## Sunday, June 3, 2012

### Deceleration by Drag Devices (and More) on Mars

The “classic” landing scenario on Mars dates all the way back to Viking in 1976. After hypersonic entry is done, the vehicles “pops out” at around Mach 2.5, ready for ribbon chute deployment of one type or another. On Earth, that’s up around 40,000 feet (13 km); on Mars, it’s substantially closer to the surface. That’s because, even with the lower gravity, the thin “air” of Mars is far less effective as a hypersonic entry decelerator.

This hypersonic decelerator effect is more-or-less proportional to density, and to speed squared during entry. Entry speeds on Mars are less than those on Earth, due to the lower gravity. But, “air” densities are very, very far lower. The density effect “outweighs” the gravity effect by quite a lot.

Once the chutes are out, it becomes a simple drag-to-weight ratio question, all other things being equal. On Earth, densities are far higher, but so is weight. The surface density on Mars varies with temperature, but is around 0.6% of that here. These densities reduce essentially with atmospheric scale height, but the overall density argument still applies.

The surface gravity on Mars is 38% of that here. Drag/weight for a chute then scales as density/gravity, for 0.6%/38% = about 2%. So chutes are simply far less effective as decelerators on Mars, as measured in the same velocity range, from about Mach 2.5 to impact.

Drag/weight for a parachute system varies directly proportional to chute blockage area, drag coefficient, air density, and velocity squared, and inversely with load weight. Kinematically, we’d like the terminal velocity of a chute-and-load system on Mars to be similar to that here, around 20-30 mph (30-50 kph). That leaves only density and weight, which ratio at about 2% for mars. See Figure 1 below.

In other words, for the same kinematic effect, the payload retarded by a chute on Mars can be about 2% of what we would expect for the same size chute on Earth. (It can be more if you can accept a higher terminal velocity, except that on Mars, you cannot. You are already too close to the surface when the atmospheric entry hypersonics are over.)

For ribbon chutes, which work at moderate supersonic Mach numbers all the way down subsonic, the drag coefficient is in the range of 1 to 1.5 and is a function of Mach number, peaking near Mach 1.1-ish. It is not substantially different with ballutes, or any other inflated structures. To get the drag to support the load at a reasonable terminal velocity, the decelerator must be around 50 times larger on Mars than on Earth. This is one real whopper of a landing design problem!

The only way out of this dilemma that I see, is a combination of aero decelerator braking, and rocket retro thrust braking, done simultaneously. That is something we have not yet attempted on Mars.

But, it is something “we” (but not all of “us”) have attempted here on Earth: parachute extraction of battle tanks from aircraft, followed by parachute descent, and last-second rocket braking to a touchdown, parachute still attached. The Russians did this, somewhere around 1960. It does offer a way to make aero-decelerators more effective, at the expense of extra rocket fuel and hardware.

Here’s another thought: augment the hypersonic aero-braking with rocket retro thrust during atmospheric entry. Once that is over, deploy aerodynamic decelerators, but maintain rocket retro thrust. Increase rocket braking thrust sharply for the final last seconds to touchdown. The main problem is axial retro plume aerodynamic stability, which can induce control moments on the craft beyond its capability to cope. Nozzle cant angle might help resolve that, however. See Figure 2 below.

There are many questions and problems with this overall retro-thrust concept, but it actually does offer a possible way to land very large payloads on Mars, assisted by aero drag retardation. See Figure 3 (added as an update 6-10-12).

Figure 1 – How Aero-Decelerators Work

Figure 2 – Stability Problems with Hypersonic Retro Thrust Might Be solved by Cant Angle

Figure 3 - Combined Chute and Rocket Braking