How much more stressful would the Indian Ocean reentry have been for Starship IFT-3?

Question

The flight tests were originally supposed to end in the Pacific, but was switched to a steeper trajectory that ended in the Indian Ocean for IFT-3. Assuming all had gone perfectly, how much harder would this have been in terms of temperature and deceleration, compared to the original trajectory?

Answer 1

Jonathan McDowell posted "Starship marginal-orbit insertion" and said, "I infer a -55 x 235 orbit". It might be possible to figure out the difference in the initial speed at the start of re-entry by using some orbital mechanics, but you would need to guess the orbit for the original trajectory. If you edit your question to provide an assumption for the "originally planned" orbit, then it might be possible to estimate the difference in speed and steepness of the reentry.

Even if we know the differences in speed and steepness, we still need to consider how those numbers translate into differences in temperature and deceleration. It's a bit easier to produce estimates for "stagnation heating" and "aerodynamic pressure". Converting numbers for stagnation heating into temperature is a fairly complex heat transfer problem. But if you're trying to get a sense of how the speed translates into "how much harder", then this slide deck will help to explain that. It shows that the...

As energy is $1/2mv^2$, if the 'v' is increased by factor $X$, then the spacecraft's kinetic energy is increased by $X^2$. A well-designed thermal protection system (TPS) will cause most of that energy to be converted into heat that is absorbed by the atmosphere. Most of the rest is absorbed by the thermal mass of the heat shield tiles (ideally) to prevent the structure of the spacecraft from becoming too hot. While numbers for Starship are not publicly available, number for the Space Shuttle might be representative. For example, a reference for the Space Shuttle that said...

Black tiles work by reflecting about 90 percent of the heat they’re exposed to back into the atmosphere, while the tiles’ interior absorbs the rest. The tiles’ interiors radiate absorbed heat so slowly that after landing, the tiles take hours to cool.

Another reference says...

Roughly 15% of the Orbiter's empty weight is TPS.

As the steepness of the reentry path doesn't really change the total amount of kinetic energy that needs to be converted into heat and dispersed or absorbed, we can say that there is at least a $V^2$ relationship between speed and total heat energy that should be fairly independent of reentry angle.

All this is to say that the $V^2$, $V^3$, and $V^8$ terms in the above equations should help illustrate that more speed makes the reentry problem much harder, and less speed makes the reentry problem a lot easier.

Changing the reentry angle, on the other hand, changes the rate at which the air density, '${\rho}$', will increase during descent. From the formulas above, stagnation heating increases with ${\rho}^{0.5}$ or ${\rho}^{1.2}$, which are much smaller effects than the velocity effects. That is to say, the velocity effects dominate the reentry angle effects. So, the original trajectory would have had a higher velocity; therefore, it would have been harder than the trajectory that ended over the Indian Ocean.