1 00:00:00,970 --> 00:00:03,940 We'd like to analyze a physics problem called 2 00:00:03,940 --> 00:00:08,860 the rocket problem, in which we have a continuous mass 3 00:00:08,860 --> 00:00:10,180 transfer. 4 00:00:10,180 --> 00:00:11,570 Let's see what we mean by that. 5 00:00:11,570 --> 00:00:14,260 Let's consider a rocket launching. 6 00:00:14,260 --> 00:00:18,330 So here's a classic example of a rocket that's launching off 7 00:00:18,330 --> 00:00:20,080 in a gravitational field. 8 00:00:20,080 --> 00:00:24,550 As a rocket starts at rest, it starts to burn fuel, 9 00:00:24,550 --> 00:00:26,850 and ejects the fuel backwards. 10 00:00:26,850 --> 00:00:29,470 And as the ejected fuel backwards 11 00:00:29,470 --> 00:00:32,409 is producing a thrust on the rocket-- and you can see, 12 00:00:32,409 --> 00:00:34,420 as the rocket starts to rise-- it's 13 00:00:34,420 --> 00:00:39,200 starting to accelerate upwards as fuel is ejected backwards. 14 00:00:39,200 --> 00:00:42,760 Now, this process goes on for some time interval. 15 00:00:42,760 --> 00:00:45,190 And a lot of fuel is ejected backwards 16 00:00:45,190 --> 00:00:46,660 during that time interval. 17 00:00:46,660 --> 00:00:50,080 In this particular case, we could analyze the situation 18 00:00:50,080 --> 00:00:52,180 from the ground frame of the rocket-- 19 00:00:52,180 --> 00:00:54,070 from the ground frame of the earth-- 20 00:00:54,070 --> 00:00:56,560 and the fuel is ejected backwards 21 00:00:56,560 --> 00:00:59,810 relative to the rocket, rocket moves up forward. 22 00:00:59,810 --> 00:01:03,310 So how do we analyze this situation? 23 00:01:03,310 --> 00:01:06,670 If we just looked at the rocket when the rocket initially 24 00:01:06,670 --> 00:01:10,930 launched, the mass of the rocket is the dry mass of the rocket 25 00:01:10,930 --> 00:01:12,550 plus all the fuel in there. 26 00:01:12,550 --> 00:01:14,200 It's at rest. 27 00:01:14,200 --> 00:01:17,890 In a state much later on, the mass of the rocket 28 00:01:17,890 --> 00:01:23,260 has decreased and the speed has increased. 29 00:01:23,260 --> 00:01:27,039 But we can't just compare this initial to this final state, 30 00:01:27,039 --> 00:01:29,050 because throughout this process, there 31 00:01:29,050 --> 00:01:31,190 was a continuous mass transfer. 32 00:01:31,190 --> 00:01:33,920 Fuel was ejected backwards. 33 00:01:33,920 --> 00:01:37,660 So how do we analyze these types of continuous mass transfer 34 00:01:37,660 --> 00:01:38,560 problems? 35 00:01:38,560 --> 00:01:42,250 And the key is, again, identifying states. 36 00:01:42,250 --> 00:01:45,460 So, for a continuous mass transfer problem, 37 00:01:45,460 --> 00:01:49,180 we always have to choose a reference frame. 38 00:01:49,180 --> 00:01:51,610 So here, in our example, we were choosing 39 00:01:51,610 --> 00:01:53,710 the ground frame of the earth. 40 00:01:53,710 --> 00:01:57,940 And then, we want to identify two states. 41 00:01:57,940 --> 00:02:01,230 We don't want to identify the initial and the final states. 42 00:02:01,230 --> 00:02:04,810 We want to pick two arbitrary states that are separated 43 00:02:04,810 --> 00:02:06,830 by a small time interval. 44 00:02:06,830 --> 00:02:10,389 So we can take a state at time t. 45 00:02:10,389 --> 00:02:14,800 And then our second state occurs at a small time afterwards, 46 00:02:14,800 --> 00:02:17,800 time t plus delta t. 47 00:02:17,800 --> 00:02:22,270 And what we want to do is now represent the state changes 48 00:02:22,270 --> 00:02:24,820 by momentum diagrams. 49 00:02:24,820 --> 00:02:28,000 And that will lead us to a differential equation, which 50 00:02:28,000 --> 00:02:29,650 is called the rocket equation. 51 00:02:29,650 --> 00:02:33,310 And that's how we'll analyze all continuous mass transfer 52 00:02:33,310 --> 00:02:34,920 problems. 53 00:02:34,920 --> 00:02:37,000 In applying the momentum principle, 54 00:02:37,000 --> 00:02:39,400 our first thing is just to choose a reference frame. 55 00:02:39,400 --> 00:02:41,440 So in our example, we're going to choose 56 00:02:41,440 --> 00:02:44,470 the ground frame of the earth as a reference frame. 57 00:02:44,470 --> 00:02:48,220 Now, the next step is the key to these processes. 58 00:02:48,220 --> 00:02:51,490 Instead of choosing some initial state where the rocket is 59 00:02:51,490 --> 00:02:54,340 at rest and some final state much later on, 60 00:02:54,340 --> 00:02:56,260 we want to choose two states that 61 00:02:56,260 --> 00:02:59,590 are separated by a small time interval, delta t. 62 00:02:59,590 --> 00:03:03,070 So our first state, at time t. 63 00:03:03,070 --> 00:03:08,020 And our second state, we choose at time t plus delta t. 64 00:03:08,020 --> 00:03:10,360 Now, the next step is that we have 65 00:03:10,360 --> 00:03:12,610 to identify what is our system. 66 00:03:12,610 --> 00:03:16,300 And so in this case, our system is 67 00:03:16,300 --> 00:03:19,840 going to be the mass of the rocket at time t. 68 00:03:19,840 --> 00:03:23,050 That mass is the dry mass of the rocket 69 00:03:23,050 --> 00:03:27,100 plus all the fuel that's still in the rocket at time t. 70 00:03:27,100 --> 00:03:31,400 And this mass will stay the same between times t and t 71 00:03:31,400 --> 00:03:34,240 plus delta t, even though some of the mass 72 00:03:34,240 --> 00:03:38,620 has been ejected downward. 73 00:03:38,620 --> 00:03:42,100 So our next step is to look at that mass conservation 74 00:03:42,100 --> 00:03:43,829 of the system.