Rocket action

[Roger Freedman <airboy@PHYSICS.UCSB.EDU>]

> Date:    Wed, 1 Dec 1999 19:28:51 -0600
> From:    Greg Kifer <gsanda@TFS.NET>
> Subject: Rocket action
> MIME-Version: 1.0
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>
> I remember, but don't know from where, hearing that a rocket can go no
> faster forward than the exhaust gases leave the exhaust nozzle.  Is this
> correct?  If so, can someone explain the whys and wherefores to a simple
> high school teacher like me.  Thanks in advance.
>
> Greg Kifer
> Olathe North High School
> Olathe, KS  66061

Dear Greg,

I'm assuming that your question is about a rocket that starts from rest, so
the question can be restated as "If a rocket starts from rest, can its
speed v ever exceed the speed v_ex of the exhaust relative to the rocket?"
The answer is to this question is "Yes, and by a lot!" (In other words, the
original statement in your message just isn't so.)

The final speed v will exceed the exhaust speed v_ex if the initial
combined mass m_0 of the rocket and the propellant is greater than the
final mass m of the rocket by a factor of at least e = 2.71828 . . ., that
is, if m_0/m > e. As an example, the main engines of the Space Shuttle use
a mixture of liquid hydrogen and liquid oxygen, which can produce exhaust
speeds of a mere 4.6 km/s. Yet the Space Shuttle is able to achieve an
orbital speed of about 8 km/s! (I'm ignoring how gravity slows the
spacecraft down as it climbs.)

Here's a detailed explanation, adapted from a textbook of my intimate
acquaintance: Young and Freedman, UNIVERSITY PHYSICS, 10th edition (Addison
Wesley Longman, 2000):

Consider a rocket fired in outer space, where there is no gravitational
force and no air resistance. We choose our x-axis to be along the rocket's
direction of motion. Suppose that at a time t, the rocket's mass is m and
the x-component of its velocity relative to our coordinate system is v. The
x-component of total momentum at this instant is P = mv. In a short time
interval dt, the mass of the rocket changes by an amount dm. This is a
negative quantity because the rocket's mass m decreases with time. During
dt, a positive mass -dm of burned fuel is ejected from the rocket. Let v_ex
be the exhaust speed of this material relative to the rocket; the burned
fuel is ejected opposite the direction of motion, so its x-component of
velocity relative to the rocket is -v_ex. The x-component of velocity
v_fuel of the burned fuel relative to our coordinate system is then

        v_fuel = v + (-v_ex) = v - v_ex,

and the x-component of momentum of the ejected mass (-dm) is

        (-dm)v_fuel = (-dm)(v - v_ex).

At the end of the time interval dt, the x-component of velocity of the
rocket and unburned fuel has increased to v + dv, and its mass has
decreased to m + dm (remember that dm is negative). The rocket's momentum
at this time is

        (m + dm)(v + dv).

Thus the total x-component of momentum P of rocket plus ejected fuel at
time t + dt is

        P = (m + dm)(v + dv) + (-dm)(v - v_ex).

        According to our initial assumption, the rocket and fuel are an
isolated system. Thus momentum is conserved, and the total x-component of
momentum P of the system must be the same at time t and at time t + dt:

        mv = (m + dm)(v + dv) + (-dm)(v - v_ex).

This can be simplified to

        m dv = -dm v_ex - dm dv.

We can neglect the term (-dm dv) because it is a product of two small
quantities and thus is much smaller than the other terms. Dropping this
term, dividing by dt, and rearranging, we find

        m(dv/dt)  =  -v_ex(dm/dt) .

Now dv/dt is the acceleration of the rocket, so the left side of this
equation (mass times acceleration) equals the net force F, or thrust, on
the rocket,

        F = -v_ex(dm/dt) .

The thrust is proportional both to the relative speed v_ex of the ejected
fuel and to the mass of fuel ejected per unit time, -dm/dt. (Remember that
dm/dt is negative because it is the rate of change of the rocket's mass.)
        The x-component of acceleration of the rocket is

        a =  (dv/dt  = F/m = -(v_ex/m) (dm/dt) .

This is positive because v_ex is positive (remember, it's the exhaust
*speed*) and dm/dt is negative. The rocket's mass m decreases continuously
while the fuel is being consumed. If vex and dm/dt are constant, the
acceleration increases until all the fuel is gone.
        An effective rocket burns fuel at a rapid rate (large -dm/dt) and
ejects the burned fuel at a high relative speed (large v_ex). If the
exhaust speed vex is constant, we can integrate the expression for a to
find a relation between the velocity v at any time and the remaining mass
m. At time t = 0, let the mass be m_0 and the velocity v_0. Then we rewrite
the acceleration equation as

        dv = - v_ex (dm/m) .

We change the integration variables to v' and m' so we can use v and m as
the upper limits (the final speed and mass):

        dv' = - v_ex (dm'/m') .

Then we integrate both sides, using limits v_0 to v and m_0 to m, and take
the constant v_ex outside the integral. When we do this, we get

        v - v0 = - v_ex ln (m/m_0)  =  v_ex ln (m0/m) .

The ratio m0/m is the original mass divided by the mass after the fuel has
been exhausted. In practical spacecraft, this ratio is made as large as
possible to maximize the speed gain, which means that the initial mass of
the rocket is almost all fuel. The final velocity of the rocket will be
greater in magnitude (and is often much greater) than the relative speed
v_ex if ln (m_0/m) >1, that is, if m_0/m > e = 2.71828 . . .

For some useful classroom exercises using this physics, see

http://tommy.jsc.nasa.gov/~woodfill/SPACEED/CDSEH/math50.html

Cheers,
Roger Freedman

+++++++++++++++++++++++++++++++++++++++++++++++++++++++

Roger A. Freedman
Department of Physics and College of Creative Studies
University of California, Santa Barbara

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