Rockets

Rockets in free space

Problem:

A rocket starts in free space by emitting mass with a constant speed with respect to the body of the rocket.  What fraction of the initial mass remains when the rocket's momentum has its maximum value?

Solution:

• Concepts:
  Motion in one dimension, Newton's 2nd law
• Reasoning:
  F = dp/dt = 0, dp = p(t + dt) - p(t)
  By Newton's 2nd law dp/dt is equal to the external force F = 0.
• Details of the calculation:
  Let m₀ be the initial mass of the rocket and v' the speed with which it is emitted with respect to the rocket body.
  m(dv/dt) = -(dm/dt)v'.
  dv = -v'dm/m,  v(m) = -v'ln(m/m₀) if v(m₀) = 0.
  p(m) = mv(m) = -v'mln(m/m₀),  dp/dm = 0 --> ln(m/m₀) = -1, m/m₀ = 1/e = 0.3678.

Problem:

A rocket with initial mass m₀ starts from rest and travels in a straight line in a gravity-free environment.  It burns its fuel at a constant rate km₀, and exhausts the burned gases at a speed v' relative to the rocket shell of mass m₁.  Find the maximum momentum of the rocket.

Solution:

• Concepts:
  Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t)
• Reasoning:
  The system consists of the rocket body and the fuel.  The speed of a small amount of fuel |dm| decreases by -v' in the time interval dt.  In the same time interval the speed of the rocket and the remaining fuel increases by dv.  The total momentum change of the system in the time interval dt therefore is
  dp = p(t + dt) - p(t) = (m - |dm|)(v + dv) + |dm|(v - v') - mv = mdv - |dm|v'.
  (Only first order terms are retained.)
  By Newton's 2nd law dp/dt is equal to the external force F = 0.
• Details of the calculation:
  0 = m(dv/dt) - |dm/dt|v'.
  Here |dm/dt| = km₀.  Therefore dv/dt = km₀v'/m.
  At t = 0 m = m₀.
  For t > 0 and m > m₁ we have m = m₀ - km₀t.  Therefore
  dv/dt = km₀v'/(m₀ - km₀t) = kv'/(1 - kt).
  At time t_c the fuel is used up. 
  m₁ = m₀ - km₀t_c, t_c = (1-m₁/m₀)/k.
  For t < t_c, dv/dt = - kv'/(1 - kt).
  Let t' = 1 - kt.  Then dv/dt' = -v'/t'. 
  v(t') = -v'ln(t'/t₀').
  v(t) = -v'ln(1 - kt).
  The momentum of the rocket at time t is p = m₀v'(kt - 1)ln(1 - kt )
  The function (kt - 1)ln(1 - kt ) has a maximum when ln(1 - kt) = -1, kt = 1 - e^-1 = 0.632.
  So if kt_c = (1 - m₁/m₀) < 0.632, then the maximum momentum of the rocket is
  p = -m₁v'ln(m₁/m₀) at time t = t_c, otherwise the maximum momentum of the rocket (shell and remaining fuel) is p = 0.368m₀v' at time t = 0.632/k.

Rockets in a gravitational field

Problem:

A rocket has an initial mass of m₀ and a constant fuel burn rate of k.  What is the minimum exhaust velocity that will allow the rocket to lift off from earth immediately after firing?

Solution:

• Concepts:
  Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t)
• Reasoning:
  Choose a coordinate axis with the positive direction upward.  The system consists of the rocket body and the fuel.  The speed of small amount of fuel dm changes by -v' in the time interval dt.  In the same time interval the speed of the rocket and the remaining fuel increases by dv.  The total momentum change of the system in the time interval dt therefore is dp = p(t+dt) - p(t) = (m-|dm|)(v+dv) + |dm|(v-v') - mv = mdv - |dm|v' (only first order terms are retained). 
• Details of the calculation:
  By Newton's 2nd law dp/dt is equal to the external force F = -mg.
  -mg = m(dv/dt) - (|dm/dt|)v'.
  Here |dm/dt| = -dm/dt = k.  Therefore dv/dt = -g + kv'/m.
  At t = 0 we have m = m₀.  We need dv/dt > 0 for the rocket to get off the ground, therefore we need  v' > m₀g/k.

Problem:

A rocket of initial mass m₀ is shot vertically upward.  Assume the motion occurs under constant gravitational acceleration g, and that the initial velocity of the rocket on the surface of the earth is zero.  The rocket expels 1/100 of its initial mass per second for 50 seconds.  The exhaust velocity is 2000 m/s relative to the rocket.  What is the maximum height reached by the rocket?  Neglect friction.

Solution:

• Concepts:
  Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t)
• Reasoning:
  Choose a coordinate axis with the positive direction upward.  The system consists of the rocket body and the fuel.  The speed of small amount of fuel dm decreases by -v' in the time interval dt.  In the same time interval the speed of the rocket and the remaining fuel increases by dv.  The total momentum change of the system in the time interval dt therefore is
  dp = p(t + dt) - p(t) = (m + dm)(v + dv) - dm(v - v') - mv = mdv + dmv' (only first order terms are retained). 
  Note: dm is negative.
  By Newton's 2nd law dp/dt is equal to the external force F = -mg.
• Details of the calculation:
   -mg = m(dv/dt) + (dm/dt)v'
  Here v' = 2000 m/s and  dm/dt = - km₀ with k = 1/(100 s).  We write
  dv/dt = -g + km₀v'/m.
  For 0 < t < 50 s we have m = m₀ - km₀t.  Therefore
  dv/dt = -g + km₀v'/(m₀ - km₀t) = -g + kv'/(1 - kt).
  Let t' = 1 - kt.  Then dv/dt' = g/k - v'/t'.  We can integrate to find v(t'₁).
  ∫₀^v(t1')dv = ∫_t0'^t1'(g/k)dt' - ∫_t0'^t1'(v/t')dt' = (g/k)(t₁' - t₀') - v'ln(t₁'/t₀').
  v(t₁) = -gt₁ - v'ln(1 - kt₁)
  At time t₁ the height of the rocket is H(t₁) = ∫₀^tv(t) dt.
  h(t₁) = -gt₁²/2 + (v'/k)[(1 - kt₁) ln(1 - kt₁) + kt₁].
  At t₁ = 50 s we have (with g = 9.81 m/s²)
  v(t₁) = -g*50 s - (2000 m/s)ln(0.5) = 896 m/s,
  h(t₁) = -g(50 s)²/2 + (2*10⁵ m)[(0.5)ln(0.5) + 0.5] = 18422 m
  For t > t₁ the rocket is coasting in a gravitational field.  It will climb until its speed has decreased to zero.
  Δh = ½v(t₁)²/g = 40918 m.  The total height reached is h_total = h(t₁) + Δh = 50340 m.

Problem:

A rocket of initial mass m₀ ejects fuel at a constant rate km₀ and at a velocity v' relative to the rocket shell of mass m₁.
(a)  Show that the minimum rate of fuel consumption that will allow the rocket to rise at once is k = g/v' where g is the gravitational acceleration. 
(b)  Find the greatest speed achieved and the greatest height reached under that condition.

Solution:

• Concept:
  Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t)
• Reasoning:
  Choose a coordinate axis with the positive direction upward.  The system consists of the rocket body and the fuel.  The speed of small amount of fuel |dm| decreases by -v' in the time interval dt.  In the same time interval the speed of the rocket and the remaining fuel increases by dv.  The total momentum change of the system in the time interval dt therefore is
  dp = p(t + dt) - p(t) = (m - |dm|)(v + dv) + |dm|(v - v') - mv = mdv - |dm|v' (only first order terms are retained). 
  By Newton's 2nd law dp/dt is equal to the external force F = -mg.
• Details of the calculation:
   (a)  -mg = m(dv/dt) - |dm/dt|v'.
  Here |dm/dt| = km₀.  Therefore dv/dt = -g + km₀v'/m.
  At t = 0 m = m₀.  We need dv/dt ≥ 0 for the rocket to get off the ground, therefore we need  k ≥ g/v'.
  (b)  For t > 0 and m > m₁ we have m = m₀ - km₀t.  Therefore
  dv/dt = -g + km₀v'/(m₀ - km₀t) = -g + kv'/(1 - kt).
  At time t_c the fuel is used up. 
  m₁ = m₀ - km₀t_c, t_c = (1 - m₁/m₀)/k = (1 - m₁/m₀)v'/g under the conditions stated in the problem.
  For t < t_c, dv/dt = -g + kv'/(1 - kt).
  Let t' = 1 - kt.  Then dv/dt' = g/k - v'/t'.  We can integrate to find v(t').
  ∫₀^v(t1')dv = ∫_t0'^t1'(g/k)dt' - ∫_t0'^t1'(v/t')dt' = (g/k)(t₁' - t₀') - v'ln(t₁'/t₀').
  t₁' = 1 - kt₁, t₀' = 1.  v(t₁) = -gt₁ - v'ln(1 - kt₁) = -gt₁ - v'ln(1 - gt₁/v').
  For t < t_c, v(t) =  -gt  - v'ln(1 - gt/v').
  At t₁ = t_c, v(t_c) = -v'(1 - m₁/m₀) - v' ln(m₁/m₀).  This is the greatest speed of the rocket.
  At time t_c the height of the rocket is
  h(t_c) = ∫₀^tcv(t) dt = -gt_c²/2 + (v'²/g)∫_1'^1-gtc/v'ln(t')dt' = -gt_c²/2 + (v'²/g)(t'ln(t') - t')|₁^m1/m0.
  h(t_c) = (v'²/g)[(m₁/m₀)ln(m₁/m₀) - ½(m₁²/m₀²) + ½].
  For t > t_c the rocket is coasting in a gravitational field.  It will climb until its speed has decreased to zero.
  Δh = ½v(t_c)²/g.  The total height reached is h_total = h(t_c) + Δh.