mass m kept at P. The shells with radius ≤ r make up a sphere of radius r for which the point P lies on the surface. This smaller sphere therefore exerts a force on a mass m at P as if its mass Mr is concentrated at the centre. Thus the force on the mass m at P has a magnitude Gm (M r ) F = (8.9) r2 We assume that the entire earth is of uniform 4π 3 density and hence its mass is M E = RE ρ 3 where ME is the mass of the earth RE is its radius and ρ is the density. On the other hand the 4π mass of the sphere Mr of radius r is ρ r 3 and Fig. 8.8 (a) g at a height h above the surface of the 3 earth. hence 4π r3 ME r 3 its distance from the centre of the earth is (RE + F = Gm ρ 2 = Gm 3 r R E3 r 2 h ). If F (h) denoted the magnitude of the force G m ME on the point mass m , we get from Eq. (8.5) : = r (8.10) RE 3 If the mass m is situated on the surface of GM E m F (h ) = (8.13) earth, then r = RE and the gravitational force on (R E + h )2 it is, from Eq. (8.10) The acceleration experienced by the point M Em F =G (8.11) mass is F (h )/ m ≡ g(h ) and we get R E2 GRAVITATION 191 F (h ) GM E Thus the force on the point mass is g(h ) = =. (8.14) m (R E + h )2 F (d) = G Ms m / (RE – d ) 2 (8.17) This is clearly less than the value of g on the Substituting for Ms from above , we get GM E F (d) = G ME m ( RE – d ) / RE3 (8.18) 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. 11. A geostationary (geosynchronous communication) satellite moves in a circular orbit in the equatorial plane at a approximate distance of 4.22 × 104 km from the Earth’s centre. 200 PHYSICS 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. (8.1) and T2 = KS R3. The constant KS is the same for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (8.38)]. 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. GRAVITATION 201 EXERCISES 8.1 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 ? 8.2 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 for mula –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. 8.3 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 ? 8.4 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. 8.5 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. 8.6 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. 8.7 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? 8.8 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. 8.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem. 8.10 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 8.12) (i) a, (ii) b, (iii) c, (iv) 0. Fig. 8.12 202 PHYSICS 8.11 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. 8.12 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). 8.13 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. 8.14 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 ? 8.15 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 ? 8.16 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 ? 8.17 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. 8.18 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. 8.19 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. 8.20 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). 8.21 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 ? Additional Exercises 8.22 As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0×1024 kg, radius = 6400 km. 8.23 A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun = 2×1030 kg). 8.24 A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2×1030 kg; mass of mars = 6.4×1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 ×108 km; G = 6.67×10-11 N m2 kg–2. 8.25 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4×1023 kg; radius of mars = 3395 km; G = 6.67×10-11 N m2 kg–2.