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Why can't we feel the Earth's revolution?
Earth's gravitational pull on ISSWhy don't we feel the speed of earth movingWhy can't gravity repel things?Why can't we see the supermassive black holes supposedly at the center of galaxies?Why don’t we feel the Earth rotating beneath us?Why can't supermassive black holes merge? (or can they?)
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I googled it and checked a few Q&A and there's only things about "Earth's rotation". But why can't we feel the revolution?
They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution?
At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and minus blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?
gravity astrophysics
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
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$begingroup$
I googled it and checked a few Q&A and there's only things about "Earth's rotation". But why can't we feel the revolution?
They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution?
At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and minus blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?
gravity astrophysics
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
$endgroup$
7
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Speak for yourself. I feel quite dizzy, thank you very much. Also, the fact that we don't all float off into space (I can feel the floor below me as I type), appears to be evidence of something.
$endgroup$
– Strawberry
Jun 13 at 8:50
2
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The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
$endgroup$
– Alchimista
Jun 14 at 12:31
1
$begingroup$
You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
$endgroup$
– jean
Jun 14 at 13:26
add a comment
|
$begingroup$
I googled it and checked a few Q&A and there's only things about "Earth's rotation". But why can't we feel the revolution?
They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution?
At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and minus blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?
gravity astrophysics
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
$endgroup$
I googled it and checked a few Q&A and there's only things about "Earth's rotation". But why can't we feel the revolution?
They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution?
At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and minus blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?
gravity astrophysics
gravity astrophysics
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
asked Jun 12 at 8:28
dolcodolco
2151 gold badge1 silver badge6 bronze badges
asked Jun 12 at 8:28
dolcodolco
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2151 gold badge1 silver badge6 bronze badges
7
$begingroup$
Speak for yourself. I feel quite dizzy, thank you very much. Also, the fact that we don't all float off into space (I can feel the floor below me as I type), appears to be evidence of something.
$endgroup$
– Strawberry
Jun 13 at 8:50
2
$begingroup$
The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
$endgroup$
– Alchimista
Jun 14 at 12:31
1
$begingroup$
You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
$endgroup$
– jean
Jun 14 at 13:26
add a comment
|
7
$begingroup$
Speak for yourself. I feel quite dizzy, thank you very much. Also, the fact that we don't all float off into space (I can feel the floor below me as I type), appears to be evidence of something.
$endgroup$
– Strawberry
Jun 13 at 8:50
2
$begingroup$
The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
$endgroup$
– Alchimista
Jun 14 at 12:31
1
$begingroup$
You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
$endgroup$
– jean
Jun 14 at 13:26
7
7
$begingroup$
Speak for yourself. I feel quite dizzy, thank you very much. Also, the fact that we don't all float off into space (I can feel the floor below me as I type), appears to be evidence of something.
$endgroup$
– Strawberry
Jun 13 at 8:50
$begingroup$
Speak for yourself. I feel quite dizzy, thank you very much. Also, the fact that we don't all float off into space (I can feel the floor below me as I type), appears to be evidence of something.
$endgroup$
– Strawberry
Jun 13 at 8:50
2
2
$begingroup$
The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
$endgroup$
– Alchimista
Jun 14 at 12:31
$begingroup$
The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
$endgroup$
– Alchimista
Jun 14 at 12:31
1
1
$begingroup$
You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
$endgroup$
– jean
Jun 14 at 13:26
$begingroup$
You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
$endgroup$
– jean
Jun 14 at 13:26
add a comment
|
4 Answers
4
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Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment.
edited Jun 13 at 10:10
mcalex
1031 bronze badge
answered Jun 12 at 10:13
Steve LintonSteve Linton
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Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
20
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This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
6
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
1
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Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
add a comment
|
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You do, but it's too small to really notice
First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.
Think about driving a car, or riding in an airplane. Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".
However, When you take a sharp turn, or take off from the runway, you definitely feel something. That's the acceleration. It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.
More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.
Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.
Acceleration is quite noticeable, depending on its magnitude. Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.
So how large is the acceleration keeping Earth in orbit? About 0.0059 m/s².
What about the acceleration of Earth's rotation? A ever so slightly larger 0.0339 m/s².
Small wonder that it seems like you can't feel these forces!
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
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A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
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– PM 2Ring
Jun 13 at 8:49
2
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@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
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– Moyli
Jun 13 at 14:26
1
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@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
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– Moyli
Jun 13 at 16:50
2
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
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– Acccumulation
Jun 13 at 17:03
3
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@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
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– PM 2Ring
Jun 14 at 3:30
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show 15 more comments
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The problem with how you're looking at it is that velocities don't cause or result from forces, but accelerations do. Think of Newton's 2nd law, $F = m a$. Circular motion is motion at constant speed but changing direction, this changing direction is a type of acceleration because velocity is a vector (has direction) and acceleration is change in velocity.
We do actually feel a difference due to this acceleration - if you think about a sphere rotating around an axis, the points on near the intersection of that axis with the surface are rotating slower than anywhere else, and so if you measure the local gravitational acceleration $g$ at the poles you get a slightly larger value than you do at the equator (about 0.3%). This is because the rotation acts in opposition to the force of gravity due to the earth's mass.
There is also a very small effect from the earth's orbit around the sun. In this case, the force due to acceleration is exactly the same as the force that holds the earth in orbit, so you can just look at the gravitational effects as done in this answer on physics.se, which came up with a value of around 26 parts per billion, or 0.0000026%. The interesting thing is that you become lighter both when the sun is overhead, and when it's directly on the opposite side of the earth to you.
answered Jun 12 at 21:25
llamallama
2736 bronze badges
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As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
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– PM 2Ring
Jun 13 at 9:08
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I will give a biologically motivated answer to this question:
Feeling the Earth rotation carries no meaning to us. It is always approximately the same and we will blend out such background information and concentrate on news that are really important to us: Is there a danger approaching? Is the some food to gain? What are our peers doing?
Because there was no evolutionary pressure to feel the Earth rotation, we did not develop a sense for it. The fact that it is weak and really difficult to measure adds to this. We need an apparatus like Foucault's pendulum to see its effect.
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
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1
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It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
1
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I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
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– Eric Towers
Jun 14 at 3:30
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4 Answers
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4 Answers
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Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment.
edited Jun 13 at 10:10
mcalex
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answered Jun 12 at 10:13
Steve LintonSteve Linton
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Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
20
$begingroup$
This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
6
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
1
$begingroup$
Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
add a comment
|
$begingroup$
Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment.
edited Jun 13 at 10:10
mcalex
1031 bronze badge
answered Jun 12 at 10:13
Steve LintonSteve Linton
5,9241 gold badge9 silver badges32 bronze badges
$endgroup$
8
$begingroup$
Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
20
$begingroup$
This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
6
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
1
$begingroup$
Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
add a comment
|
$begingroup$
Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment.
edited Jun 13 at 10:10
mcalex
1031 bronze badge
answered Jun 12 at 10:13
Steve LintonSteve Linton
5,9241 gold badge9 silver badges32 bronze badges
$endgroup$
Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment.
edited Jun 13 at 10:10
mcalex
1031 bronze badge
answered Jun 12 at 10:13
Steve LintonSteve Linton
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edited Jun 13 at 10:10
mcalex
1031 bronze badge
edited Jun 13 at 10:10
mcalex
1031 bronze badge
edited Jun 13 at 10:10
mcalex
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1031 bronze badge
answered Jun 12 at 10:13
Steve LintonSteve Linton
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answered Jun 12 at 10:13
Steve LintonSteve Linton
5,9241 gold badge9 silver badges32 bronze badges
answered Jun 12 at 10:13
Steve LintonSteve Linton
5,9241 gold badge9 silver badges32 bronze badges
5,9241 gold badge9 silver badges32 bronze badges
8
$begingroup$
Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
20
$begingroup$
This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
6
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
1
$begingroup$
Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
add a comment
|
8
$begingroup$
Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
20
$begingroup$
This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
6
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
1
$begingroup$
Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
8
8
$begingroup$
Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
$begingroup$
Which in turn is a reflection of just how weak gravity is.
$endgroup$
– Pieter Geerkens
Jun 12 at 19:49
20
20
$begingroup$
This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
$begingroup$
This answer is misleading in that it is not the size of the speeds but any acceleration (change velocity in its speed or direction) that is related to an applied force. When traveling in a circle a force needs to be applied to change the direction of motion. The answers by llama and ap55 are better. However I like the "the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity". At constant velocity any forces are all balanced out (net force = zero, F=0)
$endgroup$
– TazAstroSpacial
Jun 13 at 6:31
6
6
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
$begingroup$
Galileo, not Einstein.
$endgroup$
– Carsten S
Jun 13 at 22:56
1
1
$begingroup$
Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
$begingroup$
Yeah it didn't take Einstein to realize this...
$endgroup$
– Mehrdad
Jun 13 at 23:09
add a comment
|
$begingroup$
You do, but it's too small to really notice
First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.
Think about driving a car, or riding in an airplane. Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".
However, When you take a sharp turn, or take off from the runway, you definitely feel something. That's the acceleration. It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.
More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.
Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.
Acceleration is quite noticeable, depending on its magnitude. Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.
So how large is the acceleration keeping Earth in orbit? About 0.0059 m/s².
What about the acceleration of Earth's rotation? A ever so slightly larger 0.0339 m/s².
Small wonder that it seems like you can't feel these forces!
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
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A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
$endgroup$
– PM 2Ring
Jun 13 at 8:49
2
$begingroup$
@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
$endgroup$
– Moyli
Jun 13 at 14:26
1
$begingroup$
@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
$endgroup$
– Moyli
Jun 13 at 16:50
2
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
$endgroup$
– Acccumulation
Jun 13 at 17:03
3
$begingroup$
@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
$endgroup$
– PM 2Ring
Jun 14 at 3:30
|
show 15 more comments
$begingroup$
You do, but it's too small to really notice
First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.
Think about driving a car, or riding in an airplane. Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".
However, When you take a sharp turn, or take off from the runway, you definitely feel something. That's the acceleration. It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.
More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.
Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.
Acceleration is quite noticeable, depending on its magnitude. Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.
So how large is the acceleration keeping Earth in orbit? About 0.0059 m/s².
What about the acceleration of Earth's rotation? A ever so slightly larger 0.0339 m/s².
Small wonder that it seems like you can't feel these forces!
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
$endgroup$
2
$begingroup$
A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
$endgroup$
– PM 2Ring
Jun 13 at 8:49
2
$begingroup$
@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
$endgroup$
– Moyli
Jun 13 at 14:26
1
$begingroup$
@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
$endgroup$
– Moyli
Jun 13 at 16:50
2
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
$endgroup$
– Acccumulation
Jun 13 at 17:03
3
$begingroup$
@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
$endgroup$
– PM 2Ring
Jun 14 at 3:30
|
show 15 more comments
$begingroup$
You do, but it's too small to really notice
First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.
Think about driving a car, or riding in an airplane. Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".
However, When you take a sharp turn, or take off from the runway, you definitely feel something. That's the acceleration. It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.
More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.
Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.
Acceleration is quite noticeable, depending on its magnitude. Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.
So how large is the acceleration keeping Earth in orbit? About 0.0059 m/s².
What about the acceleration of Earth's rotation? A ever so slightly larger 0.0339 m/s².
Small wonder that it seems like you can't feel these forces!
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
$endgroup$
You do, but it's too small to really notice
First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.
Think about driving a car, or riding in an airplane. Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".
However, When you take a sharp turn, or take off from the runway, you definitely feel something. That's the acceleration. It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.
More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.
Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.
Acceleration is quite noticeable, depending on its magnitude. Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.
So how large is the acceleration keeping Earth in orbit? About 0.0059 m/s².
What about the acceleration of Earth's rotation? A ever so slightly larger 0.0339 m/s².
Small wonder that it seems like you can't feel these forces!
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
answered Jun 13 at 3:04
ap55ap55
4611 gold badge1 silver badge3 bronze badges
4611 gold badge1 silver badge3 bronze badges
2
$begingroup$
A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
$endgroup$
– PM 2Ring
Jun 13 at 8:49
2
$begingroup$
@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
$endgroup$
– Moyli
Jun 13 at 14:26
1
$begingroup$
@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
$endgroup$
– Moyli
Jun 13 at 16:50
2
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
$endgroup$
– Acccumulation
Jun 13 at 17:03
3
$begingroup$
@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
$endgroup$
– PM 2Ring
Jun 14 at 3:30
|
show 15 more comments
2
$begingroup$
A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
$endgroup$
– PM 2Ring
Jun 13 at 8:49
2
$begingroup$
@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
$endgroup$
– Moyli
Jun 13 at 14:26
1
$begingroup$
@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
$endgroup$
– Moyli
Jun 13 at 16:50
2
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
$endgroup$
– Acccumulation
Jun 13 at 17:03
3
$begingroup$
@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
$endgroup$
– PM 2Ring
Jun 14 at 3:30
2
2
$begingroup$
A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
$endgroup$
– PM 2Ring
Jun 13 at 8:49
$begingroup$
A particle in orbit is in freefall, and it doesn't feel the centripetal force. But on an extended body there is the tidal force. So standing on the Earth's surface we don't feel the full force of the Sun's gravity, we just feel the difference between the strength of the Sun's gravity on the centre of the Earth vs its strength at our location. Similar remarks apply to the tidal force due to the Moon.
$endgroup$
– PM 2Ring
Jun 13 at 8:49
2
2
$begingroup$
@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
$endgroup$
– Moyli
Jun 13 at 14:26
$begingroup$
@Neyt Correct me if I'm wrong, but it seems that you're comparing the size of an ant (in mm) to the magnitude of acceleration (in mm/s²) but that doesn't make much sense – they are completely unrelated. The ant feels the same acceleration as every other object on Earth regardless of their size.
$endgroup$
– Moyli
Jun 13 at 14:26
1
1
$begingroup$
@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
$endgroup$
– Moyli
Jun 13 at 16:50
$begingroup$
@Neyt "Small wonder" is a saying, it doesn't have anything to do with physical size. And the answer is still no, neither size or weight has anything to do with acceleration.
$endgroup$
– Moyli
Jun 13 at 16:50
2
2
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
$endgroup$
– Acccumulation
Jun 13 at 17:03
$begingroup$
@PM2Ring Depends on what you mean by "feel". You are being pulled towards the sun by the gravity. However, if you try to measure this force by standing on a scale, the scale is also being pulled towards the sun. Since the acceleration is the same for both, the sun's gravity doesn't cause anything to be registered on the scale.
$endgroup$
– Acccumulation
Jun 13 at 17:03
3
3
$begingroup$
@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
$endgroup$
– PM 2Ring
Jun 14 at 3:30
$begingroup$
@Draco18s The Earth is not a point. The point at the centre of the Earth is in freefall around the Sun. The rest of the Earth is subject to a tidal force from the Sun because it's not at the centre point.
$endgroup$
– PM 2Ring
Jun 14 at 3:30
|
show 15 more comments
$begingroup$
The problem with how you're looking at it is that velocities don't cause or result from forces, but accelerations do. Think of Newton's 2nd law, $F = m a$. Circular motion is motion at constant speed but changing direction, this changing direction is a type of acceleration because velocity is a vector (has direction) and acceleration is change in velocity.
We do actually feel a difference due to this acceleration - if you think about a sphere rotating around an axis, the points on near the intersection of that axis with the surface are rotating slower than anywhere else, and so if you measure the local gravitational acceleration $g$ at the poles you get a slightly larger value than you do at the equator (about 0.3%). This is because the rotation acts in opposition to the force of gravity due to the earth's mass.
There is also a very small effect from the earth's orbit around the sun. In this case, the force due to acceleration is exactly the same as the force that holds the earth in orbit, so you can just look at the gravitational effects as done in this answer on physics.se, which came up with a value of around 26 parts per billion, or 0.0000026%. The interesting thing is that you become lighter both when the sun is overhead, and when it's directly on the opposite side of the earth to you.
answered Jun 12 at 21:25
llamallama
2736 bronze badges
$endgroup$
$begingroup$
As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
$endgroup$
– PM 2Ring
Jun 13 at 9:08
add a comment
|
$begingroup$
The problem with how you're looking at it is that velocities don't cause or result from forces, but accelerations do. Think of Newton's 2nd law, $F = m a$. Circular motion is motion at constant speed but changing direction, this changing direction is a type of acceleration because velocity is a vector (has direction) and acceleration is change in velocity.
We do actually feel a difference due to this acceleration - if you think about a sphere rotating around an axis, the points on near the intersection of that axis with the surface are rotating slower than anywhere else, and so if you measure the local gravitational acceleration $g$ at the poles you get a slightly larger value than you do at the equator (about 0.3%). This is because the rotation acts in opposition to the force of gravity due to the earth's mass.
There is also a very small effect from the earth's orbit around the sun. In this case, the force due to acceleration is exactly the same as the force that holds the earth in orbit, so you can just look at the gravitational effects as done in this answer on physics.se, which came up with a value of around 26 parts per billion, or 0.0000026%. The interesting thing is that you become lighter both when the sun is overhead, and when it's directly on the opposite side of the earth to you.
answered Jun 12 at 21:25
llamallama
2736 bronze badges
$endgroup$
$begingroup$
As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
$endgroup$
– PM 2Ring
Jun 13 at 9:08
add a comment
|
$begingroup$
The problem with how you're looking at it is that velocities don't cause or result from forces, but accelerations do. Think of Newton's 2nd law, $F = m a$. Circular motion is motion at constant speed but changing direction, this changing direction is a type of acceleration because velocity is a vector (has direction) and acceleration is change in velocity.
We do actually feel a difference due to this acceleration - if you think about a sphere rotating around an axis, the points on near the intersection of that axis with the surface are rotating slower than anywhere else, and so if you measure the local gravitational acceleration $g$ at the poles you get a slightly larger value than you do at the equator (about 0.3%). This is because the rotation acts in opposition to the force of gravity due to the earth's mass.
There is also a very small effect from the earth's orbit around the sun. In this case, the force due to acceleration is exactly the same as the force that holds the earth in orbit, so you can just look at the gravitational effects as done in this answer on physics.se, which came up with a value of around 26 parts per billion, or 0.0000026%. The interesting thing is that you become lighter both when the sun is overhead, and when it's directly on the opposite side of the earth to you.
answered Jun 12 at 21:25
llamallama
2736 bronze badges
$endgroup$
The problem with how you're looking at it is that velocities don't cause or result from forces, but accelerations do. Think of Newton's 2nd law, $F = m a$. Circular motion is motion at constant speed but changing direction, this changing direction is a type of acceleration because velocity is a vector (has direction) and acceleration is change in velocity.
We do actually feel a difference due to this acceleration - if you think about a sphere rotating around an axis, the points on near the intersection of that axis with the surface are rotating slower than anywhere else, and so if you measure the local gravitational acceleration $g$ at the poles you get a slightly larger value than you do at the equator (about 0.3%). This is because the rotation acts in opposition to the force of gravity due to the earth's mass.
There is also a very small effect from the earth's orbit around the sun. In this case, the force due to acceleration is exactly the same as the force that holds the earth in orbit, so you can just look at the gravitational effects as done in this answer on physics.se, which came up with a value of around 26 parts per billion, or 0.0000026%. The interesting thing is that you become lighter both when the sun is overhead, and when it's directly on the opposite side of the earth to you.
answered Jun 12 at 21:25
llamallama
2736 bronze badges
answered Jun 12 at 21:25
llamallama
2736 bronze badges
answered Jun 12 at 21:25
llamallama
2736 bronze badges
answered Jun 12 at 21:25
llamallama
2736 bronze badges
2736 bronze badges
$begingroup$
As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
$endgroup$
– PM 2Ring
Jun 13 at 9:08
add a comment
|
$begingroup$
As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
$endgroup$
– PM 2Ring
Jun 13 at 9:08
$begingroup$
As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
$endgroup$
– PM 2Ring
Jun 13 at 9:08
$begingroup$
As John Rennie explains in the linked answer on Physics.SE, the force we feel from the Sun is the tidal force which arises because we're on the surface of the Earth, not at its centre.
$endgroup$
– PM 2Ring
Jun 13 at 9:08
add a comment
|
$begingroup$
I will give a biologically motivated answer to this question:
Feeling the Earth rotation carries no meaning to us. It is always approximately the same and we will blend out such background information and concentrate on news that are really important to us: Is there a danger approaching? Is the some food to gain? What are our peers doing?
Because there was no evolutionary pressure to feel the Earth rotation, we did not develop a sense for it. The fact that it is weak and really difficult to measure adds to this. We need an apparatus like Foucault's pendulum to see its effect.
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
$endgroup$
1
$begingroup$
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
1
$begingroup$
I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
$endgroup$
– Eric Towers
Jun 14 at 3:30
add a comment
|
$begingroup$
I will give a biologically motivated answer to this question:
Feeling the Earth rotation carries no meaning to us. It is always approximately the same and we will blend out such background information and concentrate on news that are really important to us: Is there a danger approaching? Is the some food to gain? What are our peers doing?
Because there was no evolutionary pressure to feel the Earth rotation, we did not develop a sense for it. The fact that it is weak and really difficult to measure adds to this. We need an apparatus like Foucault's pendulum to see its effect.
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
$endgroup$
1
$begingroup$
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
1
$begingroup$
I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
$endgroup$
– Eric Towers
Jun 14 at 3:30
add a comment
|
$begingroup$
I will give a biologically motivated answer to this question:
Feeling the Earth rotation carries no meaning to us. It is always approximately the same and we will blend out such background information and concentrate on news that are really important to us: Is there a danger approaching? Is the some food to gain? What are our peers doing?
Because there was no evolutionary pressure to feel the Earth rotation, we did not develop a sense for it. The fact that it is weak and really difficult to measure adds to this. We need an apparatus like Foucault's pendulum to see its effect.
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
$endgroup$
I will give a biologically motivated answer to this question:
Feeling the Earth rotation carries no meaning to us. It is always approximately the same and we will blend out such background information and concentrate on news that are really important to us: Is there a danger approaching? Is the some food to gain? What are our peers doing?
Because there was no evolutionary pressure to feel the Earth rotation, we did not develop a sense for it. The fact that it is weak and really difficult to measure adds to this. We need an apparatus like Foucault's pendulum to see its effect.
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
answered Jun 13 at 16:44
jknappenjknappen
4813 silver badges10 bronze badges
4813 silver badges10 bronze badges
1
$begingroup$
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
1
$begingroup$
I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
$endgroup$
– Eric Towers
Jun 14 at 3:30
add a comment
|
1
$begingroup$
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
1
$begingroup$
I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
$endgroup$
– Eric Towers
Jun 14 at 3:30
1
1
$begingroup$
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
$begingroup$
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
$endgroup$
– jpa
Jun 13 at 18:41
1
1
$begingroup$
I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
$endgroup$
– Eric Towers
Jun 14 at 3:30
$begingroup$
I know Foucault's pendula are capable of detecting (latitude adjusted) rotation. But revolution? Really? How does that compare with the wobble in period caused by ambient thermal cycling of the pendulum's length (or the room's dimensions, since one may make a pendulum out of Invar or its friends, but one does not spend the money to make such a room)?
$endgroup$
– Eric Towers
Jun 14 at 3:30
add a comment
|
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Speak for yourself. I feel quite dizzy, thank you very much. Also, the fact that we don't all float off into space (I can feel the floor below me as I type), appears to be evidence of something.
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– Strawberry
Jun 13 at 8:50
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The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
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– Alchimista
Jun 14 at 12:31
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You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
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– jean
Jun 14 at 13:26