Assignment 11

Problem 1:

Show that the second Kepler law, the radius vector of a planet covers equal areas in equal times, follows directly from the angular momentum conservation law.

Solution:

• Concepts:
  Kepler's second law, conservation of angular momentum
• Reasoning:
  Consider a particle moving in a central potential U = U(r).
  The motion is in a plane.  Choose this plane to be the x-y plane.  Then
  L = T - U = ½m[(dr/dt)² + r²(dφ/dt)²] - U(r).
  E = T + U = ½m[(dr/dt)² + r²(dφ/dt)²] + U(r) = constant, energy is conserved.
  M = p_φ = ∂L/∂(dφ/dt) = mr²(dφ/dt) = constant, angular momentum is conserved.
• Details of the calculation:
  Areal velocity = dA/dt = ½r(rdφ/dt) = ½r²(dφ/dt) = constant.
  The area swept out per unit time is constant for all central potentials.

Problem 2:

A satellite is in a circular orbit of radius r around an airless spherical planet of radius R.  An asteroid of equal mass falls radially towards the planet, starting at zero velocity from a very large distance.  The satellite and the asteroid collide inelastically and stick together, moving in a new orbit that just misses the planet's surface.  What was the radius r of the satellite's original circular orbit in terms of R?

Solution:

• Concepts:
  Kepler orbits
• Reasoning:
  Since M >> m, assume the CM is the center of the planet.  For a single object moving in a central potential the energy E and the angular momentum L are conserved.  The motion is in a plane. For an object with mass m' in a circular orbit about a fixed center with mass M we have
  T = ½m'v² = ½GMm'/r,  U = -GMm'/r,  E = -½GMm'/r, 
  p = m'v = (GMm'²/r)^½,  L = rp = (GMm'²r)^½.
  In an inelastic collision momentum and angular momentum are conserved.
• Details of the calculation:
  Just before the collision:
  For the orbiting mass m, E₁ = -½GMm/r,  T₁ = ½GMm/r,  p₁ = (GMm²/r)^½,  L₁ = (GMm²r)^½.
  For the falling mass m, E₂ = 0,  T₂ = GMm/r,  p₂ = (2GMm²/r)^½,  L₂ = 0.

  Just after the collision:
  Momentum and angular momentum are conserved in the totally inelastic collision.  The momenta of the two colliding objects are perpendicular to each other.  Therefore the magnitude of the momentum of the combined mass is p = (p₁² + p₂²)^½,  p² = 3GMm²/r.
  The magnitude of the angular momentum of the combined object is  L = (GMm²r)^½.
  Its energy is E = T + U = p²/4m - 2GMm/r = -(5/4)GMm/r.
  
  From now on we have motion of a single object in a central potential.
  Energy and angular momentum are conserved.
  At the distance of closest approach L = p_maxr_min = p_maxR = (GMm²r)^½, p_max²R² = GMm²r.
  E = p_max²/(4m) - 2GMm/R = -(5/4)GMm/r.
  GMm²r/(4mR²) - 2GMm/R = -(5/4)GMm/r.
  r² - 8Rr + 5R² = 0,  r = 4R ± √(11)R.
  The perihelion of the orbit is at R, so r > R.
  r =  4R + √(11)R.

Problem 3:

Two particles with the same mass m orbit a massive planet of mass M >> m and radius R.  The orbit of one of the particles is circular, with radius 4R.  The other particle's orbits in the same plane, but its orbit is elliptical with semi-major axis 8R and r_min = 4R.  The angular momenta of the two particles point in opposite directions.  The particles collide and stick together, forming a new object.
(a)  What is the semi-major axis of the orbit of the new object in terms of R?
(b)  Does the new object collide with the planet?

Solution:

• Concepts:
  Kepler orbits
• Reasoning:
• For Kepler orbits:
  All orbits with the same semi-major axis have the same total energy per unit mass and the same period.
  Let us write down all energies per unit mass m.
  For a circular orbit of radius r we have v²/r = GM/r²,  T = ½GM/r, U = -GM/r,  E = -½GM/r,  r = ½GM/|E|.
• Details of the calculation:
  (a)  For the circular orbit E = -½GM/(4R),  for the elliptical orbit  E' = -½GM/(8R) = E/2.
  
  The kinetic energy of the particle in the circular orbit is T = GM/(8R) = -E and its speed is v = (GM/(4R))^½.
  
  At r = r_min the particle in the elliptical orbit has kinetic energy T' = E' - U(4R) =  E/2 + GM/(4R) = E/2 - 2E = -3E/2.
  T'/T = 3/2,  v'/v = √(3/2) is the ratio of the speeds.
  The collision conserves momentum.  mv' - mv = 2mv'',  v'' = (v' - v)/2 = (√(3/2) - 1)v/2.
  
  For the new object:
  T''(per unit mass) = ½v''² = (5/2 - 2√(3/2))v²/8  = T(5/2 - √6)/4  = -E(5/2 - √6)/4 = -0.0126E.
  U''(per unit mass) = -GM/(2R) = 2E,  E'' = 1.987 E. 
  R'' = 4R/1.987 = 2.013 R is the semi-major axis of the orbit of the new object.
  
  (b)  r'_max = 4R,  r''_min = 2 R'' - 4R = 0.0262 R.
  The new object collides with the planet.

Problem 3:

A particle of mass m moves under the action of a central force whose potential energy function is U(r) = kr³,  k > 0. 
(a)  For what energy and angular momentum will the orbit be a circle of radius a about the origin?  What is the period of this circular motion? 
(b)  If the particle is slightly disturbed from this circular motion, what will be the period of small radial oscillations about r = a?

Solution:

• Concepts:
  A particle moving in a central potential
  U = U(r),  F = f(r)(r/r),  f(r) = -∂U/∂r.
  In a central potential the energy E and the angular momentum M are conserved.
  E = ½m(dr/dt)² + M²/(2mr²) + U(r) = ½m(dr/dt)² + U_eff(r),
  with U_eff(r) = M²/(2mr²) + U(r).
  md²r/dt² = -∂U_eff(r)/∂r.
• Reasoning:
  A potential energy function U(r) is given.  The particle moves under the action of a central force.
• Details of the calculation:
  (a)  For a circular orbit with radius a we need r = a, dr/dt = 0, d²r/dt² = 0.
  For a circular orbit at with radius a we therefore need ∂U_eff(r)/∂r|_a = 0.  
  ∂U_eff(r)/∂r =  3kr² - M²/(mr³).
  ∂U_eff(r)/∂r|_a = 0 --> 3ka² = M²/(ma³),  a⁵  =  M²/(3mk),  M = (3kma⁵)^½.
  E = U_eff(a) = ka³ + M²/(2ma²) = ka³ + 3kma⁵/(2ma²) = 5ka³/2.
  The period for the circular motion is τ = 2π/ω, ω = dΦ/dt = M/ma² = (3ka/m)^½.
  τ = 2π/(3ka/m)^½.
   
• (b)  For a stable orbit we need a restoring force.  Let r = a + ρ, r >> ρ.
  We need md²ρ/dt² = -αρ.
  md²ρ/dt² = -∂U_eff(ρ)/∂ρ = -(∂U_eff/∂ρ)|_ρ=0 - (∂²U_eff/∂ρ²)|_ρ=0ρ (series expansion).
  We need ∂²U_eff/∂ρ² > 0 at ρ = 0, or ∂²U_eff/∂r² > 0 at r = a.
  ∂U_eff(r)/∂r =  3kr² - M²/(mr³).
  ∂²U_eff(r)/∂r² = 6kr + 3M²/(mr⁴).
  ∂²Ueff/∂ρ²|_r=a = 6ka + 3M²/(ma⁴) = 6ka² + 3kma⁵/(ma⁴) = 9ka² > 0.
  The orbit with radius a is stable, α = 9ka².  The period of small radial oscillation about a is
  τ = 2π/ω_r,  ω_r = (α/m)^½ = (9ka²/m)^½.  τ = 2π/(9ka²/m)^½.

Problem 5:

Consider a particle of mass m moving in the xy plane.  The potential energy function is
U(x,y) = (k/2)(x² + y²), with k a positive constant.
(a)  Find the equations of motion.
(b)  Are there circular orbits?  If yes, do they all have the same period?
(c)  Is the total energy conserved?

Solution:

• Concepts:
  Central force problem in 2 dimensions
• Reasoning:
  We can use the Lagrangian formalism or work in Cartesian coordinates and use Newton's second law.
• Details of the calculation:
  Use the Lagrangian formalism.  Choose polar coordinates (r, θ).
  (a)  T = ½m[(dr/dt)² + r²(dθ/dt)²],   U(r) = kr²/2
  L = T - U
  ∂L/∂((dr/dt)) = mdr/dt,  d/dt(∂L/∂(dr/dt)) = md²r/dt²,  ∂L/∂r = mr(dθ/dt)² - kr.
  md²r/dt² - mr(dθ/dt)² + kr = 0.
  ∂L/∂((dθ/dt)) = mr²(dθ/dt),  ∂L/∂θ = 0.
  mr²(dθ/dt) = M = constant.  The angular momentum is conserved.
  The equation of motion for θ is dθ/dt = M/(mr²).
  md²r/dt² - M²/(mr³) + kr = 0. 
  The equation of motion for r is d²r/dt² - M²/(m²r³) + kr/m = 0. 
  (b)  circular orbit:  d²r/dt² = 0,  dr/dt = 0,  r = R,  M²/m = kR⁴
  ω = (dθ/dt) = (k/m)^½ independent of R, T = 2π/ω does not depend on R.
  (c)  If the Lagrangian does not explicitly depend on time and the generalized coordinates do not explicitly depend on time, then H = T + U = E and the energy is a constant of motion.  This is the case here.
  
  We can also work in Cartesian coordinates and use Newton's second law.
  (a)  F_x = -∂U/∂x = -kx,  F_y = -∂U/∂y = -ky,
  d²x/dt² = -(k/m)x,  d²y/dt² = -(k/m)y.
  (b)  Circular orbit: x² + y² = R².  For a circular orbit of radius R we need  
  d²x/dt² =  -ω²x, d²y/dt² =  -ω²y,  ω = (k/m)^½.
  Circular orbits are possible.
  ω = (dθ/dt) = (k/m)^½ independent of R, T = 2π/ω does not depend on R.
  (c)  E = T + U,  dE/dt = dT/dt + dU/dt
  = (dT/dv_x)(dv_x/dt) + (dT/dv_y)(dv_y/dt) + (dU/dx)(dx/dt) + (dU/dy)(dy/dt)
  = mv_xa_x + mv_ya_y  - F_xv_x - F_yv_y =  v_x(ma_x - F_x) + v_y(ma_y - F_y) = 0 using Newton's second law.
  Energy is conserved.