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

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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?










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  • Highly related: space.stackexchange.com/questions/12011/…
    – Organic Marble
    8 hours ago














up vote
14
down vote

favorite












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?










share|improve this question









New contributor




Alexander Klauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • Highly related: space.stackexchange.com/questions/12011/…
    – Organic Marble
    8 hours ago












up vote
14
down vote

favorite









up vote
14
down vote

favorite











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?










share|improve this question









New contributor




Alexander Klauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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?







iss reentry soyuz-spacecraft aerobraking velocity






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edited 15 mins ago









Nathan Tuggy

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Alexander Klauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.











  • Highly related: space.stackexchange.com/questions/12011/…
    – Organic Marble
    8 hours ago
















  • Highly related: space.stackexchange.com/questions/12011/…
    – Organic Marble
    8 hours ago















Highly related: space.stackexchange.com/questions/12011/…
– Organic Marble
8 hours ago




Highly related: space.stackexchange.com/questions/12011/…
– Organic Marble
8 hours ago










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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 in the air; atmospheric deceleration of the airframe does almost all of the work, because it's "free" apart from the heat shielding.






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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
    10 hours ago






  • 2




    @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
    – MindS1
    9 hours ago






  • 8




    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
    9 hours ago







  • 1




    @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
    6 hours ago






  • 1




    @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
    – ArtOfCode
    3 hours ago

















up vote
14
down vote













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. 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.






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    2 Answers
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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 in the air; atmospheric deceleration of the airframe does almost all of the work, because it's "free" apart from the heat shielding.






    share|improve this answer






















    • 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
      10 hours ago






    • 2




      @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
      – MindS1
      9 hours ago






    • 8




      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
      9 hours ago







    • 1




      @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
      6 hours ago






    • 1




      @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
      – ArtOfCode
      3 hours ago














    up vote
    25
    down vote













    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 in the air; atmospheric deceleration of the airframe does almost all of the work, because it's "free" apart from the heat shielding.






    share|improve this answer






















    • 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
      10 hours ago






    • 2




      @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
      – MindS1
      9 hours ago






    • 8




      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
      9 hours ago







    • 1




      @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
      6 hours ago






    • 1




      @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
      – ArtOfCode
      3 hours ago












    up vote
    25
    down vote










    up vote
    25
    down vote









    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 in the air; atmospheric deceleration of the airframe does almost all of the work, because it's "free" apart from the heat shielding.






    share|improve this answer














    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 in the air; atmospheric deceleration of the airframe does almost all of the work, because it's "free" apart from the heat shielding.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 11 hours ago

























    answered 11 hours ago









    Russell Borogove

    72.7k2224309




    72.7k2224309











    • 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
      10 hours ago






    • 2




      @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
      – MindS1
      9 hours ago






    • 8




      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
      9 hours ago







    • 1




      @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
      6 hours ago






    • 1




      @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
      – ArtOfCode
      3 hours ago
















    • 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
      10 hours ago






    • 2




      @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
      – MindS1
      9 hours ago






    • 8




      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
      9 hours ago







    • 1




      @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
      6 hours ago






    • 1




      @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
      – ArtOfCode
      3 hours ago















    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
    10 hours ago




    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
    10 hours ago




    2




    2




    @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
    – MindS1
    9 hours ago




    @geoffc Should be noted that due to drastically reduced gravity the orbital velocity around Mars and the Moon is significantly lower.
    – MindS1
    9 hours ago




    8




    8




    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
    9 hours ago





    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
    9 hours ago





    1




    1




    @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
    6 hours ago




    @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
    6 hours ago




    1




    1




    @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
    – ArtOfCode
    3 hours ago




    @zwol Kinetic energy is 0.5mv^2, so 88.9 GJ.
    – ArtOfCode
    3 hours ago










    up vote
    14
    down vote













    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. 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.






    share|improve this answer
























      up vote
      14
      down vote













      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. 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.






      share|improve this answer






















        up vote
        14
        down vote










        up vote
        14
        down vote









        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. 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.






        share|improve this answer












        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. 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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 11 hours ago









        Steve Linton

        4,3311531




        4,3311531




















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