Gravitation Chapter 9 PDF

60 168 Gravitation 8.1 Introduction. E3 N ewton at the age of twenty-three is said to have seen an apple falling down from tree in his orchid. This was the year 1665. He started thinking about the role of earth's attraction in the motion of moon and other heavenly bodies. ID By comparing the acceler...

60 168 Gravitation 8.1 Introduction. E3 N ewton at the age of twenty-three is said to have seen an apple falling down from tree in his orchid. This was the year 1665. He started thinking about the role of earth's attraction in the motion of moon and other heavenly bodies. ID By comparing the acceleration due to gravity due to earth with the acceleration required to keep the moon in its orbit around the earth, he was able to arrive the Basic Law of U Gravitation. 8.2 Newton's law of Gravitation. D YG Newton's law of gravitation states that every body in this 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 their centres. The direction of the force is along the line joining the particles. Thus the magnitude of the gravitational force F that two particles of masses m 1 and m 2 m m separated by a distance r exert on each other is given by F  1 2 2 r A B   FG r m1 F 12 2 U or m 1m 2 F 21 m2 r Vector form : According to Newton's law of gravitation  Gm 1 m 2 ST F 12  r2 r̂21   Gm 1 m 2 r3 r 21   Gm 1 m 2 r 21 | r 21 | 3 r̂12 = unit vector from A to B r̂21 = unit vector from B to Here negative sign indicates that the direction of F 12 is opposite to that of r̂21. F 21  Similarly  Gm 1 m 2 r2  Gm 1 m 2  Gm 1 m 2  Gm 1 m 2 r̂  r  r12 12 12 r2 r3 3 | r12 | r̂21 [rˆ12  rˆ21 ] A, F12 = gravitational force exerted on body A by body B F 21 = gravitational force  It is clear that F 12 = – F 21. Which is Newton's third law of motion. exerted on body B by body A Here G is constant of proportionality which is called 'Universal gravitational constant'. Gravitation 169 If m 1  m 2 and r  1 then G  F i.e. universal gravitational constant is equal to the force of attraction between two bodies each of unit mass whose centres are placed unit distance apart. Important points 60 (i) The value of G in the laboratory was first determined by Cavendish using the torsional balance. E3 (ii) The value of G is 6.67×10–11 N–m2 kg–2 in S.I. and 6.67×10–8 dyne- cm2-g–2 in C.G.S. system. (iii) Dimensional formula [M 1 L3 T 2 ]. (iv) The value of G does not depend upon the nature and size of the bodies. ID (v) It also does not depend upon the nature of the medium between the two bodies. (vi) As G is very small hence gravitational forces are very small, unless one (or both) of the masses is huge. U 8.3 Properties of Gravitational Force. D YG (1) It is always attractive in nature while electric and magnetic force can be attractive or repulsive. (2) It is independent of the medium between the particles while electric and magnetic force depend on the nature of the medium between the particles. (3) It holds good over a wide range of distances. It is found true for interplanetary to inter atomic distances. (4) It is a central force i.e. acts along the line joining the centres of two interacting bodies. U (5) It is a two-body interaction i.e. gravitational force between two particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles is the resultant of forces due to individual particles i.e. ST F  F 1  F 2  F 3 ........ While nuclear force is many body interaction (6) It is the weakest force in nature : As Fnuclear > F electromagnetic > F gravitational. (7) The ratio of gravitational force to electrostatic force between two electrons is of the order of 10 43. (8) It is a conservative force i.e. work done by it is path independent or work done in moving a particle round a closed path under the action of gravitational force is zero. 170 Gravitation (9) It is an action reaction pair i.e. the force with which one body (say earth) attracts the second body (say moon) is equal to the force with which moon attracts the earth. This is in accordance with Newton's third law of motion. Note :  The law of gravitation is stated for two point masses, therefore for any two 60 arbitrary finite size bodies, as shown in the figure, It can not be applied as there is not unique value for the separation. m2 m1 m2 E3 m1 r r=? ID But if the two bodies are uniform spheres then the separation r may be taken as the distance between their centres because a sphere of uniform mass behave as a point mass for any point lying outside it. Problem 1. U Sample problems based on Newton’s law of gravitation The gravitational force between two objects does not depend on D YG (a) Sum of the masses (c) Gravitational constant masses Solution : (a) F  Problem 2. [RPET 2003] (b) Product of the masses (d) Distance between the Gravitatio nal constant  product of the masses. (Distance between the masses) 2 Mass M is divided into two parts xM and (1 – x)M. For a given separation, the value of x for U which the gravitational attraction between the two pieces becomes maximum is 1 2 ST (a) Solution : (a) Gravitational force F  (b) Problem 3. (c) 1 (d) 2 Gm 1m 2 GxM (1  x )M GM 2   2 x (1  x ) r2 r2 r For maximum value of force  3 5  d  GM 2 x dF (1  x )  0 0   2 dx  r dx  d (x  x 2 )  0  1  2 x  0  x  1 / 2 dx The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth (a) Is the same (b) Is smaller (c) Is greater Solution : (a) Earth and moon both exerts same force on each other. (d) Varies with its phase Gravitation 171 Problem 4. Three identical point masses, each of mass 1kg lie in the x-y plane at points (0, 0), (0, 0.2m) and (0.2m, 0). The net gravitational force on the mass at the origin is (a) 1.67  10 9 (ˆj  ˆj)N (b) 3.34  10 10 (ˆi  ˆj)N (c) 1.67  10 9 (ˆi  ˆj)N (d) 3.34  10 10 (ˆi  ˆj)N FAC  60 Solution : (a) Let particle A lies at origin, particle B and C on y and x -axis respectively G m A m B ˆ 6.67  10 11  1  1 ˆ i  i  1.67  10 9 ˆi N 2 rAB (0.2) 2 Y B FAB E3 Similarly FAB  1.67  10 9 ˆj N (0, 0)  Net force on particle A F  F AC  F AB  1.67  10 9 (ˆi  ˆj) N (0.2, C0) A FAC X Four particles of masses m, 2m, 3m and 4m are kept in sequence at the corners of a square ID Problem 5. (0, 0.2) of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be 24 m 2 G a2 (b) 6m 2G a2 (c) U (a) 4 2 Gm 2 a2 (d) Zero D YG Solution : (c) If two particles of mass m are placed x distance apart then force of attraction Gmm F x2 (Let) Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experience four forces FPA D 4 m Gmm F  force at point P due to particle A  x2 G 2mm x2 U Similarly FPB   2 F , FPC  ST  Gmm x 2 4 2 Gm 2 a2 2 2 x2  3 F and FPD  D P m G 4 mm x2 FPC FPA  4F A m FPB 2 B m F net  F PA  F PB  F PC  F PD  2 2 F Hence the net force on P  F net  2 2 G3mm 3 C m FP Gm 2 (a / 2 ) 2 [x  a  half of the diagonal of the square] 2. 8.4 Acceleration Due to Gravity. The force of attraction exerted by the earth on a body is called gravitational pull or gravity. We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate. 172 Gravitation The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g. Consider a body of mass m is lying on the surface of earth then gravitational force on the F  GMm R2 …..(i) Where M = mass of the earth and R = radius of the earth. 60 body is given by E3 If g is the acceleration due to gravity, then the force on the body due to earth is given by Force = mass  acceleration F = mg …..(ii) GMm From (i) and (ii) we have mg  R2 g GM R2  g G R2  g R Earth …..(iii) 4 [As mass (M) = volume ( R 3 ) × density 3 D YG ()] 4 3   R   3   m U  mg ID or 4 GR 3 …..(iv) Important points GM 4  GR it is clear that its value depends upon the mass 3 R2 radius and density of planet and it is independent of mass, shape and density of the body placed U (i) From the expression g  ST on the surface of the planet. i.e. a given planet (reference body) produces same acceleration in a light as well as heavy body. (ii) The greater the value of (M / R 2 ) or R, greater will be value of g for that planet. (iii) Acceleration due to gravity is a vector quantity and its direction is always towards the centre of the planet. (iv) Dimension [g] = [LT–2] (v) it’s average value is taken to be 9.8 m/s2 or 981 cm/sec2 or 32 feet/sec2, on the surface of the earth at mean sea level. (vi) The value of acceleration due to gravity vary due to the following factors : (a) Shape of the earth, (b) Height above the earth surface, (c) Depth below the earth surface and (d) Axial rotation of the earth. Gravitation 173 Sample problems based on acceleration due to gravity (a) 5 Re 18 (b) 1 Re 6 Solution : (a) Acceleration due to gravity g  (c)  1 5     6 3  Rm  5 Re 18 ID D02 Solution : (c) We know g  Re gm  1 5 and e   m ge 6 3 [As (b) 4 mGM 0 / GM R 2 GM  (D / 2) 2 D02 (c) 4 GM 0 / D02  [MP PMT 1987; DPMT 2002] (d) GmM 0 / D02 4 GM D2 D YG If mass of the planet  M 0 and diameter of the planet  D0. Then g  4 GM 0 D02. The moon's radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is g (b) 5 g (a) 4 U Solution : (b) Acceleration due to gravity g  ST gmoon  gearth  Problem 9. 2 3 A spherical planet far out in space has a mass M 0 and diameter D0. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to (a) GM 0 / Problem 8. 1 (d) U Problem 7. R m  g m   e    Re  g e   m 3 Re 18 g  R 4  GR  g   R or m  m. m ge  e Re 3 (given)]  60 Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the   5 ratio of densities of earth (m ) and moon (  e ) is  e   then radius of moon Rm in terms  m  3 of R e will be [MP PMT 2003] E3 Problem 6. GM R2  [MP PMT 1997; RPET 2000; MP PET 2000, 2001] g (c) 6 (d) gmoon M R2  1  4   moon. 2earth     gearth M earth Rmoon  80  1  g 8 2 16 g . 80 5 If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would[IIT-JEE 1981; CPMT 1981; MP PMT 1996, 97; Roor (a) Decrease by 2% Solution : (c) We know g  (b) Remain unchanged (c) Increase by 2% (d) 1 [As R decreases, g increases] R2 So % change in g  2 (% change in R )  2  1%  2%  acceleration due to gravity increases by 2%. Problem 10. Mass of moon is 7.34  10 22 kg. If the acceleration due to gravity on the moon is 1.4m/s2, the radius of the moon is (G  6.667  10 11 Nm 2 / kg 2 ) (a) 0.56  10 4 m (b) 1.87  10 6 m (c) 1.92  10 6 m (d) 1.01  10 8 m 174 Gravitation Solution : (b) We know g  GM R2 GM  g  R 6.67  10 11  7.34  10 22  1.87  10 6 m. 1.4 Problem 11. A planet has mass 1/10 of that of earth, while radius is 1/3 that of earth. If a person can throw a stone on earth surface to a height of 90m, then he will be able to throw the stone on that planet to a height [RPMT 1994] (b) 40m (c) 100m M planet  R earth  2 g earth M earth  R planet R If a stone is thrown with velocity u from the surface of the u2 H  2g H planet H earth  gearth g planet  H planet  GM  g planet  2 2    1   3   9  10  1  10  planet then maximum height E3 Solution : (c) Acceleration due to gravity g  (d) 45m 60 (a) 90m 10 10  90 = 100 metre.  H earth  9 9 ID Problem 12. The radii of two planets are respectively R1 and R 2 and their densities are respectively 1 and  2. The ratio of the accelerations due to gravity at their surfaces is 1 R 12 : 2 (b) g1 : g2  R1 R2 : 1  2 R 22 (c) g1 : g 2  R1  2 : R 2  1 U (a) g 1 : g 2  (d) g1 : g 2  R1 1 : R 2  2 4  GR 3 D YG Solution : (d) Acceleration due to gravity g   g 1 : g 2  R1  1 : R 2  2. 8.5 Variation in g Due to Shape of Earth. Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The equatorial radius is about 21 km longer than polar radius, from GM g 2 R ge  GM R e2 U At equator ST At poles gp  gp Rp......(i) Re GM R p2 ge.....(ii) R p2 ge From (i) and (ii)  2 gp Re Since R equator  R pole  g pole  g equator and g p  g e  0.018 ms 2 Therefore the weight of body increases as it is taken from equator to the pole. Sample problems based on variation in g due to shape of the earth Problem 13. Where will it be profitable to purchase 1 kg sugar (by spring balance) (a) At poles (b) At equator (c) At 45° latitude (d) At 40°latitude Gravitation 175 Solution : (b) At equator the value of g is minimum so it is profitable to purchase sugar at this position. Problem 14. Force of gravity is least at (a) The equator (b) The poles (c) A point in between equator and any pole (d) None of these Solution : (a) 60 8.6 Variation in g With Height. Acceleration due to gravity at the surface of the earth GM R2.....(i) g E3 g Acceleration due to gravity at height h from the surface of the earth GM (R  h)2.....(ii) =g R2 r2 2 O ID  R  g'  g   R h From (i) and (ii).....(iii).....(iv) [As r = R + h] D YG Important points g R U g'  h r (i) As we go above the surface of the earth, the value of g decreases because g   1. r2 (ii) If r   then g   0 , i.e., at infinite distance from the earth, the value of g becomes zero. (iii) If h  R i.e., height is negligible in comparison to the radius then from equation (iii) we get 2 U h  R   g   g   g1   R R h  2 2h    g 1   R  [As h  R ] ST (iv) If h  R then decrease in the value of g with height : Absolute decrease g  g  g   Fractional decrease Percentage decrease 2hg R g g  g  2 h   g g R g 2h  100 %   100 % g R Sample problems based on variation in g with height Problem 15. The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth) 176 Gravitation (a) g 9 g  R  1  R       g R h R R  2 9   2 Solution : (a) g 3 (b) (c) 2  g  g 4 (d) g g. 9 60 Problem 16. The height of the point vertically above the earth's surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R) (a) 8R (b) 9R (c) 10 R (d) 20R  2 g  R   g  100 R  h  R 1  h  9R.  R  h 10 2 E3  R  Solution : (b) Acceleration due to gravity at height h is given by g  g   R h earth is (R = radius of earth) (b) 16N (c) 32N   R  R  W  W  W R  R  h   R 2  U (a) 28N ID Problem 17. At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of 2 D YG Solution : (c) Weight of the body at height R, (d) 72N 2  2    W  2   4 W  4  72  32 N.  9 9 3   Problem 18. If the distance between centres of earth and moon is D and the mass of earth is 81 times the mass of moon, then at what distance from centre of earth the gravitational force will be zero [RPET 1996] (a) D/2 (b) 2D/3 (c) 4D/3 (d) 9D/10 Solution : (d) If P is the point where net gravitational force is zero then FPA  FPB ST U Gm 1m Gm 2m  x2 (d  x )2 By solving x  m1 A m1 d d P FPA x So x  d–x m1  m 2 For the given problem d  D , m 1  earth, m 2  moon and m1  81m 2  m 2  m1 D m1  m 2  m1 D m1  m1 81  D 1 1 9  9D 10 8.7 Variation in g With Depth. Acceleration due to gravity at the surface of the earth m2 FPB m1 81 B Gravitation 177 g GM 4  GR R2 3 …..(i) g d Acceleration due to gravity at depth d from the surface of the earth From (i) and (ii) g r O 4 G(R  d ) 3 …..(ii) 60 g  R P d  g   g 1   R  E3 Important points (i) The value of g decreases on going below the surface of the earth. From equation (ii) we get g   (R  d ). ID So it is clear that if d increase, the value of g decreases. (ii) At the centre of earth d  R  g   0 , i.e., the acceleration due to gravity at the centre of earth becomes zero. U (iii) Decrease in the value of g with depth dg R D YG Absolute decrease g  g  g   Fractional decrease Percentage decrease g g  g  d   g g R g d  100 %   100 % g R (iv) The rate of decrease of gravity outside the earth ( if h  R ) is double to that of inside U the earth. Sample problems based on variation in g with depth ST Problem 19. Weight of a body of mass m decreases by 1% when it is raised to height h above the earth's surface. If the body is taken to a depth h in a mine, change in its weight is[KCET 2003; MP PMT 2003] (a) 2% decrease (b) 0.5% decrease (c) 1% increase (d) 0.5% increase Solution : (b) Percentage change in g when the body is raised to height h , g 2h  100  100 %   1% g R Percentage change in g when the body is taken into depth d, g d h  100 %   100 %   100 % g R R [As d  h ]  Percentage decrease in weight  1  2h  1  100   1 %   0. 5 %.  2 R  2 178 Gravitation g is (R = radius of 4 Problem 20. The depth at which the effective value of acceleration due to gravity is the earth) [MP PET 2003] (b) 3R 4 R 2 (c) R 4 60 (a) R (d) E3 g d d 3R    g1    d  Solution : (b) g  g 1    4 R R 4   Problem 21. Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine 100 km below the earth's surface (Given R = 6400km) Solution : (a) Acceleration 63  9.66 m / s 2. 64 to gravity at depth d, 100  1  d    g   g 1    g 1    9.8 1  64  6400 R       U  9.8  due (d) 3.10 m / s 2 (c) 5.06 m / s 2 ID (b) 7.64 m / s 2 (a) 9.66 m / s 2 Problem 22. The depth d at which the value of acceleration due to gravity becomes 1 times the value n D YG at the surface, is [R = radius of the earth] (a) R n n 1 (b) R    n  (c)  n  (d) R   n 1 R n2 d g d d 1   n 1 Solution : (b) g  g 1     g 1     1   d   R R n R R n      n  8.8 Variation in g Due to Rotation of Earth. U As the earth rotates, a body placed on its surface moves along the circular path and hence ST experiences centrifugal force, due to it, the apparent weight of the body decreases.  P r Since the magnitude of centrifugal force varies with the mg  Fc mg latitude of the place, therefore the apparent weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body. If the body of mass m lying at point P, whose latitude is , then due to rotation of earth its apparent weight can be given by m g   mg  Fc or  m g   (mg )2  (Fc )2  2mg Fc cos(180  ) m g   (mg )2  (m  2 R cos )2  2mg m  2 R cos  ( cos ) [As Fc  m  2 r  m  2 R cos  ] Gravitation 179 By solving we get g   g   2 R cos 2  Note :  The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of earth makes with equatorial plane. It is denoted by . 60  For the poles   90 o and for equator   0 o Important points  g pole  g E3 (i) Substituting   90 o in the above expression we get g pole  g   2 R cos 2 90 o …..(i) i.e., there is no effect of rotational motion of the earth on the value of g at the poles.  g equator  g   2 R ID (ii) Substituting   0 o in the above expression we get g eqator  g   2 R cos 2 0 o …..(ii) U i.e., the effect of rotation of earth on the value of g at the equator is maximum. D YG From equation (i) and (ii) g pole  g equator  R 2  0.034 m / s 2 (iii) When a body of mass m is moved from the equator to the poles, its weight increases by an amount m(g p  g e )  m  2 R (iv) Weightlessness due to rotation of earth : As we know that apparent weight of the body decreases due to rotation of earth. If  is the angular velocity of rotation of earth for which a body U at the equator will become weightless g   g   2 R cos 2  [As   0 o for equator] 0  g   2 R cos 2 0 o ST   g 2R   g R or time period of rotation of earth T  2   2 Substituting the value of R  6400  10 3 m  1 rad  1.25  10 3 800 sec R g and and g  10 m / s 2 we get T  5026.5 sec  1.40 hr. 180 Gravitation Note :  This time is about 1 times the present time period of earth. Therefore if earth 17 starts rotating 17 times faster then all objects on equator will become weightless.  If earth stops rotation about its own axis then at the equator the value of g m 2 R. 60 increases by  2 R and consequently the weight of body lying there increases by  After considering the effect of rotation and elliptical shape of the earth, acceleration due to gravity at the poles and equator are related as  g p  g e  0.052 m / s 2 E3 g p  g e  0.034  0.018 m / s 2 Sample problems based on variation in g due to rotation of the earth Problem 23. The angular velocity of the earth with which it has to rotate so that acceleration due to (a) 2.5×10–3rad/sec ID gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms– 2 ) [EAMCET 2000] (b) 5.0×10–1rad/sec (c) 10  10 1 rad / sec (d) 7.8  10 2 rad / sec  2R 4  g   D YG  0  g   2 R cos 2 60 o  U Solution : (a) Effective acceleration due to gravity due to rotation of earth g  g   2 R cos 2    4g g 2 rad 2  R R 800 sec [As g  0 and   60 o ] 1 rad.  2.5  10  3 400 sec Problem 24. If earth stands still what will be its effect on man's weight (a) Increases (b) Decreases (c) Remains same (d) None of these U Solution : (a) When earth stops suddenly, centrifugal force on the man becomes zero so its effective weight increases. Problem 25. If the angular speed of earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly ST (a) 1.5 hours (b) 8 hours Solution : (a) Time period for the given condition T  2 (c) 18 hours (d) 24 hours R  1.40 hr  1.5 hr nearly. g 8.9 Mass and Density of Earth. Newton’s law of gravitation can be used to estimate the mass and density of the earth. As we know g   M gR 2 GM M  , so we have G R2 9.8  (6.4  10 6 )2  5.98  10 24 kg  10 25 kg 6.67  10 11 Gravitation 181 and as we know g    4 3g GR , so we have   3 4GR 3  9.8  5478.4 kg / m 3 4  3.14  6.67  10 11  6.4  10 6 60 8.10 Inertial and Gravitational Masses. (1) Inertial mass : It is the mass of the material body, which measures its inertia. If an external force F acts on a body of mass mi, then according to Newton’s second law of F  m i a or m i  E3 motion F a ID Hence inertial mass of a body may be measured as the ratio of the magnitude of the external force applied on it to the magnitude of acceleration produced in its motion. Important points U (i) It is the measure of ability of the body to oppose the production of acceleration in its motion by an external force. (ii) Gravity has no effect on inertial mass of the body. D YG (iii) It is proportional to the quantity of matter contained in the body. (iv) It is independent of size, shape and state of body. (v) It does not depend on the temperature of body. (vi) It is conserved when two bodies combine physically or chemically. (vii) When a body moves with velocity v , its inertial mass is given by m0 v2 1 2 c , where m0 = rest mass of body, c = velocity of light in vacuum, U m ST (2) Gravitational Mass : It is the mass of the material body, which determines the gravitational pull acting upon it. If M is the mass of the earth and R is the radius, then gravitational pull on a body of mass m g is given by F GMm g R 2 or m g  F F  2 (GM / R ) E Here m g is the gravitational mass of the body, if E  1 then m g  F Thus the gravitational mass of a body is defined as the gravitational pull experienced by the body in a gravitational field of unit intensity, (3) Comparison between inertial and gravitational mass (i) Both are measured in the same units. 182 Gravitation (ii) Both are scalars (iii) Both do not depends on the shape and state of the body (iv) Inertial mass is measured by applying Newton’s second law of motion where as gravitational mass is measured by applying Newton’s law of gravitation. (4) Comparison between mass and weight of the body Mass (m) 60 (v) Spring balance measure gravitational mass and inertial balance measure inertial mass. Weight (W) It is the attractive force exerted by earth on any body. Its value does not change with g Its value changes with g. Its value can never be zero for any material particle. At infinity and at the centre of earth its value is zero. Its unit is kilogram and its dimension is [M]. Its unit is Newton or kg-wt and dimension are ID E3 It is a quantity of matter contained in a body. [ MLT 2 ] It is determined by a spring balance. U It is determined by a physical balance. It is a scalar quantity. It is a vector quantity. D YG Sample problems based on inertial and gravitational mass Problem 26. Gravitational mass is proportional to gravitational (a) Field Solution : (d) (b) Force (c) Intensity (d) All of these Problem 27. The ratio of the inertial mass to gravitational mass is equal to (a) 1/2 (c) 2 (d) No fixed number U Solution : (b) (b) 1 [CPMT 1978] 8.11 Gravitational Field. ST The space surrounding a material body in which gravitational force of attraction can be experienced is called its gravitational field. Gravitational field intensity : The intensity of the gravitational field of a material body at any point in its field is defined as the force experienced by a unit mass (test mass) placed at that point, provided the unit mass (test mass) itself does not produce any change in the field of the body. So if a test mass m at a point in a gravitational field experiences a force F then I  Important points F m Gravitation 183 (i) It is a vector quantity and is always directed towards the centre of gravity of body whose gravitational field is considered. (ii) Units : Newton/kg or m/s2 (iii) Dimension : [M0LT–2] it then by Newton’s law of gravitation F  GMm r2 Test mass F GMm / r 2  m m M Source point GM I 2 r  m r E3 then intensity of gravitational field I  60 (iv) If the field is produced by a point mass M and the test mass m is at a distance r from ID (v) As the distance (r) of test mass from the point mass (M ) , increases, intensity of gravitational field decreases 1 GM ;  I 2 2 r r M U I (vi) Intensity of gravitational field I  0 , when r  . I1 M1 (vii) Intensity at a given point (P) due to the combined effect of different point masses can be calculated by vector sum of different intensities I2 2 D YG I3 M 3 Inet  I1  I 2  I 3 ........ (viii) Point of zero intensity : If two bodies A and B of different masses m 1 and m 2 are d distance apart. ST U Let P be the point of zero intensity i.e., the intensity at this point is equal and apposite due to two bodies A and B and if any test mass placed at this point it will not experience any force. For point P I1  I 2  0 By solving x  m1 d m1  m 2  Gm 1 x2  Gm 2 (d  x )2 and (d  x )  0 d A m1 x P I1 I2 m2 d m1  m 2 (ix) Gravitational field line is a line, straight or curved such that a unit mass placed in the field of another mass would always move along this line. Field lines for an isolated mass m are radially inwards. d–x m B m2 184 Gravitation GM GM and also g  2  I  g 2 r R (x) As I  Thus the intensity of gravitational field at a point in the field is equal to acceleration of test mass placed at that point. (1) Intensity due to uniform solid sphere On the surface r=R Inside the surface r>R GM I r2 GM I R2 (2) Intensity due to spherical shell Outside the surface r=R r>R I O I GM 2 R I I=0 D YG 2 r r=R R Inside the surface r