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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?

Nathan Tuggy
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Alexander Klauer
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  • Highly related: https://space.stackexchange.com/questions/12011/how-could-a-90-m-s-delta-v-be-enough-to-commit-the-space-shuttle-to-landing/12014#12014 – Organic Marble Oct 11 '18 at 19:12

2 Answers2

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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.

gibson
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Russell Borogove
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  • Which is why landing on Mars is harder, much thinner atmosphere, and on the moon is even harder. (Moon has to be all propulsive). – geoffc Oct 11 '18 at 16:35
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    @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower. – MindS1 Oct 11 '18 at 18:11
  • @MindS1 Agreed. NOte however Russel's point. Of the 7700 m/s only 360 m/s is cancelled by something OTHER than atmospheric drag. So absence of atmosphere is really painful to mission planning. – geoffc Oct 11 '18 at 18:14
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    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
  • Should be noted that anything which we intend to de-orbit onto the Moon or Mars is not going to be at orbital velocity for those bodies when it's time to start decelerating it. It'll be way above that - minimum value would seem to be no less than the value of Earth's escape velocity. – Beanluc Oct 11 '18 at 20:41
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    @Beanluc, minimum value is considerably lower than that. To a first approximation, the minimum velocity for a non-orbital landing on the Moon is roughly the Moon's escape velocity, while that for Mars is too complicated for me to do in my head. (Orbital mechanics is tricky, and frequently counter-intuitive.) – Mark Oct 11 '18 at 21:16
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    @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ. – ArtOfCode Oct 12 '18 at 00:00
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    @zwol, yes, though only a small fraction of that is absorbed by the capsule - the rest ends up in the atmosphere. – Aethernaught Oct 12 '18 at 00:16
  • @ArtOfCode Doh! – zwol Oct 12 '18 at 00:17
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    @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 02:05
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    Useless fact: The final 3 m/s impact is equivalent to falling from a height of 46 cm. – JollyJoker Oct 12 '18 at 12:05
  • Here's what Earth escape velocity has to do with landing something on another body: The item has to leave Earth first. That kinetic energy doesn't magically disappear. – Beanluc Oct 12 '18 at 17:12
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    @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
  • Sigh. It's MOMENTUM. When the thing arrives at Mars, it's going to be moving way faster than Mars's escape velocity. Period. That's my point. Potential energy in Earth's gravity well, potential energy in Mars's gravity well, same difference. De-orbiting gently means bleeding off >11m/s, we aren't magically at the 5 m/s which is Mars's escape velocity when we arrive there and start the procedure. – Beanluc Oct 12 '18 at 17:21
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    @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
  • I didn't say minimum LANDING velocity. At all. I was responding to MindS who seemed to be saying that minimum orbital velocity at the remote body was all that needed to be accounted for when determining how much energy would need to be dissipated to achieve a soft landing. To quote: "Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower." My point is that you don't get that lower orbital velocity for free, orbital velocity starts at inter-body velocity before you start reducing it. – Beanluc Oct 12 '18 at 17:38
  • Read again because I was very very clear about what velocity your traveling body would be at at the time you start decelerating it. In what way NONE does that imply that we're going to land without decelerating from that very high velocity. – Beanluc Oct 12 '18 at 17:40
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    @Beanluc: The "inter-body velocity", as you call it, can be arbitrarily low. This is what jamesqf was referring to as "however fast you were going at the midpoint". The only drawback to making it very low is that it increases mission duration, which has other limits. So we agree on that point. But it still has nothing to do with Earth escape velocity, which is what you keep asserting. And MindS was not wrong -- when landing from an orbit around the body in question, the inter-body velocity is essentially identical to the orbital velocity. – Dave Tweed Oct 12 '18 at 17:46
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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.

Steve Linton
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