# Why can't we deorbit to earth more gently? [duplicate]

Atmospheric re-entry from orbit seems such a violent affair with tons of speed and energy having to be shed, which necessitates heat shields and what not.

I do understand that the velocity of the craft has to slow down to de-orbit but does it have to be done in such a rapid fashion?

Is it possible for a slower deorbit that doesn't require a heat shield? What would this take? Larger wings on the orbital craft? Big engines? Large parachutes?orbit

• Sep 12, 2016 at 1:40
• All that energy has to go somewhere. At the upper edge of the atmosphere, there is no air to carry it away; at best it can only be radiated. To radiate at a high enough rate probably requires the vehicle surface to get really hot. Also, temperature equates to particle velocity, so the vehicle's re-entry velocity equates to a temperature - a high one. So any surface in contact with the atmosphere is going to get hot regardless. The solutions we already know about seem to be optimal. Sep 12, 2016 at 4:17
• This is not a duplicate of the 'Moon landings vs Earth landings' question because that deals entire with the physics of the moon landing.
– user10509
Sep 12, 2016 at 7:44

When de-orbiting, all the horizontal velocity needs to be transferred into another form of energy. In a normal reentry it is converted to heat. If a much larger aerodynamic surface is used, the heat is dispersed over a larger area. There have been suggestions to use balloons (called a Ballute) to increase surface area for reentry. In a very shallow angle or skip reentry, the heat is dispersed over longer time. A space plane can do this. Retro rocket landings on earth could avoid reentry heat, but it would probably require atomic rockets to have enough energy.

If you are not going to use atmospheric re-entry the only other option is to use engines to slow down, and that would be incredibly expensive.

Imagine if you wanted to land on a body with 1G but no atmosphere, from LEO. Only a rocket will get you down safely in this case. In order to land from LEO you would need a rocket as big as the one that put you up there in the first place. To put a rocket that size up there would require a significant number of launches. It would be fabulously expensive to do it that way when the atmosphere lets us do it practically for free.

Anyway, it's not actually a problem in the first place - atmospheric re-entry is not particularly violent compared to launch, both are between 3-5G for most human rated systems. It's not a problem that needs solving.

Larger parachutes would not help. At the speed from the orbit, large (and also small!) parachutes would be destroyed at the upper edge of the atmosphere. Parachutes may be used only if the craft is slowed down by the heat shield to subsonic speed, about 0.2 km/s. But the speed in the orbit is much higher, about 7.8 km/s, therefore more than 97 % of the initial speed must be decelerated without parachutes.

• A ballute is a combination of a parachute and balloon. It is theorized that it can be used at high velocities in the thin upper atmosphere for atmospheric braking. Here is a paper about using one for lunar return which is 40% faster than re-entry from earth orbit. ssdl.gatech.edu/papers/mastersProjects/ClarkI-8900.pdf Sep 13, 2016 at 0:05
• @Johnny Robinson: But in this paper the ballute is used only for deceleration from 11,2 km/s down to 7.8 km/s.
– Uwe
Sep 13, 2016 at 10:53