### 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 tc the fuel is used up.
m1 = m0 - km0tc, tc = (1-m1/m0)/k.
For t < tc, 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 ktc = (1 - m1/m0) < 0.632, then the maximum momentum of the rocket is
p = -m1v'ln(m1/m0) at time t = tc, 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).
0v(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) = ∫0tv(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 htotal = 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 tc the fuel is used up.
m1 = m0 - km0tc, tc = (1 - m1/m0)/k = (1 - m1/m0)v'/g under the conditions stated in the problem.
For t < tc, dv/dt = -g + kv'/(1 - kt).
Let t' = 1 - kt.  Then dv/dt' = g/k - v'/t'.  We can integrate to find v(t').
0v(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 < tc, v(t) =  -gt  - v'ln(1 - gt/v').
At t1 = tc, v(tc) = -v'(1 - m1/m0) - v' ln(m1/m0).  This is the greatest speed of the rocket.
At time tc the height of the rocket is
h(tc) = ∫0tcv(t) dt = -gtc2/2 + (v'2/g)∫1'1-gtc/v'ln(t')dt' = -gtc2/2 + (v'2/g)(t'ln(t') - t')|1m1/m0.
h(tc) = (v'2/g)[(m1/m0)ln(m1/m0) - ½(m12/m02) + ½].
For t > tc the rocket is coasting in a gravitational field.  It will climb until its speed has decreased to zero.
Δh = ½v(tc)2/g.  The total height reached is htotal = h(tc) + Δh.