WEBVTT
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We've seen the rocket
equation, and we've
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studied the example when
there's no external forces.
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Now, let's consider an example
when the external force is not
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0.
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Well, one of our
examples is just a rocket
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taking off in a
gravitational field.
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And so when we want to focus
on the external forces,
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let's draw our
rocket at time, t.
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And now, consider-- what are the
force diagrams on this rocket?
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Well, we have a
gravitational force, mg.
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So we're going to
choose j hat up.
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So now, what does our
rocket equation look like?
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We wrote the exhaust
velocity as minus mu j hat.
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And so if we write this now as
a vector equation in the j hat
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component, we get minus mg.
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We have the rate that the
rocket mass is changing,
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and this is in our-- we
have this as-- remember,
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dm dr/dt is going
to be negative.
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So there's a minus.
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And over here, we have another--
this is u because u is minus u.
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And finally, we have
m rocket dvr/dt.
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Now, once again, we're
going to try to integrate--
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this is actually equal.
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And we're going to try to
integrate this equation
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to find a solution for the
velocity as a function of time
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when all the fuel is burned.
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So what we do is we'll
multiply through by dt,
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and we have minus dmr
times u equals mr dvr.
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And this is mass of the rocket.
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And now, let's divide through
by mass of the rocket,
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so we have minus g minus
dmr over mru equals dvr.
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And once again, we can
integrate this equation
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from some initial time
to some final time.
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I'll just denote that by
i initial and i final,
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and what we see here is we have
minus g times t final minus t
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initial because
we're integrating
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with respect to time.
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Here, we're integrating mass,
so that's minus mu natural log.
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And I'm going to write this
as mr final over mr initial.
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And over here, we have vr
final minus vr initial.
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And let's just
remove that time, t.
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And so now, here is
our rocket equation.
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Now, what we notice
here is-- let's look
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at some rocket taking off.
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So suppose that, at
10 initial equals 0,
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we have the special condition
that vr initial equals 0
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and mr initial equals the
mass of the rocket-- dry mass.
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That's all.
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That's not counting the fuel.
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Plus the total mass of the fuel.
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And the final-- at t final, when
the engine turns off-- we're
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going to try to find
this velocity at t final.
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And our condition
for the final mass
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is all the fuel has been
burned, so this is just
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the dry mass of the rocket.
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And so what we get is
we get minus g t final.
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Now, I'm going to
switch the sign here,
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so we have-- if
I invert the log,
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that will give a minus sign.
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And I have plus u log of mr
of d plus mass of the fuel.
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That's the initial
mass of the rocket.
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Divided by just the
dry mass, and that's
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equal to vr at t final.
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So one of the interesting
questions that you might ask
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is-- how fast should
you burn the fuel,
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and what will that have to
do with the final speed?
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So we can see from
this expression
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that, if you burn the fuel for
a very long period of time--
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so t final is big--
then this piece will
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diminish the final velocity.
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So if you want the fastest speed
after you've burned the fuel,
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you want to burn the fuel so
that t final is as short as
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possible.
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We have the shortest
burn time, will give us
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the largest velocity when
all the fuel has been burned.