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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:
- the spacecraft (Soyuz) engine,
the atmosphere:
a. descent module only,
b. descent module with parachutes deployed,
- 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
edited 14 mins ago
Nathan Tuggy
3,18242135
asked 12 hours ago
Alexander Klauer
1714
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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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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:
- the spacecraft (Soyuz) engine,
the atmosphere:
a. descent module only,
b. descent module with parachutes deployed,
- 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
edited 14 mins ago
Nathan Tuggy
3,18242135
asked 12 hours ago
Alexander Klauer
1714
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
add a comment |Â
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:
- the spacecraft (Soyuz) engine,
the atmosphere:
a. descent module only,
b. descent module with parachutes deployed,
- 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
edited 14 mins ago
Nathan Tuggy
3,18242135
asked 12 hours ago
Alexander Klauer
1714
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:
- the spacecraft (Soyuz) engine,
the atmosphere:
a. descent module only,
b. descent module with parachutes deployed,
- 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
iss reentry soyuz-spacecraft aerobraking velocity
edited 14 mins ago
Nathan Tuggy
3,18242135
asked 12 hours ago
Alexander Klauer
1714
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.
edited 14 mins ago
Nathan Tuggy
3,18242135
asked 12 hours ago
Alexander Klauer
1714
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.
edited 14 mins ago
Nathan Tuggy
3,18242135
edited 14 mins ago
Nathan Tuggy
3,18242135
edited 14 mins ago
Nathan Tuggy
3,18242135
3,18242135
asked 12 hours ago
Alexander Klauer
1714
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.
asked 12 hours ago
Alexander Klauer
1714
asked 12 hours ago
Alexander Klauer
1714
1714
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.
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.
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
add a comment |Â
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
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
25
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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.
answered 11 hours ago
Russell Borogove
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
 |Â
show 5 more comments
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.
answered 11 hours ago
Steve Linton
4,3311531
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 11 hours ago
Russell Borogove
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
 |Â
show 5 more comments
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.
answered 11 hours ago
Russell Borogove
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
 |Â
show 5 more comments
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.
answered 11 hours ago
Russell Borogove
72.7k2224309
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.
answered 11 hours ago
Russell Borogove
72.7k2224309
edited 11 hours ago
answered 11 hours ago
Russell Borogove
72.7k2224309
answered 11 hours ago
Russell Borogove
72.7k2224309
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
 |Â
show 5 more comments
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
 |Â
show 5 more comments
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.
answered 11 hours ago
Steve Linton
4,3311531
add a comment |Â
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.
answered 11 hours ago
Steve Linton
4,3311531
add a comment |Â
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.
answered 11 hours ago
Steve Linton
4,3311531
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.
answered 11 hours ago
Steve Linton
4,3311531
answered 11 hours ago
Steve Linton
4,3311531
answered 11 hours ago
Steve Linton
4,3311531
answered 11 hours ago
Steve Linton
4,3311531
4,3311531
add a comment |Â
add a comment |Â
Alexander Klauer is a new contributor. Be nice, and check out our Code of Conduct.
Alexander Klauer is a new contributor. Be nice, and check out our Code of Conduct.
Alexander Klauer is a new contributor. Be nice, and check out our Code of Conduct.
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Highly related: space.stackexchange.com/questions/12011/…
– Organic Marble
8 hours ago