Mars Dogleg - Phobos refueling station

RobS · 2005-04-01 14:32:03

Well, let's start with an opposition where the distance between Mars and Earth is average, such as the May 22, 2016 opposition.

I am assuming that refueling in Mars orbit is possible and that aerocapture into orbit on arrival is possible. That means a delta-v for trans-Mars injection from Earth orbit of as much as 5 km/sec is reasonable; maybe even 6 or 7 km/sec (though the fuel load gets pretty big, even with hydrogen-oxygen; lox-augmented nuclear thermal could handle 7 km/sec, though). Ditto for trans-Earth injection. So the question is, would this work:

trans-Earth injection from Mars: Dec. 1, 2015

Earth arrival: 1 April 2016 (122 days in transit)

trans-Mars injection: 22 April 2016 (this is just a month before opposition, so maybe we have to move everything earlier, but while that will decrease delta-v for the outboud leg, it probably increases the delta-v of the inbound leg)

Mars arrival: 22 Sept. 2016 (123 days in transit)

OR, would you like to try the reverse:

trans-Mars injection from Earth: Nov. 1, 2015

Mars arrival: 2 March 2016 (120 days in transit; is the arrival speed too high for aerobraking, though?)

trans-Earth injection: 2 April 2016

Earth arrival: 1 Aug. 2016 (122 days in transit)

My intuition (for what it's worth) tells me we should start from Earth earlier and arrive at Mars earlier than for flights starting at Mars and heading first to Earth. But maybe I am wrong.

If you can get these to work, the next question is: how do we tweak the various departure and arrival dates to minimize the delta-v? Should start sooner or later? Should we make one leg a longer flight and the other leg a shorter flight?

Very interesting questions, when flights to Mars become more common, expensive tourism becomes possible, and settlement gradually becomes possible.

-- RobS

Hop · 2005-04-02 14:16:25

trans-Earth injection from Mars: Dec. 1, 2015
delta vee 12.134 km/sec
Earth arrival: 1 April 2016 (122 days in transit)
delta vee 27.295 km/sec
Here's a pic
http://clowder.net/hop/etc./RobS.jpg]ht … ./RobS.jpg

r1 is position vector of departure point and r2 is position vector of destination point. c is vector from r1 to r2. These 3 are sides of a Lambert Space Triangle.

Given a certain time between departure and destination, there are usually two elliptical paths that pass through departure and destination points.

My more elaborate model for Earth and Mars orbits is a 1000 sided polygon where each vertice has (x,y,z,t) coordinates. Each orbit must start at time of perihelion.

I was able to find departure and arrival points close to the dates you mentioned. This particular spreadsheet uses AU and years most of the time.

I pasted these coordinates in a lambtrans spreadsheet.
http://clowder.net/hop/etc./lambtrans.x … btrans.xls
The magnitude of the delta v vectors in kilometers/sec are over on the right hand side (column AG), ignor rows 14, 17, 20, 23.

Some of the delta vees are quite high.

This ugly spreadsheet is still a beta version, so don't take it as the gospel truth.

RobS · 2005-04-04 01:59:07

Sorry I didn't see this sooner! Thanks, I'll look at it in detail tomorrow.

-- RobS

RobS · 2005-04-05 21:23:19

Thank you again for your time and effort. The data are not encouraging, that's for sure. If you are in the mood, calculate this one, if you can:

trans-Earth injection from Mars: Oct. 15, 2015

Earth arrival: 15 Mar. 2016 (151 days in transit)

trans-Mars injection: 1 April 2016

Mars arrival: 1 Oct. 2016 (183 days in transit)

This leaves Mars earlier, leaves Earth earlier, spends alonger time in transit, and reduces the time at Earth rather severely.

-- RobS

Hop · 2005-04-08 16:08:52

Thank you again for your time and effort. The data are not encouraging, that's for sure. If you are in the mood, calculate this one, if you can:
trans-Earth injection from Mars: Oct. 15, 2015
Earth arrival: 15 Mar. 2016 (151 days in transit)
trans-Mars injection: 1 April 2016
Mars arrival: 1 Oct. 2016 (183 days in transit)
This leaves Mars earlier, leaves Earth earlier, spends alonger time in transit, and reduces the time at Earth rather severely.
-- RobS

The first one I'm having problems with. I believe the starting and ending position vectors will be on opposite sides of the transfer ellipse's axis. Will have to study that some more.

The second scenario:
trans-Mars injection: 1 April 2016
Mars arrival: 1 Oct. 2016 (183 days in transit)

is more Hohmann like. 1st delta vee is 5.12 km/sec and the second burn is 3.94 km/sec. Not bad!

I've been wanting to make this spreadsheet for a long time. Thanks for providing a nudge. If I can perfect this, it will be a very useful implement in my tool belt.

RobS · 2005-04-08 16:22:06

Yes, the second one is very Hohmann-like. it's also similar to Mars Direct. What's the delta-v of 3.94 km/sec; Mars orbit insertion? Into what orbit?

- RobS

Hop · 2005-04-08 18:00:19

The 3.94 is leaving transfer for Mars. Mars gravity isn't considered.

3.94 km/sec is Vinfinity of a hyperbolic trajectory. Let's say you set this hyperbola's periapsis at 300 km above Mars surface. Then the hyperbola's periapsis velocity will be 6.22 km/sec.

A 1.74 burn at this periapsis would be enough for capture to an elliptical orbit with a 20062 km apoapsis (Deimos altitude). Then at Deimos a 1.28 burn would match velocities with the moon.

So two burns totalling 3.02 km/sec would land you on Deimos.

Hop · 2005-04-08 18:32:53

Here's some studies of the Lambert space triangle:
http://clowder.net/hop/lambert/Lambert. … mbert.html
http://clowder.net/hop/lambert2/lambert … bert2.html
http://clowder.net/hop/lambert3/lambert … bert3.html

Given departure and destination points, a focus, there are usually two ellipses with semimajor axis a that pass through both points. (although sometimes there's just one ellipse and sometimes none. The Lambert.html illustrates those possibilities)

At the bottom of the lambert3.html are 4 possible paths from departure to destination points if your ellipse has a specific semi-major axis. Each path takes a different time. I've been using the top possibility: the path is the short part of an ellipse's circumference where both starting and ending points are on the same side of ellipse's axis.

RobS · 2005-04-08 23:23:32

Thank you for the diagrams, but I apologize that I really don't know how to read them. I know next to nothing about orbital mechanics.

Have you ever seen

http://www.marsacademy.com/mmdc/trajdes … rajdes.htm

It's a beta-test version of a program that plots trajectories from Earth to Mars. You chose the year and the departure date and the length of the trip; it gives you the delta-v. It tends to be quirky, but I suppose it's more or less accurate. The big problem is that it assumes that both orbits are circular. It starts with 1999, so I suppose it's set up with the opposition of that year (about April, I suppose) in mind. But I think it then calculates oppositions based on circular orbits, with the result that oppositions don't occur quite when they should. For example, 2003's opposition was late August and the MER rovers left for Mars in early June and early July, respectively, for six month trips to Mars. But if you try flights to Mars of any length in those months it says there is no solution to the problem. On the other hand, back up to mid May and launch them to Mars, and you get low delta-vees. In fact, the program says you can launch to Mars as late as May 26 on a 129-day trip; after that and you can't get to Mars at all. that would be true if opposition occurred earlier in the year than it really did, i.e., early July instead of late August. And you would expect an error of that sort if it is calculating opposition based on a circular orbit.

-- RobS

Hop · 2005-04-09 12:38:11

The sun, earth at departure, and Mars at destination form 3 corners of a Lambert triangle. r1 is vector from sun to departure position, r2 vector from sun to destination position and c is chord from r1 to r2.

Green ellipse is transfer orbit with axis indicated by a green horizontal line.

Notice how Earth and Mars are both on the same side of the axis (the top half?) This is how my Lambert spreadsheet is set up and I'm guessing also Mars Academy's applet.

With a Hohmann orbit departure earth and destination mars are separated by 180 degrees. That puts the departure and destination points right on the boundary -- another degree and they'd be on opposite sides of the transfer ellipse's axis.

I tested this. Using my spreadsheet based on circular orbits I get a Hohmann launch window of May 19, 2003 with about a 258 day trip time. I input this info into their applet - no solution. Then I changed trip time to 254 days and their picture shows a very Hohmann like orbit.

Which leads me to believe that their applet, like my lambert spreadsheet, isn't set up to handle departure and destination points on opposite sides of transfer ellipse's axis.

Hop · 2005-04-09 12:50:34

If chord c is long and time short, you need to haul ass to get from point A to B. If you have to go too fast (solar escape velocity or higher), the transfer orbit is no longer an ellipse, but a parabola or hyperbolic orbit about the sun. These solutions lie outside of Lambert's approach.

RobS · 2005-04-09 14:58:33

Very interesting! I've played with their program for over a year. Sometimes you can get a solution that shows the trajectory crossing the orbit of Mars twice, because that's the only way to get there in the time specified I guess. I've wondered how accurate the results are.

-- RobS

SpaceNut · 2019-11-18 16:50:27

This one goes with the new topic Solving Mars mission docking with Phobos

SpaceNut · 2020-03-02 20:19:28

The question for ION matching of a greater mass for the ability to not only get to the moon but now what will we do once there to make fuel and to make it a way station that man can use.
Of course rendevousing with an in orbit mars station to refuel and get the ship home would seem to be possible.
What we know is that we can spin basalt for encasing the fragile moons shell and for mining a place for man to be protected while building up the somewhat automated base.

tahanson43206 · 2020-03-02 21:22:58

For SpaceNut re #39

I like the way you've adapted the cocoon discussion from Calliban's asteroid topic for the Phobos case.

While the moon is MUCH larger, it has the distinct advantage of (small but sufficient) gravity to permit practical use of a thread deploying robot. It would take a lot of trips around the moon to start to make a meaningful layer of thread, but there is plenty of time.

(th)

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#26 2005-04-01 14:32:03

Re: Mars Dogleg - Phobos refueling station

#27 2005-04-02 14:16:25

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#28 2005-04-04 01:59:07

Re: Mars Dogleg - Phobos refueling station

#29 2005-04-05 21:23:19

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#30 2005-04-08 16:08:52

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#31 2005-04-08 16:22:06

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#32 2005-04-08 18:00:19

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#33 2005-04-08 18:32:53

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#34 2005-04-08 23:23:32

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#35 2005-04-09 12:38:11

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#36 2005-04-09 12:50:34

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#37 2005-04-09 14:58:33

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#38 2019-11-18 16:50:27

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#39 2020-03-02 20:19:28

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#40 2020-03-02 21:22:58

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