10
$\begingroup$

I am seeing a lot of references to the “ballistic reentry” mode of the crew return vehicle in relation to the abort during the powered ascent stage. In partiulat, the recent crewed Soyuz MS-10 abort is said by Scott Manley in his video to have experienced a ballistic reentry. A table of space accidents on Wikipedia also suggest that a mission abort is correlated with the ballistic reentry: Soyuz 33 and the above mentioned Soyuz MS-10 are listed as having had a ballistic reentry due to a pre-orbital mission abort. 2 is not a large sample, but I suspect there is a relation.

I have only a superficial understanding of what exactly is this mode of anomalous descent. A section in the Wikipedia article on the reentry process explains that a normal reentry mode uses some aerodynamic lift to reduce acceleration on descent (lift = upward component of the aerodynamic force = acceleration less that the $1 g$ of a free fall). What I do not understand is how does that translate to an increased maximum deceleration in excess of $10 g$ experienced by the vehicle in this “ballistic” mode. At a later stage of descent perhaps?.

My probably superficial understanding that a ballistic trajectory is the one that is closer to one a body would follow absent the atmosphere, or at the least experiencing the least possible drag (as it is not possible to experience a truly ballistic descent, i. e. under the force of gravity alone, except in a vacuum. Is this related to the craft initial attitude only, or are there other factors at play, such as low initial descent altitude and very low orbital velocity?

I am a physicist, albeit an academic deserter, so I can not only easily take a load of math, but would rather even prefer it!


Update: S. Manley posted a video on 2018-10-23 with a sensible explanation and Kerbal simulations of a ballistic and several modes of aerodynamic reentry, concluding with one reminiscent of the MS-10 booster early cut-off and abort. While Kerbal simulations must not be regarded as in any way precise, they still exhibit notable differences in descent duration and peak deceleration.

This update was made long since the accepted answer by @GdD has been posted, and as such does not reflect the context in which that answer was provided.

$\endgroup$
  • 1
    $\begingroup$ Reentry is the process of transforming the high orbital speed into heat. Entering the atmosphere at hypersonic speed causes a lot of drag and this force causes a high decceleration. A pure ballisitic mode is possible only on a planet without an atmosphere. $\endgroup$ – Uwe Oct 16 '18 at 8:22
  • 1
    $\begingroup$ Ballistic here just means that the descent is uncontrolled, i.e. the capsule behaves like a dumb projectile. It doesn't have anything to do with not considering air resistance (artillery wouldn't be very accurate if you ignore that). In general you want to stay in the upper regions of the atmosphere longer, because the lower density leads to more gentle deceleration. $\endgroup$ – Almoturg Oct 16 '18 at 10:50
  • $\begingroup$ @Almoturg, you are right, I must have been thinking about ballistics in a too abstract way. But artillery is where the word originated, after all! $\endgroup$ – kkm Oct 17 '18 at 16:50
  • $\begingroup$ In general, ballistic reentry is Soyuz's "safe mode" - the glide requires specific initial conditions and active stabilization. Ballistic reentry is a passive mode - the capsule stabilizes itself due to its shape, not active controls. It's harder on the crew, but there's practically nothing that could fail - generally any non-catastrophic instability in any reentry will reduce it to ballistic. $\endgroup$ – SF. Oct 25 '18 at 2:30
11
$\begingroup$

When you're in orbit you have velocity roughly parallel to the atmosphere, a flat-ish shape angled properly can "fly" across it in a lifting entry where you can dissipate energy in the thinner upper atmosphere before you hit thicker atmosphere.

The first stage of a rocket uses most of its energy to gain altitude, getting above the thickest part of the atmosphere where the upper stages won't have to fight air friction to get the payload to orbital velocity. When the Soyuz aborted the second stage hadn't fired yet, most of its velocity was upwards, so it traveled an arc, coming down at high speed with a mostly downward component. It was going up like a cannonball and then came down like a cannonball, essentially, with limited opportunity to bleed off speed before hitting thicker air.

$\endgroup$
  • $\begingroup$ Thank you, I just want to note how amazing it's to think about the effect. A quick back of an envelope calculation (for an orbit height $\ll$ Earth radius, i. e. constant $g$) shows you have ≈15 times more kinetic than the potential energy on a 200 km orbit (unless I'm dead wrong with my math, but I cannot see where). This gliding maneuver helps dissipate not only you potential energy, but also 15 times more, in a more gradual fashion than if you fall straight down. Wow. Just wow! $\endgroup$ – kkm Oct 17 '18 at 16:43
  • $\begingroup$ GdD, I just posted a footer update to the Q with a reference to a new video from S. Manley with qualitative (Kerbal) simulations of different modes of reentry. While my understanding is that it aligns closely with your answer, I believe I must alert you since you wrote your A to my original Q w/o this update, so that the integrity of the context of your answer is not compromised in any way. I'll remove the update if you find it inappropriate; no reason needs to be given--just let me know please! $\endgroup$ – kkm Oct 24 '18 at 17:32
1
$\begingroup$

When reentering aerodynamically the time to decelerate is increased. This will mean that the highest g-force experienced will be reduced. When the reentry is totally ballistic, it is less controlled, and will usually in fact be faster, which means that the peak acceleration is increased.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.