# What eliminates the velocity when occupants return from ISS to earth, and how much?

The ISS has an orbital velocity of ~28000 km/h; the velocity $$v$$ relative to the landing site of the descent module is probably even higher than that most of the time. Once the occupants have landed, their velocity relative to the landing site is zero.

My first question is: what is it that eliminates the velocity between detaching from the ISS and arrival? Three things come to mind:

1. the spacecraft (Soyuz) engine,
2. the atmosphere:

a. descent module only,

b. descent module with parachutes deployed,

3. the earth itself (final impact).

Anything else?

My second question is: how much does each of these modes contribute (measured in $$\Delta v/v$$ or $$(\Delta v/v)^2$$)? For the sake of the occupants' prolonged joy in space travel, I figure 3. has the smallest impact (pun intended), but how do the others relate to each other?

Nearly all the velocity is cancelled by atmospheric deceleration of the descent module, before its parachutes are deployed.

ISS orbital velocity is around 7700 m/s. An initial retro-burn of the Soyuz engines, of something like 115 m/s magnitude, is sufficient to lower the perigee of orbit into the uppermost part of the atmosphere. The orbital module and service module are then separated from the descent module. Once the descent module starts to enter the atmosphere, air resistance slows it, which further lowers the orbit, bringing the capsule into denser atmosphere, which slows it further, and so on.

The Soyuz parachutes deploy starting at ~240 m/s (first drogue chutes to bring the capsule down to ~90 m/s then the mains to reach a 6 m/s descent rate). Just before touchdown, small solid rockets are fired for the final deceleration, producing another 3 m/s of ∆v.

Thus, of the 7700m/s initial velocity, only about 360 m/s is cancelled via parachutes, reentry burn, and final retrorockets; 7340m/s (95%) of the deceleration is done by the descent module moving through the atmosphere.

(I shamelessly stole correct figures from Steve Linton's answer.)

This breakdown applies generally for all crewed spacecraft, though American capsules didn't have the final braking rockets, and the space shuttle touched down at ~100 m/s horizontal velocity, without deploying parachutes while airborne; atmospheric deceleration of the airframe does almost all of the work, because it's "free" apart from the heat shielding.

• This is also the fundamental reason why reentry heating is so much of a problem. The Soyuz reentry module is ~3000 kg unloaded, so aerobraking is converting more than 3000kg*(7700m/s)^2 = 177 gigajoules of kinetic energy into heat. (Wolfram Alpha helpfully informs me this is roughly as much energy as would be released by burning 5100 liters of jet fuel.) – zwol Oct 11 '18 at 18:22
• @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ. – ArtOfCode Oct 12 '18 at 0:00
• @Beanluc: Earth escape velocity really has little to do with landing on the Moon or Mars. For the Moon, you wind up going pretty slow at the gravitational midpoint (I mean where the gravity of Earth and Moon are equal, which of course is much closer to the Moon), then accelerate under the Moon's gravity. So final speed is ~= lunar escape velocity + however fast you were going at the midpoint. Likewise for Mars: you spend most of the trip at whatever the Hohmann transfer orbit's velocity is, then gain Mars' escape velocity on the approach. – jamesqf Oct 12 '18 at 2:05
• @Beanluc: Not magically, but it does get converted to potential energy while climbing out of the gravity well. – Dave Tweed Oct 12 '18 at 17:17
• @Beanluc: Sorry, your intuition is simply wrong on this. Reread jamesqf's most recent comment. The Apollo LEM had no trouble landing on the moon, but it never could have landed on Earth, even if it had no atmosphere. You keep saying the minimum landing velocity is the Earth's escape velocity, but Earth's escape velocity has literally NOTHING to do with it. How would you explain our ability to rendezvous with smaller interplanetary bodies like comets and asteroids, or even the ISS itself? The same rules apply everywhere. – Dave Tweed Oct 12 '18 at 17:33

The process is described here, which answers nearly all of your question. The reentry burn removes about 120 m/s of velocity from the capsule (that's your 1) and the final impact is 15 miles per hour (about 6 m/s). That's your 3. That leaves about 7.5 km/s for part 2. The only remaining question is the split between 2a and 2b, ie the velocity when the parachute opens. This source gives that. Firstly 2b is given as 240 m/s, leaving 7.25 km/s for 2a. Finally, a set of small rockets fire just before landing and reduce the 6 m/s to 3 m/s.