Gravitation Theory PDF

Gravitation 389 Chapter...

Gravitation 389 Chapter 8 Gravitation Gm1m 2  Gm1m 2 Gm m Introduction F12  2 r̂21  3 r21  1 2 r21 r r | r21 | 3 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 Here negative sign indicates that the direction of F 12 is opposite to that thinking about the role of earth's attraction in the motion of moon and other heavenly bodies. of r̂21. Gm1m 2  Gm1m 2  Gm m Similarly F 21  2 r̂12  3 r12  1 2 r12 r r | r12 | 3 Gm1m 2  r̂21 [ rˆ12  rˆ21 ] r2  It is clear that F12 = – F 21. Which is Newton's third law of motion. Here G is constant of proportionality which is called 'Universal gravitational constant'. Fig. 8.1 due to gravity due to earth with the By comparing the acceleration If m1  m 2 and r  1 then G  F acceleration required to keep the moon in its orbit around the earth, he was able to arrive the Basic Law of Gravitation. i.e. universal gravitational constant is equal to the force of attraction between two bodies each of unit mass whose centres are placed Newton's law of Gravitation unit distance apart. (i) The value of G in the laboratory was first determined by Newton's law of gravitation states that every body in this universe Cavendish using the torsional balance. 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 (ii) The value of G is 6.67×10 N–m kg in S.I. and 6.67×10 dyne- –11 2 –2 –8 distance between their centres. The direction of the force is along the line cm -g 2 – t–––– in C.G.S. system. 2 –2 joining the particles. (iii) Dimensional formula [M 1 L3 T 2 ]. Thus the magnitude of the gravitational force F that two particles of (iv) The value of G does not depend upon the nature and size of the masses m 1 and m 2 are separated by a distance r exert on each other is bodies. m1 m 2 A B (v) It also does not depend upon the nature of the medium between given by F    r2 m1 F 12 F 21 m2 the two bodies. m1m 2 (vi) As G is very small, hence gravitational forces are very small, or F  G r unless one (or both) of the mass is huge. r2 Fig. 8.2 Vector form : According to Newton's law of gravitation Properties of Gravitational Force  390 Gravitation (1) It is always attractive in nature while electric and magnetic force or F = mg …(ii) m can be attractive or repulsive. GMm (2) It is independent of the medium between the particles while From (i) and (ii) we have mg  mg R2 R electric and magnetic force depend on the nature of the medium between the particles. Earth GM (3) It holds good over a wide range of distances. It is found true for  g 2 …(iii) R Fig. 8.4 interplanetary to inter atomic distances. (4) It is a central force i.e. acts along the line joining the centres of G 4 3  two interacting bodies.  g  R   R2 3  (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 4 principle of superposition is valid i.e. force on a particle due to number of [As mass (M) = volume ( R 3 ) × density ()] 3 particles is the resultant of forces due to individual particles i.e. 4 F  F1  F 2  F 3 ........  g GR …(iv) 3 While nuclear force is many body interaction (6) It is the weakest force in nature : As F > F >F. GM 4 nuclear electromagnetic gravi ta tional (i) From the expression g  2 GR it is clear that its  (7) The ratio of gravitational force to electrostatic force between two R 3 value depends upon the mass radius and density of planet and it is electrons is of the order of 10 43. independent of mass, shape and density of the body placed on the surface (8) It is a conservative force i.e. work done by it is path independent of the planet. i.e. a given planet (reference body) produces same acceleration or work done in moving a particle round a closed path under the action of in a light as well as heavy body. gravitational force is zero. (9) It is an action reaction pair i.e. the force with which one body (say (ii) The greater the value of (M / R 2 ) or R, greater will be value earth) attracts the second body (say moon) is equal to the force with which of g for that planet. moon attracts the earth. This is in accordance with Newton's third law of motion. (iii) Acceleration due to gravity is a vector quantity and its direction is always towards the centre of the planet. Note :  The law of gravitation is stated for two (iv) Dimension [g] = [LT ] –2 point masses, therefore for any two arbitrary finite size bodies, as shown in the figure, It can not be applied as there is not unique value for the (v) it’s average value is taken to be 9.8 m/s or 981 cm/sec or 32 2 2 separation. feet/sec , on the surface of the earth at mean sea level. 2 (vi) The value of acceleration due to gravity vary due to the m1 m2 m1 m2 following factors : (a) Shape of the earth, (b) Height above the earth surface, (c) Depth below the earth surface and (d) Axial rotation of r the earth. r=? Fig. 8.3 Variation in g Due to Shape of Earth 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. Earth is elliptical in shape. It is gp flattened at the poles and bulged out at Acceleration Due to Gravity the equator. The equatorial radius is Rp The force of attraction exerted by the earth on a body is called about 21 km longer than polar radius, ge gravitational pull or gravity. GM Re from g  2 We know that when force acts on a body, it produces acceleration. R Therefore, a body under the effect of gravitational pull must accelerate. Fig. 8.5 GM The acceleration produced in the motion of a body under the effect At equator ge  Re2 of gravity is called acceleration due to gravity, it is denoted by g. …(i) Consider a body of mass m is lying on the surface of earth then gravitational force on the body is given by GM At poles g p  …(ii) R p2 GMm F  …(i) R2 R p2 ge From (i) and (ii)  2 Where M = mass of the earth and R = radius of the earth. gp Re If g is the acceleration due to gravity, then the force on the body Since Requator  R pole due to earth is given by Force = mass  acceleration  g pole  gequator and g p  ge  0.018 ms 2  Gravitation 391 Therefore the weight of body increases as it is taken from equator to So it is clear that if d increase, the value of g decreases. the pole. (ii) At the centre of earth d  R  g  0 , i.e., the acceleration due to gravity at the centre of earth becomes zero. Variation in g With Height (iii) Decrease in the value of g with depth Acceleration due to gravity at the surface of the earth dg Absolute decrease g  g  g  R GM g …(i) g g  g  d R2 Fractional decrease   g g R Acceleration due to gravity at height h from the surface of the earth g d Percentage decrease  100%   100% g g R GM h (iv) The rate of decrease of gravity outside the earth ( if h  R ) is g'  …(ii) (R  h)2 r g double to that of inside the earth. R 2 Variation in g Due to Rotation of Earth  R  O From (i) and (ii) g'  g  …(iii) As the earth rotates, a body placed  Rh  on its surface moves along the circular path Fig. 8.6 and hence experiences centrifugal force, due P R2 to it, the apparent weight of the body r Fc =g …(iv)  r2 decreases. mg mg  [As r = R + h] Since the magnitude of centrifugal force varies with the latitude of the place, (i) As we go above the surface of the earth, the value of g decreases therefore the apparent weight of the body 1 varies with latitude due to variation in the because g   2. magnitude of centrifugal force on the body. Fig. 8.8 r If the body of mass m lying at point P, whose latitude is , then due (ii) If r   then g  0 , i.e., at infinite distance from the earth, to rotation of earth its apparent weight can be given by m g  mg  Fc the value of g becomes zero. (iii) If h  R i.e., height is negligible in comparison to the radius or m g  (mg )2  (Fc )2  2mg Fc cos(180  ) then from equation (iii) we get  m g  (mg )2  (m  2 R cos )2  2mg m  2 R cos  ( cos ) 2 2  R   h  2h  [As Fc  m  2r  m  2 R cos  ] g  g   g1    g 1    R  h   R   R By solving we get g  g   2 R cos 2  [As h  R ] (iv) If h  R then decrease in the value of g with height : 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 2hg of earth makes with equatorial plane. It is denoted by . Absolute decrease g  g  g  R  For the poles   90 o and for equator   0 o g g  g  2h (i) Substituting   90 o in the above expression we get Fractional decrease   g g R g pole  g   R cos 90 2 2 o g 2h  g pole  g …(i) Percentage decrease  100%   100% g R i.e., there is no effect of rotational motion of the earth on the value Variation in g With Depth of g at the poles. Acceleration due to gravity at the surface of the earth (ii) Substituting   0o in the above expression we get GM g  2  GR 4 …(i) geqator  g   R cos 0 2 2 o R 3 Acceleration due to gravity at depth d from the surface of the earth  gequator  g   2 R …(ii) 4 g i.e., the effect of rotation of earth on the value of g at the equator g  G(R  d ) …(ii) d 3 P g is maximum. R r  d From equation (i) and (ii) From (i) and (ii) g   g 1    R O g pole  gequator  R 2  0.034 m / s 2 (iii) When a body of mass m is moved from the equator to the (i) The value of g decreases on going below the surface Fig. 8.7of the earth. poles, its weight increases by an amount From equation (ii) we get g  (R  d ).  392 Gravitation m(g p  ge )  m  2 R F  m ia or m i  F a (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 Hence inertial mass of a body may be measured as the ratio of the velocity of rotation of earth for which a body at the equator will become magnitude of the external force applied on it to the magnitude of weightless acceleration produced in its motion. g  g   2 R cos 2  (i) It is the measure of ability of the body to oppose the production of acceleration in its motion by an external force.  0  g   2 R cos 2 0 o [As   0 o for equator] (ii) Gravity has no effect on inertial mass of the body.  g  R  0 2 (iii) It is proportional to the quantity of matter contained in the body. g   (iv) It is independent of size, shape and state of body. R (v) It does not depend on the temperature of body. 2 R (vi) It is conserved when two bodies combine physically or or time period of rotation of earth T   2  g chemically. (vii) When a body moves with velocity v , its inertial mass is given Substituting the value of R  6400  10 3 m and g  10m / s 2 by we get m0 1 rad m , where m = rest mass of body, c = velocity of light   1.25  10  3 and T  5026.5 sec  1.40 hr. 0 800 sec v2 1 2 c Note :  This time is about 1 17 times the present time in vacuum, period of earth. Therefore if earth starts rotating 17 times faster then all (2) Gravitational Mass : It is the mass of the material of body, which objects on equator will become weightless. determines the gravitational pull acting upon it.  If earth stops rotation about its own axis then at the If M is the mass of the earth and R is the radius, then gravitational equator the value of g increases by  R and consequently the weight of 2 pull on a body of mass m g is given by body lying there increases by m  2 R. GMm g F F F or m g    After considering the effect of rotation and elliptical R2 GM /R 2 I shape of the earth, acceleration due to gravity at the poles and equator are related as Here m g is the gravitational mass of the body, if I  1 then g p  ge  0.034  0.018m / s 2  g p  ge  0.052m / s 2 mg  F Mass and Density of Earth Thus the gravitational mass of a body is defined as the gravitational pull experienced by the body in a gravitational field of unit intensity, Newton’s law of gravitation can be used to estimate the mass and density of the earth. (3) Comparison between inertial and gravitational mass (i) Both are measured in the same units. GM gR 2 As we know g  , so we have M  (ii) Both are scalar. R2 G (iii) Both do not depend on the shape and state of the body 9.8  (6.4  10 6 )2  M  5.98  10 24 kg  10 25 kg (iv) Inertial mass is measured by applying Newton’s second law of 6.67  10 11 motion where as gravitational mass is measured by applying Newton’s law 4 3g of gravitation. and as we know g  GR , so we have   3 4GR (v) Spring balance measure gravitational mass and inertial balance measure inertial mass. 3  9.8    5478.4 kg / m 3 (4) Comparison between mass and weight of the body 4  3.14  6.67  10 11  6.4  10 6 Mass (m) Weight (W) Inertial and Gravitational Masses It is a quantity of matter It is the attractive force exerted (1) Inertial mass : It is the mass of the material of the body, which contained in a body. by earth on any body. measures its inertia. Its value does not change with g Its value changes with g. If an external force F acts on a body of mass m, then according to i Its value can never be zero for At infinity and at the centre of Newton’s second law of motion any material particle. earth its value is zero. Its unit is kilogram and its Its unit is Newton or kg-wt and  Gravitation 393 dimension is [M]. dimension are [ MLT 2 ] Gm1 Gm 2 For point P, I1  I2  0   0 It is determined by a physical It is determined by a spring x2 (d  x )2 balance. balance. m1 d m2 d It is a scalar quantity. It is a vector quantity. By solving x and (d  x )  m1  m 2 m1  m 2 Gravitational Field (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 The space surrounding a material body in which gravitational force line. Field lines for an isolated mass m are radially inwards. 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 m unit mass (test mass) itself does not produce any change in the field of the body. GM GM So if a test mass m at a point in a gravitational field experiences a (x) As I  2 and also  Fig.g8.12  Ig r R2 force F then Thus the intensity of gravitational field at a point in the field is equal to acceleration of test mass placed at that point. F I  Gravitational Field Intensity for Different Bodies m (i) It is a vector quantity and is always directed towards the centre (1) Intensity due to uniform solid sphere of gravity of body whose gravitational field is considered. R (ii) Units : Newton/kg or m/s 2 (iii) Dimension : [M LT ] 0 –2 (iv) If the field is produced by a point mass M and the test mass I m is at a distance r from it then by Newton’s law of gravitation GM/R 2 GMm F , then intensity of gravitational field r r2 O r=R Test mass Fig. 8.13 F GMm / r 2 Outside the surface On the surface Inside the surface I  r m m m r>R r=R rR r=R rR r=R r