Starship to Mars: faster than Hohmann transfer trajectories

[InterestedEngineer]

So they are assuming that atmosphere arrival speed for Mars needs to be south of 7km/sec.   I can't recall the numbers, but others here have calculated max arrival speeds, and I seem to recall it was on the order of 9km/sec before you couldn't brake without multiple passes.  Anyone remember?

At a 9km/s, 70km periapse (slightly different from entry speed, but not by much), you need 2G of downward lift, which (I think) comes out to be 4.5G of inertial forces on the crew, assuming L/D=0.5.  Not very nice after 3 months of microgravity

Probably still doable though, right?

Depends on the heating model.  As lift goes up, so does drag.  As drag goes up, peak heating goes up non-linearly.

I've never seen anybody discussing 9km/s entry speeds at Mars.  7500m/s, yes, but that only shaves off a couple of days from the 7000m/s speed.

can you give us the Gs for inertial acceleration at Mars for the 7000m/sec and 7500m/sec case for EDL?

[Vultur]

So they are assuming that atmosphere arrival speed for Mars needs to be south of 7km/sec.   I can't recall the numbers, but others here have calculated max arrival speeds, and I seem to recall it was on the order of 9km/sec before you couldn't brake without multiple passes.  Anyone remember?

At a 9km/s, 70km periapse (slightly different from entry speed, but not by much), you need 2G of downward lift, which (I think) comes out to be 4.5G of inertial forces on the crew, assuming L/D=0.5.  Not very nice after 3 months of microgravity

Probably still doable though, right?

Depends on the heating model.  As lift goes up, so does drag.  As drag goes up, peak heating goes up non-linearly.

I've never seen anybody discussing 9km/s entry speeds at Mars.  7500m/s, yes, but that only shaves off a couple of days from the 7000m/s speed.

.

Is the heating worse at Mars than entering 9 km/s at Earth would be?

I imagine Starship should ultimately be able to come back from the Moon, at a higher entry velocity than that.

[TheRadicalModerate]

can you give us the Gs for inertial acceleration at Mars for the 7000m/sec and 7500m/sec case for EDL?

and

Depends on the heating model.  As lift goes up, so does drag.  As drag goes up, peak heating goes up non-linearly.

I've never seen anybody discussing 9km/s entry speeds at Mars.  7500m/s, yes, but that only shaves off a couple of days from the 7000m/s speed.

.

Is the heating worse at Mars than entering 9 km/s at Earth would be?

I imagine Starship should ultimately be able to come back from the Moon, at a higher entry velocity than that.

I think I messed this up up-thread.  Let me try again.

Let's build a simple, stupid model.  We'll assume that the Starship instantly falls to the periapse altitude, then stays there all the way through peak heating.  It does this by applying some amount of lift, either positive (up) or negative (down).  Note that this model is only going to tell us what the initial lift is, because it'll change as the Ship slows down.

Let's assume that we want to kill as much speed as possible at some optimal altitude.  From the IFT-5 and -6 flight data, that appears to be about 70km on Earth.

It'll be a different altitude on Mars, one preferably with the same atmospheric density.  (We actually need a higher density, but go with me for a bit--it's a complicated problem.)  I went into my atmosphere models (which are sketchy this high) and found it was about 57km.  Note that I'm just assuming that the ship magically appears at periapse and stays there.  As the speed drops, the amount of lift required will also drop, reducing acceleration on the crew.

We can write an equation for the condition where Starship maintains altitude.  Let's define "up" as positive:

μ/r² - v²/r = L/m

The first term is acceleration from gravity, the second is the centrifugal acceleration that's trying to fling the Starship back into space, and the acceleration from lift is what balances everything out.

That's far from the whole story, though.  If we assume that Starship has a constant lift to drag ratio, then the Starship and crew won't feel acceleration from gravity or the centrifugal acceleration, but they will from both the lift and the drag.  If we define the lift to drag as LD (we've mostly been using LD=0.5), then the total acceleration on the Starship structure and crew is:

totalAccel = sqrt((L/m)² + (D/m)²)

But D = L/LD, so:

totalAccel = sqrt(L²(1/LD²+1))

Note that, at L/D = 0.5, totalAccel = sqrt(5L²).

Both the pure lift acceleration and the acceleration on the crew and vehicle are plotted below.  Note that the crew acceleration for Earth has a minimum, because the slower entry speeds need upward lift, while the faster ones need downward lift.

But this is still wrong, and here's where I'm not sure how to compute things.

If we have the same speed and density, then we should be able to generate the same amount of lift.  That's why we chose the Mars altitude as 57km--because that's where the density of the martian atmosphere is the same as Earth's at 70km, the apparent Starship sweet spot.

But we need more lift at the 57km altitude, for a given entry speed.  But we can't just conjure that extra lift out of nowhere.  We can get more lift in one of three ways:

1) If we aren't exactly at the optimal L/D, we can fiddle with the angle of attack until we are.  But Starship is essentially a flying brick, so there's going to be pretty narrow range of functional AoA's.

2) Go faster.  But that's the exact thing we want to avoid, because it generates more centrifugal acceleration, which requires even more lift.  (Heating rate also increases as the cube of the speed.)

3) Most likely, we need to go even lower, at the same speed, to get denser air.  This increases the negative lift we need by a little bit, because the centrifugal acceleration increases faster than the gravitational acceleration does, but we're only changing the radius by small percentage.  (Mars is small, but it's not small in relation to the altitudes we're using.) The extra lift is almost negligible.

But denser air at the same velocity is going to heat up more.  It's not terrible:  The heating rate is proportional to the square root of the density.  But the heating rate that melts the vehicle is down there somewhere.

To do a better solution, we'd need to find the density that gives us the extra (negative) lift we need, reverse lookup that density to find the altitude, and then iterate until the centrifugal acceleration converges.  Then you have to decide if you can live with that heating rate.  If not, you can't enter that fast.

Bottom line:  The charts below are probably fairly accurate in terms of G forces on the crew and vehicle, but the altitude will be lower, and peak heating will be higher. 

Which answers Vultur's question--hopefully correctly this time--about whether the heating's worse:  yes, it is, because you need more negative lift, which means you need denser air, which increases heating.

Hopefully, this gives you an idea of why you can't use the same entry speeds on Mars as you can on the Earth.

PS:  Crap.  Old charts.  Updating the attachments.

[InterestedEngineer]

3) Most likely, we need to go even lower, at the same speed, to get denser air.  This increases the negative lift we need by a little bit, because the centrifugal acceleration increases faster than the gravitational acceleration does, but we're only changing the radius by small percentage.  (Mars is small, but it's not small in relation to the altitudes we're using.) The extra lift is almost negligible.

But denser air at the same velocity is going to heat up more.  It's not terrible:  The heating rate is proportional to the square root of the density.  But the heating rate that melts the vehicle is down there somewhere.

To do a better solution, we'd need to find the density that gives us the extra (negative) lift we need, reverse lookup that density to find the altitude, and then iterate until the centrifugal acceleration converges.  Then you have to decide if you can live with that heating rate.  If not, you can't enter that fast.

Bottom line:  The charts below are probably fairly accurate in terms of G forces on the crew and vehicle, but the altitude will be lower, and peak heating will be higher. 

Thanks, exactly what I was hoping for!

If we line up the G-forces for the Moon Return to the Mars Entry, the Mars Entry as a max velocity of 6.5km/sec if we limit our selves to the same G-force as a Moon Return entry of 11km/sec.

Less than I would have expected.  Ouch.

It seems like we have enough information to examine this, if we take the benchmark of Starship as return-from-Moon at 11km/sec, we can solve these two equations where we set F to be equivalent for our Mars and Earth entries and solve for p

F ∝ pv²
q_dot ∝ p^0.5 x v³

 p_earth * v_earth² =  p_mars * v_mars²

p_mars / p_earth = v_earth² / v_mars²

which is 11² / 6.5² ~= 3.  We need three times the density on Mars to get the same force (and thus same acceleration).

at 60km, 0.0001 kg/m³ means we need 0.0003 kg/m³  on Mars

so relative q_dot on Earth is 13.31 and relative q_dot on Mars is 4.76, or 2.8x less heating on Mars

if you plug in a constant force or pv² = constant, you get density is inversely proportional to velocity squared, and if you plug that into the q_dot proportional equation you get q_dot ∝ v²

in other words we can do entry on Mars faster than 6.5km/s, but the limit won't be heating it'll be G-load.

It's a shorter soak on Mars in terms of G-load and heating, so that might help us go a little faster.

3Gs seems reasonable, given that on launch our astronauts experience that, so that corresponds to 7.25km/sec.

[InterestedEngineer]

Now if we had a TMI velocity vs periapse velocity on Mars graph or equation, we could figure out how much extra deltaV can be applied to the TMI vs the engine brake at Mars once we've exceeded about 7.25km/sec periapse velocity at Mars.

[TheRadicalModerate]

Now if we had a TMI velocity vs periapse velocity on Mars graph or equation, we could figure out how much extra deltaV can be applied to the TMI vs the engine brake at Mars once we've exceeded about 7.25km/sec periapse velocity at Mars.

That I have.  But the TMI velocity isn't exactly what you need.

There are two kinds of non-hohmann transfers to consider:

1) Ones where the perihelion = Earth's orbit and the aphelion is > Mars's orbit.  (Note that aphelion = Mars's orbit is a hohmann orbit.)

2) Ones where the perihelion < Earth's orbit, and the aphelion >= Mars's orbit.

If you want the most bang for your delta-v, it's always best to use type 1.  However, a moment's thought about the geometry will convince you that your arrival angle with respect to Mars will always be maximized by this.  (Angle is between the planet's orbit and the vehicle's orbit.)  That increases the arrival v∞, because you can only reduce the arrival speed by a factor of cos(arrivalAngle) of the planet's speed.

Type 2, with the lower perihelion, means that you're departing Earth's orbit at some angle, which reduces the efficiency of the departure delta-v, but reduces the arrival v∞ as well.  TANSTAAFL.

I'll try to come up with a sharable model that won't drive somebody isn't me insane.  It does not have a Lambert solver in it, so it doesn't specify departure and arrival times, and doesn't deal with the vagaries of planetary eccentricity or non-coplanar orbits.  It kinda works like this:

1) Use the departure v∞ and a departure angle, which lets you calculate the semi-major axis, angular momentum, and eccentricity. 

2) Then you can use the orbit equation to determine your departure (r = 1AU) and arrival (r = 1.524AU) true anomalies, from which you can derive time of flight. 

3) At your arrival, you have to compute the angle between Mars's orbit and the transfer orbit, which gives you your arrival v∞.  Note: higher angles (measured between the trajectory and Mars's orbital path) result in higher v∞.

4) Pick a periapse altitude and convert it to a Mars-relative orbit radius. You can use that and the arrival v∞ to compute the speed at periapse.  Then it's on to reentry.

[TheRadicalModerate]

so relative q_dot on Earth is 13.31 and relative q_dot on Mars is 4.76, or 2.8x less heating on Mars

You can probably go higher than 2G (what the 12km/s Earth entry says).  I'd think they'd be OK at 3G.  And I'm surprised that qdot doesn't get you first.

Your math looks OK, but I didn't grok the fullness. (Seems I'm having a Heinlein cliche day today...)  I had a slightly different plan in mind for determining an equivalent altitude to generate the needed lift.  I'll post it when it's done.

[Robotbeat]

"15 full Starship launches per crew Starship".

That's 3000t of fuel, they only need 1500t of fuel, so they are assuming only 100t of fuel to LEO per launch.

Maybe in 2026 window, but after that?  That's too much pessimism.

1500t of fuel with 100t of cargo is a MR of 7, or by odd coincidence 7km/sec of delta V.

They are using 4.6km/sec on the TMI.   So where is the other 2.4km/sec going?  I'm pretty sure the landing on Mars is < 1km/sec of deltaV.

I also don't think 4.6km/sec is a 90 day transfer.  I seem to recall that was on the order of 120 days, but it's been a while since I dug through those calculations.

it depends on the opportunity. For the same fuel load, transit time can vary by a factor of 2 depending on which synod you pick.

[sdsds]

The paper that started this has a good pork-chop plot that shows the reasonable opportunities don't occur every synod. It's the red/orange parts that matter for this discussion. The blue tails are just a distraction.

[InterestedEngineer]

so relative q_dot on Earth is 13.31 and relative q_dot on Mars is 4.76, or 2.8x less heating on Mars

You can probably go higher than 2G (what the 12km/s Earth entry says).  I'd think they'd be OK at 3G.  And I'm surprised that qdot doesn't get your first.

Your math looks OK, but I didn't grok the fullness. (Seems I'm having a Heinlein cliche day today...)  I had a slightly different plan in mind for determining an equivalent altitude to generate the needed lift.  I'll post it when it's done.

Let's go by intuition.  At the same deceleration, it takes the same time to go from 11km/sec to 10km/sec as it does from 8km/sec to 7km/sec

The former requires shedding for a 150t vehicle 9 - 7.5 = 1.5TJ.   The latter requires 4.8 - 3.7 = 1.1TJ.

so intuitively it makes sense there's less heat when slowing down the same amount at a lower velocity.

Thus it's limited by G-loading.  For reference

Shuttle: 3G (on launch)
Falcon-9/Dragon: 4.5G (launch, reentry is less)
Starship: 3G (on launch)

so somewhere in the 3-3.5 range is likely the limit.

[InterestedEngineer]

The paper that started this has a good pork-chop plot that shows the reasonable opportunities don't occur every synod. It's the red/orange parts that matter for this discussion. The blue tails are just a distraction.

Vinf (which I think means sun-frame deltaV) can go to 15km/sec with a GTO-refueled Oberth burn.

So the chart doesn't even go high enough.  Which is quite common with those charts, as for very hyperbolic orbits to Pork chop math goes haywire.  They'd need ultraviolet to cover that, possibly soft x-ray

I prefer Radical Moderates' approach.  The Lambert solver was deisgned for 1970s computers, we have much more computational resources, one can just brute-force it these days, and get less wonky results.

(It helps to take derivatives with respect to radius instead of respect to time, but it's so ingrained in everyone to take derivatives against time it occurs to extremely few people to do so)

[TheRadicalModerate]

The paper that started this has a good pork-chop plot that shows the reasonable opportunities don't occur every synod. It's the red/orange parts that matter for this discussion. The blue tails are just a distraction.

Vinf (which I think means sun-frame deltaV) can go to 15km/sec with a GTO-refueled Oberth burn.

No.  It's an Earth-relative measure of velocity at infinity.  v² = vInfinity² + vEscape².  You convert this to delta-v from your parking orbit by substituting v = vParking + deltaV and solving for deltaV.

To convert to heliocentric velocity from v∞, use vHelio = vEarthHelio(29,783m/s average) + cos(departureAngle) * v∞.¹

I prefer Radical Moderates' approach.  The Lambert solver was deisgned for 1970s computers, we have much more computational resources, one can just brute-force it these days, and get less wonky results.

Well, 18th century computers.

The wonky results don't happen as much in-plane.  It's when you start doing extreme out-of-plane maneuvers that things get weird--and suboptimal.  But you need something capable of dealing with a broken-plane maneuver to do better.

Just to be clear:  my model can't capture any of the delta-v changes and ToF differences caused by out-of-plane maneuvers and Earth and Mars eccentricity.  For that you need an ephemeris and at least a Lambert solver.

__________
¹Note that the "Sun-frame dV" label on the paper's porkchop plot is a terrible name.  It's really just v∞.  They could have put it in C3 (= v∞²) and saved a lot of confusion.

[sdsds]

Confession: for me the realization that broken-plane trajectories were probably better has been a stumbling block. Even a single impulsive maneuver has made finding optimal launch opportunities seem intractable. I would love to have code where: (a) it finds an optimal transfer and (b) I can be confident the result is correct. And that's without concern about how much computation is involved, so long as it's something that only takes a few days on an Amazon EC2 instance.

Maybe the bigger stumbling block though is the realization that single-impulse mid-course maneuvers are themselves probably sub-optimal. At least I assume the electric propulsion people will advocate for even better constant-thrust non-planar trajectories....

[TheRadicalModerate]

Maybe the bigger stumbling block though is the realization that single-impulse mid-course maneuvers are themselves probably sub-optimal. At least I assume the electric propulsion people will advocate for even better constant-thrust non-planar trajectories....

Electric propulsion is never going to be delta-v-efficient.  But it can be prop-efficient at the proper specific impulse and thrust.

[sdsds]

[...] At least I assume the electric propulsion people will advocate for even better constant-thrust non-planar trajectories....

Electric propulsion is never going to be delta-v-efficient.  But it can be prop-efficient at the proper specific impulse and thrust.

Yes, I'm doing lots of hand-waving because when considering "better trajectories" I don't have an objective function to optimize. In other words, I don't know what's better than what.

If you don't know where you are going, any road will get you there.

[Vultur]

so relative q_dot on Earth is 13.31 and relative q_dot on Mars is 4.76, or 2.8x less heating on Mars

You can probably go higher than 2G (what the 12km/s Earth entry says).  I'd think they'd be OK at 3G.  And I'm surprised that qdot doesn't get your first.

Your math looks OK, but I didn't grok the fullness. (Seems I'm having a Heinlein cliche day today...)  I had a slightly different plan in mind for determining an equivalent altitude to generate the needed lift.  I'll post it when it's done.

Let's go by intuition.  At the same deceleration, it takes the same time to go from 11km/sec to 10km/sec as it does from 8km/sec to 7km/sec

The former requires shedding for a 150t vehicle 9 - 7.5 = 1.5TJ.   The latter requires 4.8 - 3.7 = 1.1TJ.

so intuitively it makes sense there's less heat when slowing down the same amount at a lower velocity.

Thus it's limited by G-loading.  For reference

Shuttle: 3G (on launch)
Falcon-9/Dragon: 4.5G (launch, reentry is less)
Starship: 3G (on launch)

so somewhere in the 3-3.5 range is likely the limit.

But haven't some crew vehicles pulled much higher Gs?

Also, are we talking peak G load or sustained? For peak, I think New Shepard is fairly high and they fly very elderly passengers.

[TheRadicalModerate]

But haven't some crew vehicles pulled much higher Gs?

Also, are we talking peak G load or sustained? For peak, I think New Shepard is fairly high and they fly very elderly passengers.

The crew will have spent 3-4 months in microgravity before landing. Unlike a return to Earth, they’ll have no ground support staff to help them adjust to gravity after they land.

The good news is that they’re only in 1/3G. But they’ll want to be pretty conservative about crew health.

[meekGee]

But haven't some crew vehicles pulled much higher Gs?

Also, are we talking peak G load or sustained? For peak, I think New Shepard is fairly high and they fly very elderly passengers.

The crew will have spent 3-4 months in microgravity before landing. Unlike a return to Earth, they’ll have no ground support staff to help them adjust to gravity after they land.


The good news is that they’re only in 1/3G. But they’ll want to be pretty conservative about crew health.

Butch Cassidy and the Sundance Kid:  "But I can't swim!   Don't worry, the fall will kill you anyway..."

Same situation, fewer horses.

[InterestedEngineer]

Confession: for me the realization that broken-plane trajectories were probably better has been a stumbling block. Even a single impulsive maneuver has made finding optimal launch opportunities seem intractable. I would love to have code where: (a) it finds an optimal transfer and (b) I can be confident the result is correct. And that's without concern about how much computation is involved, so long as it's something that only takes a few days on an Amazon EC2 instance.

Maybe the bigger stumbling block though is the realization that single-impulse mid-course maneuvers are themselves probably sub-optimal. At least I assume the electric propulsion people will advocate for even better constant-thrust non-planar trajectories....

Maybe given RTN coordinate system a try.

It drives me crazy that most papers I see try to use the cartesian coordinate system.

In a rotating frame, that makes things far more difficult.

[Twark_Main]

I'm skeptical of Lambert solver results for deeply (high deltaV) hyperbolic orbits, they always seem to behave funny when I try them.

Me too.  The reason for funny behavior is that they do dumb things to match inclinations.

The hyperbolic trajectories have problems because of convergence of the actual algorithm, not matching inclinations. Matching inclinations is mostly an issue for near-Hohmann transfers.

https://hopsblog-hop.blogspot.com/2013/01/deboning-porkchop-plot.html

However, the things they do are smarter than my model, which assumes everything is in-plane.

Ironically, the first thing all Lambert solvers do is convert the problem to planar.  Three points makes a plane, after all. 

Broken-plane is almost certainly the way to go, but I have no idea how to compute them.

You're gonna laugh when I tell you the answer...   

They use a regular lambert solver, and they pick a plausible initial guess burn location and time, and then they simply "hill climb" in 4D space from that point to find the optimum broken-plane trajectory.