Citing the Wikipedia article:

The orbit's perigee and apogee had been selected to cause reentry due to orbital decay within 10 days (the limit of the life support system function) in the event of retrorocket malfunction. However, the actual orbit differed from the planned and would not have allowed descent until 20 days.

The obvious case where Gagarin would be stuck in orbit is the retro thruster not firing. But from miscellaneous sources ;-) I know one needs very little delta-V to perform a retrograde burn at the apogee that drops perigee enough for aerodynamic forces to grow enough to deorbit the craft. Assuming inability to start the main retro engine, could the nitrogen thrusters of Vostok-1 have provided enough delta-V to reduce orbital decay below the critical 10 days?

  • $\begingroup$ I downloaded several papers to try to write an answer to this question but haven't got to it yet. While there may be a "standard atmosphere model" of density vs altitude that goes up high enough, I think the answer here will still need to contain a "it depends a lot on what the sun was doing that week" caveat. $\endgroup$
    – uhoh
    Commented Jan 16, 2017 at 15:42
  • $\begingroup$ In light of this answer, I guess not! $\endgroup$
    – uhoh
    Commented Jan 16, 2017 at 16:18

1 Answer 1


Not likely. For early manned missions, rendezvous wasn't a requirement, so thrusters provided only for rotation, not for translation. According to this reference:

In each of the two [redundant] sections of 8 [thrusters] there were 2 thrusters each for pitch and yaw and 4 for roll.

Depending on how the thrusters were offset from the craft's center of gravity, it's possible that some alternating pattern of thruster use would give some translational force, but that would be terribly inefficient.

Even if the thrusters were arranged for translation, they wouldn't have been able to de-orbit the craft. There was a total of 20kg of gas carried out of a spacecraft mass of 4.7 tons; nitrogen gas specific impulse is around 73 seconds. According to the rocket equation this yields only about 3 m/s worth of delta-V. At a guess that might cut a few hours off the natural orbital decay time.

It does seem a little strange to provide 10 days of consumables for what was supposed to be a two hour flight, rather than spending the mass on some sort of backup retrofire solution. The follow-on Voskhod design did have a backup, solid-fuel retrorocket.

  • $\begingroup$ How much would perigee of 169km drop with apogee of 327km retrograde burn of 3m/s? $\endgroup$
    – SF.
    Commented Jan 16, 2017 at 16:03
  • 2
    $\begingroup$ As a rule of thumb, 2fps delta-v in LEO changes altitude by about a mile, so 3 m/s would amount to some 8km decrease in perigee. $\endgroup$ Commented Jan 16, 2017 at 16:09
  • $\begingroup$ It's not nearly so bad. "Apogee 327 km, perigee 181."src - that was only declassified in 1996. Wikipedia lists probably the planned/official ones, 327/169. That means only 12km off. 8km is still short of it, but I believe even venting the fuel into space without combustion could buy the missing 1.5m/s. $\endgroup$
    – SF.
    Commented Jan 16, 2017 at 18:12
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – SF.
    Commented Jan 16, 2017 at 19:33
  • 1
    $\begingroup$ So, no risk that Yuri Gagarin would defect to the West by deorbiting early. $\endgroup$
    – LocalFluff
    Commented Jan 19, 2017 at 23:15

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