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I ran Google AI again, using the 2031 departure trajectory. It says hyperbolic approach velocity is 4.77 km/s.
"for a spacecraft departing earth orbit on February 23 2031 toward mars with 176 day transit what is approach velocity"
Then what is atmospheric entry velocity? 6.86 to 6.89 km/s depending on specific altitude defined for entry interface.
"assuming hyperbolic approach velocity of 4.77km/s to Mars, and aerocapture with apoapsis of 80km above surface, what is atmospheric entry velocity"
So the specific dates and trajectory reduce atmospheric entry velocity a little. And using 4.90 km/s is slightly higher exit velocity. So using Google the higher entry velocity we need to shed 1.99 km/s. That's a little better.
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Rob:
I updated the analysis and the spreadsheet to be more realistic and defendable. It's available from the drop box by means of links here on the forums, in the GW postings thread, if nowhere else. Tom put it up there.
The problem with running other people's software while trying to do orbit mission planning is not knowing what the results really mean, and/or not knowing garbage from good stuff. Computers "happily" process bad data in garbage, and it looks the same as when they process good data into good stuff.
I suggest that you use whatever 2-body orbit "thingie" you prefer to define the trajectory of your ship with respect to the sun. Ship, sun: 2-body. The textbook formulas give you that. The speed and direction of the Earth and Mars with respect to the sun are well known. Planet, sun: 2-body. Textbook formulas.
The problem is the 3-body effect when the ship is near Earth with the sun acting on both, and near Mars with the sun acting on both. ship, planet, sun, 3-body: there are NO textbook formulas for that, only finite-element computer codes can do it "right". And the amateur using somebody else's code often gets into trouble. I know I do, and did.
What I outlined in post 22 above was a magnitude-only close approximation that gets you ship speed (but NOT direction) near the Earth at departure, and near Mars at arrival. It's arithmetic for Hohmann transfers (because the ship and planet velocity vectors are parallel where the orbits touch tangentially). It's vector math for faster trajectories, because the ship and planet velocity vectors are NOT parallel where the orbits cross!
But you can easily get the speed of the ship on its 2-body orbit at the planet distance from the sun, with a textbook formula! I put that into the orbits spreadsheet in the drop box linked from the forums. It may or may not be in the NASA online orbits calculator. The angle between them is something you must figure from the local slopes or angles of the ship trajectory and the planet orbit, and then difference. My spreadsheet produces the plot data in a table that lets you figure that out.
GW
Last edited by GW Johnson (2026-09-25 09:07:29)
GW Johnson
McGregor, Texas
"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"
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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.
Last edited by HG Muller (Yesterday 03:32:19)
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Dr. Muller
I do have a crude spreadsheet tool for estimating aerobraking. It is in the tools topic as a free download.
It uses a best guess for how far around the planet one must go while aerobraking, along with how much velocity you must lose to drag, to get an average speed around that path (which is also your speed at periapsis), and therefore a max deceleration gee at the periapsis of that path.
If you know your ballistic coefficient, you can use that plus the max deceleration gee at periapsis to estimate the max drag force you must have. If you know the drag coefficient, you can use that with your periapsis speed estimate, to find the density of the atmosphere that must exist at your periapsis point. That density, used with a simple exponential scale height model of density versus altitude, tells you just how deep into the atmosphere your periapsis must be.
During entry, peak heating always occurs prior to peak deceleration gees and at a significantly higher speed than the speed at peak gees, but not a large difference in altitude (because the angle must be shallow to be survivable at all). So as a rule-of-thumb estimate, I use the average of speed at entry and speed at periapsis as the peak heating speed. I put that and the periapsis density into the rough correlation equations for convective and plasma radiation heating rates, to estimate stagnation heating rates for those two effects.
I put all this into the aerobraking spreadsheet tool. It actually works pretty good. I even have some rules of thumb in it for spreading those stagnation heating rates around to other locations as realistically-reduced values. Given some surface emissivities, the tool even estimates refractory surface radiation temperatures at those sam locations.
You need only know what planet (for the atmosphere model), what speed at entry interface, how much speed you need to lose in order to capture, your vehicle's ballistic coefficient, and the hypersonic drag coefficient that went into that ballistic coefficient. Make the best guess you can, for how far around the plant your aerobraking path will go, in the atmosphere. A quarter turn (about 1.5 radian) is a good starting guess for aerobraking aerocapture.
GW
GW Johnson
McGregor, Texas
"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"
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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.)
Last edited by HG Muller (Yesterday 12:26:41)
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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.
Last edited by HG Muller (Yesterday 13:40:17)
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Well, my guess for the wrap-around angle was just empirical, based on what I have seen. You can always run a spread of angles and see how much they affect the answers. As near as I can tell, it's not the strongest effect, once you are anywhere in the ballpark. It is the heating rates, pressures, and refractory re-radiation temperatures that help you select the right kinds of materials for your heat protection. Ablatives depend upon heating rate and pressure. Refractories depend upon servce temperature and pressure.
My aerocapture spreadsheet tool is for crude estimates, to be made by people who do not have (or have access to) an trajectory program that includes both thrusts and atmospheric drag, at orbital-class speeds. Most people do not have that. I do not have such.
I come from the slide rule days, when almost everything was done by hand, pencil-and-paper, running numbers with slide rules (later, scientific calculators). Only those things absolutely too complex to do that way, were run on the main frame computer, which for most of my defense career was an enormous thing in a room cooled with an enormous air conditioner, fed batch input by punch cards in steel trays, and with about a 3-hour to 3-day turnaround time for each run, the majority of which bombed out. All the while with management yelling at you, that you were delaying payroll! There were no desktops or laptops in those days! There were no word processors, no spreadsheet software, and no powerpoint to make slides with.
That is when I learned how to do those things that I can do. I did do some scientific programming to automate some of the worst stuff, but the only languages I knew are long dead now. Things like Fortran IV (NOT Fortran 77 or newer; I learned on Fortran II in college, over half a century ago), and some of the advanced BASIC's like QuickBASIC 4.5. I still have access to QuickBASIC 4.5 on my laptop here, but it takes a couple of gamer shells to run it, and to print out the results. It was a Microsoft product that only ran under DOS operating systems, not Windows! The gamer shells get around that. My ramjet cycle analysis codes are written in it, about 3000 lines each.
I was forced out of defense work after 20 years, by the huge drawdown that took place after the fall of the Soviet Union. I went into a second career of mostly teaching, plus some civil engineering work. I taught math, physics, and engineering at levels from high school to university. The civil engineering included both fire protection and environmental. That went on for another 20 years. I have been fully retired for 11 years now.
This last year I did some paid consulting in ramjet propulsion work. But that's about all, other than about 20 years custom-building farm implements that kill prickly pear cactus out of farm and ranch pastures, with no pickup and disposal, and no chemicals. I invented that totally by accident, but it worked, and I have built and sold over 60 of them, in my shop out here on the farm. About half a ton each.
GW
Last edited by GW Johnson (Yesterday 14:06:00)
GW Johnson
McGregor, Texas
"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"
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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.
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The drag equation used from incompressible to hypersonic is D = CD q A, where A is the reference area used for CD. It is q that beyond about Mach 0.5 to 0.8 can no longer be measured accurately with a pitot-static tube. It is simply calculated as 0.5 density velocity-squared, even hypersonic. The drag coefficient has a constant subsonic value up to around Mach 0.8, then there is a transonic drag rise to a value about twice the subsonic value, peaking in the Mach 1 to 1.1 range. On the supersonic side there is a decrease to values comparable to (and sometimes lower than) the subsonic value, by around Mach 3 or 4. From about Mach 5 on up, the supersonic/hypersonic drag coefficient value is almost constant (the shape of the shock field and flow pattern around the object no longer changes significantly with speed). For really blunt objects, that constancy of hypersonic patterns and drag coefficients starts nearer Mach 3 than Mach 5.
The lift force from a blunt object like a capsule is quite real, but not large at all compared to the drag, quite unlike airplanes. The capsule flies at an angle of attack not large enough to cause its lateral wall to become parallel to the flow on the one side, because of flow reattachment and much higher heating rates on that one side. During entry with most capsule shapes, lift/drag is only near 0.1 at most. You roll the angled capsule about the oncoming velocity vector axis to point that lift vector in the direction you want, to fine-control your entry path. Once past peak gees, you are slow enough that dynamic pressure is low, so that drag and lift are low, and this type of control no longer works. The trajectory inherently droops downward from there, and really droops once you are only supersonic again.
You have to be very careful with extended surfaces of any kind in hypersonic flight through the atmosphere, even up high in the thin air. There are all sorts of different geometries, and those that shed a bow shock wave which impinges upon an adjacent portion, like two parallel nacelles, one upon the other, will incur something called "shock-impingement heating". When that occurs, the normal supersonic convection heating rates get multiplied by a factor in the neighborhood of 7 or higher. Not the driving temperatures, just the heating rates. The material so affected will come to full thermal equilibrium in a second or so, to something pretty near the flow stagnation temperature, which above Mach 5 to 6 is usually higher than any known materials can survive, especially the metals.
That last occurred on the X-15 flight that reached 4520 mph at about 100,000 feet, the Mach 6.67 flight Pete Knight made in the X-15A-2. That flight had a scramjet test article mounted on the ventral fin stub, instead of the usual ventral fin. The scramjet spike shock impinged upon the ventral fin stub and on the underside of the tail section, nearly destroying the stub, and came close to cutting the tail section off the airplane. This effect is quite real, and quite dangerous! And it is exactly why no hypersonic vehicle can ever have a shape resembling the SR-71. The shocks off the engine inlet spikes would cut the wings off.
GW
Last edited by GW Johnson (Today 08:17:36)
GW Johnson
McGregor, Texas
"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"
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