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Two Bodies Central Force Problem Lecture 1

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Two Bodies Central Force Problem Lecture 1 Chapter 4 Lecture 1 Two body central Force Problem Akhlaq Hussain 4.1 Introduction ▪ One of the most important problems of classical mechanics is to understand the motion of a body moving under the influence of a central force field. ▪ Force which is always directed towards the centre or line joining two bodies ▪ The motion of the planets around the sun. 2 ▪ Motion of satellites around the earth ▪ Motion of two charge particles with respect to each other ▪ In this chapter, we study the two- body problem, which is reduced to an equivalent one-body problem. 3 The motion of a particle in central force field can be classified as; 1) Bound motion The distance between two bodies never exceeds a finite limit, e.g. motion of planets around the sun. 2) Unbound motion The distance between two particles or bodies is infinite at initial and final stage. The bodies move from infinite distance and approach to interact in close proximity and finally move far from each other to an infinite distance. For example, scattering of alpha particles by gold nuclei as studied by Rutherford. ▪ It is always possible to reduce the motion of two bodies to that of an equivalent single-body problem. ▪ The exact solution and understanding of two bodies motion problem is possible. ▪ However, the presence of the third body complicates the situation and an exact solution to the problem become an impossibility. ▪ Therefore, on must adopt the approximate methods to solve the many bodies problem. ▪ We can always reduce many body systems to a two-body problem either by neglecting the effects of the others or by some other screening methods, where the effects of the other bodies don’t play prominent role. ▪ Such as the motion of a planets around the sun, where the effect due to the presence of other planets is neglected. However, we will restrict ourselves to the two bodies problem only. Consider the motion of two particles. Let F(ext) be the total external force acting on the system. Let Fint be the total internal force due to the interaction between two particles. Total external force will be 풆풙풕 풆풙풕 풆풙풕 푭 = 푭ퟏ + 푭ퟐ (4.1.1) Further according to the Newton’s 3rd law 풊풏풕 풊풏풕 푭ퟏퟐ = −푭ퟐퟏ ` (4.1.2) Action and reaction forces. If the action and reaction forces are same Why only apple falls for earth? The equations of motion can be written as Force on Particle 1 풆풙풕 풊풏풕 푚1풓ሷ ퟏ = 푭ퟏ + 푭ퟏퟐ (4.1.3) Force on Particle 2 풆풙풕 풊풏풕 푚2풓ሷ ퟐ = 푭ퟐ + 푭ퟐퟏ (4.1.4) Total force on system of Particles 푭풆풙풕 = 푀푹ሷ (4.1.5) Total mass of the system 푀 = 푚1 + 푚2 (4.1.6) Position vector of the centre of mass of the system is 푚 풓 +푚 풓 ▪ 푹 = 1 ퟏ 2 ퟐ (4.1.7) 푚1+푚2 Position vector of particle 1 relative to particle 2 be 풓 = 풓ퟏ − 풓ퟐ (4.1.8) 풓ퟏ = 풓 + 풓ퟐ (4.1.9) Putting in Eq. (4.1.9) in Eq. (4.1.7) 푚1풓 풓ퟐ = 푹 − (4.1.10) 푚1+푚2 Similarly, Eq. (4.1.8) can be written as 풓ퟐ = 풓ퟏ − 풓 (4.1.11) Putting in equation (4.1.11) in equation (4.1.7) 푚2풓 풓ퟏ = 푹 + (4.1.12) 푚1+푚2 Multiplying Eq. (4.1.3) by m2 & Eq. (4.1.4) by m1 and subtracting, 풆풙풕 풆풙풕 풊풏풕 풊풏풕 푭ퟏ 푭ퟐ 푚1푚2 풓ሷ ퟏ − 풓ሷ ퟐ = 푚2푭ퟏퟐ − 푚1푭ퟐퟏ + 푚1푚2 − 푚1 푚2 풊풏풕 풊풏풕 Dividing the above equation by (푚1 +푚2) and using 푭ퟏퟐ = −푭ퟐퟏ 풆풙풕 풆풙풕 푚1푚2 (푚1+푚2) 풊풏풕 푚1푚2 푭ퟏ 푭ퟐ 풓ሷ ퟏ − 풓ሷ ퟐ = 푭ퟏퟐ + − (푚1+푚2) (푚1+푚2) (푚1+푚2) 푚1 푚2 풆풙풕 풆풙풕 풊풏풕 푭ퟏ 푭ퟐ ⇒ 휇 풓ሷ ퟏ − 풓ሷ ퟐ = 푭ퟏퟐ + 휇 − (4.1.13) 푚1 푚2 Where 휇 is reduce mass of the system. 푚 푚 1 1 휇 = 1 2 = + (4.1.14) (푚1+푚2) 푚1 푚2 Special case If no external force is acting 풆풙풕 풆풙풕 푭ퟏ = 푭ퟐ = ퟎ (A) equation (4.1.13) will be reduced to 풊풏풕 ⇒ 휇 풓ሷ ퟏ − 풓ሷ ퟐ = 푭ퟏퟐ 풊풏풕 ⇒ 휇풓ሷ = 푭ퟏퟐ (4.1.15)a If the forces produce same acceleration 풆풙풕 풆풙풕 푭ퟏ 푭ퟐ = ⇒ 풓ሷ ퟏ = 풓ሷ ퟐ (B) 풎ퟏ 풎ퟐ The condition B is realized if centre producing the external forces is at a considerable distance from the system and the force due to it on any mass is proportional to that of the mass. Such as gravitational force. In Earth-moon mutual motion, force due to the sun is assumed such that it satisfy the condition mentioned in Eq. B. Equation will be reduced to 풊풏풕 ⇒ 휇 풓ሷ ퟏ − 풓ሷ ퟐ = 푭ퟏퟐ 풊풏풕 ⇒ 휇풓ሷ = 푭ퟏퟐ (4.1.15)b Eq. (4.1.15)b represent motion of a particle of mass equal 휇 and moving 풊풏풕 under the action of force 푭ퟏퟐ . The reduction is equivalent to replace the system of two bodies by a mass 휇 and considering the acceleration produced is due to the internal force. 풊풏풕 풆풙풕 ሷ Eq. (4.1.15)a ( 휇풓ሷ = 푭ퟏퟐ ) together with Eq. (4.1.5) ( 푭 = 푀푹 ) represents the motion of a two body system under the action of internal and external forces as long as the conditions mentioned in equations A & B are valid. If the internal forces are attractive and these are the only forces acting on the system, the two bodies move around the centre of mass which acts as centre of force. i.e. directed towards the centre. Condition on mass If the mass of one of the particles is extremely large as compared to that of the other, say m1 >> m2, then the reduced mass is simply 푚1푚2 푚1푚2 휇 = = 푚2 (푚1+푚2) 푚1(1+ ൗ푚1) 푚2 푚2 ⇒ 휇 = 푚2 as Τ푚1 ≈ 0 (1+ ൗ푚1) ⇒ 휇 = 푚2 In this case the centre of mass of the system coincides with the centre of mass of the heavier body. This approximation is equivalent to neglecting the recoil of mass m1. This is used in Bohr’s theory of hydrogen atom and motion of satellites around the earth. It can be assumed for the motion of earth around the Sun. Since mass m1>>m2, acceleration in mass m1 풊풏풕 푭ퟐퟏ 풂ퟏ = ≈ ퟎ 풎ퟏ acceleration in mass m2 풊풏풕 푭ퟐퟏ 풂ퟐ = > ퟎ 풎ퟐ That’s is why “An apple appears to fall towards the earth and not the earth towards the apple”. Lagrangian of the System If 푈 풓, 풓ሶ is the function of “풓”and higher derivative of “풓ሶ”. Then Lagrangian of the system can be written as 퐿 = 푇 푹ሶ , 풓ሶ − 푈 풓, 풓ሶ (4.1.16) 1 ퟐ ′ 1 ퟐ ′ Where 푇 푹ሶ , 풓ሶ = Τ2 푀푹ሶ + 푇 = Τ2 (푚1+푚2)푹ሶ + 푇 (4.1.17) 1 1 And 푇′ = 푚 풓ሶ ′ ퟐ + 푚 풓ሶ ′ ퟐ (4.1.18) 2 1 1 2 2 ퟐ ′ Where 풓ퟏ = 풓ퟏ − 푹 ′ 푚1풓ퟏ+푚2풓ퟐ 푚2 풓ퟏ−풓ퟐ ⇒ 풓1= 풓ퟏ − = 푚1+푚2 푚1+푚2 ′ 푚2 ⇒ 풓1= 풓 (4.1.19) 푚1+푚2 ′ Similarly, 풓2 = 풓ퟐ − 푹 ′ 푚1풓ퟏ+푚2풓ퟐ 푚1 풓ퟏ−풓ퟐ ⇒ 풓ퟐ= 풓ퟐ − = − 푚1+푚2 푚1+푚2 ′ 푚1 ⇒ 풓ퟐ= − 풓 (4.1.20) 푚1+푚2 Lagrangian of the System Therefore, the kinetic energy from Eq (4.1.18) can be written as ퟐ ퟐ ′ 1 푚2 1 푚1 푇 = 푚1 풓ሶ + 푚2 − 풓ሶ 2 푚1+푚2 2 푚1+푚2 ′ 1 푚1푚2 ퟐ ⇒ 푇 = 푚2 + 푚1 2 풓ሶ 2 푚1+푚2 1 푚 푚 ⇒ 푇′ = 1 2 풓ሶ ퟐ (4.1.21) 2 푚2+푚1 The Lagrangian of the system can be written as; 퐿 = 푇 푹ሶ , 풓ሶ − 푈 풓, 풓ሶ 1 ퟐ 1 푚1푚2 ퟐ 퐿 = Τ2 푚1 +푚2 푹ሶ + 풓ሶ − 푈 풓, 풓ሶ 2 푚1+푚2 1 퐿 = 1Τ 푀푹ሶ ퟐ + 휇풓ሶ ퟐ − 푈 풓, 풓ሶ (4.1.22) 2 2 Where M is the total mass of the system and 휇 is the reduce mass of the system. 4.2 Properties of central Force 4.2.1a Under the central force, the angular momentum of the particle is conserved a. In cartesian coordinates The Torque on the system (if any) can be written as; 푵 = 풓 × 푭 (4.2.1) and the angular momentum of the body is l= 풓 × 푷 (4.2.2) 풅l We know that; = 푵 (4.2.3) 풅풕 Since the force acting on the body is central force and always directed towards the line joining the body with the centre therefore 푵 = 풓 × 푭 = 푟푟Ƹ × 퐹푟푟Ƹ = 푟퐹푟 푟Ƹ × 푟Ƹ = 0 (4.2.4) 풅l ⇒ = 푵 = ퟎ ⇒ l = 퐶표푛푠푡푎푛푡 (4.2.5) 풅풕 Eq.(4.2.4) & (4.2.5) suggests that the total torque “푵” acting on the system will be zero and angular momentum “푳” of the body will be constant. 4.2 Properties of central Force b. In Polar coordinates 푭 = 퐹푟푟Ƹ + 퐹휃휃መ (4.2.6) And similarly, the torque acting on a particle in polar coordinates is 푵 = 풓 × 푭 = 푟푟Ƹ × 휇푟ሷ − 푚푟휃ሶ 2 푟Ƹ + 휇푟휃ሷ + 2휇푟ሶ휃ሶ 휃መ ⇒ 푵 = 푟 휇푟ሷ − 휇푟휃ሶ 2 푟Ƹ × 푟Ƹ + 푟 휇푟휃ሷ + 2휇푟ሶ휃ሶ 푟Ƹ × 휃መ ⇒ 푵 = 0 + 푟 휇푟휃ሷ + 2휇푟ሶ휃ሶ 푟Ƹ × 휃መ ⇒ 푵 = 휇푟2휃ሷ + 2휇푟푟ሶ휃ሶ 푛ො where 푛ො is ⏊ to both 푟Ƹ and 휃መ 푑 ⇒ 푵 = 휇푟2휃ሶ 푛ො (4.2.7) 푑푡 For Radial force, the angular part of the force is zero 푑l 푵 = = 0 ⇒ l = 휇푟2휃ሶ = 퐶표푛푠푡푎푛푡 (4.2.8) 푑푡 Note: Also l= 풓 × 푷 = 풓 × 휇풗 = 풓 × 휇푟흎 = 풓 × 휇푟휽ሶ ⇒ 푳 = 휇푟2휃ሶ 4.2 Properties of central Force 4.2.2 The path of a particle moving under the central force must be a Plane Consider the central force 푭 = 퐹푟푟Ƹ (4.2.9) Taking cross product with radius vector of above equation 풓 × 푭 = 풓퐹푟 푟Ƹ × 푟Ƹ = 0 푑풗 ⇒ 풓 × 푭 = 풓 × 휇 = 0 푑푡 푑풗 푑 ⇒ 풓 × 휇 = 휇 풓 × 풗 = 0 푑푡 푑푡 푑 ⇒ 풓 × 풗 = 0 (4.2.10) 푑푡 Integrating above equation 풓 × 풗 = 풒 = 푐표푛푠푡푎푛푡 (4.2.11) Since the vector “풒” is perpendicular to both “풓” and “풗”is zero 풓 ∙ 풓 × 풗 = 풓 ∙ 풒 = 0 Therefore, the particle is in Plane.
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