More Problems

Problem 1:

A satellite is launched into a circular orbit of radius R around Earth.  A second satellite is launched into an orbit of radius 1.01 R.  Find the ration of the periods of the two satellites.

Solution:

• Concepts:
  Kepler's third law
• Reasoning:
  T² ∝ r³,  T = period, R = radius of orbit.
• Details of the calculation:
  (T₂/T₁)² = (R₂/R₁)³ = (1.01)³ = (1 + 0.01)³ ~ 1 + 0.03.
  T₂/T₁ = (1 + x)^½ ~ 1 + ½x = 1.015.

Problem 2:

Consider a particle of mass m moving in a central potential with potential energy U(r).   Assume U(r) --> 0 as r --> ∞.
(a)  Write down the Lagrangian for this system and determine the second-order equations of motion in polar coordinates. 
(b)  What condition must U(r) fulfill, so that a circular orbit of radius b is possible?
(c)  If a circular orbit of radius b is possible, when is it a stable orbit?

Solution:

• Concept:
  Motion in a central potential,  Lagrange's equations, stable orbits
• Reasoning:
  We are asked to write down the Lagrangian, and investigate the stability of a circular.
• Details of the calculation:
  (a)  L = T - U = ½m[(dr/dt)² + r²(dΦ/dt)²] - U(r).
  The generalized coordinates are r and Φ.  The coordinate Φ is cyclic.
  Lagrange's equations yield the equations of motion.
  (d/dt)[mr²(dΦ/dt)] = 0.  mr²(dΦ/dt) = M = constant.
  md²r/dt² - M/(mr³) = -dU/dr,
  or md²r/dt² = -dU_eff/dr, with U_eff(r) = M²/(2mr²) + U(r).
  
  (b)  For a circular orbit of radius b we need md²r/dt²|_b = 0, or dU/dr|_b = M²/mb³.
  Since M² > 0, we need dU/dr > 0 for r = b.  We cannot have a purely repulsive potential.
  
  (c)  For a stable orbit we need a restoring force.  Let r = b + 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 = b.
  
  ∂²U_eff/∂r² = 3M²/mr⁴ + d²U/dr².
  ∂²U_eff/∂r²|_r=b = 3M²/mb⁴ + d²U/dr²|_{r = b} > 0.

Problem 3:

Two earth-like planets, each with mass m = 6*10²⁴ kg, orbit each other in a circular orbit once every 50 days.  Find the distance between their centers in km.

Solution:

• Concepts:
  Kepler's third law, relative motion
• Reasoning:
  The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass μ moving in a central potential U(r). 
  Here U(r) =  Gm²/r,  F(r) = Gm²/r²,  r = distance between their centers.
• Details of the calculation:
  μv²/r  = Gm²/r².  v² = Gm²/(μr).  μ = m/2.  v² = 2Gm/r.
  (2πr/T)² = 2Gm/r,  r = [T²Gm/(2π²)]^1/3.
  T = 50*24*3600 s,  r = 7.23*10⁵ km.
  
  Another approach:
  The two planets orbit their CM.  Let r' be the distance of each particle from the CM, v' the speed of the particle with respect to the CM.  Then
  mv'²/r' = Gm²/(2r')²,  v'² = Gm/(4r') = (2πr'/T)².
  r'³ = d³/8 = GmT²/(16π²).  Here  d = distance between the centers of the planets.
  d = [T²Gm/(2π²)]^1/3.