Math needed for 5-week flight from Earth to Mars

[Staiduk]

Thanks, dgatsoulis - I'm gratified to know that while I have little understanding of the mechanics; I have at least enough knowledge to be able to write a plausible-sounding paragraph detailing Princess Astoria's flight - one that you have been able to flesh out (magnificently, I might add ) with solid mathematics I could only hope to apply.

I have no idea if the point will ever come up in the novel but I'm glad the numbers are there if it does.

Cheers!

Edit: Dgatsoulis; I'm just curious: What software did you use to a)perform the calculations and b) produce your excellent graph?
I'm fairly confident that given your superb knowledge of mathematics (a knowledge I can only respect and envy); the only software you needed was a pocket calculator; but I'd really like to know how you made that lovely graph. It explained things so well to me - a man who has little understanding of mathematics.

Thanks again!

 

[dgatsoulis]

Thanks, dgatsoulis - I'm gratified to know that while I have little understanding of the mechanics; I have at least enough knowledge to be able to write a plausible-sounding paragraph detailing Princess Astoria's flight - one that you have been able to flesh out (magnificently, I might add ) with solid mathematics I could only hope to apply.

I have no idea if the point will ever come up in the novel but I'm glad the numbers are there if it does.

Cheers!

No problem, glad you found this stuff useful. If I may add, eventhough the two burn solution may seem cheaper for the transfer time you are mentioning in your story, it also seems unnecessary. If you have a ship with the capability to get to Mars in five weeks (a cruise liner for that matter), 2-3 of km/s of spare ΔV aren't that difficult to come up with.

Also, a Low Earth Orbit would make more sense as a starting point, for ease of access from Earth and also radiation protection. But you can easily "handwave" that out with some "anti-cosmic ray/radiation" material or coating for the ship and the station.
It's not the backround technical details that make a story good to read. Like I said before, even Ray Bradbury got his science wrong sometimes; who cares!?

Edit: Dgatsoulis; I'm just curious: What software did you use to a)perform the calculations and b) produce your excellent graph?
I'm fairly confident that given your superb knowledge of mathematics (a knowledge I can only respect and envy); the only software you needed was a pocket calculator; but I'd really like to know how you made that lovely graph. It explained things so well to me - a man who has little understanding of mathematics.

Thanks again!

Thanks for the kind words, but I'd hardly call my understanding of mathematics superb. The equations involved in those calculations are not that difficult and a working knowledge of highschool level maths (my level) can get you a long way, before you start needing to look things up.

This is is where I first starting reading on orbital mechanics about a year and a half ago, but I've been an orbinaut for about 7-8 years. So I had an intuitive understanding of how interplanetary travel works, before learning the actual equations. That link above and the example problems were very helpful in understanding orbital mechanics.

As for the software itself all you need is a calculator and some pen and paper. The default windows calculator is pretty handy but I prefer to use this one because I am used to the notation.

The graph was created with a google spreadsheet where I plugged in the equations for the three different transfer orbits and
then the different values for the transfer ΔV variable.
Here is a link for that particular calculation and graph. I only wrote down column A and the equations that "spit out" columns B C and D and then converted the data to a graph.

Hope this helps,

 

[Staiduk]

It does indeed; as do your links.
Huh - I'm wondering if 48 is too late an age to start learning math. I'll cruise around a little bit and let you know.

LOVED Mr. Bradbury's talk, by the way. Thanks for that!

 

[RGClark]

Short answer is yes. Starting from a 40000x40000km orbit around Earth, for a 5 week transfer to Mars, you have significant savings by making two burns; One retrograde to drop to a low perigee and a second one to escape. For comparison here is the ΔV required for all three cases:

1. Single burn from a 40000x40000km orbit: 14.73 km/s

2. Single burn from a 200x200km orbit: 12.61 km/s

3. Two burns. To drop from 40000x40000 to 40000x200km you need 1.47km/s. Then -at periapsis- you need an additional 10.09 km/s. Total ΔV for two burns is 1.47+10.09 = 11.56 km/s

...

The lower required delta-V from LEO to make the 35 day requirement is coming from the higher orbital velocity there.
This would seem to suggest departing from Mars orbit would not provide as much a speed boost. So how much would the required delta-v be for leaving Mars to return in only 35 days?

Bob Clark

 

[RGClark]

Nice calculation dgatsoulis. I really don't like the 6 to 8 month travel times that have been suggested for manned flights to Mars, either for the first astronauts to be sent or for the Mars colonists to travel there. So I've been investigating delta-v's for shorter travel times.
...

Seeing the ISS astronauts pulled out of the Soyuz like geriatrics after returning from a six months mission onboard the ISS is not a pleasant sight.

Chris Hadfield in for a long recovery after ISS mission.
Bone density, muscle mass and blood pressure all impacted by mission.
Lauren Strapagiel
Published: May 14, 2013, 1:11 pm

We watched on Monday night as Hadfield, Tom Marshburn and Roman Romanenko were pulled out of the Soyuz capsule and carried to waiting chairs. Experiencing the weight of gravity for the first time in months, and having just endured a wild ride through the atmosphere, the crew were in no condition to walk on their own.
http://o.canada.com/2013/05/14/chris-hadfield-in-for-a-long-recovery-after-iss-mission/

That's why I say trips to Mars should not be the 6 to 8 months duration of a Hohmann trajectory flight. You don't want the astronauts to have to wait days to weeks to recover before they can walk on the Martian surface, assuming they will even recover in the same length of time in the reduced gravity of Mars. To shorten the flight, we need increased departure speeds. Using chemical propellant that requires a large amount of propellant.
That would be much easier to get from lunar-derived propellant stored in depots in cislunar space rather than having to haul it up from Earth's huge gravity well.
That is the importance of returning to the Moon. By using the propellant we can get from there, it makes flights to other destinations much easier.


Bob Clark

 

[MattBaker]

That's why I say trips to Mars should not be the 6 to 8 months duration of a Hohmann trajectory flight.

6 to 8 months travel time does not necessarily mean 6 to 8 month microgravity.
In my opinion artificial gravity will become a very, maybe the most, important factor in human exploration of other planets.
Reasons to fund CERN: Find and create gravitons...OK, maybe in 200 years.

That would be much easier to get from lunar-derived propellant stored in depots in cislunar space rather than having to haul it up from Earth's huge gravity well.
That is the importance of returning to the Moon. By using the propellant we can get from there, it makes flights to other destinations much easier.

I'm not an expert like dgatsoulis salute on the subject of delta-V but I think you're overestimating the effect of a moon launch. Also, you'd send your propellant depots to a Moon orbit to launch to Mars from there? Why not use LEO as an assembly place? Easier to access, also no atmosphere, cheaper.

Also, propellants on the Moon? No methane nor huge H2O (yes, there might be some but I'd prefer to not have a base in a steep crater, safety reasons and that boring stuff). And else? Helium-3 fusion? If you like your spacecraft having a reactor hotter than the center of the sun, sure. Oh, we're not exactly good in that technology? Dammit.

 

[Staiduk]

Well; the argument is a little beyond me - I'm a writer, not a mathematician - but the whole point of the Princess Astoria is that she spins herself to provide artificial gravity for her passengers and crew. It might be an SF concept but one which serves well as the location of a murder to be solved.

 

[RGClark]

I'm not an expert like dgatsoulis salute on the subject of delta-V but I think you're overestimating the effect of a moon launch. Also, you'd send your propellant depots to a Moon orbit to launch to Mars from there? Why not use LEO as an assembly place? Easier to access, also no atmosphere, cheaper.

Also, propellants on the Moon? No methane nor huge H2O (yes, there might be some but I'd prefer to not have a base in a steep crater, safety reasons and that boring stuff). And else? Helium-3 fusion? If you like your spacecraft having a reactor hotter than the center of the sun, sure. Oh, we're not exactly good in that technology? Dammit.

Actually, there is a huge amount of H2O there in the permanently shadowed craters. That is why the propellant in the cislunar depots should be derived from lunar H2O.
The increase in delta-v from a lunar or L1/L2 launch is just an additional advantage of launching from there.
As an indication of how much an advantage a virtually unlimited supply of fuel could be, I estimated that if we used a Saturn V sized vehicle launched from L2, we could make the Mars trip in just two weeks from chemical propulsion alone. A problem though is that it is able to make the trip in such a short time by having a high departure speed. Then we would have an even larger problem slowing down by aerobraking than there is now for large vehicles, a problem currently unsolved.


Bob Clark

 

[thepenguin]

"The space elevator will be built approximately ten years after people stop laughing at it" - Arthur C. Clarke

 

[RGClark]

Dgatsoulis, did you include the effect of various departure angles in your calculation? The Earth has a 30 km/s orbital speed around the Sun. If you launch from LEO at a tangential direction to the Earth's orbit then you have an additional 30 km/s speed added on to your departure speed. The disadvantage though is the geometry. This means you have a longer distance to reach Mars' orbit.
This compares to a shorter distance if you launch in a radial direction towards Mars. But the radial direction gives you zero boost from the Earth's orbital speed around the Sun. To see the different distances draw two concentric circles and note there is a longer distance to travel at the tangential direction to the inner circle compared to the radial direction.
So the question is what is the optimal intermediate angle that will add on part of the Earth's orbital speed but won't be too far from the radial direction so as to make the distance too long.

Bob Clark

 

[dgatsoulis]

So the question is what is the optimal intermediate angle that will add on part of the Earth's orbital speed but won't be too far from the radial direction so as to make the distance too long.

Bob Clark

Look at pic #2 on post #8 of this thread. For that year (2287), minimum dV solution for 35 day journey is V∞ = 17.17 km/s. ("oV" in the pic, under the "Escape Vector").
The angle between the tangent of Earth's orbit and the tangent of the outbound trajectory is 23.608° ("EjA" in the pic, right below the departure date). The blue line between the "Earth at Departure" and the "Mars at Arrival" arrows is the distance.
The arrival at Mars is at an angle of 36.259°.(InA, right below the the EjA).

Compare to pic #1, where you have a larger distance, slightly smaller angle and same time interval between departure and arrival. Look at the oV.

 

[RGClark]

I assume that by departing from LEO the shortest time would be at a similar angle but with the departure delta-v at ca. 10 km/s in accordance with the calculation in post #17?


Bob Clark

 

[RGClark]

As an indication of how much an advantage a virtually unlimited supply of fuel could be, I estimated that if we used a Saturn V sized vehicle launched from L2, we could make the Mars trip in just two weeks from chemical propulsion alone.

Mars rover confirms dangers of space radiation.
Future manned missions to Mars will need internal shielding and advanced propulsion systems to shorten transit times, minimizing exposure to space radiation, scientists say.
by William Harwood May 30, 2013 3:06 PM PDT

Chris Moore, deputy director of advanced exploration systems at NASA headquarters, said shorter transit times and improved shielding will be needed to protect future deep space crews.
"To get really fast trip times to cut down on radiation exposure we'd probably need nuclear thermal propulsion, and we're working with the U.S. Department of Energy to look at various types of fuel elements for these rockets," Moore said.
"But it's a long-range technology development activity and it will probably be many years before that is ready. But it is part of our design reference mission architecture for sending humans to Mars.... That could probably cut the (one-way) trip time down to around 180 days."

http://news.cnet.com/8301-11386_3-57586958-76/mars-rover-confirms-dangers-of-space-radiation/

An expensive and far off development using nuclear propulsion that is already controversial and would still only make the travel time 6 months(!)
This is a big reason why I argue for getting the propellant from the Moon. Then we would have virtually unlimited amount of propellant to drastically cut the travel time, no new, expensive, (potentially) dangerous, far off propulsion systems required.

Then a manned Mars mission is simply dependent on setting up a propellant production base on the Moon. Since as I argue manned/cargo lunar flights can be done at costs of a few hundred million per flight, making multiple flights per year possible, constructing such as base and therefore mounting a Mars mission can be done in less than a decade.

Bob Clark

 

[dgatsoulis]

I assume that by departing from LEO the shortest time would be at a similar angle but with the departure delta-v at ca. 10 km/s in accordance with the calculation in post #17?


Bob Clark

AFAIK the angle would be exactly the same.

 

[RGClark]

AFAIK the angle would be exactly the same.

Here's another brain teaser for you. Suppose we wanted to use the speed boost we can get by the Oberth effect this time from both the Earth and the Moon. We would depart from L2 at a time when the Moon, Earth, and Mars are on an approximate straight-line.
We would head towards the Moon from L2, skimming the Moon's surface, firing our rockets at closest approach, heading towards Earth, getting the lunar Oberth boost. Then skimming past Earth we would fire our rockets again, getting an additional Oberth boost there.
A couple of questions: suppose you have fixed amount of delta-v you can supply from your rockets, then how should you break up the burns to wind up with the highest final speed?
Secondly, after you get the Oberth boost from the Moon, you'll have some final speed from that, then to calculate the Oberth effect from the Earth you need to know the speed at periapsis. Should you just add Earth's escape velocity to the final speed from the lunar encounter?

I did a sample calculation using the equations on the Wikipedia Oberth effect page. It is in effect the same equations you used in post #17.
I took the Moon's escape velocity as 2.4 km/s and the Earth's as 11 km/s. I set the max rocket delta-v available as 10 km/s. For this test calculation I split up the rocket burns as 2.4 km/s at the Moon, and 7.6 km/s at Earth.
Then to calculate Oberth at Earth, for the speed at periapsis I just added the Earth's escape velocity to the speed obtained after the lunar (Oberth) encounter.
I was surprised I was able to get up to 23 km/s this way.

Bob Clark

Edit: I tried the calculation again and this time I got 19.9 km/s. Also, the calculation does not take into account the orbital velocity of the Moon, so it should be regarded as just a tentative calculation.

 

[Deleted Orbinaut]

This compares to a shorter distance if you launch in a radial direction towards Mars. But the radial direction gives you zero boost from the Earth's orbital speed around the Sun. To see the different distances draw two concentric circles and note there is a longer distance to travel at the tangential direction to the inner circle compared to the radial direction.

1) You can't really launch away from the Earth completely radially. The Earth is travelling 29.8 km/s in its orbit around the sun. To zero out that orbital velocity, you'd have to have at least that much excess velocity when inside the Earth's sphere of influence, which means an Earth-relative velocity of something like 40 km/s when near the Earth.

2) If you do a Hohmann transfer to Mars (tangential departure/tangential arrival), the transfer time is fixed at about 8.5 months (plus or minus a few days due to the relative eccentricities of the orbits of Earth and Mars.) The two ways you can make this time faster are:
(a) Leave Earth tangentially going faster than Hohmann speeds. This means that you'll intercept Mars's orbit prior to reaching aphelion (and again, symmetrically after aphelion.)
(b) Do (a), but also add some radial velocity. This would have the effect of making your transfer orbit more or less eccentric. A less eccentric orbit could intercept Mars's orbit sooner, shortening the transfer time. Ironically, to make your orbit less eccentric, you'd actually want a radial velocity component that points away from, not toward the sun.

 

[RGClark]

Mars rover confirms dangers of space radiation.
Future manned missions to Mars will need internal shielding and advanced propulsion systems to shorten transit times, minimizing exposure to space radiation, scientists say.
by William Harwood May 30, 2013 3:06 PM PDT

http://news.cnet.com/8301-11386_3-57586958-76/mars-rover-confirms-dangers-of-space-radiation/

An expensive and far off development using nuclear propulsion that is already controversial and would still only make the travel time 6 months(!)
This is a big reason why I argue for getting the propellant from the Moon. Then we would have virtually unlimited amount of propellant to drastically cut the travel time, no new, expensive, (potentially) dangerous, far off propulsion systems required.
Then a manned Mars mission is simply dependent on setting up a propellant production base on the Moon. Since as I argue manned/cargo lunar flights can be done at costs of a few hundred million per flight, making multiple flights per year possible, constructing such as base and therefore mounting a Mars mission can be done in less than a decade.

Further on the radiation issue for manned flights to Mars:

Manned mission to Mars an unlikely proposition.
Current limits on exposure to radiation make chances of flight in near future pretty slim.
Sep. 22, 2013
Written by Todd Halvorson FLORIDA TODAY

It’s “the elephant in the room,” NASA Chief Astronaut Robert Behnken recently told a National Academy of Sciences committee.
“We’re talking about a lot of ionizing radiation, almost a guarantee for cancer, and you are really close to the edge of the range for lethal exposure,” said Kristin Shrader-Frechette, a University of Notre Dame professor and a specialist in ethical issues that arise in scientific research and technology development. “If we can’t get shorter transit times in space, and we can’t get better shielding, then we really can’t do (a Mars) spaceflight.”

http://www.floridatoday.com/article...6/Manned-mission-Mars-an-unlikely-proposition


A near term solution is already apparent: lunar derived propellant depots.


Bob Clark

 

[RGClark]

...
I did a sample calculation using the equations on the Wikipedia Oberth effect page. It is in effect the same equations you used in post #17.
I took the Moon's escape velocity as 2.4 km/s and the Earth's as 11 km/s. I set the max rocket delta-v available as 10 km/s. For this test calculation I split up the rocket burns as 2.4 km/s at the Moon, and 7.6 km/s at Earth.
Then to calculate Oberth at Earth, for the speed at periapsis I just added the Earth's escape velocity to the speed obtained after the lunar (Oberth) encounter.
I was surprised I was able to get up to 23 km/s this way.

Bob Clark

Edit: I tried the calculation again and this time I got 19.9 km/s. Also, the calculation does not take into account the orbital velocity of the Moon, so it should be regarded as just a tentative calculation.

After a more careful calculation I don't believe you can do better by using two "Oberth burns", one at the Moon the other at the Earth. The reason is the propellant you used up in the burn at the Moon could be put to better use if the entire burn was just at the Earth. This is because the higher escape velocity at the Earth would give a higher Oberth boost.

BTW, Dgatsoulis, what would be the arrival speed at Mars for the scenario you mentioned in post #17?

Bob Clark

 

[dgatsoulis]

19.53 km/s.

Remember that that's the encounter velocity, not the velocity at periapsis.
You can calculate the velocity at periapsis with:

[math] V_{pe} = \sqrt{V_{enc}^2 + V_{esc}^2} [/math]

where [math]V_{esc}[/math] is the local escape velocity, aka the escape velocity for the periapsis altitude you choose.

[math]V_{esc} = \sqrt{\frac{2GM_{planet}}{R_{planet}+alt}}[/math]

 

[Ripley]

...The Living module came out a little more nautical than I intended but you get the general idea...

Looks like something drawn by Jules Verne.