1 00:00:04,340 --> 00:00:06,350 We've seen the rocket equation, and we've 2 00:00:06,350 --> 00:00:08,870 studied the example when there's no external forces. 3 00:00:08,870 --> 00:00:15,920 Now, let's consider an example when the external force is not 4 00:00:15,920 --> 00:00:17,150 0. 5 00:00:17,150 --> 00:00:19,310 Well, one of our examples is just a rocket 6 00:00:19,310 --> 00:00:21,380 taking off in a gravitational field. 7 00:00:21,380 --> 00:00:24,560 And so when we want to focus on the external forces, 8 00:00:24,560 --> 00:00:28,970 let's draw our rocket at time, t. 9 00:00:28,970 --> 00:00:32,689 And now, consider-- what are the force diagrams on this rocket? 10 00:00:32,689 --> 00:00:37,190 Well, we have a gravitational force, mg. 11 00:00:37,190 --> 00:00:39,860 So we're going to choose j hat up. 12 00:00:39,860 --> 00:00:42,590 So now, what does our rocket equation look like? 13 00:00:42,590 --> 00:00:47,650 We wrote the exhaust velocity as minus mu j hat. 14 00:00:47,650 --> 00:00:52,490 And so if we write this now as a vector equation in the j hat 15 00:00:52,490 --> 00:00:56,090 component, we get minus mg. 16 00:00:56,090 --> 00:01:02,240 We have the rate that the rocket mass is changing, 17 00:01:02,240 --> 00:01:06,740 and this is in our-- we have this as-- remember, 18 00:01:06,740 --> 00:01:10,200 dm dr/dt is going to be negative. 19 00:01:10,200 --> 00:01:11,750 So there's a minus. 20 00:01:11,750 --> 00:01:18,080 And over here, we have another-- this is u because u is minus u. 21 00:01:18,080 --> 00:01:21,725 And finally, we have m rocket dvr/dt. 22 00:01:24,920 --> 00:01:28,570 Now, once again, we're going to try to integrate-- 23 00:01:28,570 --> 00:01:31,460 this is actually equal. 24 00:01:31,460 --> 00:01:33,830 And we're going to try to integrate this equation 25 00:01:33,830 --> 00:01:37,850 to find a solution for the velocity as a function of time 26 00:01:37,850 --> 00:01:40,340 when all the fuel is burned. 27 00:01:40,340 --> 00:01:45,630 So what we do is we'll multiply through by dt, 28 00:01:45,630 --> 00:01:54,080 and we have minus dmr times u equals mr dvr. 29 00:01:54,080 --> 00:01:56,630 And this is mass of the rocket. 30 00:01:56,630 --> 00:02:00,080 And now, let's divide through by mass of the rocket, 31 00:02:00,080 --> 00:02:11,330 so we have minus g minus dmr over mru equals dvr. 32 00:02:11,330 --> 00:02:16,070 And once again, we can integrate this equation 33 00:02:16,070 --> 00:02:18,350 from some initial time to some final time. 34 00:02:18,350 --> 00:02:22,700 I'll just denote that by i initial and i final, 35 00:02:22,700 --> 00:02:27,560 and what we see here is we have minus g times t final minus t 36 00:02:27,560 --> 00:02:29,270 initial because we're integrating 37 00:02:29,270 --> 00:02:30,680 with respect to time. 38 00:02:30,680 --> 00:02:35,329 Here, we're integrating mass, so that's minus mu natural log. 39 00:02:35,329 --> 00:02:43,579 And I'm going to write this as mr final over mr initial. 40 00:02:43,579 --> 00:02:49,310 And over here, we have vr final minus vr initial. 41 00:02:49,310 --> 00:02:54,590 And let's just remove that time, t. 42 00:02:54,590 --> 00:02:58,190 And so now, here is our rocket equation. 43 00:02:58,190 --> 00:02:59,990 Now, what we notice here is-- let's look 44 00:02:59,990 --> 00:03:02,370 at some rocket taking off. 45 00:03:02,370 --> 00:03:07,700 So suppose that, at 10 initial equals 0, 46 00:03:07,700 --> 00:03:13,460 we have the special condition that vr initial equals 0 47 00:03:13,460 --> 00:03:18,650 and mr initial equals the mass of the rocket-- dry mass. 48 00:03:18,650 --> 00:03:19,970 That's all. 49 00:03:19,970 --> 00:03:21,980 That's not counting the fuel. 50 00:03:21,980 --> 00:03:26,450 Plus the total mass of the fuel. 51 00:03:26,450 --> 00:03:32,690 And the final-- at t final, when the engine turns off-- we're 52 00:03:32,690 --> 00:03:36,440 going to try to find this velocity at t final. 53 00:03:36,440 --> 00:03:38,960 And our condition for the final mass 54 00:03:38,960 --> 00:03:42,140 is all the fuel has been burned, so this is just 55 00:03:42,140 --> 00:03:44,750 the dry mass of the rocket. 56 00:03:44,750 --> 00:03:49,510 And so what we get is we get minus g t final. 57 00:03:49,510 --> 00:03:53,540 Now, I'm going to switch the sign here, 58 00:03:53,540 --> 00:03:59,630 so we have-- if I invert the log, 59 00:03:59,630 --> 00:04:01,790 that will give a minus sign. 60 00:04:01,790 --> 00:04:11,180 And I have plus u log of mr of d plus mass of the fuel. 61 00:04:11,180 --> 00:04:13,340 That's the initial mass of the rocket. 62 00:04:13,340 --> 00:04:18,320 Divided by just the dry mass, and that's 63 00:04:18,320 --> 00:04:24,140 equal to vr at t final. 64 00:04:24,140 --> 00:04:27,170 So one of the interesting questions that you might ask 65 00:04:27,170 --> 00:04:34,580 is-- how fast should you burn the fuel, 66 00:04:34,580 --> 00:04:37,430 and what will that have to do with the final speed? 67 00:04:37,430 --> 00:04:39,530 So we can see from this expression 68 00:04:39,530 --> 00:04:43,290 that, if you burn the fuel for a very long period of time-- 69 00:04:43,290 --> 00:04:46,220 so t final is big-- then this piece will 70 00:04:46,220 --> 00:04:48,320 diminish the final velocity. 71 00:04:48,320 --> 00:04:51,830 So if you want the fastest speed after you've burned the fuel, 72 00:04:51,830 --> 00:04:54,620 you want to burn the fuel so that t final is as short as 73 00:04:54,620 --> 00:04:55,430 possible. 74 00:04:55,430 --> 00:04:57,830 We have the shortest burn time, will give us 75 00:04:57,830 --> 00:05:03,577 the largest velocity when all the fuel has been burned.