Fast starship flight to Mars with a full tank of bi-fuel

[KILYAV]

I have calculated that a fully fueled starship (1200 tons) can reach Mars in 85 days, which will significantly reduce the impact of space radiation on the health of astronauts.

There is no formula editor on the forum, so I wrote them in a separate file.

[geza]

Thanks for calculating! Entry velocity at Mars can be an issue.

[Yggdrasill]

It does confirm what the NASA Ames Research Center Trajectory Browser shows.

[Danirode]

Around 16km/s arriving at Mars, and orbital speed for low mars orbit is what, ~3 km/s?
They have to loose 13km/s to enter  a orbit. That's a lot of energy for aerocapture/aerobraking. Can SS survive that? The acceleration would be pretty high on the first pass.

[Yggdrasill]

Around 16km/s arriving at Mars, and orbital speed for low mars orbit is what, ~3 km/s?
They have to loose 13km/s to enter  a orbit. That's a lot of energy for aerocapture/aerobraking. Can SS survive that? The acceleration would be pretty high on the first pass.

16 km/s is probably a bit much. If they limit it to 11 km/s that still gives some opportunities for a fast transit, and you would just have to live with longer transit times in the years where the Mars window isn't optimal.

I think it's also very likely they will start out quite conservative, and as they get more and more data on the performance of the heat shield and tweak the design, they can do faster and faster approaches.

[1]

Around 16km/s arriving at Mars, and orbital speed for low mars orbit is what, ~3 km/s?
They have to loose 13km/s to enter  a orbit. That's a lot of energy for aerocapture/aerobraking. Can SS survive that? The acceleration would be pretty high on the first pass.

Well, Mars escape is ~5km/s, so in principal they'd "only" have to bleed off 11km/s in one shot rather than 13. Not sure if that's enough of difference to change your overall point though. That's still a lot of velocity to scrub.

[KILYAV]

There is a very simple formula to find out if it is possible to brake around Mars at a speed of 16.5 km / s. This is the formula for centripetal acceleration:

a = V^2 / R_M = 16.663^2 / 3,396.2 = 0.082 km/c^2 = 81.76 m/c^2

If the starship can move, for a while, with an overload of more than 8G, then it will be able to slow down through aerobraking.

[Yggdrasill]

I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.

[eriblo]

Here is a simplistic simulation of a 210 t Starship aerocapturing at mars from 16 km/s. It has an effective area (C_D*A) of 640 m² and a lift-to-drag-ratio (L/D) of 0.3 corresponding to the ~70° angle of attack usually specified (these are from  older calculations done on the "Tintin" 3 fin version but should very similar). It does the whole pass nose down, using the lift to stay in the atmosphere until captured into an ~10 h orbit.

I could not find an easily implemented model for the radiative heating at these speeds but maximum total heat flux would likely be about one order of magnitude larger than a direct lunar entry on Earth.

[Yggdrasill]

Nice.

Yeah, I think 13G is a little high for comfort... How much better would it be at 11 km/s?

[eriblo]

Nice.

Yeah, I think 13G is a little high for comfort... How much better would it be at 11 km/s?

Better, although I think NASA likes to use a 5 g limit. Heating is now probably ~2X lunar entry.

[KILYAV]

I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.

Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.

With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.

[KILYAV]

Here is a simplistic simulation of a 210 t Starship aerocapturing at mars from 16 km/s. It has an effective area (C_D*A) of 640 m² and a lift-to-drag-ratio (L/D) of 0.3 corresponding to the ~70° angle of attack usually specified (these are from  older calculations done on the "Tintin" 3 fin version but should very similar). It does the whole pass nose down, using the lift to stay in the atmosphere until captured into an ~10 h orbit.

I could not find an easily implemented model for the radiative heating at these speeds but maximum total heat flux would likely be about one order of magnitude larger than a direct lunar entry on Earth.

Is it possible in this simulation to change the angle of attack in order to thus reduce the maximum acceleration near the surface of Mars and increase it with distance from the surface. Make a "slide".

What model of the atmosphere of Mars do you use, how does density vary with altitude?

[eriblo]

I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.

Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.

With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.

The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.

[eriblo]

Here is a simplistic simulation of a 210 t Starship aerocapturing at mars from 16 km/s. It has an effective area (C_D*A) of 640 m² and a lift-to-drag-ratio (L/D) of 0.3 corresponding to the ~70° angle of attack usually specified (these are from  older calculations done on the "Tintin" 3 fin version but should very similar). It does the whole pass nose down, using the lift to stay in the atmosphere until captured into an ~10 h orbit.

I could not find an easily implemented model for the radiative heating at these speeds but maximum total heat flux would likely be about one order of magnitude larger than a direct lunar entry on Earth.

Is it possible in this simulation to change the angle of attack in order to thus reduce the maximum acceleration near the surface of Mars and increase it with distance from the surface. Make a "slide".

What model of the atmosphere of Mars do you use, how does density vary with altitude?

Yes, although I have not implemented any guidance so it is all trial end error. However, the figures above are for constant maximum lift down and should therefore be the shallowest/lowest g-force/lowest heating trajectories possible - They are hyperbolic until the last seconds before exiting the atmosphere. As you can see the acceleration peak is at the beginning of a period of several minutes at roughly constant altitude.

This constant maximum negative lift trajectory is of course unrealistic and any real entry would have to aim slightly deeper in order to have some margin to modulate the lift. The entry angle window between skipping out into deep space and supersonic lithobraking in the 16 km/s simulation is << 1 arc second...

I used a digitized version of the attached density profile below from here: https://www.researchgate.net/figure/5-Comparison-of-Nominal-Atmospheric-Density-versus-Height-for-Earth-and-Mars-NASA_fig8_265093506 and double checked it with MAVEN data (it should be accurate to well within the natural fluctuations).

[KILYAV]

I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.

Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.

With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.

The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.

You are using too “perfect” aerodynamic bodies. Other body shapes can give much better drag coefficients.

For example, a triangle with angles 45-90-45. Front bottom corner 45, rear bottom corner 90, rear top corner 45. At hypersonic speeds, such a figure should reflect most of the atmospheric flow upward, with the deceleration being equal to centrifugal acceleration.

The magnitude of the centrifugal acceleration decreases very quickly and is equal to 8G only at a speed of 16.5 km / s, after the speed drops to 11.7 km / s, the acceleration drops to 4G.

[KILYAV]

I used a digitized version of the attached density profile below from here: https://www.researchgate.net/figure/5-Comparison-of-Nominal-Atmospheric-Density-versus-Height-for-Earth-and-Mars-NASA_fig8_265093506 and double checked it with MAVEN data (it should be accurate to well within the natural fluctuations).

I am not saying that your calculations are wrong, I am asking you to try to calculate yourself.

[eriblo]

I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.

Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.

With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.

The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.

You are using too “perfect” aerodynamic bodies. Other body shapes can give much better drag coefficients.

For example, a triangle with angles 45-90-45. Front bottom corner 45, rear bottom corner 90, rear top corner 45. At hypersonic speeds, such a figure should reflect most of the atmospheric flow upward, with the deceleration being equal to centrifugal acceleration.

The magnitude of the centrifugal acceleration decreases very quickly and is equal to 8G only at a speed of 16.5 km / s, after the speed drops to 11.7 km / s, the acceleration drops to 4G.

I do not quite follow...? Higher lift is certainly possible with a more aerodynamic shape but I thought we were discussing Starship?

I took the lift and drag values at an angle of attack (AoA) of 70° modeled here: https://www.researchgate.net/publication/334611503_3DoF_simulation_model_and_specific_aerodynamic_control_capabilities_for_a_SpaceX%27s_Starship-like_atmospheric_reentry_vehicle
Thus is for an older version of Starship but the values are probably similar as most of the drag and lift are generated by the body. Starship is certainly able to generate more lift at a lower AoA but I have not seen any mention of this despite the fact that it would greatly help with g-forces and heat flux. This suggest that they are limited by stability or local heating to an AoA around 70° and a L/D of about 0.3.

[eriblo]

I used a digitized version of the attached density profile below from here: https://www.researchgate.net/figure/5-Comparison-of-Nominal-Atmospheric-Density-versus-Height-for-Earth-and-Mars-NASA_fig8_265093506 and double checked it with MAVEN data (it should be accurate to well within the natural fluctuations).

I am not saying that your calculations are wrong, I am asking you to try to calculate yourself.

I think something is getting lost in translation or quoting. I ran the (fairly simple) simulation I presented myself and would hope you are not requiring me to get my own measurements of Mars atmospheric density!?

[KILYAV]

I used a digitized version of the attached density profile below from here: https://www.researchgate.net/figure/5-Comparison-of-Nominal-Atmospheric-Density-versus-Height-for-Earth-and-Mars-NASA_fig8_265093506 and double checked it with MAVEN data (it should be accurate to well within the natural fluctuations).

I am not saying that your calculations are wrong, I am asking you to try to calculate yourself.

I think something is getting lost in translation or quoting. I ran the (fairly simple) simulation I presented myself and would hope you are not requiring me to get my own measurements of Mars atmospheric density!?

Translation spoils everything.
I asked for the data to try and build the "optimal" trajectory myself.

[KILYAV]

I can't see how that is accurate. This is assuming the Starship is aerodynamically controlled to follow a circular trajectory around Mars, and has enough control authority to do so. But that's not the case. It will be going through thorugh the atmosphere in a hyperbolic trajectory, which then degrades into a parabolic trajectory, which then degrades into a suborbital parabolic trajectory, right? Also you have to subtract the Mars gravitational acceleration.

Moving at an acceleration of more than 8 G, the starship will be able to fly at the same height above the surface and decelerate as much as necessary.

With 8G acceleration, the starship will be able to "follow" the "bend" of the planet all the time, remaining at the same altitude, which will allow it to decelerate as much as necessary.

The problem is that for a L/D ratio of 0.3 you need 3.3 g of drag for every 1 g of lift. So the 8 g circular trajectory implies a total of ~28 g... Fortunately, the peak is much less due to the lower but still significant drag as it descends into the atmosphere.

You are using too “perfect” aerodynamic bodies. Other body shapes can give much better drag coefficients.

For example, a triangle with angles 45-90-45. Front bottom corner 45, rear bottom corner 90, rear top corner 45. At hypersonic speeds, such a figure should reflect most of the atmospheric flow upward, with the deceleration being equal to centrifugal acceleration.

The magnitude of the centrifugal acceleration decreases very quickly and is equal to 8G only at a speed of 16.5 km / s, after the speed drops to 11.7 km / s, the acceleration drops to 4G.

I do not quite follow...? Higher lift is certainly possible with a more aerodynamic shape but I thought we were discussing Starship?

I took the lift and drag values at an angle of attack (AoA) of 70° modeled here: https://www.researchgate.net/publication/334611503_3DoF_simulation_model_and_specific_aerodynamic_control_capabilities_for_a_SpaceX%27s_Starship-like_atmospheric_reentry_vehicle
Thus is for an older version of Starship but the values are probably similar as most of the drag and lift are generated by the body. Starship is certainly able to generate more lift at a lower AoA but I have not seen any mention of this despite the fact that it would greatly help with g-forces and heat flux. This suggest that they are limited by stability or local heating to an AoA around 70° and a L/D of about 0.3.

You are correct, I assume that the "Martian Express" in its shape will be much more like a Boeing X-51 than a starship.

The form of the "express" should not create lift, but press it to the surface so that it does not spill out of the atmosphere ahead of time, before it has time to decelerate to the "10 o'clock" orbit.

[tbellman]

You are correct, I assume that the "Martian Express" in its shape will be much more like a Boeing X-51 than a starship.

The form of the "express" should not create lift, but press it to the surface so that it does not spill out of the atmosphere ahead of time, before it has time to decelerate to the "10 o'clock" orbit.

That's why Starship will enter with the nose pointing downwards, so the lift force will keep the ship in the atmosphere.  It might be more correct to say that the angle of attack will be minus 70 degrees.

[Twark_Main]

The form of the "express" should not create lift, but press it to the surface

"Lift," by definition, is nothing more than the component of aerodynamic force that's at right angles to the flight direction (ie the direction of the vehicle's motion with respect to the free-stream fluid flow).  Thus, downward lift is still considered "lift."

It might be more correct to say that the angle of attack will be minus 70 degrees.

The angle of attack is positive 70 degrees because the nose is up. Negative 70 degrees would be nose down.

What we're seeing is a positive 70 degree angle of attack combined with a 180 degree bank angle.