# Powered atmospheric re-entry

1. Apr 11, 2010

### awygle

This is something I've wondered for some time. Our spaceships need all this ablative heat shielding because they slam into the atmosphere at ludicrous speeds, thereby generating a ton of friction and heat. Wouldn't it make more sense to slow down first?

There is an orbit where the speed of the spaceship keeps it moving at the rotational velocity of the Earth times its height above sea level - geosynchronous orbit. But orbital speed is determined by centripetal force. So could a spacecraft use its engines to exert a force so that it was going in an orbit at the height at which atmospheric drag becomes a problem moving at $$\varpi_{E}*(R_{E} + h_{d})$$, where $$h_{d}$$ is the height at which atmospheric drag becomes noticeable? In this situation, the difference in velocity between the atmosphere and the ship would be very low, and the usual massive amounts of friction and heat would not be generated.

Is this plausible? If so, why don't spacecraft use this method?

Last edited: Apr 11, 2010
2. Apr 11, 2010

### Theheretic

Don't things in geosynchronous orbit fly at somewhere around 15-20,000mph which perhaps would be too prohibitive to slow such a speed down, i.e. perhaps would require too much thrust for too long of a period of time where they do not have enough fuel for such a thing...just an idea..I actually don't know for certain.

3. Apr 11, 2010

### willem2

This method is not used because it would cost an enormous amount of fuel. They need to lose about 7 km/s of speed. You'd have to start with a 2 million kg ship in orbit to get 70000 kg back on the ground. (the spaceshuttle uses 2 600,000 kg solid rocket boosters and a tank with more than 700,000 kg of fuel to get a 70,000 kg orbiter into space)
While the speed is slower in geosynchronous orbit, a spaceship has more energy. Even if you decelerate to get to an orbit that just strikes the atmosphere, the ship will still go faster than a low altitude orbit)

4. Apr 12, 2010

### Thecla

Why not have parachutes deployed before re-entry. In space they will hang limp, as the capsule enters the atmosphere , they will fill-up and slow the capsule down.

5. Apr 12, 2010

### PaulS1950

At orbital speeds any drag chute would burn up before it could slow down enough to make a difference. (specs of space dust burn up in the atmosphere before they can slow down) It would take as much fuel for a non-orbital re-entry as it does to get the ship into orbit. Actually to get that kind of controlled re-entry it would take considerably more fuel to get it into orbit in the first place. It is a lot cheaper to replace heat shielding than it is to attempt to use fuel to control the re-entry speeds.

6. Apr 12, 2010

### Lunar_Lander

Exactly the same thing (Parachutes deployed in Space) was explained by Auguste Piccard in Between Earth and Sky. He went on to propose a glider shaped vehicle that would barely touch the outer atmosphere very often, doing a kind of aerobreaking. As the book was written in 1946, I can imagine that he did not think of heat-shields, thus the very slight aerobreaking he proposed.

7. Apr 16, 2010

### awygle

I'm trying to work out the mathematics of this, just for fun/practice.
What I thought I'd do was calculate what the accelerations/velocities/jerks would have to be to a) slow the ship down and b) keep it moving in the same orbit. For that to happen it would have to be the case that
$$\frac{(v_{tan}(t))^{2}}{r}=a_{c}(t) \forall t \in [0,T]$$
where T is the total time the deceleration takes, right? And
$$v_{tan}(T)=\varpi_{E}*r$$ ?
From there it seemed like there were any number of arbitrary functions that could make this work, but that choosing a function for velocity would fix the acceleration function, and vice versa. So to make it (hopefully) easy, I chose $$v_{tan}(t)=v_{0}+a_{x}t$$ for some constant acceleration $$a_{x}$$.
This, obviously, gives the following function for $$a_{c}(t)$$:
$$a_{c}(t)=\frac{(v_{0}+a_{x}t)^2}{r}$$

Am I doing OK so far? I thought next I might try to calculate the delta v function, but I wouldn't want to waste time doing that if there's something wrong with the above...

Last edited: Apr 16, 2010