Title Med-RM_Phy_SP-2_Ch-8-Gravitation.pdf ✅

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Chapter Contents KEPLER’S LAWS (1) Law of orbits : Each planet revolve around the sun in an elliptical orbit with the Sun at one of the foci of the ellipse. Sun Aphelion Perihelion 2a F2 B A F1 ea 2b (2) Law of areas : Line joning sun and planet sweeps out equal area in equal time interval i.e. Areal speed of the planet is constant. AB AB    Area of portion SAB = Area of portion SAB A A S t B t + dt B ( ) t dt  t v1 2 v 1 2 2 dA L vr dt m   = constant vr v r 11 2 2  Angular momentum ‘L’ about the Sun for all planets is constant. (3) Law of periods : The square of time taken (T) for a planet to complete one revolution about the sun is proportional to cube of semi-major axis (a) of the elliptical orbit of the planet. i.e., T 2  a3 Chapter 8 Gravitation Kepler’s laws Newton’s law of Gravitation Gravitational field intensity ( ) I Relation between Acceleration due to Gravity (g) and Gravitational Constant (G) Gravitational potential energy Gravitational potential Escape speed Earth satellite Energy of orbiting satellite Shape of Trajectories Corresponding to Different Velocities Geostationary and polar satellite Binary star system
2 Gravitation NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Note : Eccentricity 2 2 1 b e a   Aphelion Perihelion S At aphelion Kinetic energy minimum Potential energy maximum At perihelion Kinetic energy maximum Potential energy minimum 1 1 min GM e v a e          1 1 max GM e v a e          Example 1 : The distance of two planets from the sun are 1013 m and 1012 m respectively. What is the ratio of their time periods? Solution : As T2  R3 ...(By Kepler’s third law)  2 3 1 1 2 3 2 2 T R T R   3 2 13 1 2 12 2 10 10 T T         2 1 3 2 10 T T         1 2 10 10 1 T T  Example 2 : A geostationary satellite is orbiting the earth at a height 6R above the surface of earth, where R is radius of earth. What will be the time period of another satellite at a height 2.5 R from the surface of earth? Solution : As T2  R3 ...(By Kepler’s third law)  2 3 1 1 2 3 2 2 T R T R  or 2 3 2 2 1 1           T R T R  3 2 2 2 1 3.5 7 R T T R         2 2 1 2 8 T T   T2 = 1 2 2 T T2 = 24 2 2 ...(∵ T1 = 24 hours)  T2 = 6 2 hours
NEET Gravitation 3 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 NEWTON’S LAW OF GRAVITATION The force of attraction between two point masses is directly propotional to product of their masses and inversely propotional to the square of distance between them. 1 2 2  Gm m F r m1 r m2 Where G is universal gravitation constant. G = 6.67 × 10–11 2 2 Nm kg Newton’s Law in Vector Form Force on m1 due to m2 1 2 12 12 3 12 | |   Gm m F r r m1 m2 F12 12 r Force on m2 due to m1 1 2 21 3 21 21 | |   Gm m F r r m1 m2 F21 21 r Principle of Superposition of Gravitational Force If number of masses placed in any region, then the resultant gravitational force on any one of them is the vector sum of gravitational forces exerted by all the other point masses. If masses m1, m2, m3 .... mn are placed in any region. Then gravitational force on m1 is 1 12 13 1 ... FF F F  n Example 3 : If the masses of two spherical bodies are quadrupled and the distance between their centres is doubled, then how many times the force of gravitation between them will be changed? Solution : 1 2 1 2 Gm m F r  1 2 2 2 Gm m F R  Here, M1 = 4m1, M2 = 4m2 and R = 2r  1 2 2 2 (4 )(4 ) (2 ) Gm m F r  = 1 2 2 16 4 Gm m r  F2 = 4F1 Hence, the gravitational force will increase to four times the initial magnitude.
4 Gravitation NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 GRAVITATIONAL FIELD INTENSITY ( ) l The intensity of the gravitational field at any point P is defined as the gravitational force per unit mass at that point. F I m The intensity of gravitational field is simply called “gravitational field”. Its SI unit is newton per kilogram (N/kg) and the dimensional formula is [LT –2]. It is a vector quantity. (i) Gravitational field intensity of point mass (m) at distance 'r' 2 Gm I r  (ii) Gravitational field intensity of uniform spherical shell at distance 'r' from its centre (a) If r < R [where R is radius of shell] I = 0 (b) If r R r O R GM R2 I 2 Gm I r  (iii) Gravitational field intensity of uniform solid sphere at distance 'r' from its centre (a) If r < R 3 GMr I R  (b) If r  R r O R GM R2 I 2 Gm I r  Example 4 : Three equal masses 2m each are placed at the vertices an equilateral triangle PQR (i) What is the force acting on a mass m placed at the centroid G of the triangle? (ii) What is the force on mass m if the mass at the vertex P is quadrupled? Take PG = QG = RG = 1 m 30° 2m 2m y x Q R P m G 2m Solution : (i) The angle between GR and the positive x-axis is 30°. Similarly the angle between GQ and negative x-axis is 30°. Now, (2 ) ˆ 1 GP G mm F j   (i.e., along positive y-axis)