[notact]

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]

I tried what you said in the most realistic simulator we have, Kerbal Space Program, assuming like you that a gentler approach would be better. And I learned that no, that most certainly is not better.

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]

What happens if you do a geostationary deorbit? I.e straight down.

[stevage]

You mean use fuel to remove all horizontal velocity? Yes, that works fine and puts much less stress on the exterior, but it's a gigantic amount of fuel, almost as much as it took to get into orbit in the first place.

[HPsquared]

Coriolis force. If a geostationary satellite was instantaneously accelerated straight down towards the Earth, its path (from Earth's perspective) would not be straight down as its original circumferential velocity at geostationary orbit (pi * geostationary orbit diameter * Earth radial velocity) would be "too high" when the altitude is lower.

To continue moving "straight down" its angular velocity would need to be constant, which means as its altitude decreases the circumferential velocity would need to decrease. But gravity only pulls down, there's no force to accelerate it in that direction. So therefore it appears to curve off to the side.

[magicalhippo]

How would you do that?

Keep in mind that the object orbiting is already falling. Orbiting earth is literally "falling around the earth", compared to "falling down to earth" which we are more familiar with from throwing rocks and whatnot.

So to go "straight down" either it would need to orbit the sun (instead of the earth) and have its orbit intersect that of the earth, like the meteors we're worried about, or it would need to do a very strong deceleration burn.

[pricechild]

You still need to reduce your horizontal velocity. (For "geo-stationary", think "really high and really, really fast")

Either you do that with atmospheric drag, or a huge amount of fuel. The weight of heat protection is much lower and more efficient than the fuel option.

[macintux]

https://what-if.xkcd.com/58/

[dotnet00]

You need to balance peak heating and heating duration. Shallower reentry means lower peak heating, but higher heating duration. Steeper entry means higher peak heating, but lower heating duration.

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]

Scott Manley has a good video looking at this question and goes into a bit of a dive into the physics and engineering issues involved

https://www.youtube.com/watch?v=5kl2mm96Jkk

[CarVac]

To get a gentler reentry, you need a greater lift-to-drag ratio.

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]

Exactly! Re-entry is the transition from orbital dynamics to aerodynamics. If you want the transition from orbit to flight to occur at a lower speed, then you need to be able to produce lift equal to your weight at that speed, at the altitude where you will hit that speed.

[panick21_]

That's one of the reason why space planes were preferred for so long. Bleeding of speed while skipping along the atmosphere and then coming in for landing.

[GMoromisato]

Not sure if you got an answer already, but the reason is gravity. Gravity pulls the ship down, so you have to move very quickly horizontally so that the curvature of the Earth starts falling away from you. For LEO that's something like 25,000 kph.

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]

If you want to do slow star-trek style landings, you need star-trek level tech. Namely, propulsion tech that doesn't exist.

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]

A positive way of framing it is that atmospheric recently is free. If the Earth didn’t have an atmosphere it would take just as big a velocity change to land as it does to get into orbit and getting to be orbit would be as hard as an interplanetary flight. It's worse than it sounds because the rocket equation has a logarithm in it...

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.

[thehappypm]

No atmosphere would be much easier.

There's enough energy in a Tesla battery to for the Tesla to reach escape velocity. If you could simply drive at max acceleration (and the car didn't fall apart, and the tires continued to have grip, and a million other reasons why this is impossible) eventually you'd reach escape velocity and still have some percentage left.

In a more realistic sense, a long railgun type system would be very practical in a no-atmosphere environment, and then not being subject to the tyranny of the rocket equation, you could launch whatever you wanted. Enough fuel to decelerate is no problem.

[Denvercoder9]

> There's enough energy in a Tesla battery to for the Tesla to reach escape velocity.

No, it's not even remotely close. A Model S weighs around 2000kg and has a battery of 100 kWh. That's √((100 kWh)/(1/2*2000kg)) = 600m/s of delta-v. Escape velocity for Earth is 11.2km/s, almost a factor 20 more.

[thehappypm]

My math was off!

This nerd sniped me a LOT, I’m wondering if it’s possible for a chemical battery to reach orbital velocity (not escape).

An idealized Tesla would just be its battery (500kg) perfectly dumping energy to mechanical forward speed. Cutting 3/4 of the weight gets you closer to the delta-v you need, but youre still off by a factor of 5. Though orbital velocity, and leaving from the equator and gaining that speed, means you only need to get up to ~7.2 km/s. Still only a third of the way there.

Maybe you could split your battery into chunks, and expel them once they’re expended?

[PaulHoule]

Seems challenging to get decelerated by a rail gun, coil gun or such at the destination. You get one chance to get caught by it otherwise you crash and die.

[thehappypm]

Use a giant spring! When you land, aim for the spring, and charge it for the next launch!

[jccooper]

An airless Earth would be extremely hard to land on. You could easily get to quite low altitude, but orbital velocity at sea level is almost 8km/2... faster than at "orbital altitude".

Once you hit an orbit intersecting the ground, you have to scrub all your speed in whatever that amount of time is, which is gonna be short. It's basically an orbital launch in reverse.

[garaetjjte]

If you are coming at higher speed eg. from the moon, then it's possible to slow down to get reentry equivalent to low earth orbit one. But you can't really slow down much more because you would just plunge into atmosphere at steeper angle. Some vehicles utilize skip reentry trajectories, where it does high altitude pass through atmosphere and then goes in second time: https://en.wikipedia.org/wiki/Non-ballistic_atmospheric_entr...

[bagels]

It would take a tremendous amount of fuel to do what you're imagining, probably to the point of making the craft impossible to build with current technology.

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]

I'm not sure why people are misunderstanding my question as "Why not bring more fuel and burn the rockets in reverse". I am simply asking: why not reenter the atmosphere at a shallower angle, spreading the atmospheric braking friction over a longer period of time, which I'd expect would allow more time for the accumulated heat to radiate away before it becomes catastrophic.

[jaggederest]

What makes you think they aren't already taking the shallowest possible descent?

Once you start touching the atmosphere, it very quickly becomes deterministic. There are a limited number of descent profiles that actually get you to the ground, and believe it or not, starship as far as I can tell is actually taking a "shallow angle" and spreading the atmospheric braking friction over the largest possible time. A steeper entry would melt every conceivable material

[pixl97]

Gravity is one you are still being pulled down.

The other is at too shallow of angle at high speed you bounce off like skipping a stone off the surface of a lake.

[mrandish]

I'm no expert but I think reentering at a shallower angle results in "bouncing off" the atmosphere. So, even if you did it multiple times like a rock skipping on water, you'd have to have extra fuel to counter the bounce "up" and go back down for each skip. Thus, back to the same "bring more fuel/weight to orbit" problem.

[rtkwe]

Any heat you see is velocity lost to the craft will eventually hit the atmosphere again. I think the main reason is that the skip and the second reentry is way less predictable than doing the descent in a single pass so for predictability of landing agencies much prefer to do a harder more controlled reentry.

[friend_and_foe]

So, speed of reentry is directly a consequence of surface area (or energy expended by fuel to counter, in the case of the booster which does not use friction with the atmosphere to slow down) of one side. You'll produce the same amount of heat (and sound, light, but let's keep the model simple so we can understand better) no matter how fast you come in and no matter how wide your surface area is (assuming the same mass), it's just the thermal properties of the material and the surrounding environment dictate how quickly that heat dissipates, and the surface area determines how distributed the heat is, and the speed it's entering determines how quickly the heat is generated.

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]

The velocity of a spacecraft in low earth orbit is over 15,000 miles per hour. Smashing into the atmosphere is perhaps the most fuel- and cost-efficient way to slow down to a speed at which landing is possible.

[93po]

It doesn't really answer the question though. Why not descend slower so that the 15k MPH isn't meeting so much air? And bleed it off much slower so there is less heat

[avmich]

Ellipse, circle, parabola, hyperbola - all so called conic sections - are orbital trajectories; when you entering the atmosphere (which means you're technically not on a strictly circular orbit), you're initially following the part of that curve which is closest to the planet.

The curve is such that if you don't lose enough speed, you're going to start moving way from the planet.

If you're still on parabola (technically you never are, it's infinitely thin case between ellipse and hyperbola, physically not really possible) or hyperbola, you're not comping back - so if you need to get to the planet, you have to be on elliptical trajectory.

Even if you're on ellipse, you don't want that ellipse to be too elongated - e.g. the elliptical trajectory from the Earth to the Moon, which is rather close to parabolic one, takes about 4 days one way. You don't want to spend that much time when you're landing, so you need to lose enough of speed in the atmosphere. Which means you need to brake relatively aggressively.

This means there's a "reentry corridor" - not too steep, not too shallow, and the spacecraft needs to survive the reentry, and going from the Moon is harder than going from LEO because coming from the Moon the spacecraft has higher initial speed entering the atmosphere. It's still possible to balance various approaches, but you can't have (correction: it must be particularly hard to have...) zero fuel use, relatively fast landing (without long ellipses between reentries), speedy planet approach and low heating at the same time.

[verzali]

It's hard to do that. What you suggest would mean losing all your orbital speed before you hit the thicker layers of the atmosphere. You could probably do that, but you'd use a lot of fuel to decelerate. And then you are still being accelerated downwards by gravity, so you need something to counter that, which means you need to burn fuel all the way down. All that fuel adds a lot of weight, which cuts down on the amount of useful stuff you can take with you.

[tsimionescu]

It's because slowing down from 26000 km/h to something that wouldn't cause extreme heating (say, 1000km/h) would only be possible by firing thrusters in the opposite direction. Otherwise, any contact with a medium that could slow you down would lead to the same extreme heating.

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]

If you're reentering from say the moon, you don't really have a choice. If you reenter too slowly, you won't end up landing at all, and will skip out the other side. Or you would have to do multiple passes, which would take days to weeks.

[tocs3]

They use the atmosphere to help slow the ship down. It takes most of the tank of fuel to get up there and moving so fast. It would take most a tank to slow down. So, they would need about double the fuel plus some for landing.

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]

I understand the atmosphere is used to slow the vehicle - it's basically free brakes that you don't have to carry with you. I never suggested using rockets in reverse to slow the vehicle down. What I am asking is, instead of effectively standing on the breaks and generating enormous amounts of friction in a short period of time, why can't the vehicle ease onto the breaks and spread the friction out over time so it can be more safely dissipated (via a more shallow reentry angle).

[ta1243]

The shallower the angle the less energy you lose, but you are still losing altitude.

At some point you lose enough energy that your speed drops enough that your altitude starts dropping significantly. You can't lose the energy without losing altitude, and once you lose altitude you start losing energy whether you like it or not

I think what you are wondering is "can I stay in the thin atmosphere bleeding X Joules of energy for 50 minutes until most of the energy has gone rather than entering more steeply and bleeding 10X Joules for 5 minutes"

However once you lose energy, you lose your altitude, and as you lose altitude the atmosphere thickens and you start very quickly losing 5X, 10X, 20X joules every minute.

[phkahler]

See lift to drag ratio. To get enough lift to maintain altitude you need a certain amount of drag. At those speeds the drag causes the heating while still not producing enough lift to stay up.

[gitfan86]

If you had extremely big light weight wings it would help, but the materials that can do that don't do well when heated up

[panick21_]

You also run into issues of what do with the wings on the way up. You can't just put huge ass wings on that thing. You likely need it deploy-able.

And then the wings would also survive the flip and vertical landing. Or if you want to land like a plane, then you also need landing gear.

So there is really no way to add wings without adding a huge amount of mass. You are building a completely new thing.

There are some super cool mega-space planes designed in the 70s (I think). But of course these were never built or even tested. I remember they had some overlapping metal heat shields and a big ass delta wing. They would also start vertically and use air breathing engines.

[bagels]

They already use a shallow angle. There's just a lot of energy involved. As soon as the drag kicks in, the angle gets steeper and steeper on its own as the drag slows the craft down.

[notact]

I guess this sorta makes sense - the slightest slowdown starts to deorbit the vehicle, at which point a particular descent rate becomes difficult to maintain?

[bagels]

Its more than double.

[baq]

It's mind-boggingly more than double. The rocket equation takes no prisoners.

[chasd00]

Slowing down from Mach 20-something takes a huge amount of energy in its own right.

[robertsdionne]

You are not riding the atmosphere down, the atmosphere is riding you.