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