0:00:00.000,0:00:20.679
36C3 preroll music
0:00:20.679,0:00:25.929
Herald: OK, so the next talk for this[br]evening is on how to get to Mars and all
0:00:25.929,0:00:31.890
in very interesting ways. Some of them[br]might be really, really slow. Our next
0:00:31.890,0:00:36.640
speaker has studied physics and has a PhD[br]in maths and is currently working as a
0:00:36.640,0:00:41.180
mission planner at the German Space[br]Operations Center. Please give a big round
0:00:41.180,0:00:50.142
of applause to Sven.[br]Sven: Thank you.
0:00:50.142,0:00:52.551
Hello and welcome to[br]"Thrust is not an option: How to get a
0:00:52.551,0:00:56.910
Mars really slow". My name is Sven. I'm a[br]mission planner at the German Space
0:00:56.910,0:01:01.380
Operations Center, which is a part of the[br]DLR, the Deutsches Zentrum für Luft- und
0:01:01.380,0:01:05.190
Raumfahrt. And first of all, I have to[br]apologize because I kind of cheated a
0:01:05.190,0:01:11.461
little bit in the title. The accurate[br]title would have been "Reducing thrust: How
0:01:11.461,0:01:16.990
to get to Mars or maybe Mercury really[br]slow". The reason for this is that I will
0:01:16.990,0:01:22.750
actually use Mercury as an example quite[br]a few times. And also we will not be able
0:01:22.750,0:01:29.000
to actually get rid of all the maneuvers[br]that we want to do. So the goal of this
0:01:29.000,0:01:34.550
talk is to give you an introduction to[br]orbital mechanics to see what we can do.
0:01:34.550,0:01:37.860
What are the techniques that you can use[br]to actually get to another planet, to
0:01:37.860,0:01:44.000
bring a spacecraft to another planet and[br]also go a few more, go a bit further into
0:01:44.000,0:01:49.950
some more advanced techniques. So we will[br]start with gravity and the two body
0:01:49.950,0:01:54.900
problem. So this is the basics, the[br]underlying physics that we need. Then we
0:01:54.900,0:01:59.110
will talk about the two main techniques[br]maybe to get to Mars, for example, the
0:01:59.110,0:02:04.530
Hohmann-transfer as well as gravity[br]assists. The third point will be an
0:02:04.530,0:02:08.520
extension of that that's called a planar[br]circular restricted three body problem.
0:02:08.520,0:02:14.710
Sounds pretty complicated, but we will see[br]in pictures what it is about. And then we
0:02:14.710,0:02:21.920
will finally get a taste of certain ways[br]to actually be even better, be even more
0:02:21.920,0:02:26.190
efficient by looking at what's called[br]ballistic capture and the weak stability
0:02:26.190,0:02:32.770
boundary. All right, so let's start. First[br]of all, we have gravity and we need to
0:02:32.770,0:02:36.300
talk about a two body problem. So I'm[br]standing here on the stage and I'm
0:02:36.300,0:02:41.660
actually being well accelerated downwards,[br]right? The earth actually attracts me. And
0:02:41.660,0:02:47.537
this is the same thing that happens for[br]any two bodies that have mass. OK. So they
0:02:47.537,0:02:51.730
attract each other by gravitational force[br]and this force will actually accelerate
0:02:51.730,0:02:56.620
the objects towards each other. Notice[br]that the force actually depends on the
0:02:56.620,0:03:03.720
distance. OK. So we don't need[br]to know any details. But in principle, the
0:03:03.720,0:03:11.310
force gets stronger the closer the objects[br]are. OK, good. Now, we can't really
0:03:11.310,0:03:17.080
analyze this whole thing in every[br]detail. So we will make a few assumptions.
0:03:17.080,0:03:23.090
One of them will be that all our bodies,[br]in particular, the Sun, Earth will
0:03:23.090,0:03:27.470
actually be points, OK? So we will just[br]consider points because anything else is
0:03:27.470,0:03:32.680
too complicated for me. Also, all our[br]satellites will actually be just points.
0:03:32.680,0:03:38.569
One of the reasons is that, in principle,[br]you have to deal with the attitude of the
0:03:38.569,0:03:42.620
satellites. For example, a solar panel[br]needs to actually point towards the sun,
0:03:42.620,0:03:47.300
but of course that's complicated. So we[br]will skip this for this talk. Third point
0:03:47.300,0:03:51.790
is that none of our planets will have an[br]atmosphere, so there won't be any
0:03:51.790,0:03:58.760
friction anywhere in the space. And the[br]fourth point is that we will mostly
0:03:58.760,0:04:03.730
restrict to movement within the plane. So[br]we only have like two dimensions during
0:04:03.730,0:04:11.350
this talk. And also, I will kind of forget[br]about certain planets and other masses
0:04:11.350,0:04:16.070
from time to time. Okay. I'm mentioning[br]this because I do not want you to go home
0:04:16.070,0:04:20.590
this evening, start planning your own[br]interplanetary mission, then maybe
0:04:20.590,0:04:24.720
building your spacecraft tomorrow,[br]launching in three days and then a week
0:04:24.720,0:04:31.030
later I get an e-mail: "Hey, this [br]didn't work. I mean, what did you tell me?"
0:04:31.030,0:04:35.680
OK. So if you actually want to do this at[br]home, don't try this just now but please
0:04:35.680,0:04:40.539
consult your local flight dynamics department, [br]they will actually supply with the necessary
0:04:40.539,0:04:46.410
details. All right. So what's the two body[br]problem about? So in principle we have
0:04:46.410,0:04:51.229
some body - the Sun - and the spacecraft[br]that is being attracted by the Sun. Now,
0:04:51.229,0:04:55.520
the Sun is obviously much heavier than a[br]spacecraft, meaning that we will actually
0:04:55.520,0:05:01.699
neglect the force that the spacecraft[br]exerts on the Sun. So instead, the Sun
0:05:01.699,0:05:06.319
will be at some place. It [br]might move in some way, or a
0:05:06.319,0:05:12.309
planet. But we only care about a[br]spacecraft, in general. Furthermore,
0:05:12.309,0:05:16.469
notice that if you specify the position[br]and the velocity of a spacecraft at some
0:05:16.469,0:05:23.487
point, then the gravitational force will[br]actually determine the whole path of the
0:05:23.487,0:05:31.370
spacecraft for all time. OK. So this path[br]is called the orbit and this is what we
0:05:31.370,0:05:34.930
are talking about. So we want to determine[br]orbits. We want to actually find ways how
0:05:34.930,0:05:44.129
to efficiently change orbits in order to[br]actually reach Mars, for example. There is
0:05:44.129,0:05:51.380
one other thing that you may know from[br]your day to day life. If you actually take
0:05:51.380,0:05:55.680
an object and you put it high up and you[br]let it fall down, then it will accelerate.
0:05:55.680,0:06:01.879
OK. So one way to actually describe this[br]is by looking at the energy. There is a
0:06:01.879,0:06:05.620
kinetic energy that's related to movement,[br]to velocity, and there is a potential
0:06:05.620,0:06:10.680
energy which is related to this[br]gravitational field. And the sum of those
0:06:10.680,0:06:17.901
energies is actually conserved. This means[br]that when the spacecraft moves, for
0:06:17.901,0:06:23.370
example, closer to the Sun, then its[br]potential energy will decrease and thus
0:06:23.370,0:06:28.909
the kinetic energy will increase. So it[br]will actually get faster. So you can see
0:06:28.909,0:06:32.550
this, for example, here. We have [br]two bodies that rotate around their
0:06:32.550,0:06:37.550
center of mass. And if you're careful, if[br]you're looking careful when they actually
0:06:37.550,0:06:43.229
approach each other, then they are quite a[br]bit faster. OK. So it is important to keep
0:06:43.229,0:06:48.249
in mind. All right, so how do spacecrafts[br]actually move? So we will now actually
0:06:48.249,0:06:55.210
assume that we don't use any kind of[br]engine, no thruster. We just cruise along
0:06:55.210,0:07:00.180
the gravitational field. And then there[br]are essentially three types of orbits that
0:07:00.180,0:07:04.360
we can have. One of them are hyperbolas.[br]So this case happens if the velocity is
0:07:04.360,0:07:10.759
very high, because those are not periodic[br]solutions. They're not closed. So instead,
0:07:10.759,0:07:15.819
our spacecraft kind of approaches the Sun[br]or the planet in the middle and the center
0:07:15.819,0:07:21.210
from infinity. It will kind of turn,[br]it will change its direction and then it
0:07:21.210,0:07:27.990
will leave again to infinity. Another[br]orbit that may happen as a parabola, this
0:07:27.990,0:07:33.180
is kind of similar. Actually, we won't[br]encounter parabolas during this talk. So I
0:07:33.180,0:07:38.029
will skip this. And the probably most[br]common orbit that we all know are
0:07:38.029,0:07:44.509
ellipses. In particular circles because,[br]well, we know that the Earth is actually
0:07:44.509,0:07:49.449
moving around the sun approximately in a[br]circle. OK. So those are periodic
0:07:49.449,0:07:56.869
solutions. They are closed. And in[br]particular, they are such that if a
0:07:56.869,0:08:00.789
spacecraft is on one of those orbits and[br]it's not doing anything, then it will
0:08:00.789,0:08:09.120
forever stay on that orbit, OK, in the two[br]body problem. So now the problem is we
0:08:09.120,0:08:13.069
actually want to change this. So we need[br]to do something. OK. So we want to change
0:08:13.069,0:08:17.589
from one circle around the Sun, which [br]corresponds to Earth orbit, for example, to
0:08:17.589,0:08:21.509
another circle around the Sun, which [br]corresponds to Mars orbit. And in order to
0:08:21.509,0:08:27.319
change this, we need to do some kind of[br]maneuver. OK. So this is an actual picture
0:08:27.319,0:08:33.360
of a spacecraft. And what the spacecraft[br]is doing, it's emitting some kind of
0:08:33.360,0:08:40.500
particles in some direction. They have a[br]mass m. Those particles might be gases or
0:08:40.500,0:08:48.100
ions, for example. And because these gases[br]or these emissions, they carry some mass,
0:08:48.100,0:08:53.160
they actually have some momentum due to[br]conservation of momentum. This means that
0:08:53.160,0:08:58.050
the spacecraft actually has to accelerate[br]in the opposite direction. OK. So whenever
0:08:58.050,0:09:03.980
we do this, we will actually accelerate[br]the spacecraft and change the velocity and
0:09:03.980,0:09:12.660
this change of velocity as denoted by a[br]delta v. And delta v is sort of the basic
0:09:12.660,0:09:16.980
quantity that we actually want to look at[br]all the time. OK. Because this describes
0:09:16.980,0:09:26.009
how much thrust we need to actually fly[br]in order to change our orbit. Now,
0:09:26.009,0:09:32.440
unfortunately, it's pretty expensive to,[br]well, to apply a lot of delta v. This is
0:09:32.440,0:09:37.339
due to the costly rocket equation. So the[br]fuel that you need in order to reach or to
0:09:37.339,0:09:45.850
change your velocity to some delta v this[br]depends essentially exponentially on the
0:09:45.850,0:09:52.740
target delta v. So this means we really[br]need to take care that we use as few
0:09:52.740,0:10:00.490
delta v as possible in order to reduce the[br]needed fuel. There's one reason for
0:10:00.490,0:10:04.990
that is... we want to maybe reduce[br]costs because then we need to carry
0:10:04.990,0:10:10.009
less fuel. However, we can also actually[br]think the other way round if we actually
0:10:10.009,0:10:16.769
use less fuel than we can[br]bring more stuff for payloads, for
0:10:16.769,0:10:24.170
missions, for science experiments. Okay.[br]So that's why in spacecraft mission
0:10:24.170,0:10:28.399
design we actually have to take care of[br]reducing the amount of delta v that is
0:10:28.399,0:10:34.269
spent during maneuvers. So let's see, what[br]can we actually do? So one example of a
0:10:34.269,0:10:41.500
very basic maneuver is actually to, well,[br]sort of raise the orbit. So imagine you
0:10:41.500,0:10:48.100
have a spacecraft on a circular orbit[br]around, for example, Sun here. Then you
0:10:48.100,0:10:52.269
might want to raise the orbit[br]in the sense that you make it more
0:10:52.269,0:10:57.410
elliptic and reach higher altitudes. For[br]this you just accelerate in the direction
0:10:57.410,0:11:00.680
that you're flying. So you apply some[br]delta v and this will actually change the
0:11:00.680,0:11:08.029
form of the ellipse. OK. So it's a very[br]common scenario. Another one is if you
0:11:08.029,0:11:12.370
approach a planet from very far away, then[br]you might have a very high relative
0:11:12.370,0:11:18.570
velocity such that with respect to the[br]planet, you're on a hyperbolic orbit. OK.
0:11:18.570,0:11:22.540
So you would actually leave the planet.[br]However, if this is actually your
0:11:22.540,0:11:26.840
target planet that you want to reach, then[br]of course you have to enter orbit. You
0:11:26.840,0:11:31.290
have to somehow slow down. So the idea[br]here is that when you approach
0:11:31.290,0:11:37.449
the closest point to the planet,[br]for example, then you actually slow down.
0:11:37.449,0:11:41.830
So you apply delta v in sort of in the[br]opposite direction and change the orbit to
0:11:41.830,0:11:45.709
something that you prefer, for example an[br]ellipse. Because now you will actually
0:11:45.709,0:11:54.760
stay close to the planet forever. Well, if[br]relative it would a two body problem. OK,
0:11:54.760,0:12:02.230
so. Let's continue. Now, we actually want[br]to apply this knowledge to well, getting,
0:12:02.230,0:12:08.829
for example, to Mars. Let's start with [br]Hohmann transfers. Mars and Earth both
0:12:08.829,0:12:16.589
revolve around the Sun in pretty much[br]circular orbits. And our spacecraft starts
0:12:16.589,0:12:21.220
at the Earth. So now we want to reach[br]Mars. How do we do this? Well, we can fly
0:12:21.220,0:12:27.270
what we just said. So we accelerate [br]when we are at the Earth orbit,
0:12:27.270,0:12:36.810
such that our orbit touches the Mars orbit[br]on the other side. OK. So this gives us
0:12:36.810,0:12:40.990
some amount of delta v we have to apply.[br]We need to calculate this. I'm not going
0:12:40.990,0:12:47.939
to do this. Then we actually fly around[br]this orbit for half an ellipse. And once
0:12:47.939,0:12:53.139
we have reached the Mars orbit, then we[br]can actually accelerate again in order to
0:12:53.139,0:12:59.680
raise other side of the Ellipse until that[br]one reaches the Mars orbit. So with two
0:12:59.680,0:13:04.839
maneuvers, two accelerations, we can[br]actually change from one circular orbit to
0:13:04.839,0:13:09.960
another one. OK. This is the basic idea of[br]how you actually fly to Mars. So let's
0:13:09.960,0:13:16.339
look at an animation. So this is the orbit[br]of the InSight mission. That's another Mars
0:13:16.339,0:13:25.199
mission which launched and landed last[br]year. The blue circle is the Earth and the
0:13:25.199,0:13:33.130
green one is Mars. And the pink is[br]actually the satellite or the probe.
0:13:33.130,0:13:40.381
You can see that, well, it's flying in[br]this sort of half ellipse. However, there
0:13:40.381,0:13:47.339
are two... well, there's just one problem,[br]namely when it actually reaches Mars, Mars
0:13:47.339,0:13:51.779
needs to be there. I mean, that sounds[br]trivial. Yeah. But I mean, imagine you fly
0:13:51.779,0:13:57.449
there and then well, Mars is somewhere[br]else, that's not good. I mean this happens
0:13:57.449,0:14:05.439
pretty regularly when you begin playing a[br]Kerbal Space Program, for example.
0:14:05.439,0:14:11.050
So we don't want to like play around[br]with this the whole time, we actually want
0:14:11.050,0:14:16.760
to hit Mars. So we need to take care of[br]that Mars is at the right position when we
0:14:16.760,0:14:21.779
actually launch. Because it will traverse[br]the whole green line during our transfer.
0:14:21.779,0:14:27.980
This means that we can only launch such a[br]Hohmann transfer at very particular times.
0:14:27.980,0:14:31.579
And sort of this time when you can do[br]this transfer is called the transfer
0:14:31.579,0:14:39.599
window. And for Earth-Mars, for example.[br]This is possible every 26 months. So if
0:14:39.599,0:14:44.639
you miss something, like, software's not[br]ready, whatever, then you have to wait for
0:14:44.639,0:14:53.000
another twenty six months. So, the flight[br]itself takes about six months. All right.
0:14:53.000,0:14:59.399
There is another thing that we kind of[br]neglected so far, namely when we start,
0:14:59.399,0:15:04.450
when we depart from Earth, then well[br]there's Earth mainly. And so that's the
0:15:04.450,0:15:11.009
main source of gravitational force. For[br]example, right now I'm standing here on
0:15:11.009,0:15:19.800
the stage and I experience the Earth. I[br]also experience Sun and Mars. But I mean,
0:15:19.800,0:15:24.899
that's very weak. I can ignore this. So at[br]the beginning of our mission to Mars, we
0:15:24.899,0:15:29.410
actually have to take care that we[br]are close to Earth. Then during the
0:15:29.410,0:15:34.379
flight, the Sun actually dominates the[br]gravitational force. So we will only
0:15:34.379,0:15:38.029
consider this. But then when we approach[br]Mars, we actually have to take care about
0:15:38.029,0:15:44.430
Mars. Okay. So we kind of forgot this[br]during the Hohmann transfer. So what you
0:15:44.430,0:15:49.970
actually do is you patch together[br]solutions of these transfers. Yeah. So in
0:15:49.970,0:15:55.240
this case, there are there are essentially[br]three sources of gravitational force so
0:15:55.240,0:15:59.389
Earth, Sun, Mars. So we will have three two[br]body problems that we need to consider.
0:15:59.389,0:16:04.639
Yeah. One for departing, one for the[br]actual Hohmann transfer. And then the third
0:16:04.639,0:16:09.449
one when we actually approach Mars. So[br]this makes this whole thing a bit more
0:16:09.449,0:16:14.649
complicated. But it's also nice because[br]actually we need less delta v than we
0:16:14.649,0:16:19.589
would for the basic hohmann transfer. One[br]reason for this is that when we look at
0:16:19.589,0:16:25.930
Mars. So the green line is now the Mars[br]orbit and the red one is again the
0:16:25.930,0:16:31.509
spacecraft, it approaches Mars now we can[br]actually look at what happens at Mars by
0:16:31.509,0:16:40.480
kind of zooming into the system of Mars.[br]OK. So Mars is now standing still. And
0:16:40.480,0:16:46.050
then we see that the velocity of the[br]spacecraft is actually very high relative
0:16:46.050,0:16:50.399
to Mars. So it will be on the hyperbolic[br]orbit and will actually leave Mars again.
0:16:50.399,0:16:55.270
You can see this on the left side. Right.[br]Because it's leaving Mars again. So what
0:16:55.270,0:17:00.459
you need to do is, in fact, you need to[br]slow down and change your orbit into an
0:17:00.459,0:17:04.770
ellipse. Okay. And this delta v, is that[br]you that you need here for this maneuver
0:17:04.770,0:17:12.220
it's actually less than the delta v you[br]would need to to circularize the orbit to
0:17:12.220,0:17:18.400
just fly in the same orbit as Mars. So we[br]need to slow down. A similar argument
0:17:18.400,0:17:24.640
actually at Earth shows that, well, if you[br]actually launch into space, then you do
0:17:24.640,0:17:29.780
need quite some speed already to not fall[br]down back onto Earth. So that's something
0:17:29.780,0:17:33.700
like seven kilometers per second or so.[br]This means that you already have some
0:17:33.700,0:17:38.810
speed. OK. And if you align your orbit or[br]your launch correctly, then you already
0:17:38.810,0:17:43.350
have some of the delta v that you need for[br]the Hohmann transfer. So in principle, you
0:17:43.350,0:17:52.080
need quite a bit less delta v than than[br]you might naively think. All right. So
0:17:52.080,0:17:57.280
that much about Hohmann transfer. Let's look[br]at Gravity assist. That's another major
0:17:57.280,0:18:03.530
technique for interplanetary missions. The[br]idea is that we can actually use planets
0:18:03.530,0:18:10.570
to sort of getting pulled along. So this[br]is an animation, on the lower animation
0:18:10.570,0:18:16.300
you see kind of the picture when you look[br]at the planet. So the planets standing
0:18:16.300,0:18:21.320
still and we assume that the spacecraft's[br]sort of blue object is on a hyperbolic
0:18:21.320,0:18:27.120
orbit and it's kind of making a 90 degree[br]turn. OK. And the upper image, you
0:18:27.120,0:18:32.320
actually see the picture when[br]you look from the Sun, so the planet is
0:18:32.320,0:18:38.820
actually moving. And if you look very[br]carefully at the blue object then you can
0:18:38.820,0:18:45.030
see that it is faster. So once it has[br]passed, the planet is actually faster.
0:18:45.030,0:18:52.900
Well, we can actually look at this problem.[br]So this is, again, the picture. When
0:18:52.900,0:18:56.260
Mars is centered, we have some entry[br]velocity. Then we are in this hyperbolic
0:18:56.260,0:19:03.160
orbit. We have an exit velocity. Notice[br]that the lengths are actually equal. So
0:19:03.160,0:19:08.580
it's the same speed. But just a turn[br]direction of this example. But then we can
0:19:08.580,0:19:13.410
look at the whole problem with a moving[br]Mars. OK, so now Mars has some velocity
0:19:13.410,0:19:19.610
v_mars. So the actual velocity that we see[br]is the sum of the entry and the Mars
0:19:19.610,0:19:25.870
velocity before and afterwards exit, plus[br]Mars velocity. And if you look at those
0:19:25.870,0:19:31.910
two arrows, then you see immediately that,[br]well, the lengths are different. Okay. So
0:19:31.910,0:19:37.650
this is just the whole phenomenon here. So[br]we see that by actually passing close to
0:19:37.650,0:19:43.250
such a planet or massive body, we[br]can sort of gain free delta v. Okay, so of
0:19:43.250,0:19:49.080
course, it's not. I mean, the energy is[br]still conserved. Okay. But yeah, let's not
0:19:49.080,0:19:53.550
worry about these details here. Now, the[br]nice thing is we can use this technique to
0:19:53.550,0:19:58.970
actually alter course. We can speed up. So[br]this is the example that I'm shown here.
0:19:58.970,0:20:02.790
But also, we can use this to slow down.[br]Okay. So that's a pretty common
0:20:02.790,0:20:08.160
application as well. We can use this to[br]slow down by just changing the arrows,
0:20:08.160,0:20:15.860
essentially. So just approaching Mars from[br]a different direction, essentially. So
0:20:15.860,0:20:21.960
let's look at the example. And this is[br]Bepicolombo. That's actually the reason
0:20:21.960,0:20:26.240
why I kind of changed the title, because[br]Bepicolombo is actually a mission to
0:20:26.240,0:20:32.661
Mercury. So it was launched last year.[br]It's a combined ESA/JAXA mission and it
0:20:32.661,0:20:38.390
consists of two probes and one thruster[br]centrally. So it's a through three stages
0:20:38.390,0:20:43.780
that you can see in the picture. Yeah.[br]That's a pretty awesome mission, actually.
0:20:43.780,0:20:49.930
It's really nice. But it has in[br]particular, a very cool orbit. So that's
0:20:49.930,0:20:56.627
it. What can we see here? So first of all,[br]the blue line, that's actually Earth. The
0:20:56.627,0:21:00.180
green one, that's Mercury. So that's where[br]we want to go. And we have this
0:21:00.180,0:21:07.130
intermediate turquoise one - that's[br]Venus. And well the curve is
0:21:07.130,0:21:10.790
Bepicolombo's orbit, so it looks pretty[br]complicated. Yeah, it's definitely not the
0:21:10.790,0:21:16.020
Hohmann transfer. And in fact, this[br]mission uses nine Gravity assists to reach
0:21:16.020,0:21:21.950
Mercury. And if you look at the[br]path so, for example, right now
0:21:21.950,0:21:28.690
it actually is very close to Mercury[br]because the last five or six Gravity
0:21:28.690,0:21:34.500
assists are just around Mercury or just[br]slow down. OK. And this saves a lot of
0:21:34.500,0:21:41.760
delta v compared to the standard [br]Hohmann transfer. All right. But we
0:21:41.760,0:21:45.810
want to do even better. OK. So let's now[br]actually make the whole problem more
0:21:45.810,0:21:53.830
complicated in order to hope for some kind[br]of nice tricks that we can do. OK, so now
0:21:53.830,0:21:58.550
we will talk about a planar circular[br]restricted three body problem. All right.
0:21:58.550,0:22:02.590
So in general, the three body problem just[br]means, hey, well, instead of two bodies,
0:22:02.590,0:22:07.400
we have three. OK. They pairwise attract[br]each other and then we can solve this
0:22:07.400,0:22:12.080
whole equation of motion. We can ask a[br]computer. And this is one animation of
0:22:12.080,0:22:17.490
what it could look like. So the three[br]masses and their orbits are traced and we
0:22:17.490,0:22:24.080
see immediately that we don't see anything[br]that's super complicated. There is no
0:22:24.080,0:22:29.670
way we can really... I don't know,[br]formulate a whole solution theory for a
0:22:29.670,0:22:33.650
general three body problem. That's[br]complicated. Those are definitely not
0:22:33.650,0:22:40.312
ellipses. So let's maybe go a step back[br]and make the problem a bit easier. OK. So
0:22:40.312,0:22:44.520
we will now talk about a plane or circular[br]restricted three body problem. There are
0:22:44.520,0:22:49.440
three words. So the first one is[br]restricted. Restricted means that in our
0:22:49.440,0:22:54.350
application case, one of the bodies is[br]actually a spacecraft. Spacecrafts are
0:22:54.350,0:22:58.440
much lighter than, for example, Sun and[br]Mars, meaning that we can actually ignore
0:22:58.440,0:23:05.570
the force that the spacecraft exerts on[br]Sun and Mars. Okay. So we will actually
0:23:05.570,0:23:11.740
consider Sun and Mars to be independent of[br]the spacecraft. OK. So in total, we only
0:23:11.740,0:23:18.120
have like two gravitational forces now[br]acting on a spacecraft. So we neglect sort
0:23:18.120,0:23:25.610
of this other force. Also, we will assume[br]that the whole problem is what's called
0:23:25.610,0:23:30.800
circular. So we assume that Sun and Mars[br]actually rotate in circles around their
0:23:30.800,0:23:37.081
center of mass. This assumption is[br]actually pretty okay. We will see a
0:23:37.081,0:23:42.960
picture right now. So in this graph, for[br]example, in this image, you can see that
0:23:42.960,0:23:48.680
the black situation. So this might be at[br]some time, at some point in time. And then
0:23:48.680,0:23:54.520
later on, Sun and Mars actually have moved[br]to the red positions and the spacecraft is
0:23:54.520,0:24:00.840
at some other place. And now, of course,[br]feels some other forces. OK. And also we
0:24:00.840,0:24:04.330
will assume that this problem is plane,[br]meaning again that everything takes place
0:24:04.330,0:24:12.380
in the plane. OK. So let's look at the[br]video. That's a video with a very low
0:24:12.380,0:24:19.610
frame rate, something like two images per[br]day. Maybe it's actually Pluto and Charon.
0:24:19.610,0:24:27.250
So the larger one, this is the ex-planet[br]Pluto. It was taken by New Horizons in
0:24:27.250,0:24:34.360
2015 and it shows that they actually[br]rotate around the center of mass. Yeah. So
0:24:34.360,0:24:40.270
both actually rotate. This also happens,[br]for example, for Sun and Earth or Sun and
0:24:40.270,0:24:45.250
Mars or sun and Jupiter or also Earth and[br]Moon. However, in those other cases, the
0:24:45.250,0:24:50.650
center of mass is usually contained in the[br]larger body. And so this means that in the
0:24:50.650,0:24:57.910
case of Sun-Earth, for example, the Sun[br]will just wiggle a little bit. OK. So you
0:24:57.910,0:25:04.410
don't really see this extensive rotation.[br]OK. Now, this problem is still difficult.
0:25:04.410,0:25:10.140
OK. So if you're putting out a mass in[br]there, then you don't really
0:25:10.140,0:25:15.499
know what happens. However, there's a nice[br]trick to simplify this problem. And
0:25:15.499,0:25:19.730
unfortunately, I can't do this here. But[br]maybe all the viewers at home, they can
0:25:19.730,0:25:25.080
try to do this. You can take your laptop.[br]Please don't do this. And you can rotate
0:25:25.080,0:25:34.020
your laptop at the same speed as this[br]image actually rotates. OK. Well, then
0:25:34.020,0:25:39.340
what happens? The two masses will actually[br]stand still from your point of view. OK.
0:25:39.340,0:25:45.080
If you do it carefully and don't break[br]anything. So we switch to this sort of
0:25:45.080,0:25:50.590
rotating point of view. OK, then the two[br]masses stand still. We still have the two
0:25:50.590,0:25:56.020
gravitational forces towards Sun and Mars.[br]But because we kind of look at it from a
0:25:56.020,0:26:00.670
rotated or from a moving point of view, we[br]get two new forces, those forces, you
0:26:00.670,0:26:04.890
know, the centrifugal forces, of[br]course, the one that, for example, you
0:26:04.890,0:26:11.510
have when you play with some[br]children or so, they want to be pulled in
0:26:11.510,0:26:17.440
a circle very quickly and then they start[br]flying and that's pretty cool. But here we
0:26:17.440,0:26:21.730
actually have this force acting on the[br]spacecraft. Okay. And also there is the
0:26:21.730,0:26:26.790
Coriolis force, which is a bit less known.[br]This depends on the velocity of the
0:26:26.790,0:26:31.660
spacecraft. OK. So if there is no velocity[br]in particular, then there will not be any
0:26:31.660,0:26:38.270
Coriolis force. So our new problem[br]actually has four forces. OK, but the
0:26:38.270,0:26:43.580
advantage is that Sun and Mars actually[br]are standing still. So we don't need to
0:26:43.580,0:26:51.040
worry about their movement. OK. So now how[br]does this look like? Well, this might be
0:26:51.040,0:26:55.990
an example for an orbit. Well, that looks[br]still pretty complicated. I mean, this is
0:26:55.990,0:27:01.500
something that you can't have in a two[br]body problem. It has these weird wiggles.
0:27:01.500,0:27:06.320
I mean, they're not really corners. And it[br]actually kind of switches from Sun to
0:27:06.320,0:27:10.650
Mars. Yes. So it stays close to Sun for[br]some time and it moves somewhere else as
0:27:10.650,0:27:15.650
it, it's still pretty complicated. I don't[br]know. Maybe some of you have have read the
0:27:15.650,0:27:23.490
book "The Three-Body Problem". So there,[br]for example, the two masses might be a
0:27:23.490,0:27:28.760
binary star system. OK. And then you have[br]a planet that's actually moving along such
0:27:28.760,0:27:35.710
an orbit. OK, that looks pretty bad. So in[br]particular, the seasons might be somewhat
0:27:35.710,0:27:41.960
messed up. Yeah. So this problem is, in[br]fact, in a strong mathematical sense,
0:27:41.960,0:27:47.200
chaotic. OK. So chaotic means something[br]like if you start with very close initial
0:27:47.200,0:27:51.610
conditions and you just let the system[br]evolve, then the solutions will look very,
0:27:51.610,0:27:58.560
very different. OK. And this really[br]happens here, which is good. All right. So
0:27:58.560,0:28:03.950
one thing we can ask is, well, is it[br]possible that if we put a spacecraft into
0:28:03.950,0:28:08.100
the system without any velocity, is it[br]possible that all the forces actually
0:28:08.100,0:28:12.450
cancel out. And it turns out yes, it is[br]possible. And those points are called
0:28:12.450,0:28:17.950
Lagrangian points. So if we have zero[br]velocity, there is no Coriolis force. So
0:28:17.950,0:28:23.460
we have only these three forces. And as[br]you can see in this little schematics
0:28:23.460,0:28:32.116
here, it's possible that all these forces[br]actually cancel out. Now imagine. Yeah. I
0:28:32.116,0:28:36.940
give you a homework. Please calculate all[br]these possible points. Then you can do
0:28:36.940,0:28:42.280
this. But we won't do this right here.[br]Instead, we just look at the result. So
0:28:42.280,0:28:47.880
those are the five Lagrangian points in[br]this problem. OK, so we have L4 and L5
0:28:47.880,0:28:52.150
which are at equilateral triangles with[br]Sun and Mars. Well, Sun - Mars in this
0:28:52.150,0:28:59.780
case. And we have L1, L2 and L3 on the[br]line through Sun and Mars. So if you put
0:28:59.780,0:29:05.250
the spacecraft precisely at L1 without any[br]velocity, then in relation to Sun and Mars
0:29:05.250,0:29:10.150
it will actually stay at the same position.[br]Okay, that's pretty cool. However,
0:29:10.150,0:29:15.770
mathematicians and physicists will[br]immediately ask well, OK, but what happens
0:29:15.770,0:29:21.920
if I actually put a spacecraft close to a[br]Lagrangian point? OK, so this is
0:29:21.920,0:29:28.200
related to what's called stability. And[br]you can calculate that around L4 and L5.
0:29:28.200,0:29:33.330
The spacecraft will actually stay in the[br]vicinity. So it will essentially rotate
0:29:33.330,0:29:38.980
around the Lagrangian points. So those are[br]called stable, whereas L1, L2 and L3 are
0:29:38.980,0:29:43.990
actually unstable. This means that if you[br]put a spacecraft there, then it will
0:29:43.990,0:29:50.600
eventually escape. However, this takes a[br]different amount of time depending on the
0:29:50.600,0:29:55.330
Lagrangian points. For example, if you're[br]close to L2, this might take a few months,
0:29:55.330,0:29:58.730
but if you're close to L3, this will[br]actually take something like a hundred
0:29:58.730,0:30:08.140
years or so. Okay, so those points are[br]still different. All right. Okay. So
0:30:08.140,0:30:10.950
is there actually any evidence that they[br]exist? I mean, maybe I'm just making this
0:30:10.950,0:30:14.690
up and, you know, I mean, haven't shown[br]you any equations. I could just throw
0:30:14.690,0:30:19.950
anything. However, we can actually look at[br]the solar system. So this is the inner
0:30:19.950,0:30:23.570
solar system here. In the middle you see,[br]well, the center you see the Sun, of
0:30:23.570,0:30:28.970
course. And to the lower left, there's[br]Jupiter. Now, if you imagine an
0:30:28.970,0:30:35.250
equilateral triangle of Sun and Jupiter,[br]well, there are two of them. And then you
0:30:35.250,0:30:40.920
see all these green dots there. And those[br]are asteroids. Those are the Trojans and
0:30:40.920,0:30:47.770
the Greeks. And they accumulate there[br]because L4 and L5 are stable. Okay. So we
0:30:47.770,0:30:55.140
can really see this dynamics gone on in[br]the solar system. However, there's also
0:30:55.140,0:30:59.490
various other applications of Lagrangian[br]points. So in particular, you might want
0:30:59.490,0:31:05.710
to put a space telescope somewhere in[br]space, of course, in such a way that the
0:31:05.710,0:31:11.520
Sun is not blinding you. Well, there is[br]Earth, of course. So if we can put the
0:31:11.520,0:31:18.980
spacecraft behind Earth, then we might be[br]in the shadow. And this is the Lagrangian
0:31:18.980,0:31:24.860
point L2, which is the reason why this is[br]actually being used for space telescopes
0:31:24.860,0:31:30.470
such as, for example, this one. However,[br]it turns out L2 is unstable. So we don't
0:31:30.470,0:31:35.091
really want to put the spacecraft just[br]there. But instead, we put it in an orbit
0:31:35.091,0:31:40.730
close... in a close orbit, close to L2.[br]And this particular example is called the
0:31:40.730,0:31:44.560
Halo orbit, and it's actually not[br]contained in the planes. I'm cheating a
0:31:44.560,0:31:48.030
little bit. It's on the right hand side to[br]you. And in the animation you actually see
0:31:48.030,0:31:54.110
the the orbit from the side. So it[br]actually leaves the plane the blue dot is
0:31:54.110,0:32:00.620
Earth and the left hand side you see [br]the actual orbit from the top. So
0:32:00.620,0:32:06.230
it's rotating around this place. OK. So[br]that's the James Webb Space Telescope, by
0:32:06.230,0:32:11.360
the way. You can see in the animation it's[br]supposed to launch in 2018. That didn't
0:32:11.360,0:32:19.530
work out, unfortunately, but stay tuned.[br]Another example. That's how it has become
0:32:19.530,0:32:26.200
pretty famous recently as the Chinese[br]Queqiao relay satellite. This one sits at
0:32:26.200,0:32:31.090
the Earth - Moon L2 Lagrange point. And[br]the reason for this is that the Chinese
0:32:31.090,0:32:37.650
wanted to or actually did land the Chang'e 4[br]Moon lander on the backside of the Moon.
0:32:37.650,0:32:41.560
And in order to stay in contact, radio[br]contact with the lander, they had to put a
0:32:41.560,0:32:47.640
relay satellite behind the Moon, which[br]they could still see from Earth. And they
0:32:47.640,0:33:00.100
chose some similar orbit around L2. OK. So[br]let's now go to some other more advanced
0:33:00.100,0:33:07.510
technique: ballistic capture. Right. Okay.[br]So this whole business sort of started
0:33:07.510,0:33:14.410
with a mission from the beginning of the[br]1990s, and that's the Hiten mission. So
0:33:14.410,0:33:19.890
that was a Japanese well, Moon probe[br]consisted of a probe which had a small
0:33:19.890,0:33:26.290
orbiter site which was separated, and then[br]it was supposed to actually enter orbit
0:33:26.290,0:33:31.610
around Moon. Unfortunately, it missed its[br]maneuver. So it didn't apply enough delta v
0:33:31.610,0:33:37.570
so it actually flew off. And the[br]mission was sort of lost at that point
0:33:37.570,0:33:42.430
because Hiten itself, so the main probe[br]did not have enough fuel to reach the
0:33:42.430,0:33:47.701
Moon. All right. That's, of course, a[br]problem. I mean, that's a risk you have to
0:33:47.701,0:33:53.460
take. And they were probably pretty[br]devastated. However, there were two people
0:33:53.460,0:34:00.780
from JPL, NASA, who actually heard about[br]this, Belbruno and Miller, and they were
0:34:00.780,0:34:08.260
working on strange orbits, those ballistic[br]capture orbits. And they actually found
0:34:08.260,0:34:14.609
one for the Hiten probe. They sent this to[br]the Japanese and they actually use that
0:34:14.609,0:34:23.220
orbit to get the Hiten probe to the moon.[br]And it actually arrived in October 1991.
0:34:23.220,0:34:26.450
And then it could do some[br]science, you know, maybe some
0:34:26.450,0:34:31.389
different experiments, but it actually[br]arrived there. However, the transfer took
0:34:31.389,0:34:37.070
quite a bit longer. So a normal Moon[br]transfer takes like three days or so. But
0:34:37.070,0:34:42.320
this one actually took a few months. All[br]right. And the reason for this is that it
0:34:42.320,0:34:48.600
looks pretty weird. So this is a[br]picture of the orbiter - schematic picture.
0:34:48.600,0:34:54.260
And you can see the Earth. Well, there in[br]the middle sort of. And the Moon a bit to
0:34:54.260,0:35:01.820
the left at the L2 is the Lagrangian point[br]of the Sun - Earth system. OK. So it's
0:35:01.820,0:35:07.430
pretty far out. And you can see that the[br]orbit sort of consists of two parts.
0:35:07.430,0:35:13.100
First, it leaves Earth and it flies far[br]beyond the Moon. So somewhere completely
0:35:13.100,0:35:18.910
different towards some other Lagrangian[br]point. That's really far away. Then it
0:35:18.910,0:35:24.280
does some weird things. And in the upper[br]part of picture there it actually does a
0:35:24.280,0:35:30.240
maneuver. So we apply some thrusts there[br]to be to change on a different orbit. And
0:35:30.240,0:35:36.830
this orbit led the probe directly to the[br]moon where it was essentially captured for
0:35:36.830,0:35:42.320
free. OK. So it just entered orbit around[br]the Moon. And this is, of course, not
0:35:42.320,0:35:46.470
possible in the two body problem, but we[br]may find a way for doing this in the three
0:35:46.470,0:35:56.530
body problem. OK, so what do we mean by[br]capture? Now we have to sort of think
0:35:56.530,0:36:02.320
a bit more abstractly. The idea is... [br]we have Sun and Mars and we
0:36:02.320,0:36:08.100
have a spacecraft that flies in this three[br]body problem. So the red orbit is actually
0:36:08.100,0:36:14.960
the orbit of the spacecraft. Now, at any[br]point in time, we may decide to just
0:36:14.960,0:36:20.970
forget about the Sun. OK. So instead we[br]consider the two body problem of Mars and
0:36:20.970,0:36:26.760
a spacecraft. OK. Because at this point[br]in time, the spacecraft has a certain
0:36:26.760,0:36:31.240
position relative to Mars and a certain[br]velocity. So this determines its orbit in
0:36:31.240,0:36:36.440
the two body problem. Usually it would be[br]very fast. So it would be on a hyperbolic
0:36:36.440,0:36:43.269
orbit, which is denoted by the dashed line[br]here. OK. Or a dashed curve. So usually
0:36:43.269,0:36:47.240
you happen to be in a hyperbolic orbit.[br]But of course, that orbit is only an
0:36:47.240,0:36:50.280
approximation because in the three body[br]problem, the movement is actually
0:36:50.280,0:36:57.490
different. But later on, it might happen[br]that we continue on the orbit. We can do
0:36:57.490,0:37:01.530
the same kind of construction, but just[br]looking... but just ignoring the Sun
0:37:01.530,0:37:09.710
essentially, and then we could find that[br]the spacecraft suddenly is in a elliptical
0:37:09.710,0:37:14.190
orbit. This would mean that if you[br]forgot about the Sun, then the spacecraft
0:37:14.190,0:37:19.670
would be stable and would be captured by[br]Mars. It would be there. That would be
0:37:19.670,0:37:27.210
pretty nice. So this phenomenon, when this[br]happens, we call this a temporary capture.
0:37:27.210,0:37:33.810
OK. Temporary because it might actually[br]leave that situation again later on. Now,
0:37:33.810,0:37:37.320
because the actual movement depends on the[br]three body problem, which is super
0:37:37.320,0:37:42.150
complicated. So it's possible that it[br]actually leaves again. But for that moment
0:37:42.150,0:37:46.300
at least, it's captured and we want to[br]find a way or describe some kind of
0:37:46.300,0:37:54.000
algorithm perhaps how we can find[br]this situation essentially. OK, and in a
0:37:54.000,0:38:01.451
reasonable way, and the notion for this is[br]what's called, well, n-stability, the idea
0:38:01.451,0:38:08.020
is the following: we look at the three[br]body probleme, we want to go to Mars. So we
0:38:08.020,0:38:13.250
pick a line there. And on the line we take[br]a point x, which has some distance r to
0:38:13.250,0:38:20.090
the Mars and we pick a perpendicular[br]speed, a perpendicular velocity to the
0:38:20.090,0:38:25.810
line such that this corresponds to some[br]kind of elliptic orbit in the two body
0:38:25.810,0:38:30.290
problem. Okay. So that's the dashed line.[br]But then we actually look at the problem
0:38:30.290,0:38:37.820
in the three body problem and we just[br]evolve the spacecraft. And it's following
0:38:37.820,0:38:46.080
the red orbit. It might follow the red[br]orbit. And it can happen that after going
0:38:46.080,0:38:54.200
around Mars for one time, it hits again[br]the line. Okay, then we can do the same
0:38:54.200,0:38:59.540
construction with forgetting the Sun again[br]and just look at the two body problem. And
0:38:59.540,0:39:04.990
it's possible that this point actually[br]still corresponds to an elliptic orbit.
0:39:04.990,0:39:10.870
That's somewhat interesting, right?[br]Because now this means that if we actually
0:39:10.870,0:39:17.120
hit the point x, then we can follow the[br]orbit and we know that we wrap around
0:39:17.120,0:39:24.030
Mars once and are still sort of captured[br]in the corresponding two body problem.
0:39:24.030,0:39:29.230
Okay. If we actually are able to wrap[br]around Mars twice, then we would call this
0:39:29.230,0:39:35.980
2-stable and, well, for more rotations[br]that it is n-stable. Okay, so that's good
0:39:35.980,0:39:39.170
because such an orbit corresponds to[br]something that's usable because we will
0:39:39.170,0:39:45.370
wrap around Mars n times. However, it's[br]also possible that you have an unstable
0:39:45.370,0:39:49.020
point, meaning that we again start in[br]something that corresponds to an ellipse
0:39:49.020,0:39:54.170
around Mars. But if we actually follow the[br]orbit in a three body problem, it will,
0:39:54.170,0:39:58.110
for example, not come back. It will not[br]wrap around Mars, it will go to the Sun or
0:39:58.110,0:40:03.470
somewhere else. OK. So that's that's of[br]course, not a nice point. This one's
0:40:03.470,0:40:10.480
called unstable. And then there's another[br]thing we can do. That's actually a pretty
0:40:10.480,0:40:18.280
common trick in finding orbits, etc. We[br]can instead of solving the problem in
0:40:18.280,0:40:23.970
forward time we actually go back, okay. So[br]essentially in your program you just
0:40:23.970,0:40:28.810
replace time by minus time, for example,[br]and then you just solve the thing and you
0:40:28.810,0:40:37.070
go back in the past and it's possible [br]that a point that corresponds to such
0:40:37.070,0:40:41.641
an ellipse when you go back into the past[br]and it does not wrap around, but it
0:40:41.641,0:40:47.010
actually goes to the Sun, for example, we[br]call this unstable in the past. Okay. So
0:40:47.010,0:40:56.680
that's just some random definition. [br]And we can use this. The reason for
0:40:56.680,0:41:05.080
this is we can actually kind of take these[br]concepts together and build an orbit from
0:41:05.080,0:41:13.020
that. The idea being we pick a point x[br]that is n-stable. So, for example, it
0:41:13.020,0:41:19.710
might wrap around Mars six times, some[br]number that we like. This is the blue part
0:41:19.710,0:41:24.130
here in the picture. So it wraps around[br]Mars six times. But the way we go back in
0:41:24.130,0:41:30.560
time, it actually leaves Mars or at least[br]it doesn't come back in such a way that
0:41:30.560,0:41:42.070
it's again on an ecliptic curve. So this[br]is the red part. Okay. So we can
0:41:42.070,0:41:48.390
just follow this and then we pick a point[br]y on that curve. Okay. So this one will be
0:41:48.390,0:41:57.990
pretty far away from Mars or we can choose[br]it. And then we sort of use a Hohmann
0:41:57.990,0:42:03.650
transfer to get from Earth to that point[br]y. Okay? So our orbit actually consists of
0:42:03.650,0:42:08.520
three parts now. Okay. So we have the[br]Hohmann transfer, but it's not actually
0:42:08.520,0:42:14.390
aiming for Mars. It's actually aiming for[br]the point y. There we do a maneuver
0:42:14.390,0:42:21.470
because we want to switch onto this red[br]orbit. Okay. And then this one will bring
0:42:21.470,0:42:28.780
us to the point x where we know because it[br]was constructed in this way that the
0:42:28.780,0:42:36.050
spacecraft will continue to rotate around[br]Mars for example six times. Okay. So in
0:42:36.050,0:42:43.619
particular at x there is no maneuver[br]taking place. Okay. So that's a
0:42:43.619,0:42:49.360
possible mission scenario. And the way[br]this is done then usually is you kind of..
0:42:49.360,0:42:54.280
you calculate these points x that[br]are suitable for doing this. Okay. So they
0:42:54.280,0:42:58.460
have to be stable and unstable in the past[br]at the same time. So we have to find them.
0:42:58.460,0:43:02.500
And there's a lot of numerical[br]computations involved in that. But once we
0:43:02.500,0:43:07.480
have this, you can actually build these[br]orbits. OK. So let's look at an actual
0:43:07.480,0:43:16.021
example. So this is for Earth - Mars. On[br]the left, you see, well, that the two
0:43:16.021,0:43:23.380
circular orbits of Earth, Mars, and on the[br]right you see the same orbit, but from a
0:43:23.380,0:43:28.370
point of view centered around Mars. Okay.[br]And the colors correspond to each other.
0:43:28.370,0:43:32.500
So the mission starts on the left side by[br]doing a Hohmann transfer. So that's the
0:43:32.500,0:43:35.770
black line starting at Earth and then[br]hitting the point, which is called x_c
0:43:35.770,0:43:42.290
here. So that's the y that I had on [br]the other slide. So this point y
0:43:42.290,0:43:47.930
or x_c here is pretty far away still from[br]Mars. There we do a maneuver and we switch
0:43:47.930,0:43:56.730
under the red orbit. Which brings us to[br]the point x closer to Mars, after which we
0:43:56.730,0:44:01.310
will actually start rotating round Mars.[br]And the point x is actually at the top of
0:44:01.310,0:44:08.510
this picture. Okay. And then on the right[br]you can see the orbit and it's looking
0:44:08.510,0:44:13.550
pretty strangely. And also the red[br]orbit is when we kind of the capture orbit
0:44:13.550,0:44:19.180
our way to actually get to Mars. And then[br]if you look very carefully, you can count
0:44:19.180,0:44:26.710
we actually rotate around Mars six[br]times. Okay. Now, during those six
0:44:26.710,0:44:32.170
rotations around Mars, we could do[br]experiments. So maybe that is enough for
0:44:32.170,0:44:37.000
whatever we are trying to do. OK. However,[br]if we want to stay there, we need to
0:44:37.000,0:44:44.797
actually execute another maneuver. OK. So[br]to actually stay around Mars. And I mean,
0:44:44.797,0:44:48.060
the principle looks nice but of course,[br]you have to do some calculations. We have
0:44:48.060,0:44:55.640
to find some ways to actually quantify how[br]good this is. And it turns out that there
0:44:55.640,0:45:02.410
are few parameters that you can choose, [br]in particular the target point x, this has
0:45:02.410,0:45:07.310
a certain distance that you're aiming for[br]at around Mars. And it turns out that this
0:45:07.310,0:45:14.090
procedure here, for example, is only very[br]good if this altitude, this distance r is
0:45:14.090,0:45:17.950
actually high enough. If it is high enough[br]then you can save - in principle - up to
0:45:17.950,0:45:23.720
twenty three percent of the delta v, which[br]is enormous. OK. So that would
0:45:23.720,0:45:29.050
be really good. However, in reality it's[br]not as good usually. Yeah. And for a
0:45:29.050,0:45:34.530
certain lower distances, for example, you[br]cannot save anything, so there are
0:45:34.530,0:45:40.810
certain tradeoffs to make. However, there[br]is another advantage. Remember this point y?
0:45:40.810,0:45:45.990
We chose this along this capture orbit[br]along the red orbit. And the thing is, we
0:45:45.990,0:45:51.380
can actually choose this freely. This[br]means that our Hohmann transfer doesn't
0:45:51.380,0:45:55.000
need to hit Mars directly when it's there.[br]So it doesn't need to aim for that
0:45:55.000,0:46:02.970
particular point. It can actually aim for[br]any point on that capture orbit. This
0:46:02.970,0:46:06.310
means that we have many more Hohmann[br]transfers available that we can actually
0:46:06.310,0:46:12.500
use to get there. This means that we have[br]a far larger transfer window. OK. So we
0:46:12.500,0:46:17.730
cannot just start every 26 months. But now[br]we, with this technique, we could actually
0:46:17.730,0:46:24.460
launch. Well, quite often. However,[br]there's a little problem. If you looked at
0:46:24.460,0:46:33.950
the graph carefully, then you may have[br]seen that the red orbit actually took like
0:46:33.950,0:46:39.350
three quarters of the rotation of Mars.[br]This corresponds to roughly something like
0:46:39.350,0:46:43.750
400 days. OK. So this takes a long time.[br]So you probably don't want to use this
0:46:43.750,0:46:49.060
with humans on board because they have to[br]actually wait for a long time. But in
0:46:49.060,0:46:52.890
principle, there are ways to make this[br]shorter. So you can try to actually
0:46:52.890,0:46:58.450
improve on this, but in general, it takes[br]a long time. So let's look at a real
0:46:58.450,0:47:04.610
example for this. Again, that's[br]Bepicolombo. The green dot is now Mercury.
0:47:04.610,0:47:09.870
So this is kind of a zoom of the other[br]animation and the purple line is the
0:47:09.870,0:47:20.640
orbit. And yeah, it looks strange. So the[br]first few movements around Mercury,
0:47:20.640,0:47:28.300
they are actually the last gravity assists[br]for slowing down. And then it actually
0:47:28.300,0:47:36.780
starts on the capture orbit. So now it[br]actually approaches Mercury. So this is
0:47:36.780,0:47:41.010
the part that's sort of difficult to find,[br]but which you can do with this stability.
0:47:41.010,0:47:45.609
And once the animation actually ends,[br]this is when it actually reaches the point
0:47:45.609,0:47:52.240
when it's temporarily captured. So in this[br]case, this is at an altitude of 180,000
0:47:52.240,0:47:58.090
kilometers. So it is pretty high up above[br]Mercury, but it's enough for the mission.
0:47:58.090,0:48:03.270
OK. And of course, then they do some[br]other maneuver to actually stay around
0:48:03.270,0:48:12.230
Mercury. Okay, so in the last few minutes,[br]let's have a look. Let's have a brief look
0:48:12.230,0:48:18.997
at how we can actually extend this. So I[br]will be very brief here, because while we
0:48:18.997,0:48:23.600
can try to actually make this more general[br]to improve on this, this concept is then
0:48:23.600,0:48:29.040
called the interplanetary transport[br]network. And it looks a bit similar to
0:48:29.040,0:48:36.290
what we just saw. The idea is that, in[br]fact, this capture orbit is part of a
0:48:36.290,0:48:42.950
larger well, a set of orbits that have[br]these kinds of properties that wrap around
0:48:42.950,0:48:48.520
Mars and then kind of leave Mars. And [br]they are very closely related to the
0:48:48.520,0:48:53.280
dynamics of particular Lagrangian points,[br]in this case L1. So that was the one
0:48:53.280,0:49:00.330
between the two masses. And if you[br]investigate this Lagrangian point a bit
0:49:00.330,0:49:05.530
closer, you can see, well, you can see[br]different orbits of all kinds of
0:49:05.530,0:49:10.650
behaviors. And if you understand this,[br]then you can actually try to do the same
0:49:10.650,0:49:16.880
thing on the other side of the Lagrangian[br]point. OK. So we just kind of switch from
0:49:16.880,0:49:21.440
Mars to the Sun and we do a similar thing[br]there. Now we expect to actually find
0:49:21.440,0:49:24.920
similar orbits that are wrapping around[br]the Sun and then going towards this
0:49:24.920,0:49:31.859
Lagrangian point in a similar way. Well,[br]then we already have some orbits that are
0:49:31.859,0:49:39.270
well, kind of meeting at L1. So we might[br]be able to actually connect them somehow,
0:49:39.270,0:49:45.070
for example by maneuver. And then we only[br]need to reach the orbit around Earth or
0:49:45.070,0:49:50.130
around Sun from Earth. OK. If you find a[br]way to do this, you can get rid of the
0:49:50.130,0:49:55.270
Hohmann transfer. And this way you reduce[br]your delta v even further. The problem is
0:49:55.270,0:50:00.690
that this is hard to find because these[br]orbits they are pretty rare. And of
0:50:00.690,0:50:07.320
course, you have to connect those orbits.[br]So they you approach the Lagrangian point
0:50:07.320,0:50:14.330
from L1 from two sides, but you don't[br]really want to wait forever until they...
0:50:14.330,0:50:19.630
it's very easy to switch or so, so instead[br]you apply some delta v, OK, in order to
0:50:19.630,0:50:24.490
not wait that long. So here's a picture [br]of how this might look like. Again
0:50:24.490,0:50:28.960
very schematic. So we have Sun, we[br]have Mars and in between there is the
0:50:28.960,0:50:35.240
Lagrangian point L1. The red orbit is sort[br]of an extension of one of those capture
0:50:35.240,0:50:38.230
orbits that we have seen. OK, so that[br]wraps around Mars a certain number of
0:50:38.230,0:50:45.350
times. And while in the past, for example,[br]it actually goes to Lagrangian point. I
0:50:45.350,0:50:50.780
didn't explain this, but in fact, there[br]are many more orbits around L1, closed
0:50:50.780,0:50:55.460
orbits, but they're all unstable. And[br]these orbits that are used in this
0:50:55.460,0:51:05.030
interplanetary transport network they[br]actually approach those orbits around L1
0:51:05.030,0:51:10.570
and we do the same thing on the other side[br]of the Sun now and then the idea is, OK,
0:51:10.570,0:51:15.590
we take these orbits, we connect[br]them. And when we are in the black orbit
0:51:15.590,0:51:19.070
around L1, we actually apply some[br]maneuver, we apply some delta v to
0:51:19.070,0:51:22.450
actually switch from one to the other. And[br]then we have sort of a connection of how
0:51:22.450,0:51:28.500
to get from Sun to Mars. So we just need[br]to do a similar thing again from for Earth
0:51:28.500,0:51:35.119
to this particular blue orbit around the[br]Sun. OK. So that's the general procedure.
0:51:35.119,0:51:38.050
But of course, it's difficult. And in the[br]end, you have to do a lot of numerics
0:51:38.050,0:51:44.840
because as I said at the beginning, this[br]is just a brief overview. It's not all the
0:51:44.840,0:51:50.900
details. Please don't launch your [br]own mission tomorrow. OK. So with
0:51:50.900,0:51:54.960
this, I want to thank you. [br]And I'm open to questions.
0:51:54.960,0:52:05.640
Applause
0:52:05.640,0:52:08.210
Herald: So thank you Sven for an[br]interesting talk. We have a few minutes
0:52:08.210,0:52:11.160
for questions, if you have any questions[br]lined up next to the microphones, we'll
0:52:11.160,0:52:18.400
start with microphone number one.[br]Mic1: Hello. So what are the problems
0:52:18.400,0:52:22.680
associated? So you showed in the end is[br]going around to Lagrange Point L1?
0:52:22.680,0:52:26.710
Although this is also possible for
0:52:26.710,0:52:30.140
other Lagrange points. Could you do this[br]with L2?
0:52:30.140,0:52:38.180
Sven: Yes, you can. Yeah. So in principle,[br]I didn't show the whole picture, but
0:52:38.180,0:52:43.107
these kind of orbits, they exist at L1,[br]but they also exist at L2. And in
0:52:43.107,0:52:49.080
principle you can this way sort of leave[br]this two body problem by finding similar
0:52:49.080,0:52:53.650
orbits. But of course the the details are[br]different. So you cannot really take your
0:52:53.650,0:52:58.640
knowledge or your calculations from L1[br]and just taking over to L2, you actually
0:52:58.640,0:53:03.189
have to do the same thing again. You have[br]to calculate everything in detail.
0:53:03.189,0:53:06.650
Herald: To a question from the Internet.[br]Signal Angel: Is it possible to use these
0:53:06.650,0:53:11.350
kinds of transfers in Kerbal Space[br]Program?
0:53:11.350,0:53:23.500
Sven: So Hohmann transfers, of course,[br]the gravity assists as well, but not the
0:53:23.500,0:53:28.900
restricted three body problem because the[br]way Kerbal Space Program at least the
0:53:28.900,0:53:33.450
default installation so without any mods[br]works is that it actually switches the
0:53:33.450,0:53:40.270
gravitational force. So the thing that I[br]described as a patch solution where we
0:53:40.270,0:53:46.220
kind of switch our picture, which[br]gravitational force we consider for our
0:53:46.220,0:53:50.619
two body problem. This is actually the way[br]the physics is implemented in Kerbal space
0:53:50.619,0:53:55.400
program. So we can't really do the[br]interplanetary transport network there.
0:53:55.400,0:54:00.090
However, I think there's a mod that allows[br]this, but your computer might be too slow
0:54:00.090,0:54:04.350
for this, I don't know.[br]Herald: If you're leaving please do so
0:54:04.350,0:54:07.440
quietly. Small question and question from[br]microphone number four.
0:54:07.440,0:54:12.619
Mic4: Hello. I have actually two[br]questions. I hope that's okay. First
0:54:12.619,0:54:18.289
question is, I wonder how you do that in[br]like your practical calculations. Like you
0:54:18.289,0:54:22.950
said, there's a two body problem and [br]there are solutions that you can
0:54:22.950,0:54:27.440
calculate with a two body problem. And[br]then there's a three body problem. And I
0:54:27.440,0:54:32.050
imagine there's an n-body problem all the[br]time you do things. So how does it look
0:54:32.050,0:54:37.890
when you do that? And the second[br]question is: you said that reducing delta v
0:54:37.890,0:54:47.770
about 15% is enormous. And I wonder what[br]effect does this have on the payload?
0:54:47.770,0:54:57.550
Sven: Okay. So regarding the first[br]question. So in principle, I mean, you
0:54:57.550,0:55:05.210
make a plan for a mission. So you have to[br]you calculate those things in these
0:55:05.210,0:55:08.910
simplified models. Okay. You kind of you[br]patch together an idea of what you want to
0:55:08.910,0:55:14.910
do. But of course, in the end, you're[br]right, there are actually many massive
0:55:14.910,0:55:19.250
bodies in the solar system. And because we[br]want to be precise, we actually have to
0:55:19.250,0:55:25.280
incorporate all of them. So in the end,[br]you have to do an actual numerical search
0:55:25.280,0:55:32.010
in a much more complicated n-body problem.[br]So with, I don't know, 100 bodies or so
0:55:32.010,0:55:37.800
and you have to incorporate other effects.[br]For example, the solar radiation might
0:55:37.800,0:55:43.230
actually have a little influence on your[br]orbit. Okay. And there are many effects of
0:55:43.230,0:55:48.040
this kind. So once you have a rough idea[br]of what you want to do, you need to take
0:55:48.040,0:55:53.260
your very good physics simulator for the[br]n-body problem, which actually has all
0:55:53.260,0:55:57.050
these other effects as well. And then you[br]need to do a numerical search over this.
0:55:57.050,0:56:01.410
Kind of, you know, where to start with[br]these ideas, where to look for solutions.
0:56:01.410,0:56:06.680
But then you actually have to just try it[br]and figure out some algorithm to actually
0:56:06.680,0:56:12.010
approach a solution that has to behaviors[br]that you want. But it's a lot of numerics.
0:56:12.010,0:56:16.500
Right. And the second question, can you[br]remind me again? Sorry.
0:56:16.500,0:56:23.550
Mic4: Well, the second question was in[br]reducing delta v about 15%. What is the
0:56:23.550,0:56:28.890
effect on the payload?[br]Sven: Right. So, I mean, if you need
0:56:28.890,0:56:35.750
15% less fuel, then of course you can use[br]15% more weight for more mass for the
0:56:35.750,0:56:40.450
payload. Right. So you could put maybe[br]another instrument on there. Another thing
0:56:40.450,0:56:46.119
you could do is actually keep the fuel but[br]actually use it for station keeping. So,
0:56:46.119,0:56:52.550
for example, in the James Webb telescope[br]example, the James Webb telescope flies
0:56:52.550,0:56:58.840
around this Halo orbit around L2, but the[br]orbit itself is unstable. So the James
0:56:58.840,0:57:03.980
Webb Space Telescope will actually escape[br]from that orbit. So they have to do a few
0:57:03.980,0:57:08.140
maneuvers every year to actually stay[br]there. And they have only a finite amount
0:57:08.140,0:57:13.431
of fuels at some point. This won't be[br]possible anymore. So reducing delta v
0:57:13.431,0:57:20.609
requirements might actually have increased[br]the mission lifetime by quite a bit.
0:57:20.609,0:57:25.160
Herald: Number three.[br]Mic3: Hey. When you do such a
0:57:25.160,0:57:29.869
mission, I guess you have to adjust the[br]trajectory of your satellite quite often
0:57:29.869,0:57:34.190
because nothing goes exactly as you[br]calculated it. Right. And the question is,
0:57:34.190,0:57:38.930
how precise can you measure the orbit?[br]Sorry, the position and the speed of a
0:57:38.930,0:57:43.590
spacecraft at, let's say, Mars. What's the[br]resolution?
0:57:43.590,0:57:48.300
Sven: Right. So from Mars, I'm not[br]completely sure how precise it is. But for
0:57:48.300,0:57:52.300
example, if you have an Earth observation[br]mission, so something that's flying around
0:57:52.300,0:57:58.500
Earth, then you can have a rather precise[br]orbit that's good enough for taking
0:57:58.500,0:58:04.220
pictures on Earth, for example, for[br]something like two weeks or so. So
0:58:04.220,0:58:12.080
you can measure the orbit well enough and[br]calculate the future something like two
0:58:12.080,0:58:21.140
weeks in the future. OK. So that's good[br]enough. However. Yeah. The... I can't
0:58:21.140,0:58:25.970
really give you good numbers on what the[br]accuracy is, but depending on the
0:58:25.970,0:58:30.619
situation, you know, it can get pretty[br]well for Mars I guess that's pretty
0:58:30.619,0:58:35.440
far, I guess that will be a bit less.[br]Herald: A very short question for
0:58:35.440,0:58:38.780
microphone number one, please.[br]Mic1: Thank you. Thank you for the talk.
0:58:38.780,0:58:44.540
I have a small question. As you said, you[br]roughly plan the trip using the three
0:58:44.540,0:58:50.540
body and two body problems. And are there[br]any stable points like Lagrangian points
0:58:50.540,0:58:54.060
in there, for example, four body problem?[br]And can you use them to... during the
0:58:54.060,0:58:59.530
roughly planning stage of...[br]Sven: Oh, yeah. I actually wondered
0:58:59.530,0:59:03.830
about this very recently as well. And I[br]don't know the answer. I'm not sure. So
0:59:03.830,0:59:07.180
the three body problem is already[br]complicated enough from a mathematical
0:59:07.180,0:59:12.270
point of view. So I have never actually[br]really looked at a four body problem.
0:59:12.270,0:59:18.100
However with those many bodies, there[br]are at least very symmetrical solutions.
0:59:18.100,0:59:22.210
So you can find some, but it's a different[br]thing than Lagrangian points, right.
0:59:22.210,0:59:26.440
Herald: So unfortunately we're almost out[br]of time for this talk. If you have more
0:59:26.440,0:59:29.910
questions, I'm sure Sven will be happy to[br]take them afterwards to talk. So please
0:59:29.910,0:59:33.186
approach him after. And again, a big[br]round of applause for the topic.
0:59:33.186,0:59:33.970
Sven: Thank you.
0:59:33.970,0:59:39.658
Applause
0:59:39.658,0:59:48.850
36C3 postroll music
0:59:48.850,1:00:06.000
Subtitles created by c3subtitles.de[br]in the year 2020. Join, and help us!