How would gravity/acceleration be perceived by a human orbiting Earth at sea level
Clash Royale CLAN TAG#URR8PPP
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I understand the impracticalities of this concept but humor the 'what-ifs.'
Ignoring physical obstacles and the effects of atmospheric fluctuations affecting trajectory.
Say it is possible to have a craft capable of orbiting in Earth's atmosphere just above sea level, that in no way generates lift (just powering through that atmosphere).
How would gravity be perceived by the passenger onboard? On one hand I think theyd be weightless since they are technically always falling... But I could be wrong.
Bonus: how fast would a 200kg spherical (I guess) vessel be traveling?
physics
edited 2 hours ago
Lamar Latrell
1032
asked 8 hours ago
anon
3876
add a comment |
up vote
3
down vote
favorite
I understand the impracticalities of this concept but humor the 'what-ifs.'
Ignoring physical obstacles and the effects of atmospheric fluctuations affecting trajectory.
Say it is possible to have a craft capable of orbiting in Earth's atmosphere just above sea level, that in no way generates lift (just powering through that atmosphere).
How would gravity be perceived by the passenger onboard? On one hand I think theyd be weightless since they are technically always falling... But I could be wrong.
Bonus: how fast would a 200kg spherical (I guess) vessel be traveling?
physics
edited 2 hours ago
Lamar Latrell
1032
asked 8 hours ago
anon
3876
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
7 hours ago
2
Fine then: equatorial sea level with no moon
– anon
7 hours ago
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
7 hours ago
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
5 hours ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I understand the impracticalities of this concept but humor the 'what-ifs.'
Ignoring physical obstacles and the effects of atmospheric fluctuations affecting trajectory.
Say it is possible to have a craft capable of orbiting in Earth's atmosphere just above sea level, that in no way generates lift (just powering through that atmosphere).
How would gravity be perceived by the passenger onboard? On one hand I think theyd be weightless since they are technically always falling... But I could be wrong.
Bonus: how fast would a 200kg spherical (I guess) vessel be traveling?
physics
edited 2 hours ago
Lamar Latrell
1032
asked 8 hours ago
anon
3876
I understand the impracticalities of this concept but humor the 'what-ifs.'
Ignoring physical obstacles and the effects of atmospheric fluctuations affecting trajectory.
Say it is possible to have a craft capable of orbiting in Earth's atmosphere just above sea level, that in no way generates lift (just powering through that atmosphere).
How would gravity be perceived by the passenger onboard? On one hand I think theyd be weightless since they are technically always falling... But I could be wrong.
Bonus: how fast would a 200kg spherical (I guess) vessel be traveling?
physics
physics
edited 2 hours ago
Lamar Latrell
1032
asked 8 hours ago
anon
3876
edited 2 hours ago
Lamar Latrell
1032
asked 8 hours ago
anon
3876
edited 2 hours ago
Lamar Latrell
1032
edited 2 hours ago
Lamar Latrell
1032
edited 2 hours ago
Lamar Latrell
1032
1032
asked 8 hours ago
anon
3876
asked 8 hours ago
anon
3876
asked 8 hours ago
anon
3876
3876
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
7 hours ago
2
Fine then: equatorial sea level with no moon
– anon
7 hours ago
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
7 hours ago
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
5 hours ago
add a comment |
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
7 hours ago
2
Fine then: equatorial sea level with no moon
– anon
7 hours ago
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
7 hours ago
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
5 hours ago
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
7 hours ago
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
7 hours ago
2
2
Fine then: equatorial sea level with no moon
– anon
7 hours ago
Fine then: equatorial sea level with no moon
– anon
7 hours ago
1
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
7 hours ago
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
7 hours ago
1
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
5 hours ago
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
10
down vote
If you're orbiting, and the rocket thrusters are off, you experience weightlessness. This is true pretty much everywhere.
It's a common misconception that earth's gravity doesn't extend beyond the atmosphere. Craft in space are weightless because they are in orbit, not because earth's gravity is really weak out there. In fact, the Hill sphere (the radius at which the earth's gravitation is no longer dominant) is about 4 times the radius of the moon's orbit. That's quite far out.
The velocity of any circular orbit can be found by $ v=sqrtfracGMr $ where G is the gravitational constant, M is the earth's mass, and r is the radius of the orbit.
Plugging in the Eath's mass and its mean radius of 6371 km gives a velocity of 7909 $ms^-1$. That's about Mach 23.
answered 7 hours ago
Ingolifs
1,133116
2
Excellent answer.
– Russell Borogove
7 hours ago
2
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
2
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
If you're orbiting, and the rocket thrusters are off, you experience weightlessness. This is true pretty much everywhere.
It's a common misconception that earth's gravity doesn't extend beyond the atmosphere. Craft in space are weightless because they are in orbit, not because earth's gravity is really weak out there. In fact, the Hill sphere (the radius at which the earth's gravitation is no longer dominant) is about 4 times the radius of the moon's orbit. That's quite far out.
The velocity of any circular orbit can be found by $ v=sqrtfracGMr $ where G is the gravitational constant, M is the earth's mass, and r is the radius of the orbit.
Plugging in the Eath's mass and its mean radius of 6371 km gives a velocity of 7909 $ms^-1$. That's about Mach 23.
answered 7 hours ago
Ingolifs
1,133116
2
Excellent answer.
– Russell Borogove
7 hours ago
2
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
2
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
|
show 2 more comments
up vote
10
down vote
If you're orbiting, and the rocket thrusters are off, you experience weightlessness. This is true pretty much everywhere.
It's a common misconception that earth's gravity doesn't extend beyond the atmosphere. Craft in space are weightless because they are in orbit, not because earth's gravity is really weak out there. In fact, the Hill sphere (the radius at which the earth's gravitation is no longer dominant) is about 4 times the radius of the moon's orbit. That's quite far out.
The velocity of any circular orbit can be found by $ v=sqrtfracGMr $ where G is the gravitational constant, M is the earth's mass, and r is the radius of the orbit.
Plugging in the Eath's mass and its mean radius of 6371 km gives a velocity of 7909 $ms^-1$. That's about Mach 23.
answered 7 hours ago
Ingolifs
1,133116
2
Excellent answer.
– Russell Borogove
7 hours ago
2
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
2
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
|
show 2 more comments
up vote
10
down vote
up vote
10
down vote
If you're orbiting, and the rocket thrusters are off, you experience weightlessness. This is true pretty much everywhere.
It's a common misconception that earth's gravity doesn't extend beyond the atmosphere. Craft in space are weightless because they are in orbit, not because earth's gravity is really weak out there. In fact, the Hill sphere (the radius at which the earth's gravitation is no longer dominant) is about 4 times the radius of the moon's orbit. That's quite far out.
The velocity of any circular orbit can be found by $ v=sqrtfracGMr $ where G is the gravitational constant, M is the earth's mass, and r is the radius of the orbit.
Plugging in the Eath's mass and its mean radius of 6371 km gives a velocity of 7909 $ms^-1$. That's about Mach 23.
answered 7 hours ago
Ingolifs
1,133116
If you're orbiting, and the rocket thrusters are off, you experience weightlessness. This is true pretty much everywhere.
It's a common misconception that earth's gravity doesn't extend beyond the atmosphere. Craft in space are weightless because they are in orbit, not because earth's gravity is really weak out there. In fact, the Hill sphere (the radius at which the earth's gravitation is no longer dominant) is about 4 times the radius of the moon's orbit. That's quite far out.
The velocity of any circular orbit can be found by $ v=sqrtfracGMr $ where G is the gravitational constant, M is the earth's mass, and r is the radius of the orbit.
Plugging in the Eath's mass and its mean radius of 6371 km gives a velocity of 7909 $ms^-1$. That's about Mach 23.
answered 7 hours ago
Ingolifs
1,133116
answered 7 hours ago
Ingolifs
1,133116
answered 7 hours ago
Ingolifs
1,133116
answered 7 hours ago
Ingolifs
1,133116
1,133116
2
Excellent answer.
– Russell Borogove
7 hours ago
2
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
2
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
|
show 2 more comments
2
Excellent answer.
– Russell Borogove
7 hours ago
2
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
2
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
2
2
Excellent answer.
– Russell Borogove
7 hours ago
Excellent answer.
– Russell Borogove
7 hours ago
2
2
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration.
– anon
7 hours ago
2
2
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless.
– Ingolifs
7 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag.
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
@anon double-checking, is that what you meant?
– uhoh
5 hours ago
|
show 2 more comments
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Sea level is not at a constant distance from earth’s center of gravity.
– Paul
7 hours ago
2
Fine then: equatorial sea level with no moon
– anon
7 hours ago
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
7 hours ago
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
5 hours ago