Why are re-entry speeds so high?

Actually, if I’m understanding the notes and math I’ve seen correctly, a slower re-entry would actually mean more heat load, not less. The heat which the thermal shield system has to deal with is determined by the speed at which the ship is hitting the atmosphere and not the deceleration rate. Spending a long time decelerating means that you’re exposed to that heat for a longer period of time, and end up having to deal with more heating than a faster deceleration would. For minimal heat load you actually want to decelerate as quickly as you can.

The Apollo and Soyuz capsules have a much simpler heat shield system than the Shuttles, in part because they decelerate much more quickly. They subject the astronauts to much higher G forces during re-entry, and have a simple structure and very little manuevering capability. The Shuttle flies a slow, gentle, high-heat-load reentry path. The shuttle’s structure wouldn’t survive the high G loads of an apollo-style fast reentry. It was designed to return fragile cargo from orbit, which requires a more gentle return to the ground, and also has a much greater ability to manuever during reentry.

You know, this is why people get all excited about the idea of a Space Elevator – you can re-enter without having to use atmospheric friction to slow your descent. In fact, you can use the energy you gain by descending into batteries and get back the nergey you required to get up to orbital height (minus the friction and inefficiency costs)

I don’t think that is true…or possible, if I am understanding you correctly. The shuttle is in a stable orbit. To break orbit, it must decelerate with it’s thrusters. This will drop it into a lower and lower orbit until it eventually hits the atmosphere. When it does reach the outer atmosphere, it is still going at orbital speeds of 1000s of MPH at which point it uses friction to slow it and become a big glider.

I suppose if fuel capacity wasn’t a factor, it could essentially “hover” to a gyosynchronus position (basically moving at the same speed as the atmosphere now) in whatever orbitit is in and then gradually lower itself into the atmosphere. Is that what you are talking about?

Think of the atmosphere like diving into the water. The shuttle is going like 17,000mph. If it just plunges in at a steep angle, it’s basically like doing a belly flop from height. Hurts right? If it doesn’t come in at enough of an angle, it skips off like a rock and back out into space.

I’m not sure I understand your point. In theory, if the shuttle had unlimited fuel, it could use its thrusters to completely kill its orbital speed before it hit the atmosphere. It is true that as soon as it starts losing speed its orbit will drop to the point where it’s going to hit the atmosphere, so it would either have to lose speed quickly enough beforehand or aim some part of its thrust downward so as to maintain altitude while decelerating. In practice neither is possible, once in orbit the shuttle only has enough fuel to slow itself just enough to start hitting the atmosphere.

The shuttle isn’t in a zero gravity environment. At typical Shuttle altitude, it still feels more than 90% of the gravitational field we experience at the Earth’s surface. The reason the occupants are weightless is because they’re constantly free-falling. Luckily for them, their free-fall is around the Earth, and not towards it.

If you were stationary at the top of a “space elevator” 200 miles up (the elevator is attached to the Earth’s surface), you’d only lose about 10% of your weight.

There’s a lot of gravity up there. It’s just that the Shuttle has room to keep “falling” so no net force is felt.

Here’s the thing that people often don’t understand. The space shuttle is in orbit around the earth. That is, it is falling very very fast towards the earth. However, it just happens (because we carefully arranged for this) to have enough speed sideways from the earth that by the time it has fallen X distance towards the earth, it has traveled Y distance sideways, enough so that it is still X distance away from the spherical earth.

OK, so all you have to do is slow down a bit, and you’ll start gently falling or spiraling down towards the earth, right? No. If you just try to “slow down” you don’t break your orbit, you alter your orbit. Slowing down will change your almost circular orbit into an ellipitical orbit. All orbits are always conic sections…they can be elliptical, circular (special case of elliptical), parabolic, or hyperbolic. No spirals. No straight lines.

If you were orbiting earth and fired your rockets toward the earth, you aren’t going to fall towards the earth. You will actually be INCREASING your speed. You will continue to orbit the earth, just faster, and will come back to exactly the same height once you’ve orbited once. The only way for this to work is to make your orbit so elliptical that perigee is inside the surface of the earth…that is, you crash into the earth at orbital speed.

The other trouble is that rockets require fuel to work. But adding more fuel is a problem, since every pound of fuel your rocket carries requires fuel to move it. Doubling the amount of fuel doesn’t give you twice the ability to change your velocity, since most of that extra fuel is spent just moving around the extra fuel.

Luckily for orbits around earth we have a way to brake that doesn’t require rocket fuel…friction with the earth’s atmosphere. The space shuttle converts speed into heat, and the shuttle is able to match velocity with the surface of the earth. That is a lot of energy which results in a lot of heat, but there is no way to avoid this if you are going to use the atmosphere for braking. X amount of orbital energy will produce exactly the same amount of energy in the form of heat. Of course the shuttle is able to get rid of some of that heat…some by radiating it away (the tiles are red-hot), some by convection (the hot shuttle moves through the cold atmosphere, some of the heat gets transfered from the shuttle to the atmosphere).

You’re not looking at the whole equation. A slow reentry (really slow, that is) would involve dipping down into the atmosphere, then going back into space to radiate the heat build up off the craft, dipping back down again, and then back into space, until you’ve killed most of your forward velocity.

Make that Columbia.

This only works until you’ve lost too much velocity to get out of the atmosphere. The apollo capsules did something like this returning from the moon, but they were coming in way faster than orbital velocity to being with. They could hit the atmosphere, lose a lot of speed, and still be going fast enough to climb back out to cool off before coming down again. Something reentering from low orbit won’t have this option - once you start losing speed, you won’t be able to climb back up. (At least not for long enough to radiate any significant amount of heat away.)

Only if you’re passively doing this. If you use thrusters and control surfaces, you can still bounce yourself in and out of the atmosphere for quite some time. Remember, the only reason space craft get up to 17,000+ MPH, is so that they can stay in orbit. So you could burn off quite a bit of speed and still be able to reach space later on. You just couldn’t stay there.

If you’re using thrusters to actively lift yourself out of the atmosphere, you’re just increasing the speed/altitude energy you need to burn off to land safely. Somewhat counterproductive. And using aerodynamic lift to lift youself out of the atmosphere is going to be a tricky thing - you’ll reach a point where there’s not enough air to provide lift, yet still enough to generate a lot of frictional heating. I suppose you could try to so some kind of repeated zoom climb, climbing up and then falling back, but it seems to me like this is doing things the very hard way. You’ll need a lot of lift to pull it off, and you’ll still eventually reach a point where you won’t have the energy to get high enough to coll off, yet are still going fast enough for heating to be an issue.

We are dealing with two types of energy here. There is the “potential energy” of an item lifted away from the earth’s surface, and the “kinetic energy” of an object in motion. The potential energy is proportional to the height the item is raised (a hundred miles or so in this case, I believe) and the kinetic energy is proportional to the square of the velocity.

Left alone a falling object will convert its potential energy into kinetic energy (i.e. ‘dropped’ objects accelerate).

So, in reentry some force must be applied either via thrust or friction to remove both the initial kinetic energy (slowing down from orbital speed to landing speed) and also any additional kinetic energy arising from the decrease in potential energy (i.e. reduction in height)

I haven’t done the numbers but I would guess that the kinetic energy involved in orbit is far greater than the potential energy, due to the effect of velocity squared (in KE) vs just a linear change in height (in PE)

Has anyone done the numbers? I can dig out my old Halliday and Resnick if need be.

At least that’s how I remember it from HS Physics.

The orbital velocity can be computed from Kepler’s law that the time in seconds for one orbit and the distance in meters of the satellite from the center of the earth are related by:

d[sup]3[/sup]/t[sup]2[/sup] = 1.006*10[sup]13[/sup] m[sup]3[/sup]/sec[sup]2[/sup]

The potential energy is a little more complicated. I’m sure that someone has a formula for it, but I don’t know it. However the computation is fairly straightforward and kind of fun.

The potential energy in this case is the work done in raising the object against gravity. The work is the product of the force applied multiplied by the distance moved. The force can be computed using Newton’s law F = m*a. The mass is the mass of the object, say one kilogram, and the acceleration is the acceleration of gravity at a distance, d, from the center of the earth.

The acceleration of gravity falls off as the square of the distance so it will be:

a = 9.8*r[sup]2[/sup]/(r + y)[sup]2[/sup]. r = radius of the earth in meters, y = distance in meters above the surface, 9.8 is the surface acceleration of gravity in m/sec[sup]2[/sup]

Let’s use a small increment of distance, dy, to compute an increment of work, dW, by multiplying the force m*a by the increment of distance dy.

Equation 1) dW = m9.8r[sup]2[/sup]*dy/(r + y)[sup]2[/sup]

We could compute this by letting dy = 1 meter and compute increments of work from the surface up to, say, 250 km and then sum up all these increments to get the total.

It is much easier to just let dy approach zero. In that case the sum of all those increments is the integral of Equation 1) for y going from 0 to 250000 meters.

That turns out to be 2.27*10[sup]6[/sup] joules for the potential energy of 1 kg raised to 250 km.

I get 5.8*10[sup]7[/sup] joules for the kinetic energy.

So yes, the major part of the energy at that height is kinetic.

The original question asked why it took as much fuel to get out of orbit as to get into it and part of the questioners reasoning was that gravity was working for you coming down by against you going up. In order to avoid burn up gravity is working against you coming down also.

By the way. The constant 1.00610[sup]13[/sup] is valid for the earth, or any other primary body with a mass of 5.95410[sup]24[/sup] kg.

Just for kicks I fiddled around with the energy bit some more. Here is aplot of the kinetic, potential and total orbital energy for a 1 kg mass in orbits from 30 to 10000 km. An orbit of 30 km is pretty low for earth so just pretend the primary is earth sized with no atmosphere.

The kinetic exceeds the potential energy for orbits lower than about 3200 km (roughly 2000 miles) above the surface, and above that the potential energy is greater.

There is an arithmetic error in the kinetic energy that I computed in a previous post.

Cool curve! Unless I’m mistaken (as often happens) the shuttle’s orbital height is only about 100 miles, so the kinetic energy would greatly outwiegh the potential energy, as originally guessed.

Space probes have used aerobraking to acheive/modify orbit as a fuel saving measure. IIRC the Venus probe Magellian was one of the 1st to use this to modify the orbit though it was never designed for this (and Venus has a kickass atmosphere compaired to earth). Also I think all the recent Mars missions use this, so the trend seems to be expanding, It’s a free way to change velocity, and that’s very hard to come by in space.

Also another way of looking at it, is to try to think of the difficulity of designing a underwater ‘craft’ that is dropped from the air into the water and try to prevent the huge splash and the forces involved, When you figure the fuel requirement and the design change, it would seem better to just design it to survive the splash.

One of the earliest test pilots of the shuttle was asked about its aerodynamic qualities. He replied “Better than a bathtub.”

Skipping the Shuttle in and out of atmosphere would have the additional characteristic of repeating the most dangerous maneuver of the entire mission half a dozen times. Pilots hate stuff like that. Mission planners too.

Tris

Well… you can do that with a parachute.

Automobile brakes are a better analogy, because that’s what the Shuttle’s thermal tiles are: brake pads. Automobile brakes heat up during hard braking, and you could try to solve this “problem” by designing some sort of electromagnetic or aerodynamic brake. Or you can tell the driver to slow down gradually. But neither option is very practical. It’s better to just design brake pads that can withstand the higher temperatures, and shed heat more efficiently. (The Shuttle tiles are designed to radiate away most of the heat instead of transferring the heat to the interior structure.)

Would a nuclear-powered shuttle resolve all these concerns?