1 00:00:03,630 --> 00:00:06,720 For a rocket and an external force, 2 00:00:06,720 --> 00:00:08,670 we had the rocket equation, which 3 00:00:08,670 --> 00:00:11,680 we wrote as mass of the rocket-- now remember, 4 00:00:11,680 --> 00:00:15,540 this is a function of time-- times the derivative 5 00:00:15,540 --> 00:00:18,570 of the rocket dr dt. 6 00:00:18,570 --> 00:00:21,030 And we also had this second term that 7 00:00:21,030 --> 00:00:23,700 came from the ejecting fuel. 8 00:00:23,700 --> 00:00:26,370 Now from our mass conservation equation, 9 00:00:26,370 --> 00:00:29,820 we can rewrite this equation as a differential relationship 10 00:00:29,820 --> 00:00:33,180 that the change of mass of the fuel in time 11 00:00:33,180 --> 00:00:36,410 is equal to minus the change of the rocket. 12 00:00:36,410 --> 00:00:40,500 It doesn't matter which side we put the minus sign on. 13 00:00:40,500 --> 00:00:44,280 So now I want to interpret this equation in a different way, 14 00:00:44,280 --> 00:00:47,710 and it will come back to what we mean by our system. 15 00:00:47,710 --> 00:00:52,470 Let's first off bring this other term over to this side. 16 00:00:52,470 --> 00:00:58,650 So we have plus dMr dt u. 17 00:00:58,650 --> 00:01:03,820 And that's equal to Mr dVr dt. 18 00:01:03,820 --> 00:01:07,380 Now separately, let's make this substitution again and go back 19 00:01:07,380 --> 00:01:09,420 to our fuel term. 20 00:01:09,420 --> 00:01:18,700 So that's minus dMf dt u equals Mr dVr dt. 21 00:01:18,700 --> 00:01:22,090 Now notice that over here we have mass times 22 00:01:22,090 --> 00:01:23,820 the acceleration of the rocket. 23 00:01:23,820 --> 00:01:30,310 So if we just rethought our system as simply the rocket-- 24 00:01:30,310 --> 00:01:36,180 Mr-- then we have two forces acting on the rocket. 25 00:01:36,180 --> 00:01:38,970 The external force might be the gravitational field, 26 00:01:38,970 --> 00:01:41,789 but we have a new term here which we're 27 00:01:41,789 --> 00:01:44,414 going to refer to as thrust. 28 00:01:47,130 --> 00:01:55,289 And so our thrust can be thought of as an external force 29 00:01:55,289 --> 00:02:00,390 simply on the rocket as a system. 30 00:02:00,390 --> 00:02:06,090 And that's equal to minus the f dt u. 31 00:02:06,090 --> 00:02:10,139 Now again, if I chose j-hat up-- let's 32 00:02:10,139 --> 00:02:13,500 just look at this in components-- 33 00:02:13,500 --> 00:02:18,000 then our thrust as a vector, we'll 34 00:02:18,000 --> 00:02:25,980 write it as a y-component, is equal to minus dMfuel dt. 35 00:02:25,980 --> 00:02:28,140 Now what is that relative speed? 36 00:02:28,140 --> 00:02:30,960 Well, the fuel is being ejected backwards, 37 00:02:30,960 --> 00:02:34,230 so that's minus mu j-hat. 38 00:02:34,230 --> 00:02:38,430 The external force, by the way, would be minus mg j-hat 39 00:02:38,430 --> 00:02:40,380 if it's near the surface of the Earth. 40 00:02:40,380 --> 00:02:44,070 So we have another minus u j-hat. 41 00:02:44,070 --> 00:02:48,340 And we see that we have a positive thrust 42 00:02:48,340 --> 00:02:52,560 force in the vertical direction that 43 00:02:52,560 --> 00:02:57,060 is giving us an additional force other than the gravitational 44 00:02:57,060 --> 00:02:57,840 field. 45 00:02:57,840 --> 00:03:01,080 And in Cartesian-- in our unit vectors here, 46 00:03:01,080 --> 00:03:13,830 we have minus mg plus dMfuel dt times u is equal to Mr dV 47 00:03:13,830 --> 00:03:17,790 ydt of the rocket. 48 00:03:17,790 --> 00:03:23,240 And so this is the rocket equation in components.