Orbital mechanics

John Creighton · 2004-03-15 09:04:03

In the a few recent posts there has been some discussion ([http://www.newmars.com/cgi-bin/ikonboard/ikonboard.cgi?act=Post;CODE=06;f=4;t=292;p=23]1,[http://www.newmars.com/cgi-bin/ikonboard/ikonboard.cgi?act=Post;CODE=06;f=4;t=292;p=33]2,[http://www.newmars.com/cgi-bin/ikonboard/ikonboard.cgi?act=Post;CODE=06;f=4;t=292;p=42]3) of the delta-v required to go to mars for various transit times. Until I work out the details my self, the numbers only mean so much to me. Thus, this thread is created to discuss the principles of [http://www.braeunig.us/space/orbmech.htm#intro]orbital mechanics. I can only spend so much time learning this because I should be doing reading that is more relevant to my masters (My thesis will probably be a servo motor control system). My ultimate goal with this thread is to learn enough about orbital mechanics so that I can implement a model predictive control space flight system in [http://www.mathworks.com/]MATLAB. If I construct such a control system I can then begin simulating various flight systems to see how quickly they can travel to mars and how robust they are to error. One step at a time.

SBird · 2004-03-15 16:07:26

Well, I'm a materials scientist with a background in molecular biology (don't ask ) so I'm by no means an expert on the subject. However, here's what I do know about it.

The delta V is simply a measure of the total additional kinetic energy required to get from point A to B. It doesn't really matter if that Delta V comes from a huge chemical rocket boost in the first 5 minutes of the trip or an ion engine that slowly adds it over the course of 6 months if you're in orbit. There's all sorts of little details that are left out of this such as the relative positions of A and B, directional vectors etc. However, all thsoe are factored in when calculating the initial delta V.

I've seen some of the equations for calculating detla V in various circumstances but I've never been motivated to actually work them out. For a more visceral interaction with orbital mechanics, look fora program called Orbiter by Martin Schweiger. It's a free flight sim that uses realistic orbital mechanics. It helps conceptually, IMO to be able to play around with the orbits hands on.

John Creighton · 2004-03-22 09:17:20

Orbital Intersections (Finding The Plane)

Consider the problem of finding a transfer trajectory between two orbits. The transfer trajectory will be elliptical or hyperbolic. Therefore the transfer trajectory must lie on a plane. Three points must be in the plane. The intial point, the final point and the sun. A plane is defined by the equation a1x1+a2x2+a3x3=c or in [http://mathworld.wolfram.com/Vector.html]vector form: a [http://mathworld.wolfram.com/DotProduct.html]dot x = c. Let the point on the sun be given by x_s=[x1_s x2_s x3_s] the point of intersections with the earth orbit be x_e=[x1_e x2_e x3_e] and the point of intersection with the other orbit be x_m=[x1_m x2_m x3_m]. The vector a is given by the system of equations:
a dot x_S=c
a dot x_e=c
a dot x_m=c

or in matrix form

Xa=c
where:
X=[x1_s x2_s x3_s]
[x1_e x2_e x3_e]
[x1_m x2_m x3_m].
c=[c]
[c]
[c]

Once the equation of the plane is found we can describe the transfer orbit with only two coordinates:

b1=[1 ]
[0 ]
[(c-a1)/a3]

b2=[ 0 ]
[ 1 ]
[(c-a2)/a3]

All points in the plane are equal to come constant x multiplied by b1 plus y multiplied by b2. The vectors b1 and b2 form a set of [http://mathworld.wolfram.com/VectorBasis.html]basis vectors for the plane of the transfer orbit, and x b1+y b2 is called a [http://mathworld.wolfram.com/LinearCombination.html]linear combination of the basis vectors b1 and b2.

John Creighton · 2004-03-22 11:10:03

Orbital Intersections (Finding the ellipse)

Let the point where the orbit intersects the earths orbit:
(x,y)=(x_e1,x_e2)
and the point where the transfer orbit intersects the other orbit:
(x,y)=(x_m1, x_m2)
One of the [http://mathworld.wolfram.com/Focus.html]foci of the [http://mathworld.wolfram.com/Ellipse.html]ellipse is the sun.
Lets denote this point (x_s1,xs_2).

Denote the coordinates of the other
foci by (x_f1,x_f2).

In an ellipse the sum of the [http://mathworld.wolfram.com/L2-Norm.html]distance from each foci is constant.
There fore
||(x_f1,x_f2)-(x_e1,x_e2)||+||(x_s1,x_s2)-(x_e1, x_e2)||=d
||(x_f1,x_f2)-(x_m1,x_m2)||+||(x_s1,x_s2)-(x_m1, x_m2)|| =d
rearranging:
||(x_f1,x_f2)-(x_e1,x_e2)||=d-||(x_s1,x_s2)-(x_e1, x_e2)||
||(x_f1,x_f2)-(x_m1,x_m2)||=d-||(x_s1,x_s2)-(x_m1, x_m2)||
Squaring both sides:
||(x_f1,x_f2)-(x_e1,x_e2)||^2=
d^2+||(x_s1,xs_2)-(x_e1, x_e2)||^2-2 d ||(x_s1,x_s2)-(x_e1, x_e2)||
|(x_f1,x_f2)-(x_m1,x_m2)||^2=
d^2+||(x_s1,x_s2)-(x_m1, x_m2)|| ^2-2 d ||(x_s1,x_s2)-(x_m1, x_m2)||
equivalently:
(x_f1-x_e1)^2+(x_f2-x_e2)^2=
d^2+(x_s1-x_e1)^2+(x_s2-x_e2)^2-2 d ||(x_s1,x_s2)-(x_e1, x_e2)||
(x_f1-x_m1)^2+(x_f2-x_m2)^2=
d^2+(x_s1-x_m1)^2+(x_s2-x_m2)^2-2 d ||(x_s1,x_s2)-(x_m1, x_m2)||

John Creighton · 2004-03-22 14:09:15

Numeric Solutions (Transfer Orbit of Mars Direct)

(x_f1-x_e1)^2+(x_f2-x_e2)^2=
d^2+(x_s1-x_e1)^2+(x_s2-x_e2)^2-2 d ||(x_s1,x_s2)-(x_e1, x_e2)||
(x_f1-x_m1)^2+(x_f2-x_m2)^2=
d^2+(x_s1-x_m1)^2+(x_s2-x_m2)^2-2 d ||(x_s1,x_s2)-(x_m1, x_m2)||

The above equations describe all possible says a rocket can go from, a point on earth orbit, (x_e1,x_e2) to a point in another orbit (x_m1,x_m2). There are two equations and three unknowns. The unknowns are the second focus (x_f1,x_f2) (note the first focus is the sun) and the constant d which is defined so that the sum of the distance from each focus to a point on the ellipse is d. These equations could be solved numerically or algebraically. I would like to solve the problem numerically first to put some meaning behind these equations. The first problem I would like to solve is to find the transfer orbit used in mars direct. To solve this problem I need to know the coordinates of the earth when the rocket leaves earth and the coordinates of mars when the rocket arrives in mars. Does anyone know these coordinates?

Rxke · 2004-03-22 14:34:23

Bit off-topic,
but *every* time i see the equation a1x1+a2x2+a3x3=c and extenions of that (a1x1y1+ etc i forgot...) my blood boils...

Why? Way back, when i was 18 (last year in highschool) we had that problem *very* extensively covered in mathclass (i had 9 hours maths&calculus a week, five dedicated to this particular problem... because a Belgian mathematician in the past had done some breakthrough work on it, chauvinism stuff and all that...)

BUT our teacher never told it was used for calculating orbits, he just kept mumbling about 'cone-intersections' he made it very abstract, boring stuff...

At that age i was already a total Mars-devotee, heh, but if only i knew i was learning stuff that was related to *spaceflight,* boy, my grades would've been *much* better.

Sigh....

John Creighton · 2004-03-22 15:09:46

Do you want to learn now? You sound like you have more then enough background in math.

Rxke · 2004-03-23 00:10:11

I'm 34, so it was quite a while ago... And i was *not good* in it, but strangely i loved it... (w/o knowing the 'space' connection)

I always thought the 'general' equation was a beautiful one... a*x^2+b*y^2+2*c*x*y+2*d*x+2*e*y+f=0 literally gave me some kind of religious experience, when i saw it solved first time on the blackboard. I was staring at the blackboard, totally over-awed,... And then that braindead teacher called me to attention, chiding in public me for sitting dreaming in his lesson (!) I'll *never* forget that, what a come-down!

So i'll quietly slip in at the back and just sit in and listen, in awe .

John Creighton · 2004-03-23 09:25:54

I thought I’d post some relevant links:

[http://nssdc.gsfc.nasa.gov/space/helios … 0054659957](1) At this link you can give the time and date and it will calculate the position of any planet.

[http://www.marsacademy.com/traj.htm](2) The mars academy gives some equations to help calculate delta v’s and travel times.
Unfortunately they do not derive anything.

[http://en.wikipedia.org/wiki/Interplanetary_travel](3) I thought this site was interesting. It is a free online encyclopedia.
It talks a little bit about with new computational power how some of the old
orbital transfer solutions have been reevaluated

John Creighton · 2004-03-23 11:28:13

Here are the coordinates the [http://nssdc.gsfc.nasa.gov/space/helios … 0054659957]website gave for earth:

planet = 07
resolution = 30
start_year = 2004
start_day = 1
stop_year = 2005
stop_day = 1
EARTH coordinates:

YYYY DDD AU ELAT ELON HLAT HLON HILON

2004 1 0.983 0.00 99.89 -2.94 164.57 23.89
2004 31 0.985 0.00 130.43 -5.90 129.53 54.39
2004 61 0.991 0.00 160.73 -7.21 94.48 84.87
2004 91 0.999 0.00 190.61 -6.57 59.04 114.95
2004 121 1.007 0.00 219.98 -4.23 22.92 144.37
2004 151 1.014 0.00 248.92 -0.87 346.16 173.14
2004 181 1.017 0.00 277.58 2.68 309.08 201.60
2004 211 1.015 0.00 306.21 5.57 272.11 230.15
2004 241 1.010 0.00 335.02 7.11 235.53 259.10
2004 271 1.002 0.00 4.21 6.87 199.41 288.52
2004 301 0.994 0.00 33.90 4.83 163.65 318.29
2004 331 0.987 0.00 64.06 1.47 128.14 348.31
2004 361 0.983 0.00 94.54 -2.32 92.86 18.5

SBird · 2004-03-23 11:49:43

John, this is some really good info - thanks! I'll have to pull my old linear algebra books out and brush up a bit.

Wikipedia's got quite a bit of useful info on spaceflight - their spaceship propulsion section is quite nicely put together.

Palomar · 2004-03-23 12:26:04

I'm 34, so it was quite a while ago... And i was *not good* in it, but strangely i loved it... (w/o knowing the 'space' connection)
I always thought the 'general' equation was a beautiful one... a*x^2+b*y^2+2*c*x*y+2*d*x+2*e*y+f=0 literally gave me some kind of religious experience, when i saw it solved first time on the blackboard. I was staring at the blackboard, totally over-awed,... And then that braindead teacher called me to attention, chiding in public me for sitting dreaming in his lesson (!) I'll *never* forget that, what a come-down!
So i'll quietly slip in at the back and just sit in and listen, in awe .

*Rik: I envy you, as I could only hope to have an experience like that. I'm not good at math at all; never have been.

I was considering buying a $70 DVD course regarding mathematics (supposedly somewhat explained in laymen's terms) from a reputable learning company...I forgot the name of the company. I decided -not- to pluck the $70 down, with the certainty I still probably "can't get it."

I can see this thread isn't going to be a mainstay for me, unfortunately. I don't mean to get off-topic, etc.; just wanted to respond to Rik's post.

--Cindy

John Creighton · 2004-03-23 12:57:50

Here are the coordinates of mars from the [http://nssdc.gsfc.nasa.gov/space/helios … 0054659957]website:

You submitted the following name/value pairs:

planet = 02
resolution = 60
start_year = 2004
start_day = 1
stop_year = 2006
stop_day = 1
MARS coordinates:

YYYY DDD AU ELAT ELON HLAT HLON HILON
2004 1 1.475 0.04 51.08 3.07 116.12 335.44
2004 61 1.552 1.01 82.82 0.12 16.68 7.07
2004 121 1.618 1.63 111.70 -2.61 274.37 35.82
2004 181 1.658 1.85 138.72 -4.59 170.30 62.81
2004 241 1.664 1.67 164.98 -5.56 65.57 89.14
2004 301 1.635 1.14 191.61 -5.37 321.25 115.89
2004 361 1.576 0.32 219.75 -3.94 218.39 144.09
2005 55 1.500 -0.65 250.44 -1.33 117.96 174.72
2005 115 1.428 -1.51 284.44 1.96 20.74 208.57
2005 175 1.385 -1.85 321.39 4.74 286.58 245.47
2005 235 1.391 -1.43 359.36 5.61 193.61 283.57
2005 295 1.442 -0.45 35.75 4.21 99.06 320.08
2005 355 1.517 0.61 68.99 1.47 1.18 353.27

John Creighton · 2004-03-23 13:29:28

I’m glad to see there is some interest in this topic. My section on finding the ellipse I made more complicated then it needs to be. The second focus can be found by intersecting two circles. This could be done on a piece of graph paper without the aid of a calculator or a computer. The center of one of the circles is the location of the earth. The location of the second circle is the location of the mars. The radius of the circle with earth as its center is: d-||s-e|| and the radius of the circle with mars as its center is: d-||s-m||.
Note:
||s-e|| denotes the [http://mathworld.wolfram.com/L2-Norm.html]norm (magnitude or length) of the vector s-e:
s is the [http://mathworld.wolfram.com/RadiusVector.html]position vector of the sun
e is the position vector or the earth
m is the position vector of mars.

If you would rather use a calculator then a piece of graph paper, then the intersection could be found using the [http://mathworld.wolfram.com/LawofCosines.html]law of cosines.

Also note that if:
d>max{||s-e||,||s-m||}=||s-m||
and the intersection of the two circles exists then two elliptical transfer exsit, one focus is the sun and the other focus of either of the two points of intersection. There is always a possible hyperbolic transfer orbit.

Rxke · 2004-03-23 13:50:30

Glad Wiki is up again! when i first checked it, it was down, and for a moment i thought they had troubles again...

There was an article i think about a year ago about those 'super-highway trajectories' in new scientist or sci Am...
Amazing what they came up with. Thanks to computers, of course.

The 'intersecting 2 circles', i have that on a .pdf, (section 2.2) that gives a quite nice overview of he stuff...but it's in Dutch, [http://www.luc.ac.be/scholennetwerk/wis … sneden.pdf]http://www.luc.ac.be/scholen....den.pdf (159kB, so might be interesting to give it a try, math is math, and it helps you visualising things...)

John Creighton · 2004-03-23 13:55:49

silly me. There is always an elliptical trajectory. There just isn't one for every value of d (the some of the distance from each foci).

SBird · 2004-03-23 16:56:41

Glad Wiki is up again! when i first checked it, it was down, and for a moment i thought they had troubles again...
There was an article i think about a year ago about those 'super-highway trajectories' in new scientist or sci Am...
Amazing what they came up with. Thanks to computers, of course.

I stumbled across Wikipedia a couple of months ago and it's already become indispensible to me. It puts old encyclopedias like Brittanica and Encarta to shame for technical stuff.

The zero energy trajectory stuff was really interesting. It really has potential for getting cargo to Mars for cheap. The electrotether tug idea I tossed up a few days ago could boost a mass from LEO to L1 and from there, a minimal amount of propulsion could get the cargo to MArs and drop it into a low speed aerobraking trajectory. Basically, with those two combined, you can get the LEO boost capacity of a launcher to Mars for almost free.

A Saturn V equivalent could realistically get something like > 100 metric tons of mass to Mars. That's an almost 4-fold increase in the available ERV mass compared to Mars Direct. Forget a little covered rover, you can start putting airplanes and bulldozers into a Mars mission with those kind of capacities. A 100 MT payload means you could carry 2 D9 Caterpillar bulldozers to Mars - each of which is capable of towing a tank. Or a few square km of pastic sheeting for greenhouses. Or a small factory/smelter capable of producing steel and plastics from Martian soil and air.

John Creighton · 2004-03-24 06:57:14

You're right. That near [http://en.wikipedia.org/wiki/Interplane … perhighway]zero energy transfer stuff is very interesting. I guess Hohman was wrong when he showed that the [http://en.wikipedia.org/wiki/Hohmann_transfer_orbit]Hohman transfer orbit was the most energy efficient way to travel.

John Creighton · 2004-03-24 07:24:02

A simple example (1) (Finding the second Focus)

I’d like to work out a simple example first. For now assume mars and earth are in the same plane and have circular orbits. The distance earth is from the sun is 1 AU (astronomical unit) and mars is approximately 1.5 AU. For simplicity let the initial point of the earth be at (1,0) in rectangular coordinates and the let the transfer orbit take intersect the orbit of mars at the point (-1.5,0).

Recall the second focus can be found with the intersection of the two circles:
||f-e||=d-||s-e||
||f-m||=d-||s-m||
e is the position vector of the earth (1,0)
m is the position vector of where the transfer orbit will intersect the orbit of mars (-1.5,0)
f is the second focus the transfer ellipse.
The distance between where the transfer orbit intersects the orbit of mars and earth is 1-(-1.5)=2.5.
The two circles will just touch each other when:
d-||s-e||+d-||s-m||=2.5
which implies
d=(2.5+||s-e||+||s-m||)/2=(2.5+1 +1.5)/2=2.5

When two circles just touch each other the intersection will be on a line between there two centers. More specifically it will be at d-||s-e||=1.5 away from the earth and d-||s-m||=1 away from where the transfer orbit intersects the orbit of mars. The coordinates of the intersection are therefore (1,0)+(-1.5,0)=(-1.5,0)+(1,0)=(-0.5,0)

Hence the location of the second focus is (-0.5,0)

Bill White · 2004-03-24 08:02:11

You're right. That near [http://en.wikipedia.org/wiki/Interplane … perhighway]zero energy transfer stuff is very interesting. I guess Hohman was wrong when he showed that the [http://en.wikipedia.org/wiki/Hohmann_transfer_orbit]Hohman transfer orbit was the most energy efficient way to travel.

After a quick google of [http://www.gg.caltech.edu/~mwl/publicat … tions2.htm]Martin Lo, I began to wonder whether the computations needed to build a useable "schedule" for available trajectories could be done "SETI-style" by idle personal computers?

Is the math needed to do these calculations susceptible to such a method of computation? As far as I can tell, there is very little money or computational power being devoted to locating these "tunnels" or whatever you want to call them.

= = =

Any opinion on whether this site offers a good introduction to the topic of orbital mechanics?

[http://zebu.uoregon.edu/~js/space/lectures/lec06.html]http://zebu.uoregon.edu/~js/space/lectures/lec06.html

SBird · 2004-03-24 11:50:51

You could always E-mail the good prof and ask him. I know that there's an 'under construction' page on his website that is supposed to have info about his orbit calculation software. Unless he's doing proprietary work, I'd say the odds are pretty good that he'd be interested in working with people who are willing to donate computer time. It seems as if there's an almost infinite number of trajectories to chose from and the more computers, the better our grasp of the situation.

John Creighton · 2004-03-24 12:22:46

More on A simple example (1) (Finding the second Focus)

In the previous example the transfer orbit started where earths orbit crossed the point (1,0) and arrived where the orbit of mars crossed the point (-1.5,0). The second focus was found to be at the point (-0.5,0). Because that the minimum d for which an elliptical solution will exist will be the [http://en.wikipedia.org/wiki/Hohmann_transfer_orbit]Hohmann transfer orbits (I think), when the two transfer orbit intersects the original orbit and the destination orbit are on opposite sides of the sun.

The transfer time is given by the equation,

t=pi*sqrt((ro+r2)^3/(8*Gm))
where:

r0 is the radius of the intial orbit
r2 is the radius of the final orbit
Gm is the gravitational parameter (3.986x10^5 km^3/s^2)

The [http://mathworld.wolfram.com/Apoapsis.html]apoapsis is the point of the transfer orbit furtherst away from the sun. Note this is also where the transfer orbit intersects the orbit of mars. The [http://mathworld.wolfram.com/Periapsis.html]periapsis is the point where the transfer orbit is closest to the sun. Note this is also where the transfer orbit intersects the orbit of earth.

The [http://mathworld.wolfram.com/SemimajorAxis.html]semi major axissemi major axis of the transfer orbit is half the distance between the apoapsis and periapsis.
a=||(1,0) – (-1.5,0)||/2=2.5/2

The [http://mathworld.wolfram.com/SemiminorAxis.html]semi minor b axis can be found by using simple triganomatry. At the semi minor axis each foci is d/2 away and the distance between the foci is ||(0,0)-(-0.5,0)||=0.5. The resulting triangle formed is an [http://mathworld.wolfram.com/IsoscelesTriangle.html]isocialise triangle. Half of an isocialise triangle is a [http://mathworld.wolfram.com/RightTriangle.html]right angle triangle. In this right angle triangle the hypotensis is d/2, the distance from one foci to the cener of the ellipse is 0.25. Thus the third side can be found using the [http://mathworld.wolfram.com/PythagoreanTheorem.html]Pythagorean Theorem.
b=sqrt((d/2)^2-(0.25/2)^2)=(1/2)*sqrt(6.5) ~ 1.27

Rxke · 2004-03-24 13:41:02

if the zero energy transfer calculations are distributable, it would be a good candidate for [http://boinc.berkeley.edu/]BOINC ... I bet a significant percentage of the SETI crowd would be interested... It is a much more open systen than the SETI screensaver, allowing you to essentially build your own projects, so maybe someone can look into the calculation requirements and go for it, heehee

It might be possible to distribute some of the work:

"computational and numerical methods can be brought to bear which iteratively approximate the invariant set and allow us to develop a systematic procedure to numerically construct orbits with finite prescribed itineraries."

(From section four: Numerical Construction of Orbits with Prescribed Itineraries." from [http://www.gg.caltech.edu/~mwl/publicat … eeBody.pdf]http://www.gg.caltech.edu/~mwl....ody.pdf (1.4 Mb .pdf file))

SBird · 2004-03-24 15:53:24

Distributed computing would be worthwhile if we had alot of people on the project. For right now, it sounds like he's got this LTool software that one can just run on a single PC. If it's just a few of us banging away on Earth to Mars trajectories, that's sufficient. If it's stuff like Mars-Ceres-Moon-Earth trajectories, we'll probably have to go to a more sophisticated system.

SBird · 2004-03-24 15:57:11

Distributed computing would be worthwhile if we had alot of people on the project. For right now, it sounds like he's got this LTool software that one can just run on a single PC. If it's just a few of us banging away on Earth to Mars trajectories, that's sufficient. If it's stuff like Mars-Ceres-Moon-Earth trajectories, we'll probably have to go to a more sophisticated system.

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