Tiêu đề GRAVITATION.pdf ✅

 Digital www.allendigital.in [ 167 ] What is Gravitation Gravitation is the property of matter due to which, two bodies, always attract each other. The discovery of the law of gravitation The way the law of universal gravitation was discovered is often considered as the paradigm of modern scientific technique. The major steps involved were : • The hypothesis about planetary motion given by Nicolaus Copernicus (1473–1543). • The careful experimental measurements of the positions of the planets and the Sun by Tycho Brahe (1546–1601). • Analysis of the data and the formulation of empirical laws by Johannes Kepler (1571–1630). The development of a general theory by Isaac Newton (1642–1727). Newton's Law of Gravitation It states that every particle in the universe attract all other particles with a force which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. 12 21 F F F = = g Fg  m1m2 ...(i) Fg  2 1 r ...(ii) By equation (i) & (ii)  Fg  1 2 2 m m r 1 2 g 2 Gm m F r = Universal Gravitational Constant "G" • Universal Gravitational constant is a scalar quantity. • Value of G : SI : G = 6.67 × 10–11 N–m2/kg2 ; CGS : G = 6.67 × 10–8 dyne–cm2/g2 Dimensions : [M–1L3T–2] • Its value is same throughout the universe; G does not depend on the nature and size of the bodies; it does not depend even upon the nature of the medium between the bodies. • Its value was first found out by the scientist "Henry Cavendish" with the help of "Torsion Balance" experiment. Note: This formula is given for point masses. For spherical objects (planets), distance between centers is taken. Illustration 1: Two spherical balls of mass 10 kg each are placed 100 m apart. Find the gravitational force of attraction between them. Solution: 11 1 2 13 2 2 Gm m 6.67 10 10 10 F 6.67 10 N r (100) −    − = = =  r m1 m2 FሬԦ 12 FሬԦ 21 04 Gravitation
NEET : Physics [ 168 ] www.allendigital.in  Digital Illustration 2: Two spheres of mass 1034 kg each are placed with their centres 1010m apart. Find the force of gravitation acting between them. Solution: 1 2 g 2 Gm m F r = 11 34 34 10 2 6.67 10 10 10 (10 ) −    = 37 =  6.67 10 N Illustration 3: Two solid spheres of same size of a certain metal are placed in contact with each other. Prove that the gravitational force acting between them is directly proportional to the fourth power of their radius. Solution: The mass of the spheres may be assumed to be concentrated at their centres. So 3 3 2 2 4 4 2 4 4 G R R 3 3 4 F (G )R F R (2R) 9                  = =     Illustration 4: Two particles of mass 1kg and 2kg are placed at a separation of 50cm. Assuming that the only forces acting on the particles are their mutual gravitation. Find the initial acceleration of heavier particle. Solution: we know that 1 2 g 2 Gm m F r =  11 g 2 6.67 10 2 1 F (0.5) −    =  Fg = 5.34 × 10–10 N Acceleration of heavier particle (a1) = g 1 F m  10 1 5.34 10 a 2 −  =  a1 = 2.67 × 10–10 m/s2 (1) Gravitational force is always attractive. (2) Gravitational forces are developed in the form of action and reaction pair. They obey Newton's third law of motion. (3) It is the weakest of all fundamental forces in nature. Strong Nuclear force > Electromagnetic force > Weak forces > Gravitational force N E W G    (4) To see the effect of gravitational pull practically, at least one of the bodies must be celestial. (5) It is two body interaction i.e. gravitational force between the two particles is independent of the presence or absence of other bodies or particles. (6) Force of gravitation does not depend on medium. (7) Force developed between any two objects is called gravitational force and force between Earth and any other object is called force of gravity. (8) Force of gravitation always acts along the line joining two particles, so it is a central force. Fg Fg here m1 = 2kg m2 = 1kg r = 50 cm = 0.5 m M R M R
Gravitation  Digital www.allendigital.in [ 169 ] (9) Gravitational force is conservative force, so work done by gravitational force does not depends on path. (10) If any particle moves along a closed path under the action of gravitational force then the work done by this force is always zero for round the trip. (11) For the system of two particles,  For system linear momentum of system is conserved Fnet = 0  For particles linear momentum of particle is not conserved Fnet  0 Force due to multiple particles F F F F .......... net 1 2 3 = + + + Illustration 5: Find net force on 2kg due to 3kg and 4kg particles. Solution: We know that Fg = 1 2 2 Gm m r  F1 = 11 2 6.67 10 2 3 (3) −    = 4.44 × 10–11N Similarly F2 = 11 2 6.67 10 2 4 (2) −    = 13.34 ×10–11N So Net force  Fnet = F2 – F1 = 13.34 ×10–11 – 4.44 × 10–11 = 8.9 × 10–11N Illustration 6: Find net force on any one of the particles. Solution: Here F = 2 Gmm a = 2 2 Gm a Fnet = 2 2 F F 2(F)(F)cos60 + +   Fnet = 2 2 2 1 F F 2F 2 + +  = 222 FFF + + = 3F  2 net 2 3Gm F a = m1 Fg Fg m2 Particle system 3kg 2kg 4kg 3m 2m 2 2 R A B 2ABcos = + +  For equilateral triangle  = 60° Here F1 = Force on 2kg due to 3kg mass F2 = Force on 2kg due to 4kg mass F1 2kg F2 m m m a a a F F m m m a a a
NEET : Physics [ 170 ] www.allendigital.in  Digital Illustration 7: Infinite particles each of mass m are placed at positions x = 1m, x = 2m, x = 4m ........ . Find the magnitude of gravitational force on particle at origin (x = 0) Solution: We know that, Fg = 1 2 2 Gm m r So, the gravitational force on mass m placed at x = 0 Fg = 222 2 2 2 Gm Gm Gm ............. (1) (2) (4) +++  Fg = 2 2 2 2 1 1 1 Gm ........ 1 2 4   + + +      Fg = 2 1 1 1 Gm ........ 1 4 16   + + +      Fg = 2 1 Gm 1 1 4         −   = 2 4 Gm 3       = 2 4Gm 3 Illustration 8: A particle of mass m is placed at a distance "d" from one end of mass M and length L as shown in figure. Find the gravitational force on the particle by the rod. Solution: Since Rod is not a point mass. dFg = 2 Gmdm x  F d L g 2 0 d M Gm dx L dF x +       =   = d L 2 d GMm x dx L + −   Fg = d L 1 d GMm x L 1 + −       − = d L d GMm 1 L x + −        Fg = GMm 1 1 L d L d −   −     +  Fg = GMm d (d L) L d(d L) − − +       + = GMm d(d L) + m m m m x=0 x=1 x=2 x=4 m d M,L m d dm dx M,L x from geometric progression a S 1 r  = − Here a = first term r = common ratio a = 1 1 r 4 = By unitary method L ⎯→ M L ⎯→ M L dx ⎯→ M L dx  dm.  dm = M L dx.