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We would now like to look at
some applications of the rocket

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equation, in which we have an
external force plus dmr dt,

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where m and r is the mass of
the rocket at time t times

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the relative velocity
of the exhaust velocity

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was equal to m of
r dv rocket, dt.

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And now we want to consider
some special cases.

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So our first example
will be a case when

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there is no external force.

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Now, this might be
characteristic of a rocket that

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is far away from any type
of gravitational interaction

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and is either speeding
up or slowing down.

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This is a special
case that will enable

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us to understand our problem.

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Now, we'll draw a
picture at time t.

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Just to remind ourselves,
we have mass of the rocket.

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It's moving with a
velocity v of r of t,

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and we're going to call this
our plus i hat direction.

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And then at time
t plus delta t, we

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have our fuel
delta, delta m fuel,

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which was equal to minus
delta m rocket moving

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with the relative velocity u
plus v of r of t plus delta t.

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And our rocket itself is moving
with v of r of t plus delta t.

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So in our coordinate
system, we have

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to remember that
the exhaust speed is

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relative to the rocket,
and we define that to be u.

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Since the rocket
is moving forward

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and we've chosen plus i
hat in that direction,

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it's minus u i hat, and we
can write our vectors vr

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as v of r i hat.

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So our rocket equation
in this special case

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becomes dmr dt, minus u i
hat, equals mr dvr dt i hat.

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Now, in this equation,
remember that it's

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important to recall that m
of r is a function of time.

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Unless we have an explicit model
for how the mass is ejected,

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in principle we can't solve
for the time behavior,

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but we can look at this
differential equation

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just by eliminating the time
completely by multiplying

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through by dt, in which
case we have minus dmr u

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equals m of r dvr.

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Now, this is separable
and I can integrate it,

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so I'll bring the terms minus
dmr over mr on this side,

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times u, and that's equal
to dvr on that side.

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Now, when we
integrate, we're going

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to be integrating
mass of the rocket,

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say at some initial
time, t initial,

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to mass of the rocket
at some final time.

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And over here, we're
going to be integrating

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the velocity from some initial
time to some final time.

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When both these integrals
are straightforward,

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on the left-hand side
we have a natural log,

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so we have minus u, natural log
of mrt final over mrt initial.

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And this side, we just have
the velocity difference,

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v of r at t final
minus vr at t initial.

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Now, again, we always
check our minus signs,

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but recall the mass of
the rocket is decreasing.

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So the log of a
fraction is negative,

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so we have a positive
quantity on this side.

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And that's indicating that
the velocity has increased

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on that side, and so we conclude
that the velocity of the rocket

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at its final time is equal
to the velocity of the rocket

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at its initial time.

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Now, here, I'll
keep this minus sign

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in here-- minus natural log
of mrt final over mrt initial.

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Now, in our answer,
we need to know

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what the mass of the rocket
is at the final time,

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so in general this
equation is still not

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going to give us an
explicit expression for v

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of t final, unless we
have one special case

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in which our final time is when
all the fuel has been burned.

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And our initial time is the
rocket with all the fuel there,

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and then this ratio
is a known ratio.

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And you can look at a variety
of different examples for that.

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So there is an
example of the solving

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the motion of a rocket
with no external forces.