This evening I watched a short documentary on the causes of the Space Shuttle Colombia disaster and a question occured to me. Why are re-entry speeds so high as to require “indestructible” carbon tiles built to withstand 3000+ degrees? Wouldn’t it be far simpler and cost effective to slow the shuttle down using reverse thrusters before re-entry, and glide down at a more leisurely pace than the 17,000 m.p.h. quoted by the documentary.
I am assuming that there is a damn good reason for why they don’t do this but can’t for the life of me think what it could be. I guess the extra fuel spent using reverse thrusters would add cost but surely this would be more than offset by removing the necessity to cover the shuttle in heat-resistant tiles?
The shuttle simply can’t carry enough fuel to make a meaningful difference to reentry speed. In order to decelerate from orbital speed, it would take as much fuel as it took to accelerate the shuttle to that speed in the first place. Getting the additional weight of that fuel into orbit would require about ten times as much additional fuel (and tanks, engines, etc) at liftoff. It’s simply not practical to do. Although it might seem dangerous, friction with the atmosphere is the most weight-efficient way to slow down from orbital speed.
It takes all the fuel in the shuttle’s external fuel tank, and the throwaway boosters too, to get the vehicle up to orbital speed. It’d take that same amount to slow it back down. It’s much more efficient to hit the atmosphere at high speed, and let wind resistance supply the braking force.
I can’t see how it would take an equal amount of fuel to slow the shuttle down, surely you require much more fuel to achieve the initial speed as you are accelerating the shuttle against the pull of the Earth’s gravity, whereas in space there are no (or marginally detectable) forces to overcome so less fuel is required?
It would be possible to have the shuttle skip in and out of the atmosphere to gradually slow down before reentry, but that’s a complex time consuming manuver that could potentially put the crew at greater risk than simply plunging down.
You’re confusing mass with weight. The shuttle may not weigh anything in orbit, but it has the same mass as it does on Earth, and that mass requires just enough fuel to slow it down as it does to speed it up.
So it would take the same force to accelerate an object from standing up to 1,000 m.p.h. in space as it would taking off from the “surface” of Jupiter? Apologies for being dense, my A-level physics has been locked away in some dark dingy corner of my mind.
Not exactly, because in that case you’re dealing with a massive gravity field in comparison to the Earth’s. Earth’s own gravity field is rather modest so it doesn’t add as much to the equation as Jupiter’s does. In the case of Jupiter, it would take more fuel to get it off the ground than it would on Earth, but if you wanted it to reenter Jupiter’s atmosphere at a very slow rate, it would take just as much fuel to slow it down as it did to put it up there.
Look at it like this: In a shuttle launch, the shuttle is fighting gravity to get up into space, to return to Earth at a slow rate of speed, it has to fight just as much gravity going down as it did going up. That make sense?
mittu asks a good question. The shuttle has to be launched with enough energy to achieve escape velocity, which is like 25,000 mph here at the surface, but is about 17,000 mph up where it orbits. So to get it into orbit from standing still at the surface, you have to use (25,000 mph * shuttle’s mass) of energy.
To slow it down from 17,000 mph in orbit, to say 3,000 mph to make a less dramatic re-entry, you’d need (14,000 mph * shuttle’s mass) of energy. So the surface gravity does make a difference as he notes, but the overall reason remains the same - it would still take a helluva lot of fuel to slow it down.
I think I see where I may have been getting confused now, I was assuming the only fuel needed to slow the shuttle down was fuel expended in orbit to cause the initial slow-down of the shuttle, I forgot to account for the continuing force required to prevent acceleration of the shuttle as it drops through the atmosphere.
Escape velocity would be needed if a mission to, say, the moon were planned. As the Shuttle only does orbital missions, less energy is needed (bust still a lot).
You’ve got it. On the way up you are lifting the vehicle against the force of gravity. On the way down you would be preventing gravity from accelerating it downward which is the same thing as lifting it against the force of gravity.
Don’t forget how much the mass of the shuttle (the whole stack, really) changes as the fuel burns off. To slow the shuttle by using rocket thrust against the direction of travel would take about the same amount of fuel as the current shuttle uses to take off. But to put that mass in orbit would require much, much more fuel than what it uses now. It’s not just a matter of doubling the size of the tank and boosters.
(I remember a book in college that showed performance graphs of a Saturn V rocket; weight, speed, altitude and such as a function of time, starting at liftoff. The amount of mass that thing burned off at the beginning of the flight was amazing. Googling shows that the takeoff weight was 6.1 million pounds. IIRC, it was down to about 1.5 million by the the time the first stage was finished; about two-and-a-half minutes. And if thrust stays constant, 1/4 the mass will have 4 times the acceleration. The math was hairy and the numbers were impressive.)
The “Orbiter” freeware spaceflight simulator makes a good point of that. The various spacecraft simulations allow you to specify a perpetual fuel supply, i.e., the fuel tanks are always full. However it warns you not to do this with the Atlantis simulation, because without losing the mass of the spent fuel and oxidant, the shuttle would remain too heavy to achieve orbit.
If you try the simulator, you can see how minimally capable the Shuttle is at reaching orbit.
You’re right, I was wrong. I had always had the impression that the escape velocity at a particular altitude was the same as the speed required to have a circular orbit. As I was thinking how to explain this to you, I realized that it can’t be right. It turns out that the escape velocity is 1.414 times the circular orbit velocity.
What makes you think they just plunge in? The reentry is programmed to keep the temperature within the design limits of the heat protection system. The amount of energy to be dissipated is the same regardless of angle. However the temperature of the tiles depends upon the rate at which the energy is dissipated and the tile system has some maximum temperature that it can withstand.
I would suppose that they run the tile temperature as high as is safe so as not to take forever to reenter.