__Rockets in free space__

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 m_{0}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/m_{0}) if v(m_{0}) = 0.

p(m) = mv(m) = -v'mln(m/m_{0}), dp/dm = 0 --> ln(m/m_{0}) = -1, m/m_{0}= 1/e = 0.3678.

A rocket with initial mass m_{0} starts from rest and travels in a
straight line in a gravity-free environment. It burns its fuel at a
constant rate km_{0}, and exhausts the burned gases at a speed v'
relative to the rocket shell of mass m_{1}. 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

Here |dm/dt| = km_{0}. Therefore dv/dt = km_{0}v'/m.

At t = 0 m = m_{0}.

For t > 0 and m > m_{1}we have m = m_{0}- km_{0}t. Therefore

dv/dt = km_{0}v'/(m_{0}- km_{0}t) = kv'/(1 - kt).

At time t_{c}the fuel is used up.

m_{1}= m_{0}- km_{0}t_{c}, t_{c}= (1-m_{1}/m_{0})/k.

For t < t_{c}, dv/dt = - kv'/(1 - kt).

Let t' = 1 - kt. Then dv/dt' = -v'/t'.

v(t') = -v'ln(t'/t_{0}').

v(t) = -v'ln(1 - kt).

The momentum of the rocket at time t is p = m_{0}v'(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 - m_{1}/m_{0}) < 0.632, then the maximum momentum of the rocket is

p = -m_{1}v'ln(m_{1}/m_{0}) at time t = t_{c}, otherwise the maximum momentum of the rocket (shell and remaining fuel) is p = 0.368m_{0}v' at time t = 0.632/k.

__Rockets in a gravitational field__

A rocket has an initial mass of m_{0} 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

Here |dm/dt| = -dm/dt = k. Therefore dv/dt = -g + kv'/m.

At t = 0 we have m = m_{0}. We need dv/dt > 0 for the rocket to get off the ground, therefore we need v' > m_{0}g/k.

A rocket of initial mass m_{0} 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 = - km_{0}with k = 1/(100 s). We write

dv/dt = -g + km_{0}v'/m.

For 0 < t < 50 s we have m = m_{0}- km_{0}t. Therefore

dv/dt = -g + km_{0}v'/(m_{0}- km_{0}t) = -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)(t_{1}' - t_{0}') - v'ln(t_{1}'/t_{0}').

v(t_{1}) = -gt_{1}- v'ln(1 - kt_{1})

At time t_{1}the height of the rocket is H(t_{1}) = ∫_{0}^{t}v(t) dt.

h(t_{1}) = -gt_{1}^{2}/2 + (v'/k)[(1 - kt_{1}) ln(1 - kt_{1}) + kt_{1}].

At t_{1}= 50 s we have (with g = 9.81 m/s^{2})

v(t_{1}) = -g*50 s - (2000 m/s)ln(0.5) = 896 m/s,

h(t_{1}) = -g(50 s)^{2}/2 + (2*10^{5 }m)[(0.5)ln(0.5) + 0.5] = 18422 m

For t > t_{1}the rocket is coasting in a gravitational field. It will climb until its speed has decreased to zero.

Δh = ½v(t_{1})^{2}/g = 40918 m. The total height reached is h_{total}= h(t_{1}) + Δh = 50340 m.

A rocket of initial mass m_{0} ejects fuel at a constant rate km_{0}
and at a velocity v' relative to the rocket shell of mass m_{1}.

(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

Here |dm/dt| = km_{0}. Therefore dv/dt = -g + km_{0}v'/m.

At t = 0 m = m_{0}. 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 > m_{1}we have m = m_{0}- km_{0}t. Therefore

dv/dt = -g + km_{0}v'/(m_{0}- km_{0}t) = -g + kv'/(1 - kt).

At time t_{c}the fuel is used up.

m_{1}= m_{0}- km_{0}t_{c}, t_{c}= (1 - m_{1}/m_{0})/k = (1 - m_{1}/m_{0})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)(t_{1}' - t_{0}') - v'ln(t_{1}'/t_{0}').

t_{1}' = 1 - kt_{1}, t_{0}' = 1. v(t_{1}) = -gt_{1}- v'ln(1 - kt_{1}) = -gt_{1}- v'ln(1 - gt_{1}/v').

For t < t_{c}, v(t) = -gt_{ }- v'ln(1 - gt/v').

At t_{1}= t_{c}, v(t_{c}) = -v'(1 - m_{1}/m_{0}) - v' ln(m_{1}/m_{0}). This is the greatest speed of the rocket.

At time t_{c}the height of the rocket is

h(t_{c}) = ∫_{0}^{tc}v(t) dt = -gt_{c}^{2}/2 + (v'^{2}/g)∫_{1'}^{1-gtc/v'}ln(t')dt' = -gt_{c}^{2}/2 + (v'^{2}/g)(t'ln(t') - t')|_{1}^{m1/m0}.

h(t_{c}) = (v'^{2}/g)[(m_{1}/m_{0})ln(m_{1}/m_{0}) - ½(m_{1}^{2}/m_{0}^{2}) + 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.