More Problems

Problem 1:

A satellite of mass m orbits a planet of mass M >> m in an circular orbit of radius R. 
(a)  What is the satellite's kinetic, potential, and total energy.
(b)  If an impulse opposite to the direction of motion decrease the angular momentum per unit mass of the satellite to 0.9 times its initial value, what will be r_max and r_min of the satellite's new orbit?

Solution:

• Concepts:
  Kepler orbits
• Reasoning:
  Since M >> m, assume the CM is the center of the planet.  For a single object moving in a closed orbit in an attractive 1/r potential, the energy E per unit mass is proportional to the inverse of the semi-major axis.
• Details of the calculation:
  (a)  The energy of the satellite in the circular orbit of radius R is E = -½GMm/R, its kinetic energy is T = ½GMm/R and its potential energy is U = -GMm/R.  E = T + U.
  (b)  L = mvR,  L' = 0.9 mvR = mv'R, T' = 0.81T,
  E' = T' + U = E - T + T' = E + 0.19 E = 1.19 E.
  Let a be the semi-major axis of the new orbit.
  a/R = 1/1.19,  a = R/1.19 = 0.84 R.
  r_max of the new orbit is R, and r_min of the new orbit is 0.68 R.

Problem 2:

A space station orbits Earth on a circular trajectory.  At some moment the captain decides to change the trajectory by turning on the rocket engine for a very short period of time.  During the time the engine was on, it accelerated the station in its direction of motion.  As a result, the station speed increased by a factor of α.  Provide the conditions, in terms of α, that the new trajectory is elliptic, parabolic or hyperbolic.  Justify your answers.

Solution:

• Concepts:
  Kepler orbits
• Reasoning:
  Since the mass of a satellite is negligible compared to the mass of Earth, assume the CM is the center of Earth.  For a single object moving in a closed orbit in an attractive 1/r potential, the energy E per unit mass is proportional to inverse of the semi-major axis.
  For the Kepler problem we have:
  E > 0      hyperbola
  E = 0      parabola
  E < 0      ellipse
  The energy per unit mass of the satellite in a circular orbit of radius R  is E = -½GM/R, its kinetic energy per unit mass is T = ½GM/R and its potential energy per unit mass is U = -GM/R.  E = T + U.
  To put it in a parabolic orbit with E = 0, its kinetic energy per unit mass at a distance R from the planet has to double. Its speed therefore has to increase by a factor of √2.
  If α > √2 the orbit will be hyperbolic and if 1 < α < √2 the orbit will be elliptic.

Problem 3:

A particle of mass m moves in a central potential U(r) = kr²e^-r/a, with k and a positive constants.
(a)  Write down the Lagrangian for this system and determine the second-order equations of motion in polar coordinates. 
(b)  For what values of b is a circular orbit of radius b possible?
(c)  For a circular orbit of radius a, write down the angular momentum M in terms of m, k and a.
(d)  Is a circular orbit of radius a 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] - kr²e^-r/a.
  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) = f(r),  md²r/dt² = M²/mr³ - 2kre^-r/a + (kr²/a)e^-r/a.
  or md²r/dt² = -dU_eff/dr
  (b)  For a circular orbit of radius b we need md²r/dt²|_b = 0.
  We need 2kbe^-b/a - (kb²/a)e^-b/a > 0, or 2 - (b/a) > 0, or b < 2a.
  
  (c)  For a circular orbit of radius a we have M² = kma⁴e^-1.
  
  (d)  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 =  -M²/mr³ + 2kre^-r/a - (kr²/a)e^-r/a.
  
  ∂²Ueff/∂r² = 3M²/mr⁴ + 2ke^-r/a - 2k(r/a)e^-r/a - (2kr/a)e^-r/a +  (kr²/a²)e^-r/a.
  ∂²Ueff/∂r²|_r=a = 3M²/ma⁴ - ke^-1 > 0.
  Inserting M² from above we have 3ke^-1 - ke^-1 > 0.
  The orbit is stable.