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.

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

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.

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 m^{2} 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.

Quote from: Yggdrasill on 10/23/2021 10:49 amI 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.

Quote from: eriblo on 10/23/2021 09:08 pmHere 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 m^{2} 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?

Quote from: KILYAV on 10/25/2021 09:35 amQuote from: Yggdrasill on 10/23/2021 10:49 amI 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.

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).

Quote from: eriblo on 10/25/2021 02:30 pmQuote from: KILYAV on 10/25/2021 09:35 amQuote from: Yggdrasill on 10/23/2021 10:49 amI 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.

Quote from: eriblo on 10/25/2021 02:59 pmI 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.

Quote from: KILYAV on 10/25/2021 03:14 pmQuote from: eriblo on 10/25/2021 02:59 pmI 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!?