How would gravity/acceleration be perceived by a human orbiting Earth at sea level?
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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 the 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 they'd be weightless since they are technically always falling... But I could be wrong.
Bonus: How fast would a 200 kg spherical (I guess) vessel be traveling?
physics
edited Nov 10 at 9:04
Peter Mortensen
1867
asked Nov 9 at 20:32
anon
4246
add a comment |
up vote
4
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 the 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 they'd be weightless since they are technically always falling... But I could be wrong.
Bonus: How fast would a 200 kg spherical (I guess) vessel be traveling?
physics
edited Nov 10 at 9:04
Peter Mortensen
1867
asked Nov 9 at 20:32
anon
4246
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
Nov 9 at 21:14
2
Fine then: equatorial sea level with no moon
– anon
Nov 9 at 21:20
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
Nov 9 at 22:02
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
Nov 10 at 0:02
Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m.
– Peter Mortensen
Nov 10 at 7:55
add a comment |
up vote
4
down vote
favorite
up vote
4
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 the 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 they'd be weightless since they are technically always falling... But I could be wrong.
Bonus: How fast would a 200 kg spherical (I guess) vessel be traveling?
physics
edited Nov 10 at 9:04
Peter Mortensen
1867
asked Nov 9 at 20:32
anon
4246
I understand the impracticalities of this concept, but humor the 'what-ifs.'
Ignoring physical obstacles and the effects of atmospheric fluctuations affecting the 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 they'd be weightless since they are technically always falling... But I could be wrong.
Bonus: How fast would a 200 kg spherical (I guess) vessel be traveling?
physics
physics
edited Nov 10 at 9:04
Peter Mortensen
1867
asked Nov 9 at 20:32
anon
4246
edited Nov 10 at 9:04
Peter Mortensen
1867
asked Nov 9 at 20:32
anon
4246
edited Nov 10 at 9:04
Peter Mortensen
1867
edited Nov 10 at 9:04
Peter Mortensen
1867
edited Nov 10 at 9:04
Peter Mortensen
1867
1867
asked Nov 9 at 20:32
anon
4246
asked Nov 9 at 20:32
anon
4246
asked Nov 9 at 20:32
anon
4246
4246
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
Nov 9 at 21:14
2
Fine then: equatorial sea level with no moon
– anon
Nov 9 at 21:20
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
Nov 9 at 22:02
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
Nov 10 at 0:02
Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m.
– Peter Mortensen
Nov 10 at 7:55
add a comment |
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
Nov 9 at 21:14
2
Fine then: equatorial sea level with no moon
– anon
Nov 9 at 21:20
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
Nov 9 at 22:02
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
Nov 10 at 0:02
Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m.
– Peter Mortensen
Nov 10 at 7:55
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
Nov 9 at 21:14
Sea level is not at a constant distance from earth’s center of gravity.
– Paul
Nov 9 at 21:14
2
2
Fine then: equatorial sea level with no moon
– anon
Nov 9 at 21:20
Fine then: equatorial sea level with no moon
– anon
Nov 9 at 21:20
1
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
Nov 9 at 22:02
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
Nov 9 at 22:02
1
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
Nov 10 at 0:02
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
Nov 10 at 0:02
Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m.
– Peter Mortensen
Nov 10 at 7:55
Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m.
– Peter Mortensen
Nov 10 at 7:55
add a comment |
1 Answer
1
active
oldest
votes
up vote
13
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=sqrtfrac{GM}{r} $ 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 Nov 9 at 21:21
Ingolifs
1,183116
2
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
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
Nov 9 at 21:32
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
Nov 9 at 21:53
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
Nov 9 at 23:50
2
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
|
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
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=sqrtfrac{GM}{r} $ 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 Nov 9 at 21:21
Ingolifs
1,183116
2
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
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
Nov 9 at 21:32
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
Nov 9 at 21:53
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
Nov 9 at 23:50
2
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
|
show 5 more comments
up vote
13
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=sqrtfrac{GM}{r} $ 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 Nov 9 at 21:21
Ingolifs
1,183116
2
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
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
Nov 9 at 21:32
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
Nov 9 at 21:53
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
Nov 9 at 23:50
2
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
|
show 5 more comments
up vote
13
down vote
up vote
13
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=sqrtfrac{GM}{r} $ 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 Nov 9 at 21:21
Ingolifs
1,183116
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=sqrtfrac{GM}{r} $ 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 Nov 9 at 21:21
Ingolifs
1,183116
answered Nov 9 at 21:21
Ingolifs
1,183116
answered Nov 9 at 21:21
Ingolifs
1,183116
answered Nov 9 at 21:21
Ingolifs
1,183116
1,183116
2
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
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
Nov 9 at 21:32
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
Nov 9 at 21:53
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
Nov 9 at 23:50
2
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
|
show 5 more comments
2
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
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
Nov 9 at 21:32
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
Nov 9 at 21:53
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
Nov 9 at 23:50
2
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
2
2
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
Excellent answer.
– Russell Borogove
Nov 9 at 21:25
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
Nov 9 at 21:32
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
Nov 9 at 21:32
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
Nov 9 at 21:53
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
Nov 9 at 21:53
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
Nov 9 at 23:50
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
Nov 9 at 23:50
2
2
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory...
– DJohnM
Nov 10 at 6:23
|
show 5 more comments
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Sea level is not at a constant distance from earth’s center of gravity.
– Paul
Nov 9 at 21:14
2
Fine then: equatorial sea level with no moon
– anon
Nov 9 at 21:20
1
Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/…
– Organic Marble
Nov 9 at 22:02
1
@OrganicMarble I'll have to find a copy, that looks fun!
– uhoh
Nov 10 at 0:02
Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m.
– Peter Mortensen
Nov 10 at 7:55