3. Consider a mechanical system that consists of massive particles inter- acting

Question

3. Consider a mechanical system that consists of massive particles inter- acting via Newtonian gravity. This question will be very simple once you but there are (only a few) simple modifications. We shall build up the Lagrangian in stages. have done question I above; a) Let the two particles be at positions T1(t) and T2(t), and have masses m1 and m2 respectively What is the (non-relativistic) kinetic energy? b) Now: i. According to Newton's law of gravitational attraction, what is the force F1-2 that particle 1 exerts on particle 2? ii. According to Newton's law of gravitational attraction, what is the force F2-,1 that particle 2 exerts on particle 1? iii. What, therefore, is the gravitational potential energy? c) From these two pieces of information, write down the Lagrangian for this 2-body system Note that the Lagrangian will depend on both m and m2, the time derivatives x1(t), and (t), the positions 2,(t) and x2(t), plus the proportionality constant G that appears in Newton's law of gravitational attraction. (d) Check that when you evaluate the Euler-Lagrange equations for this Lagrangian you reproduce (for each of the two particles in dividually) the 3-dimensional Newton's equations of motion for a force determined by Newton's law of gravitational attraction.

Explanation / Answer

Given that the positions of the objects are the functions of time.

Therefore, you can get the derivative of the positions with respect to time and calculate the velocity.

--> v1 = d(x1(t))/dt

--> v2 = d(x2(t))/dt

3.a) Now that you have got the velocities, you can plug these quantities in the non-relavistic kinetic energy formula. Since there are 2 particles, you can just add the kinetic energies of individual particle to each other.

--> K.E. = 1/2* m1 * (v1)2 + 1/2* m2 * (v2)2

3.b) i) & ii)

The force exerted by particle 1 on particle 2 will be same in magnitude as that of by particle 2 on particle 1. This force is given by the Newton's famous equation of gravitational attraction.

--> Fattraction = Gm1*m2/ r2    (r is the distance between these particles)

We should express r in terms of x1(t) and x2(t)--> r = |x2 - x1|        

Therefore,

|F1-->2 |= |F2-->1| = Gm1*m2/ |x2 - x1|2 = Gm1*m2/ |r|2

3.b) iii) The expression for gravitational potential enegry is :

--> Ug(r) = - Gm1*m2/ |x2 - x1|            . . . (r = |x2 - x1|)

3. c) Lagrangian (L) of the system: Lagrangian of the system is the difference between total kinetic energy and potential energy of the system. It is given as following:

L = T - U . . . (T = total K.E. and U = potential enegy)

Thus,

L = 1/2* m1 * (v1)2 + 1/2* m2 * (v2)2 + Gm1*m2/ |x2 - x1|    . . . (Remember the sign of gravitational potential enegy)

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.