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FA strong competent woman asserting that the current male leadership of the US are "man children".
I have NO idea if that sentiment is present in more than one individual, but I got the impression she is fed up with male incompetence.
I agree with that assessment, but I don't see any evidence that this is gender related. The main problem seems to be with the U.S. electorate (of which half are women, I suppose...). As long as they behave like voting cattle that is controlled by campaign donations, and can be brainwashed to the point where they are willing to elect convicted criminals and patholgical liars as their leaders, the situation is hard to remedy.
I understand the forum is in dire straits. But just for my understanding: the forum software used is a phpBB version? I run a phpBB3 forum on talkchess.com (which currently seems to suffer from DDoS attack, but I had no time to look into that yet). I think that 3 passwords are involved in this: for the MySQL database, for the forum administrator, and to log in as a Linux user on the server.
Are all these passwords lost? Did you ever have the password for the server? If the Mars Society hosts it, I can imagine they use their own server where they also have activity not concerning us. And that they likely have their own webmaster for that stuff. Who then must have a password of his own, and perhaps even super-user access to the server. Or is the Mars Society as poorly off as we are?
What happened to Mr. Burk? Did he die or disappear? Should he be considered an adversary?
OK, so the lift of a conical craft with a blunt heat shield is only small. But what about a winged craft like the space shuttle? Would that have a lower terminal velocity when falling through the atmosphere compared to a lift-less body like a sphere? And would the extra weight of the 'wings' (which will be jettisoned as part of the heat shield just before landing) reduce the propellant requirement for the final braking to touch-down speed enough to be worth it?
I als wonder what the limits of parachutes are. It is probably not necessary to land a very heavy piece of equipment in one piece. Even if the mission enters the atmosphere in one piece, it could break apart in smaller modules when becoming sub-sonic, all parts except the heat shield) having their own parachutes and retro rockets. This to ensure they all land close to each other.
When I click the link in the leading posting I get an SQL error:
General Error
SQL ERROR [ mysqli ]Access denied for user 'newmarscom'@'vps31385.dreamhostps.com' (using password: YES) [1045]
An sql error occurred while fetching this page. Please contact an administrator if this problem persists.
Well, even standard forum software can often be made to do more that you would expect, through plugin modules. Look for instance at https://talkchess.com/viewtopic.php?t=46462&start=1 , where postings can contain images of chess positions that are generated on-the-fly from a text line, or game records where you can step through the game interactively using buttons. And you wrote me that you guys 'owned' the forum software, so what the forum can and cannot do is basically up to you.
But I was not thinking about integrating something into the forum; what I had in mind was hosting some web pages with interactive content on the server. I don't know what dgital infrastructure you are running on (e.g. VPS, shared web space, ...), but the welcome page on newmars.com does look like a page that lives outside the forum database.
I haven't had time to look into this yet, so I am not sure what orbital calculations this software does. But it gave me an idea:
Wouldn't it be great if there were on-line tools associated with this forum that people could use directly by just going to a web page and putting some numbers into a form? Solving Newton's equation of motion for a several-body problem numerically isn't a very taxing task with modern computer technology. A typical smartphone is now more powerful than the world's largest supercomputer I could use for my calculatons at the start of this century, and finite-element methods can use such small time steps that even very simplistic numerical methods can produce converged results.
I have made web applets using the JavaScript programming language (which runs on the computer of the people visiting those web pages) in connection with calculation-intensive chess questions. E.g. for demonstrating how one should win an end-game with almost arbitrary unorthodox chess pieces ( https://www.chessvariants.com/page/MScheckmating-applet ), or to allow people to play chess variants of their own design against an AI ( https://www.chessvariants.com/page/MSpl … s-variants ).
I am sure I could make something in the same spirit for calculations on orbital mechanics or the rocket equation. If I did that, would it be technically possible to host these pages on the website, in a way that makes them easily accessible from the forum?
The problem we'd like to address is the feasibility of designing a Space Tug to dock with Artemis II on it's way back from the Moon, and deliver it safely to LEO.
The Artemis II Orion capsule is equipped with a defective heat shield that will put the astronaut's lives at risk if it is used. It appears that NASA is committed to using that heat shield, regardless of the risk.
A Space Tug made using SpaceX hardware could (theoretically) dock with the capsule on it's way back from the Moon, and provide retro-propulsion sufficient to find a secure orbit in LEO. The heat shield is more than capable of handling an ordinary LEO descent.
The challenge (as I understand the problem) is to find a solution that brings the capsule on a hyperbolic trajectory close to a Space Tug in an elliptical orbit such that the relative dV between the two is manageable with the quantity of fuel that can be carried in the Space Tug, taking into account that the Space Tug has to accelerate itself from LEO to attain the desired orbit, match orbit with the capsule, and then decelerate itself and the capsule into LEO.
This does not seem such a hard problem. Of course the solution depends on how much manoeuvrability is left in either spacecraft, and what their actual orbit is when the heat-shield problem manifests itself. The Artemis must at least be able to correct its course enough to avoid immediate reentry: it must raise it's perigee to well above the atmosphere.
When it then arrives there it can use most of its remaining propellant to slow down, loweing its apogee (which otherwise would be somewhere neer the Moon) and orbit time. It can then remain in that orbit until the space tug is prepared to meet it.
In the mean time the space tug would be launched, or when already in orbit, match that orbit to pass through the Artemis its perigee at the time the Artemis will pass through it. Starting from LEO of the wrong altitude you would raise the apogee while under or above the Artemis its apogee, and when you reach that apogee, fire the engine to adjust the perigee to the altitude of that of the Artemis. By selecting the apogee you can adjust the time needed to arrive at that peogee, to match the arrival of the apogee there.
When both get there, the tug accelerates to Artemis velocity, getting into the same orbit, for meeting it. If the Artemis did not have enough propellant left to significantly decelerate at perigee, both would now be in an orbit that stretches nearly to the Moon. The delta-V to get it there would be in the order of 3km/s. And to get back into LEO it would need a similar delta-V.
I never used MATLAB. When I want to calculate something I just write a program in a simple programming language, usually C. In the past years I also got some experience with JavaScript, for web applications.
I am usually not to keen on using software written by others anyway; it often takes me more time to learn how to use it than it would have taken me to write something myself that does the one thing I want to do. And very often it doesn't do exactly what you want to do anyway. And I hate to use software that I cannot change.
~3 tons for 5MW is indeed not too bad. For Isp=2,500 the exhaust velocity is about 25km/s, and the kinetic energy per kg of ejected propellant 312.5e6 J. So 5MW could eject 16g/s, for a thrust of 400N. On a 100t vessel this would give an acceleration of 0.004m/s2. After a million seconds (which is about 278hr or 11.6 days) a delta-V of 4km/s would be realized, which on the time scale of planetary travel is fast enough.
To do Isp = 10,000 requires 16 times as much energy per ejected kg, and delivers 4 time the thrust. With the same power use you would have 4 times less thrust, and to realize 4km/s would take some 45 days. This might already start to erode efficiency, as in those 45 days you might already have significantly moved away from the Sun.
OK, thanks for the explanation. This cap of 7.669km/s indeed equals the orbital velocity at an altitude of 400km, so that confirms my reasoning for the infinitesimal-thrust case.
I am not denying that ion drives in many cases can have huge advantages in terms of required mass ratio. I am just pointing out a caveat that due to the low thrust a large part of that advantage can be neutralized, by requiring much larger dV for making the same trip. Eventually increasing the Isp can always overcome this extra requirement. But high Isp is not free in other ways: the power needed to accelerate the propellant increases as the square of the Isp. So very rapidly the choice between unacceptably long burst duration or larger power source will present itself. And the increase in mass of that power source (e.g. more solar panels) must not exceed the savings in fuel mass, or it would defeat the purpose.
Let me give another example where low thrust performs rather poorly:
Suppose we do an interplanetary mission starting from the Moon, (where we build spacecraft fur this purpose), for which we have to escape from Earth with residual speed 3.8km/s. Suppose even in the low-thrust case such a dV can be achieved in a small fraction of the transit time (so it doesn't have to spiral through the solar system all the way out to the target).
You need high thrust to lift off from the Moon. So forget that part, and start from low Lunar orbit. Orbital velocity there is 1.7km/s.
With low thrust we need 1.7km/s to escape LLO, then 1km/s (the Moon's orbital velocity) to spiral away from Earth starting at the orbit of the Moon, and then 3.8km/s to set us on course to the target. In total 6.5km/s.
Now use high thrust: we need dv=0.8km/s to accelerate to 2.5km/s in LLO. That leave us enough velocity after escape from the Moon (which in itself required only 2,38km/s) to cancel the Moon's orbital velocity around Earth enough to dive towards Earth, staying just above the atmosphere. We arrive there with a velocity about 100m/s under escape (so 11.1km/s). Then we fire the engine to boost us to 11.83, a dV of 0.73km/s. After escape we are left with a velocity sqrt(11.83^2 - 11.2^2) = 3.81km/s. Total dV was 0.8 + 0.73 = 1.53km/s. For such delta-V you are still well in the range where (for Isp = 450) the propellant mass is only a fraction of the empty mass, and the exponential nature of the rocket equation hardly manifests itself.
The low-thrust scenario required more than 4 times as much delta-V. With Isp = 2500 compared to Isp = 450 you do have more advantage of 5.55, which is more. So the ion drive would still beat LOX/LH2 in terms of mass ratio. But what area of solar panels would be needed to power the ion drive hard enough that the required delta-V after Earth escape can indeed be made in the part of the interplanetary trajectory that is still close to Earth orbit? (Say within a few weeks.) Those solar panels add to the empty mass, and thus also to the required propellant mass. Having a twice better mass ratio does no good if the empty weight is also twice as large; on the contrary, since the empty mass will also have to be launched from somewhere through high thrust. Bringing 100t to the Moon with only a few tons of propellent is no fun if 90 out of that 100t have to be solar panels powering the ion drive.
Well, I would be interested to develop a code that could calculate trajectories with thrust and drag, so I could calculate launch requirements in the presence of an atmosphere. But at the moment I don't have good understanding of the drag part at hypersonic speed.
Another issue is aerodynamics. What happens if lift is involved, because the drag force does not align with the velocity? I suppose that at high-enough speed you could fly a plane even in the thin Mars atmosphere. And the speed below which you start to drop out of the air like a stone no doubt would depend on wing size. So the question is how far one could push this to keep a landing cargo in the air, to give it more opportunity to decelerate by drag than when it would follow a simple 'ballistic' trajectory (where drag always aligns with velocity), and thus save on propellant needed for the final slowing to touch-down speed.
There exist a number of drag-enhancing methods running from parachutes (light but fragile, so not suitable for heavy loads) to large metal 'flaps' that could be used to lower the speed through passive means. And the optimum design would occur when making these structures still larger adds more mass than what this would save you on propellant for the remaining deceleration.
OK, next try. I compare a circle of constant altitude with a parabola that has its focus at the center of that circle (which we take as origin). WOrk in units of Mars radii. The circle has y = sqrt(1 - x^2), the parabola y = 1 - (x^2)/2 (both going through the point (0,1)). Expanding the square root in a taylor series we get y = 1 - (x^2)/2 - (x^4)/8 - ... The difference with the circle in lowest order is (x^4)/8. Suppose this is 20km (assumed scale height of the Mars atmosphere), i.e. 1/165 of the Mars radius.
(x^4)/8 = 1/165
x^4 = 8/165 = 0.048
x = 0.47
The total length from -0.47 to + 0.47 makes the path about 1 radian, which indeed is much closer to your guess of 1.5 radian.
I suppose I forgot to account for the coriolis force in the rotating frame.
[Edit] the value x = 0.47 is too large to justify cutting off the Taylor series after 3 terms, though, so the result is unreliable. Let us do an exact calculation:
I polar coordinates (r, phi) a parabola has the equation r = 2/(cos(phi)+1). (Top now at (1,0)).
To get to r = 1 + 1/165 requires
1 + 1/165 = 166/165 = 2/(cos(phi)+1)
cos(phi)+1 = 165/84
cos(phi) = 81/84
phi = 0.268 rad
So we stay within one scale height between +/- 0.268 rad. The angle spanned by the trajectory would then be 0.536 rad. It would become less if the scale height was a smaller fraction of the planet radius.
This is for entry and exit around escape velocity, i.e. for aerocapture into a highly elliptical orbit.
The problem I have with this is that putting a guess as input into an otherwise rigorous calculation makes the result of that calculation not better than that guess. The calculation I presented above suggests that the angle you mention is 0.3 radian rather than 1.5.
It is hard to advocate a guess would be better than a calculation, unless it can be established that the calculation is flawed... (Which of course is very possible.)
That is another way to look at it. But beware: for thrusts are so low that even during a full orbit the orbit hardly changes, the actual trajectory will always remain very close to perpendicular to the radius. The trajectory will be a spiral that only moves out very little per orbit. So you will be pumping kinetic energy all the time.
Nevertheless you suffer the inefficiency that you need dV = initial orbital velocity rather than sqrt(2) - 1 = 0.41 times it. This is because orbital mechanics will continuously convert added kinetic energy to potential energy: you cannot go faster in the same circular orbit, the increased centrifugal force will make the orbit expand, and the climb against gravity for that will go at the expense of your speed.
In such a spiral orbit the speed will decrease all the time, as 1/sqrt(r), (to adjust to orbital velocity at that distance) despite the effort of your engine.
I posted it in the tools topic because GW's announcement of the posting there mentioned that his tools did not calculate the flight path through the atmosphere, but expected it as an input that the user would have to supply (in the form of an angle). So it would be of interest to potential users of the tool to know how to calculate that angle, and what a typical value would be.
I will have a look at the tools. We have Excell only on my wife's computer, though, and my only remaining laptop that could run Linux (in a virtual machine) has a broken keyboard. Things degrade fast when you have no time to remedy them... :-(
The Wikipedia gives an equation for the nodal precession rate. It decreases very fast with orbital radius (r^-3.5), and is proportional to the cosine of the orbits inclination. So higher altitude (slowing the precession) can to a certain extent be compensated by smaller inclination (as cos is a decreasing function). But there is a limit to that, as the cosine cannot grow above 1. Having low inclination defeats the purpose anyway: the nodal precession might make the orbit keep its orientation to the Sun, but it will fly through the planet's shadow anyway if the orbit doesn't pass close enough to the poles.
For a very large radious even orbits with quite low inclination do not pass through the planet's shadow cone. Except twice per year, when the nodes align with the Sun. But if the orbit is quite large, such an eclipse will last only short compared to the orbit time.
To pass over Korolev once a day you don't necessarily need an orbit time of a day. Any integer fraction of a day would also work. The Earth would rotate under an orbit with a high inclination to make Korolev pass underneath it at the time it flies over its lattitude.
An Isp of 2,500s (ion engine) results in a propellant mass of 31.5t to push a 100t payload through TLI (6.7km/s), from a circular low earth orbit.
An Isp of 450s (LOX/LH2) results in a propellant mass of 106.5t to push a 100t payload through TLI (3.2km/s, utilizing the Oberth effect), from an elliptical low earth orbit.
Sure, a better Isp can give you equal delta-V for less fuel, more delta-V for same fuels, or anything in between. I suppose the delta-V you picked to compare are just enough to reach the Moon. The Isp is so much better that you are still off better, though. So the effect I mentioned is already clearly visible: the low-thrust case required more than double the delta-V, to achieve the same. I suppose this is because it assumes a spiral trajectory; with "apogee pumping" you should not need much more than 3.2km/s even in the low-thrust case.
But now look at interplanetary travel: Suppose I use delta-V = 4km/s in LEO. This requires 147t LOX/LH2. Starting 700m/s above escape velocity we will be left with a velocity square-root 12*12 - 11.2*11.2 = 4.31km/s. The ion propulsion would have to realize that after escape, which on top of the mentioned mass ratio would require another factor 1.192, i.e. 1.315*1.192 = 1.568, so 56.8t propellant.
So the ion drive is still better, 156.8t vs 247t, or 57% heavier. But it is not nearly as much better as the more than 5x better Isp appears to promise. When you factor in economic issues, such as the price of the propellant, and the added travel time, it might be cheaper to go for the LOX/LH2.
[Edit] BTW, how did you calculate that 6.7km/s? If I calculated right the delta-V needed for escaping through a spiral trajectory is equal to the velcity in the orbit you start from, i.e. 7.9km/s. Is the missing 1.2km/s because you only have to reach the Moon? In the interplanetary examply we would have no such luck, so that the required delta-V would be 7.9+4.31 = 12.21km/s, and the mass ratio 1.63.
I have no expertise in aerodynamics, so my take on aerobraking/capture only uses some general physics knowledge, on the level of "we can approximate a horse by a sphere". I would appreciate comments on this from GW, to tell me how ridiculous this is (or maybe not).
At hypersonic speeds an object basically just sweeps up the gas mass it is flying through. This creates a gas layer of higer pressure and temperature in front of the object, between the latter and a 'shock' where pressure and temperature change discontinuously. This layer moves with the object, and is pushed against it by the onslaught of the incoming gas acting as a piston. The gas in that layer streams away sideways at sub-sonic speed, to eventually clear and flow around it.
At velocity v and gas density rho the mass swept up per second by an object with cross-section area A is A*rho*v. This mass has a momentum v times larger, A*rho*v*v. Momentum per unit of time equals force. This supposes the velocity was perpendicular to the area, i.e. a blunt object like a heat shield. For other shapes I would be inclined to only count the momentum component perpendicular to the surface.
Anyway, the drag force appears to be proportional to rho. In an atmosphere rho decreases sort of exponentially with altitude, by a factor e for each 'scale height'. On Earth the scale height is 8km, but lower gravity on Mars increases it, while higher density of CO2 compared to O2/N2 and lower temperature decreases it. But let's say it is 20km on Mars for easy arithmetic. So rho is proportional to e^(-h/20km) where h is the altitude.
Now in a non-circular orbit h varies. In a circular orbit centrifugal force exactly cancels gravitational attraction. But centrifugal acceleration in low orbit equals v*v/R (with R the Mars radius). Escape velocity is square-root 2 times orbital velocity, so if we fly by at near escape velocity the centrifugal force in the reference frame that rotates with the spacecraft is 2 times gravity. It is as if the object that skims the atmosphere at escape velocity is accelerated away from Mars by about 4m/s2. Distance covered under such constant acceleration g equals g*t*t/2, so 2*t*t (meter). This describes how we dip into the atmosphere as a function of time, h = h0 + 2*t*t.
Plugging this into the expression for density, we see that rho (and thus the drag) is proportional to e^(-2*t*t/20km), i.e. a gaussian with a full width at 1/e height of 200 sec. The gaussian has the same area as a square pulse of height 1 and width ~180s. So it is as if the maximum drag force experienced at altitude h0 will act during 180s. If we need to shed 600m/s to get captured in a single pass we need an deceleration of 600/180 = 3.33m/s2, i.e. 1/3 of a g. Survivable. A 10 ton spacecraft would need 33kN drag to decelerate this fast. If the area of its cross section is 10m2 the pressure P behind the shock should be 3kN/m2 (0.3 bar).
To get this drag at 5km/s requires rho = P/(v*v) = 3333/(5000m/s)2 = 3.333/25000 = 0.00013kg/m3. Atmospheric density at the Mars surface is about 0.013kg/m3. So we must aim for an altitude where the atmospheric pressure has dropped by a factor 100. This is the case at ln(100) = 4.6 scale heights, i.e. 92km above the Mars surface.
How hot would we get? CO2 molecules at room temperature move at about 250m/s (about sound speed), so 5km/s is 20 times faster. Since temperature is an energy it goes as the square of velocity. Som CO2 would move (on average) 5km/s at a temperature of 20*20*300K = 120,000K. That is an upper limit, because at such temperatures in reality CO2 would be smashed to atoms, which would even partly ionize, which takes a lot of energy. Which then does not go into increasing particle velocity (and thus temperature). Not sure what would happen to a heat shield exposed to a plasma of 0.3 bar of such a high temperature. But it is clear that some heat shield would be needed.
Note that the deceleration is not nearly enough to reach velocity for a circular orbit, which for Mars would be 3.55km/s. After the aerocapture pass the spacecraft would be in a very elongated elliptical orbit, the apoapsis being perhaps 18 Mars radii from the surface. The orbit time of such an orbit (which according to Kepler's laws is the 1.5th power of the semi-major axis) would be some 31 times larger than that of a low circular orbit, i.e. some 3100min, about 2 days. But we are in orbit, so many passes through the periapsis will follow, during which we can aerobrake as gently as we want to lower the apogee further.
Of course if this was the mothership (although that would likely be much more massive) we would not be interested at all in aerobraking further; only the lander would continue to do that. The mothership would use its rocket engine to slightly accelerate at apoapis (where its velocity would only be some 250m/s to start with) to bring the periapsis safely above the atmosphere, where aerobraking is so negligable that it can remain in orbit for months. By conserving its high velocity it can then be sent back towards Earth with minimal delta-V, just undoing the aerocapture speed loss.
I don't think that explanation covers the main cause. Which is that on an extended low-thrust 'burn' you are lifting propellant out of the gravitational well, while with a short high-thrust burn at perigee you all eject it at the deepest point of the well you can reach.
It is all a matter of energy conservation. Rocket propulsion is horrible because of the exponential nature of the rocket equation. But it has one strong point: the thrust it provides is independent of the velocity you already have. Which is quite unlike when you are pushing yourself against a stationary object, as everyone riding a bicycle will have experienced. Since energy gained equals force times distance traveled during application of that force, you gain far more energy from a the same burn (in terms of thrust and duration) at high speed (where you travel a lot during the burn) than at low speed. Provided the delta-V is in the direction you were flying.
Another way to look at it is that kinetic energy is proportional to v squared, and (v+dv)2 = v2 + 2*v*dv + dv2. So the energy gained through dv involves a multiplier proportional to v.
One should realize that rocket + contained propellant + already ejected propellant in the absence of drag is a closed system, conserving its total energy. Accelerating the propellant in the engines adds mechanical energy at the expense of the chemical energy that was in the propellant you burn. This chemical energy is initially converted into kinetic energy of the ejected propellant in the frame of reference of the rocket.
E.g. suppose a spacecraft with a mass of 1 ton burns propellant at a rate of 1 kg/s, ejecting the exhaust at a speed of 4km/s w.r.t. the nozzle. This delivers a thrust of 4000 N(ewton), which will accelerate the spacecraft by 4000N/1000kg = 4m/s2 (= 0.4g). A burst of 1s will then increase the velocity by 4m/s. To accelerate the propellant in the engine required m*v*v/2 = 1kg * (4km/s)2 / 2 = 8 MJ(oule).
Suppose the spacecraft is doing this in LEO, where it is already moving with 8000m/s. It's kinetic energy was 1000kg * (8km/s)2 / 2 = 32GJ. After the burst it is 1000kg * (8004 m/s)2 / 2 = 500 * (8000 ^2 + 2*4*8000 + 16) = 32 GJ + 32 MJ + 8kJ. So it went up 4 times as much (ignoring the 8 kJ) as the energy it took to accelerate the propellant in the engine! How is this possible? Where does this extra 24 MJ come from.
Well, that becomes clear when you consider what happened to that 1kg ejected propellant. Before the burn it was in the rocket, and moving at 8km/s. After ejection it only moves by 8 - 4 = 4km/s. So its kinetic energy fell from 1kg * (8km/s)2 / 2 = 32 MJ to 1kg * (4km/s)2 / 2 = 8MJ, i.e. by an amount of 24MJ. Which is exactly the mysterious extra energy gain of the spacecraft.
What we learn from this is that it is quite important where and how the ejected propellant ends up in terms of kinetic and potential energy. The less total mechanical energy is left in the ejected propellant, the more of the energy you spent on accelerating that propellant you will find back in the spacecraft. So always try to eject the propellant as deep in the gravitational well as you can, and in (or exactly opposit to) the direction you are already moving. That will gain (or lose) you the maximum amount of energy.
If it takes much longer than the orbit time to accumulate the desired delta-V because of low thrust, this becomes difficult. A single burst would last into the period where you are already climbing out of the gravitational well. The mentioned technique of perigee pumping would switch off the engine as soon as you climbed a significant distance, waiting for the next perigee passage to fire another burst. This gradually raises the apogee. (Remember the point where you actually apply a delta-V, in this case perigee, will always remain on the orbit, so only the apogee changes.) But reaching escape velocity puts an end to the fun, because there would be no subsequent return to go faster. The additional velocity needed to reach another planet will have to be created outside the Earth's gravitational well.
This is also indicated by the conversion of fly-by velocity to interplanetary cruising velocity. Starting at 11km/s in LEO leaves you (by definition of escape velocity) with zero velocity after escape. But add 100m/s, you will be left with square root (11.1^2 - 11^2) = square root 2.21 = 1.488 km/s. The delta-V was 'amplified' nearly 15 times by applying it close to Earth compared to what you would have to do if you waited until after you fully escaped Earth. That significantly raises the bar for propulsion methods that deliver such low thrust that the required velocity can only be reached long after escape. (Even though the amplification factor decreases as you need more delta-V; with 400m/s above escape it would only be 7.5, etc.)
One of the things that worry me about non-chemical propulsion technologies is that they typically have quite low thrust. It is obvious that this makes them unsuitable to lift off from a world with gravity. But it also can be a problem when manoeuvring in space. It might seem that under micro-gravity you have all the time in the world, but this is often not true. For example, when leaving LEO for an interplanetary voyage, it is important to build up all your velocity close to Earth. Gradually spiralling away until you are free from Earth, and then continuing to accelerate to reach the desired solar orbit, takes far more delta-V than a short acceleration burst just above the atmosphere. The explanation is that the latter method leaves the ejected propellant deep into Earth's gravitational well, while in the other case it ends up much less deep, or even entirely outside the well. And somehow you have to pay for all that energy left in the ejected fuel.
So the best way to leave LOE with low thrust is only fire the thrusters near perigee, when your velocity is high. If that only accelerates you enough to slightly raise the apogee, you should wait for the next orbit for a subsequent burst of thrust. But that can only accelerate you to escape velocity, after which you will not return anymore, and be left at rest compared to Earth, i.e. following the same orbit around the sun. The remaining velocity required for reaching the target, which could be an order of magnitude higher than what you would have needed just above the atmosphere, will then have to be achieved while ejecting all fuel outside of Earth's well.
So chemical propellants might still be competitive to low-thrust propulsion methods having much larger exhaust velocity.
Let me post my take on orbital mechanics:
I make frequent use of the simple formula
v*v = 2/r - 1/a
where
r = distance to center (Sun or planet)
v = velocity in units of the orbital velocity at r=1
a = the semi-major axis of the orbit
(This is just the formula for energy conservation after cancelling out the mass, kinetic energy being v*v/2, potential gravitational energy being 1/r, and the total energy being 1/(2a).)
For orbits around the Sun I take Earth distance to the Sun as unit for r and a (i.e. use Astronomical Units) and Earth's orbital velocity (~30km/s) as unit for v. (And indeed, plugging in r = a = 1 then gives v = 1.)
For planetary orbits I express r and a in planetary radii, and v in units of the orbital velocity near the surface (which is escape velocity divided by square root of 2).
E.g. for mars at the perihelion v*v = 2/1.381 - 2/(1.666 + 1.381) = 0.792, or v = 0.890, or (multiplying with 30) 26.7 km/s. An orbit stretching from Earth at perihelion to Mars at perihelion in ithe orbit's aphelion has v*v = 2/1.381 - 2/(1 + 1.381) = 0.608, or v = 0.780 (23.4km/s). So arrival at the Mars orbit occurs with a velocity difference of 3.3km/s.
The second requirement is to use such 'asymptotic' relative velocities to calculate the velocity after falling into the planet's gravitational well. For this you have to add or subtract the velocities 'according to Pythagoras', i.e. square, add/subtract, and then take the square root. Falling from infinity to the surface gives you by definition the escape velocity. For Mars the latter is 5km/s, so you would arrive near the surface at square root of 3.3*3.3 + 5*5 = 5.99 km/s. That means you have to slow at least 990m/s to get into orbit rather than just having a single fly-by.
Another example of the orbital formula:
To get in a higly elliptic Mars orbit stretching from near the surface to 19 Mars radii, you need a velocity at periapsis for which v*v = 2/1 - 2/(1 + 19) = 2 - 2/20 = 1.9, or v = 1.379. This is in units of the Mars orbital velocity, which is 5km/s / sqrt 2 = 3.55km/s. This gives 4,87 km/s. So you only have to slow down to 130m/s below escape to get into such an orbit.
Such an orbit is a good starting for aerobraking, which will lower the apoapsis, circularizing the orbit. But it is also a good place to park the mothership for the return journey; it would only have to accelerate in the by 130 + 990 m/s to travel back to Earth. (If that journey would also start from Mars' perihelion.)
Another useful tool is the conservation of angular momentum. Which is the distance to the orbited object times the velocity component perpendicular to that (the 'tangential' velocity). The orbital formula can give you the magnitude of the velocity at any point of the orbit. But only in the tips of the elliptical orbit is the full velocity perpendicular to the line to the center. At other points there also is a radial component to the velocity.
The tangential velocity can be calculated by first calculating the velocity V in one of the extreme points (e.g. perihelion), and multiply this with the radius there to get the angular momentum. This can then be divided by the radius at the desired point, to get the tangential velocity there. The radial velocity then is the square root of the total velocity squared minus the tangential velocity squared.
For example, if we have an orbit with semi-major axis 1 AU (so that its orbit time is 1 year), which stretches to Mars its perihelion. Then its velocity where it intersects Earth orbit (the point from which you can launch it) would also be 1 (= 30km/s). For the velocity at aphelion you would have v*v = 2/1.381 - 2/(1+1) = 0.448, or v = 0.669 (20.1 km/s). The angular momentum is thus 1.381*0.669 = 0.924. At r = 1 we would thus have tangential velocity 0.924/1 = 0.924. Then the radial velocity is square-root 1*1 - 0.924*0,924 = 0.382 (11.4km/s). Since Earth has hardly any radial velocity of itself (near-circular orbit) all that velocity would have to be acquired with rockets, requiring a starting speed of more than 15km/s in LEO.
I once attended a lecture on the preparation of laser-fusion targerts. This indeed mentioned the problem that to heat the fusion fuel from scratch to a temperature where fusion occurs is larger (or at least too large considering inefficiencies of harvesting the energy) than the energy released by the resulting fusion. A solution to this problem was presented, though. The targets are hollow layered spheres. The center is filled with vapor of the surrounding deuterium-tritium layer. around that there was (IIRC) a metal layer and an 'ablation layer' of polymer. The sphere is placed inside a cylinder of some heavy metal, with holes in the end caps through which lasers can be focused to illuminate the inside of the cylinder.
A strong laser pulse than evaporates the inside of the cylinder to a hot plasma that radiates soft X-rays. These X-rays then flash-evaporate the ablation layer to create a high pressure gas to drive the implosion of the sphere. This 'indirect drive' causes very even distribution of the energy on the sphere, cleanly imploding it rather than shattering it.
It is mainly the gas content that gets compressed, and the compression factor can be large. And gasses heat up under compression. This cuses a thermal spike in the center of the sphere when the gas is compressed to almost nothing, hot enough to start the 'burn wave' of fusion events, which then propagates outward through the fuel layer to heat it up entirely to fusion temperature.
I got the impression that this part of the problem is well solved; the main problem in initial-confinement fusion is the efficiency of teh driving laser. Which only converts electricity to light with a quite poor efficiency.
What I know about muon fusion:
Hydrogen molecules (H2) consist of two nuclei and two electrons. But with one electron removed the remaining electron is still able to cause a chemical bond in the H2+ ion. It is just slightly longer than the normal bond. This can be easily understood: when a negative charge is halfway between two positive charges, the repulsive force between the positive charges is only one quarter of the attarction to the negative one. So clasically the entire thing would collapse.
Quantum mechanics limits the volume in which an electron can be kept, due to Heisenberg's uncertainty principle: if you squeeze it too much, its energy becomes so uncertain that it can escape the attractive force of the nuclei, and would escape. This prevents ordinary matter from collapsing.
A muon is negative like an electron, but some 200 times heavier. The mass affects the uncertainty relation, and allows confining the electron in a 200 times smaller volume. So a hydrogen molecule where you replace one electron by a muon becomes a muonic H2+ molecule that is 200 times smaller than a normal (electronic) one. This is so small that towards electrons it basically behaves like a point with charge +1, which then binds another electron like it is a hydrogen atom.
The muonic bond is so short that there is a significant probability that the nuclei can 'tunnel' through the repulsive energy barrier caused by their electrostatic repulsion. There then is a finite chance the nuclei touch each other to within the distance over which the strong nuclear force can be felt. If these are fusable nuclei, this results in fusion.
For deuterium and tritium there is a high probability of fusion, and this would result in helium and a neutron. These share the released fusion energy 1:4, which make both speed away quite fast compared to the muon that was binding it. Fast enough almost always shake the muon off. So it can bind to another hydrogen nucleus, effectively neutralizing that, so it can approach another nucleus and cause another fusion. There is some chance the helium nucleus will carry away the muon with it, though. And the +2 charge of helium binds the muon much faster than hydrogen, so the muon will never be transferred to a hydogen nucleus again, and is lost from the chain reaction.
This limits the number of of fusions a single muon can catalize. The fact that a muon on average decays back into an electron after about 2 microseconds puts another limit on the process. Unfortunately the number of fusions per muon is not enough to gain back the energy that is spend for creating the muons, (through pair creation with the aid of accelerated particles), although the difference is not extremely large.
One has tried to improve the process by using spin-aligned nuclei in a strong magnetic field; this enhances the probability for fusion. But it was still not enough to reach break-even, and I am not aware of any ideas that could improve the process further.
(seems the previous message refused to post because it contained the word exp for the exponential function, and after replacing that by e^ it worked!)
Propellant from rock (part 2)
The following is a 'back of the envelope' estimate of how this could work out. To escape the surface of the Moon needs 2.38km/s. But that would leave you in the same orbit as the Moon around Earth. To dip back down to Earth you would have to undo most of the Moon's orbital velocity of about 1km/s as well, leaving only some 180m/s. The velocity you would have to reach to be left with the desired delta-V for that is about 2.5km/s. M_full/M_empty = e^(v/c) with c the SSME's 4.46km/s gives a mass ratio of about 1.7. So to send, say, 100 ton from the Lunar surface towards Earth, you woul need 70 tons of fuel: 60 ton of LOX, and 10 ton of LH2. The deliverd 100 ton could be 90 ton LOX, and 10 ton empty weight for the structure of tank and engines. Plugging that into the rocket equation
The 10 ton LH2 needed for launch would have to be landed on the Moon first. So the touchdown mass of the tanker will be 20 ton. The descent needs the same delta-V as the ascent, so the required fuel for landing 20 ton is 14 ton: 12 ton LOX and 2 ton LH2. So to prep the tanker for the next run, it has to retain 12 of the 90 ton LOX. It can thus swap 78 ton of LOX for 12 ton of LH2. This saves the spacecraft that filled up more than a factor 6 on the LOX mass it would otherwise have to lift into LEO. That is pretty good, as the LOX would be most of the propellant mass: instead of bringing LH2:LOX as 1:6, it brings propellant as 2:0.
Now unfortunately bringing along cryogenics on long interplanetary flights for use on arrival and for return might not be ideal. And to get into the tanker's orbit you already need a velocity very close to escape; the additional delta-V to get to Mars or the asteroid belt is in the hundreds of m/s (when applied at perigee). While you already had to do 7.9km/s to reach LEO, and an additional 3.2km/s to stretch the orbit to the Moon. It would be nice if you could fill up with LOX before you applied that 3.2 km/s. That would require the tanker to descend to LEO, though. It could use aerobraking to do that, but it would then require an extra 3.2km/s to get back to the Moon.
The rocket equation says a mass ratio of about 2. The tanker set up for the descent to the Lunar surface measured 10 (structure) + 12 (LH2) + 12 (LOX) = 34 ton. To boost that by 3.2km/s thus requires another 34 ton propellant, 5 ton LH2 and 30 ton LOX. That leaves only 78 - 30 = 48 ton LOX to trade, now for 12 + 5 = 17 ton LH2. A profit of less than a factor 3. Bad! (But still a agin...)
To improve that we need to beat the rocket equation. We can do that in the usual way. The exponential growth of the mass ratio is due to the need to use propellant to accelerate propellant, and the answer is: don't accelerate propellant when you don't need to. Just park it before applying delta-V, and pick it up later. If you need 12 tons of LOX to get to the Moon for filling up from the elongated orbit, just leave that 12 ton orbiting there while you aerobreak the tanker with its 78 ton LOX load to get to LEO. Then you only have to accelerate the 22 ton tanker + LH2 by 3.2km/s to get it back to the orbit that stretches to the Moon. This requires some 22 ton of propellant, 3 ton LH2 plus 18 LOX. (Forgive me the rounding...)
So you the tanker can trade 78 - 18 = 60 ton LOX for for 12 + 3 = 15 ton LH2. The savings on LOX mass to be brought to LEO are now a factor 4, for each ton of LOX you need to eject the expedition from LEO you only need to lift 1 ton of LH2 into that orbit. This reduces lifted propellant from 1 + 6 to 2.5 + 0 (LH2 + LOX), still more than a factor 2.
I suppose you could even do a bit better when you also left some LOX is low Moon orbit, rather than having it escape the Moon and then bring it back. But the delta-V between Lunar escape velocity and low-orbit velocity is only some 800m/s.
Propellant from rock
Water might be as precious on the Moon as gold is on Earth, so using it for propellant might not be the most prudent thing to do. Of the lighter elements only oxygen is abundant there. It exists in the form of rock, which is composed of oxides of mainly silicon, iron, calcium, aluminum, magnesium. Most of these could be very useful for in-situ resource utilization. So it is likely that in the future we will develop some smelting process to separate these metals from the oxygen. (We can discuss the chemistry later; for now I want to investigate the orbital mechanics.)
The best propellant we developed so far is the hydrogen/oxygen combination (LH2/LOX) The water to which these reacts contains H and O in a (mass) ratio 1:8. The Space Shuttle Main Engines use these in a ratio 1:6, though, because such a 'fuel rich' mix gives a higher exhaust speed (4.46km/s) due to the lower average molecular mass of the exhaust than the mix that would create the hottest exhaust. But even then, 86% by mass of the propellant is oxygen.
So my proposal (which might not be original, but which I have never seen anywhere) would be to bring the hydrogen from Earth, and get the oxygen from the Moon. This would reduce the launch mass of the fuel by a factor 7. And consequently also the mass of the launch vehicle to get it into low Earth orbit. The strategy would be to have a 'tanker craft', shuttling LOX between the Lunar surface and a very elongated elliptical orbit stretching from just above the Earth atmosphere to the orbit of the Moon. Such an orbit has a velocity very close to the Earth's escape velocity at the perigee. A spacecraft heading for deep space could first boost itself into the same orbit, fill up with LOX, and then boost itself to the velocity it needs to travel to the intended target elsewhere in the solar system. It would trade this LOX for some LH2 given to the tanker, which would also retain some of the LOX. This should be enough to let the tanker land on the Moon, with enough LH2 left to deliver the next load of LOX.
The remainder of this story could not be posted due to 'Internal Server Error' each time I tried (as part of this, or as a seperate posting).