(no title)
notact
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1 year ago
Can someone explain why reentry must be so hellish? The energy gained during the rocket burn into orbit must be bled off during reentry, and that energy is enormous. However, why must reentry occur so quickly? It seems if the descent into the atmosphere was slower, the heat shield would be able to radiate the heat energy away more effectively, thus lowering skin temperatures, and significantly reducing the engineering challenge.
magicalhippo|1 year ago
What you need to protect is on the inside of the heat shield. Heat conduction is based on temperature difference and time[1] and the conduction of the material[2]. Since the heat shield tiles have a very low thermal conductivity, it takes a long time for significant heat to pass through.
Yes a more aggressive approach will lead to a greater temperature, but it'll also provide significantly greater drag, thus the the extreme temperatures only exist for a relatively short amount of time, and thus it doesn't have time to pass through the tiles and heat up the inside.
A very shallow approach has significantly less drag, and you spend significantly longer slowing down. The temperatures might be a fair bit less, but the much longer time spent decelerating means it has a chance to make it through the heat shield tiles.
It's not entirely unlike iron meteorites which can still be cold when landing, as they only spend a brief time in the atmosphere[3] and thus don't have time to heat up.
[1]: https://en.wikipedia.org/wiki/Heat_equation#Interpretation
[2]: https://en.wikipedia.org/wiki/Thermal_conductivity_and_resis...
[3]: https://earthscience.stackexchange.com/questions/127/what-te...
ActorNightly|1 year ago
dotnet00|1 year ago
The heat shield material can handle a certain amount of heat and a certain maximum temperature before it starts to ablate away, so you're forced to thread the regime where both variables are within its tolerances.
saratogacx|1 year ago
https://www.youtube.com/watch?v=5kl2mm96Jkk
CarVac|1 year ago
To have a better hypersonic lift-to-drag ratio you need significantly more wing area, which is dead weight (and drag and a control problem) on the way up.
avhon1|1 year ago
GMoromisato|1 year ago
If you move slower, you are no longer in orbit and your trajectory will intersect the ground.
If you tried to slow down more gradually, your orbit would keep dropping until you suddenly hit the ground.
Think of it this way: orbital speed is the speed required for a ship to stay in orbit without thrust. If you had infinite thrust, you could land on the ground at any speed you wanted. But without thrust, you have to go from orbital speed to 0 in less than one orbit.
jtriangle|1 year ago
That doesn't mean that it's impossible, just means that it'd require things that don't exist yet.
Worth mentioning that, additionally, reentry heating isn't a huge problem, and you're not going to create new propulsion tech to counter it, you're just going to make better heat tiles. What you need new propulsion tech for is doing expanse type stuff, where you can accelerate for months at 1G so you essentially have artificial gravity and can get places extremely fast. If you're into sci-fi, the show/books "The Expanse" goes into what that looks like in practice fairly well.
PaulHoule|1 year ago
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
double the Δv means you square the mass ratio. The space shuttle had a mass ratio of about 16, a mass ratio of 256 would be absolutely insane.
You get this velocity change at the cost of dealing with the heat and all but a tiny fraction of that heat ends up immediately in the atmosphere.
garaetjjte|1 year ago
bagels|1 year ago
Your orbit would have to be high enough to do a burn to cancel your orbital velocity (lots of fuel), then you have to burn against gravity for a slow vertical descent (lots of fuel). The rocket equation says... you'll need a larger craft and more fuel to carry the extra fuel in to orbit. It gets pretty out of hand.
Instead of using fuel to slow down, spacecraft make a small burn to have the orbit intersect the atmosphere, and then use drag instead of fuel to slow down.
notact|1 year ago
friend_and_foe|1 year ago
So to slow down more evenly and have less heat at the max point per square inch, you need wider surface area (or you need to expend fuel firing engines in the opposite direction of travel, what both parts do at the end to slow to 0, and a problem due to the rocket equation, fuel has mass and so increases the amount of kinetic energy you must dissipate), and that means more mass and more engineering and a bigger vehicle. The goal ultimately is of course optimizing all these variables.
yalue|1 year ago
93po|1 year ago
tsimionescu|1 year ago
And of course, the thrusters you'd need would add huge complexity for the shape, and need extra fuel in the stage 2 itself, greatly reducing its cargo capacity.
stevage|1 year ago
tocs3|1 year ago
P.S. I have not done any of the math (I might be able to figure it out but it might take a week or two to figure it out).
P.S.S : Maybe if they could refuel in space efficiently (asteroid mining?) it might be worth looking at but it will be a while before I would expect anything like that. It would just be the ship.
notact|1 year ago
bagels|1 year ago
chasd00|1 year ago
robertsdionne|1 year ago