Title 7. Gravitation.pdf ✅

Gravitation 133 7 Gravitation QUICK LOOK The gravitational force between two particles is independent of the presence of other bodies or the properties of the intervening medium. Gravitational force is conservative force, therefore work done in a body form one place to another is independent of path; it depends only on initial and final positions. The mass of air bubble in material medium is negative. Newton’s law of gravitational force 1 2 2 Gm m F r = (Attractive) Figure: 7.1 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. 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 body is given by F 2 GMm R = . . . (i) Where M = mass of the earth and R = radius of the earth. Figure: 7.2 Earth 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) From (i) and (ii) we have mg 2 GMm R = ∴ 2 GM g R = . . . (iii) ⇒ 3 2 4 3 G g R R π ρ   =     [As mass (M) = volume ( 4 3 3 πR ) × density (ρ)] ∴ 4 3 g GR = πρ . . . (iv) Dimension [g] = [LT–2]. Acceleration due to gravity is a vector quantity and its direction is always towards the centre of the planet. 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. Note From the expression 2 4 3 GM g GR R = = πρ it is clear that its value depends upon the mass radius and density of planet and it is independent of mass, shape and density of the body placed on the surface of the planet. i.e. a given planet (reference body) produces same acceleration in a light as well as heavy body. The greater the value of 2 ( / ) M R or ρR, greater will be value of g for that planet. Value of g at the height h, 2 2 ( ) 1 e h e e GM g g R h h R = = +     +   , If 2 , 1 e h e h h R g g R   << = −     Value of g at depth x form earth’s surface 1 x e x g g R   = −     (assuming earth of uniform density) Value of g due to earth’s rotation of latitude λ ⇒ 2 2 cos e g g R ′ = − ω λ R x h w Figure: 7.3 Figure: 7.4 g O r Re Below earth’s surface Above the Earths surface m mg R Earth m m F F r
134 Quick Revision NCERT-PHYSICS Table 7.1: Distance vs. Value of g Distance above from surface of Earth Value of g Mass Weight 20,000 mi (38,400 km) 1 ft/s2 (0.3 m/s2 ) 70.0 kg 4.7 lb (21 N) 60,000 mi (25,600 km) 1.3 ft/s2 (0.4 m/s2 ) 70.0 kg 6.3 lb (28 N) 16,000 mi (25,600 km) 1.3 ft/s2 (0.4 m/s2 ) 70.0 kg 6.3 lb (28 N) 12,000 mi (19,200 km) 2 ft/s2 (0.6 m/s2 ) 70.0 kg 9.5 lb (42 N) 8,000 mi (12,8000 km) 3.6 ft/s2 (1.1 m/s2 ) 70.0 kg 17 lb (77 N) 4,000 ml (6,4000 km) 7.9 ft/s2 (2.4 m/s2 ) 70.0 kg 37 lb (168 N) 0 ml (0 km) 32 ft/s2 (9.80 m/s2 ) 70.0 kg 154 lb (686 N) 4,000 ml (6,400 km) Gravitational Field Strength: g F f m = Newton/kg Due to point mass m at distance r, g 2 GM f r = Due to a uniform solid sphere at external point g 2 GM f r = (towards centre of sphere) At internal point distant r form centre g 3 GMr f R = (towards centre of sphere) Due to a hollow sphere at external point g 2 GM f r = (towards center of sphere) At internal point fg = 0 (everywhere) Point of zero intensity: If two bodies A and B of different masses m1 and m2 are d distance apart. Let P be the point of zero intensity i.e., the intensity at this point is equal and Opposite due to two bodies A and B and if any test mass placed at this point it will not experience any force. Figure: 7.5 For point P, 1 2 I I + = 0 ⇒ 1 2 2 2 0 ( ) Gm Gm x d x − + = − By solving 1 1 2 m d x m m = + and 2 1 2 ( ) m d d x m m − = + Intensity due to uniform circular ring Figure: 7.6 At a point on its axis 2 2 3/ 2 ( ) GMr I a r = + At the centre of the ring I = 0 Intensity Due to Uniform Disc Figure: 7.7 At a point on its axis 2 2 2 2 1 1 GMr I ra r a   = −     + or 2 2 (1 cos ) GM I a = − θ At the Centre of the Disc I = 0 Gravitational Potential: g W V m = Joule/K Due to a point mass g GM V r = − Due to a uniform solid sphere of radius R at external point, g G M V r = − At internal point ( ) 2 2 3 3 2 g G M V R r R = − − Due to a hollow sphere, at external point g G M V r = − At internal point, g G M V R = − (same as on surface) Gravitational Potential Energy Gravitational potential energy U of a point mass m placed in the gravitational field of a point mass M is given by the work done in moving the point mass m from infinity to the point in question, i.e., distance r from the point mass M, i.e., G M G M m U mV m r r     = = = −         The change in potential energy when a point mass m is moved vertically upwards through a height h from the surface of the θ a P I r a P I r A m1 m2 x I1 d B P d–x I2
Gravitation 135 earth is given by: ( ) mgRh U R h ∆ = + When h R << : 1 / ( ) mgh U mgh R h   ∆ = ≈   +   When h R >> : 1 / ( ) mgR U mgh R h   ∆ = ≈   +   Satellite in Circular Orbit The time period of communication satellite is 24 h and its plane coincides with equatorial plane of earth. The height of communication satellite is nearly 36000 km above earth’s surface. If a satellite is close to earth’s surface such that h R << , then orbital speed 0 GM v g = 2 R g Rg R = = and periodic time, 3 3 2 2 2 2 R R R T GM R g g π π π     = = =         For earth 6 R = = × 6400 m 6.4 10 m and g = 9.8 m/s 2 Orbital speed 9 0 v = × × (6.40 10 9.8) 3 = × = 8 10 m / s 8km / s and period of revolution, 6 6.40 10 6 5 10 sec 84.6min 9.8 T × = = × = Note The bodies are weightless in the satellite because the centrifugal force on body in rotating satellite is balanced by gravitational pull of earth Elliptical Orbit: Properties of ellipse are: If a and b are semi- major and semi-minor axes, then 2 2 1 b a ε = − Maximum distance 1 r a = + (1 ) ε Minimum distance 2 r a = + (1 ) ε ⇒ max min 1 1 v v ε ε + = − Semi-latus recturn 2 1 2 2 1 2 b 2r r L a r r = = + The torque on a satellite is zero both in circular and elliptical orbits, so Angular momentum of satellite is conserved in both circular and elliptical orbits, while velocity is minimum at maximum distance (r1) and velocity is maximum at minimum distance r2. J = mvr = constant Maximum height attained by a projectile projected vertically upward 2 2 1 r H Rg v = − Escape velocity 2 2 e e e c G M v R g R = = The total energy of just escaping body is zero. Relation between orbital velocity and escape velocity near earth’s surface 0 2 e v v = For escape of an orbiting satellite near earth’s surface, its speed must be increased 0 0.414 41.4% v = Time Period: Period of revolution of a satellite is ( ) 1/ 2 3 2 R h T R g π     + =          For a satellite revolving very near to the surface of earth 1/ 2 2 84.6 R T g π   = =     min. Like orbital speed, time period also does not depend on mass, size and shape of the satellite. It depends only on the radius of circular orbit. Kinetic energy 2 0 1 2 E mv k = 1 2 2 GM GMm m R h r = ⋅ = + Potential energy p GM E m R h = − + 2 k GMm E r = − = − Total energy of the satellite 2 E E E E K = + = − k p k k = −Ek Total energy of satellite 2 GM me E r = − Binding energy of satellite 2 e B GM m E E r = − = Geostationary Satellite A satellite which appears to be stationary for a person on the surface of the earth is called geostationary satellite. It revolves in the equatorial plane form west to east with a time period of 24 hours. Its height from the surface of the earth is nearly 35600 km and radius of the circular orbit is nearly 42000 km. The orbital velocity of this satellite is nearly 3.08 km/sec. The relative velocity of geostationary satellite with respect to earth is zero. The orbit of a geostationary satellite is called as parking orbit.
136 Quick Revision NCERT-PHYSICS MULTIPLE CHOICE QUESTIONS Newton's Law of Gravitation 1. If the distance between two masses is doubled, the gravitational attraction between them: a. Is doubled b. Becomes four times c. Is reduced to half d. Is reduced to a quarter 2. The earth (mass = 24 6 10 kg) × revolves round the sun with angular velocity 7 2 10− × rad/s in a circular orbit of radius 8 1.5 10 × km.The force exerted by the sun on the earth in newtons, is: a. 25 18 ×10 b. Zero c. 39 ×1027 d. 21 36 ×10 3. Four particles of masses m, 2m, 3m and 4m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be: a. 2 2 24m G a b. 2 2 6m G a c. 2 2 4 2Gm a d. Zero 4. If the change in the value of ‘g’ at a height h above the surface of the earth is the same as at a depth x below it, then: (both x and h being much smaller than the radius of the earth) a. x h = b. x h = 2 c. 2 h x = d. 2 x h = 5. 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: a. 2 D b. 2 3 D c. 4 3 D d. 9 10 D Acceleration Due to Gravity 6. 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: a. g / 4 b. g / 5 c. g / 6 d. g / g 7. The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is: a. 2 9.8 / sec m b. 2 m sec/9.4 c. 2 m sec/980 d. 2 m sec/6.19 8. 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: a. Decrease by 2% b. Remain unchanged c. Increase by 2% d. Increase by 1% 9. The radii of two planets are respectively R1 and R2 and their densities are respectively ρ1 and ρ 2 . The ratio of the accelerations due to gravity at their surfaces is: a. 1 2 1 2 2 2 1 2 g g: : R R ρ ρ = b. 1 2 1 1 g g R : : = ρ R ρ 22 c. 1 2 1 2 2 1 g g R R : : = ρ ρ d. 1 2 1 2 1 2 g g R R : : = ρ ρ Variation in g Due to Shape of Earth 10. 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 11. If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will be: a. 2 g b. 4g c. 4 g d. 2g Variation in g with Height 12. At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of earth is: (R = radius of earth) a. 28 N b. 16 N c. 32 N d. 72 N Variation in g with Depth 13. 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: a. 2% decrease b. 0.5% decrease c. 1% increase d. 0.5% increase 14. The depth d at which the value of acceleration due to gravity becomes 1/n times the value at the surface, is: [R = radius of the earth]