1
00:00:03,570 --> 00:00:05,620
So now, we wrote
down the momentum

2
00:00:05,620 --> 00:00:07,844
of the system of our
rocket at time, t,

3
00:00:07,844 --> 00:00:10,010
and the movement of the
system of the rocket at time

4
00:00:10,010 --> 00:00:11,310
t plus delta t.

5
00:00:11,310 --> 00:00:13,720
Let's make some
simplifications in our notation

6
00:00:13,720 --> 00:00:17,840
just to make our calculations
a little more easy to read.

7
00:00:17,840 --> 00:00:19,570
So the first thing
we want to do is

8
00:00:19,570 --> 00:00:22,680
we want to say that the
mass of the rocket at time,

9
00:00:22,680 --> 00:00:24,790
t-- this is the
mass of the rocket

10
00:00:24,790 --> 00:00:27,340
plus the mass of the fuel
in the rocket at time, t.

11
00:00:27,340 --> 00:00:30,980
We're just going to
denote that by mr.

12
00:00:30,980 --> 00:00:36,820
And keep in mind that, when
you write the mass at time

13
00:00:36,820 --> 00:00:42,150
t plus delta t, we'll write
that as the mass at time

14
00:00:42,150 --> 00:00:46,760
t plus delta mr. This
is precisely how you

15
00:00:46,760 --> 00:00:48,290
define a differential.

16
00:00:48,290 --> 00:00:50,940
Even if this
quantity is negative,

17
00:00:50,940 --> 00:00:54,000
you always write the difference.

18
00:00:54,000 --> 00:00:58,620
m of r of t plus delta t
minus m of rt equal to delta

19
00:00:58,620 --> 00:01:02,100
mr. That's how we write
a differential delta

20
00:01:02,100 --> 00:01:03,600
mass of the rocket.

21
00:01:03,600 --> 00:01:06,050
So the mass of the
rocket at time, t,

22
00:01:06,050 --> 00:01:11,460
looks simpler-- m
of r plus delta mr.

23
00:01:11,460 --> 00:01:14,910
So those are two ways of
simplifying our expressions

24
00:01:14,910 --> 00:01:18,610
for the terms that are
appearing here and there.

25
00:01:18,610 --> 00:01:22,250
Now, the next thing was, recall
from our mass conservation,

26
00:01:22,250 --> 00:01:25,520
that we had the condition
that delta m fuel was

27
00:01:25,520 --> 00:01:29,270
equal to minus delta m rocket.

28
00:01:29,270 --> 00:01:32,680
So now, we'll take those
two simplifications,

29
00:01:32,680 --> 00:01:38,610
and now we'll write down our
system momentum at time, t.

30
00:01:38,610 --> 00:01:39,720
That's very simple.

31
00:01:39,720 --> 00:01:48,120
mr v of r at time t, and
we'll now write the momentum

32
00:01:48,120 --> 00:01:52,009
at time t plus delta t.

33
00:01:52,009 --> 00:01:54,460
Well, we have the
first piece, which

34
00:01:54,460 --> 00:02:03,130
is mr plus delta mr times
the velocity of the rocket

35
00:02:03,130 --> 00:02:05,840
at time t plus delta t.

36
00:02:05,840 --> 00:02:07,690
Now, be careful
here because we're

37
00:02:07,690 --> 00:02:12,270
going to use our conservation
of mass condition

38
00:02:12,270 --> 00:02:16,190
to replace delta m fuel
with minus delta m rocket,

39
00:02:16,190 --> 00:02:19,974
and we'll use our expression
for the velocity of fuel

40
00:02:19,974 --> 00:02:21,390
in the ground frame
of the rocket.

41
00:02:21,390 --> 00:02:27,480
So we have minus delta m
rocket times the velocity u

42
00:02:27,480 --> 00:02:32,480
plus v of r of t plus delta t.

43
00:02:32,480 --> 00:02:37,160
And this simplification, we'll
write equations 1 and equation

44
00:02:37,160 --> 00:02:42,240
2, will enable us to apply
the momentum principle

45
00:02:42,240 --> 00:02:44,010
in a very transparent way.

46
00:02:44,010 --> 00:02:46,230
And that's what we'll do next.