1
00:00:03,800 --> 00:00:06,100
Let's look at this
rocket sled here.

2
00:00:06,100 --> 00:00:10,310
It's gone on in the snow,
but it wants to stop.

3
00:00:10,310 --> 00:00:12,890
There are two little
devices mounted on the sled,

4
00:00:12,890 --> 00:00:14,730
and they can eject gas.

5
00:00:14,730 --> 00:00:18,380
And so the forward one
is used to eject gas

6
00:00:18,380 --> 00:00:21,410
to make the sled stop.

7
00:00:21,410 --> 00:00:25,340
We want to derive a relation
for the differential

8
00:00:25,340 --> 00:00:29,630
between the speed of the
sled and the differential

9
00:00:29,630 --> 00:00:33,050
of the mass of the rocket sled.

10
00:00:33,050 --> 00:00:35,096
But before we do that
with a rocket equation,

11
00:00:35,096 --> 00:00:36,470
we need to actually
consider what

12
00:00:36,470 --> 00:00:39,240
else we know about this system.

13
00:00:39,240 --> 00:00:44,330
Well, we know that the
dry mass has a mass of m0.

14
00:00:44,330 --> 00:00:48,680
The fuel mass is
given also as m0.

15
00:00:48,680 --> 00:00:53,690
And we know that, at time of t
equals 0, the speed of the sled

16
00:00:53,690 --> 00:00:56,270
is v0.

17
00:00:56,270 --> 00:01:00,200
We also know that, at a
later time, t plus delta t,

18
00:01:00,200 --> 00:01:05,870
we have the sled here whose
mass is now m of t plus delta t.

19
00:01:05,870 --> 00:01:10,070
And we have this low mass
parcel that has been ejected,

20
00:01:10,070 --> 00:01:11,620
so the gas.

21
00:01:11,620 --> 00:01:15,605
And that has the mass of delta
mf, so the mass of the fuel.

22
00:01:18,690 --> 00:01:22,130
We furthermore know that,
relative to the sled,

23
00:01:22,130 --> 00:01:25,550
this little gas parcel
is moving with a speed u.

24
00:01:28,630 --> 00:01:33,910
The rocket equation says that
my force-- my external force--

25
00:01:33,910 --> 00:01:35,530
has two terms.

26
00:01:35,530 --> 00:01:39,220
We have the mass of the
rocket times the acceleration

27
00:01:39,220 --> 00:01:46,930
of the sled minus
the differential here

28
00:01:46,930 --> 00:01:51,870
of the mass times the speed u.

29
00:01:51,870 --> 00:01:54,330
And these are
actually all vectors.

30
00:01:54,330 --> 00:01:59,400
And so this describes this
little gas parcel here,

31
00:01:59,400 --> 00:02:02,920
and this one describes
rocket and we

32
00:02:02,920 --> 00:02:07,350
know that v equals v i hat.

33
00:02:07,350 --> 00:02:11,080
i hat is going in the right--
in the direction of motion,

34
00:02:11,080 --> 00:02:14,950
and u equals u i hat.

35
00:02:14,950 --> 00:02:19,720
We also know that no external
forces apply to the sled,

36
00:02:19,720 --> 00:02:23,110
so that's actually 0.

37
00:02:23,110 --> 00:02:26,490
We can then turn this equation.

38
00:02:26,490 --> 00:02:28,030
We can apply all of this.

39
00:02:28,030 --> 00:02:38,270
We will get 0 equals mr
dvr/dt minus dmr/dtu.

40
00:02:41,230 --> 00:02:44,350
We can bring this one
on the other side,

41
00:02:44,350 --> 00:02:50,110
and we end up with the
relationship that we wanted

42
00:02:50,110 --> 00:02:55,570
to obtain-- namely,
an equation that

43
00:02:55,570 --> 00:03:01,985
contains the differentials
of the speed of the rocket,

44
00:03:01,985 --> 00:03:06,660
that sled, and the
differential of the mass.