<p><strong>You do, but it's too small to really notice</strong></p>

<p>First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.</p>

<p>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".</p>

<p>However, When you take a sharp turn, or take off from the runway, you definitely feel <em>something</em>.  That's the <strong>acceleration</strong>.  It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.</p>

<p>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.</p>

<p>Even if Earth is spinning at a constant speed, the <strong>spin is a change in direction, which requires acceleration</strong>.</p>

<p>Acceleration is quite noticeable, <strong>depending on its magnitude</strong>.  Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.</p>

<p>So how large is the acceleration keeping Earth in orbit?  <a href="https://van.physics.illinois.edu/qa/listing.php?id=1063&amp;t=gravitational-pull-of-the-sun" rel="noreferrer">About 0.0059 m/s².</a>
What about the acceleration of Earth's rotation?  <a href="https://van.physics.illinois.edu/qa/listing.php?id=186" rel="noreferrer">A ever so slightly larger 0.0339 m/s².</a></p>

<p>Small wonder that it seems like you can't feel these forces!</p>

CC BY-SA 4.0
2019-06-13T08:49:31.207Z
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.
PM 2Ring
CC BY-SA 4.0
2019-06-13T08:54:39.287Z
Although these tidal forces are tiny, they do affect the period of a high-precision pendulum clock, but you need a very good clock (eg an atomic clock) to show that. See http://leapsecond.com/hsn2006/
PM 2Ring
CC BY-SA 4.0
2019-06-13T13:32:38.757Z
So, since an ant is like 10mm. 0.0059 m/s² is 5.9 mm/s² (Is it? I'm not even sure because i'm bad at this...). Does an ant "feel" this acceleration?
Neyt
CC BY-SA 4.0
2019-06-13T14:26:46.283Z
@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.
Moyli
CC BY-SA 4.0
2019-06-13T14:47:23.723Z
@Moyli I don't know...It's the "Small wonder that it seems like you can't feel these forces!" that makes me wonder. Like we, human, at average 70kg, can't "feel" the 0.0059 m/s² (or the 0.0339 m/s²). If the acceleration was higher, would we feel it ? (sorry if I don't understand all of this and being a fool here)
Neyt
CC BY-SA 4.0
2019-06-13T15:25:51.367Z
"Small wonder that it seems like you can't feel these forces!" Just as importantly, **you've been feeling them all your life**.  You're as used to it as you are to gravity.
RonJohn
CC BY-SA 4.0
2019-06-13T16:50:42.360Z
@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.
Moyli
CC BY-SA 4.0
2019-06-13T17:03:30.663Z
@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.
Acccumulation
CC BY-SA 4.0
2019-06-13T17:19:40.590Z
@Acccumulation True. But if I were in freefall around the Sun, with no interference from the Earth, I wouldn't feel anything. But I'm not exactly orbiting the Sun, because my position & velocity is slightly modified by the Earth. For that matter, I can't feel the Earth's gravity either, but I can feel the normal force exerted by the ground or floor that's stopping me from accelerating towards the centre of the Earth.
PM 2Ring
CC BY-SA 4.0
2019-06-13T19:02:08.907Z
@PM2Ring The acceleration you feel towards the center of the earth is ~2000 times greater than the acceleration towards the sun. Oh, and you are in free-fall around the sun: the Earth is in free-fall around the sun and you are part of "the Earth."
Draco18s no longer trusts SE
CC BY-SA 4.0
2019-06-13T19:22:59.017Z
@Draco18s Sure, the Earth's gravitational force on me is much greater than the gravitational force on me from the Sun, I'm not disputing that. I'm not *quite* in freefall around the Sun for the reasons I mentioned earlier. Please see the leapsecond.com link I posted above.
PM 2Ring
CC BY-SA 4.0
2019-06-13T19:56:12.833Z
Proof of freefall: 1) (Presumed fact) Earth as a whole is in freefall around the sun. 2) Therfore a random chunk of soil is in freefall around the sun. 3) Therfore a dead body which will decay to a random chunk of soil is in freefall. 4) Therefore a living body (excluding those with sufficient means of aided acceleration, eg a rocket) are in freefall around the sun. QED. In order to not be in freefall around the sun you need [at least 18.25 km/s of delta-v](https://space.stackexchange.com/a/3613).
Draco18s no longer trusts SE
CC BY-SA 4.0
2019-06-14T03:30:47.727Z
@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.
PM 2Ring
CC BY-SA 4.0
2019-06-14T13:23:49.743Z
@PM2Ring Yes, that is true. A whopping 0.0059 m/s² worth. If the entire earth vanished into nothingness instantly leaving you floating in space with your current velocity with respect to the sun intact you would....continue to orbit the sun. That's close enough to "freefall" as to be indistinguishable to your nervous system.
Draco18s no longer trusts SE
CC BY-SA 4.0
2019-06-14T13:46:32.503Z
`or when the airplane circles the airport before landing, are much less likely to spill your drink.`  No, this is entirely different.  Aircraft make [coordinated turns](https://en.wikipedia.org/wiki/Coordinated_flight), and the fact that you don't spill your drink is because the pilot is carefully managing the acceleration vector to keep it pointing at the floor of the aircraft.  If the turn were flat, like in a car, everyone would be flying around the cabin like a barrel of monkeys the drinks would be pasted to the windows.
J...
CC BY-SA 4.0
2019-06-14T13:49:56.603Z
@Draco18s No, it's even smaller than that. See John Rennie's calculation in the answer linked by llama. Yes, of course I'd continue to orbit the Sun if the Earth magically disappeared. I never disputed that. Nor did I claim that my nervous system could detect the solar (or lunar) tidal force. It's hard enough to do that with a precision pendulum and a caesium atomic clock, as described in the leapsecond.com articles by Tom van Baak in my 2nd comment above.
PM 2Ring
CC BY-SA 4.0
2019-06-15T00:22:04.847Z
@J Can you elaborate on that?  If everything's weight was suddenly pointed in an unexpected direction, that would certainly cause chaos.  The intended point, though, is that the turn itself is very slow and gradual, and thus the resulting acceleration is small enough to be virtually unnoticable - regardless of what's necessary to keep your weight pointed towards the floor of the aircraft.  https://en.wikipedia.org/wiki/Standard_rate_turn
ap55
CC BY-SA 4.0
2019-06-15T10:22:55.030Z
@ap55 The point is that the acceleration induced by a standard turn in an aircraft is not negligible, it is just because that aircraft is kept oriented such that the total acceleration vector acting on the aircraft (gravity plus the acceleration of the turn) is maintained normal to the floor of the aircraft.  Look out the window next time you're in an aircraft in a standard turn - the bank angle is huge.  You see the ground out one window and the sky out the other. The g-force into the seat will increase to perhaps 1.15g in a turn. If that 0.15g was sideways instead you would really notice it.
J...
CC BY-SA 4.0
2019-06-17T19:35:52.520Z
@J That number is indeed concerningly higher than I expected (and indeed much higher than the numbers for the question's topic), although I'm still not quite convinced it's non-negligible in terms of human perception.  It's comparable to the moon's gravity, for instance.  I can't find any numbers for roundabouts to crunch at the moment, but I'm wondering how they would measure up.
ap55
CC BY-SA 4.0
2021-10-31T07:47:56.230Z
Both objects in freefall and objects on the ground are subject to the ‘centrifugal’ force simply because we live and measure things in a noninertial reference frame. At the equator, this is enough to change the measured acceleration to at least 9.79 m/s^2 from centrifugal factors alone, which isn’t enough to reasonably feel but noticeable when dealing when dealing with things where that precision matters. Furthermore, the Coriolis effect is an artifact from the fact that we live on a spinning planet, and while that may not be felt day to day, you certainly see it’s effects in hurricanes.
Justin T
CC BY-SA 4.0
2019-06-13T03:04:29.703Z
2019-06-13T03:04:29.703Z
ap55
<p>I will give a biologically motivated answer to this question:</p>

<p>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?</p>

<p>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.</p>

CC BY-SA 4.0
2019-06-13T18:41:25.697Z
It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.
jpa
CC BY-SA 4.0
2019-06-14T03:30:39.450Z
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)?
Eric Towers
CC BY-SA 4.0
2019-06-13T16:44:54.887Z
2019-06-13T16:44:54.887Z
Sir Cornflakes
<p>The problem with how you're looking at it is that <em>velocities</em> don't cause or result from forces, but <em>accelerations</em> do. Think of Newton's 2nd law, <span class="math-container">$F = m a$</span>. 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.</p>

<p>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 <span class="math-container">$g$</span> 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.</p>

<p>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 <a href="https://physics.stackexchange.com/a/121785/11645">this answer on physics.se</a>, 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.</p>

CC BY-SA 4.0
2019-06-13T09:08:09.603Z
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.
PM 2Ring
CC BY-SA 4.0
2019-06-12T21:25:54.420Z
2019-06-12T21:25:54.420Z
llama
<p>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.</p>

CC BY-SA 4.0
2019-06-12T19:49:10.440Z
Which in turn is a reflection of just how ***weak*** gravity is.
Pieter Geerkens
CC BY-SA 4.0
2019-06-13T06:31:13.270Z
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)
TazAstroSpacial
CC BY-SA 4.0
2019-06-13T22:56:33.860Z
Galileo, not Einstein.
Carsten S
CC BY-SA 4.0
2019-06-13T23:09:37.973Z
Yeah it didn't take Einstein to realize this...
user541686
CC BY-SA 4.0
2019-06-12T10:13:23.900Z
2019-06-13T10:10:38.993Z
2019-06-13T10:10:38.993Z
Steve Linton
<p>I googled it and checked a few Q&amp;A and there's only things about "Earth's rotation". But why can't we feel the revolution?</p>

<p>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?</p>

<p><a href="https://i.sstatic.net/xiWmA.png" rel="noreferrer"><img src="https://i.sstatic.net/xiWmA.png" alt="enter image description here"></a></p>

<p>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 <strong><em>minus</em></strong> blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?</p>

CC BY-SA 4.0
2019-06-13T08:50:11.193Z
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.
Strawberry
CC BY-SA 4.0
2019-06-14T12:31:39.593Z
The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.
Alchimista
CC BY-SA 4.0
2019-06-14T13:26:12.150Z
You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides
jean
<gravity><astrophysics>
Why can't we feel the Earth's revolution?
dolco
# Why can't we feel the Earth's revolution?

(Asked by: username_0 on 2019-06-12)

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?

[![enter image description here](https://i.sstatic.net/xiWmA.png)](https://i.sstatic.net/xiWmA.png)

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?

**Tags:** gravity, astrophysics

> **Comments:**
> **username_1**: 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.  
> **username_2**: The explanation you read about Earth rotation is wrong but the answers should clarify that point as well.  
> **username_3**: You also cannot feel the Moon or the Sun gravitational force, but the sea can, and you can measure the rise of the tides  

---

## 4 answers

### Answer by username_17 on 2019-06-12. Score: 25 | Accepted answer

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.

> **Comments:**
> **username_18**: Which in turn is a reflection of just how ***weak*** gravity is.  
> **username_19**: 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 username_16 and username_4 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)  
> **username_20**: Galileo, not Einstein.  
> **username_21**: Yeah it didn't take Einstein to realize this...  

---

### Answer by username_4 on 2019-06-13. Score: 58

**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.](https://van.physics.illinois.edu/qa/listing.php?id=1063&t=gravitational-pull-of-the-sun)
What about the acceleration of Earth's rotation? [A ever so slightly larger 0.0339 m/s².](https://van.physics.illinois.edu/qa/listing.php?id=186)

Small wonder that it seems like you can't feel these forces!

> **Comments:**
> **username_5**: 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.  
> **username_5**: Although these tidal forces are tiny, they do affect the period of a high-precision pendulum clock, but you need a very good clock (eg an atomic clock) to show that. See http://leapsecond.com/hsn2006/  
> **username_6**: So, since an ant is like 10mm. 0.0059 m/s² is 5.9 mm/s² (Is it? I'm not even sure because i'm bad at this...). Does an ant "feel" this acceleration?  
> **username_7**: @username_6 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.  
> **username_6**: @username_7 I don't know...It's the "Small wonder that it seems like you can't feel these forces!" that makes me wonder. Like we, human, at average 70kg, can't "feel" the 0.0059 m/s² (or the 0.0339 m/s²). If the acceleration was higher, would we feel it ? (sorry if I don't understand all of this and being a fool here)  
> **username_8**: "Small wonder that it seems like you can't feel these forces!" Just as importantly, **you've been feeling them all your life**.  You're as used to it as you are to gravity.  
> **username_7**: @username_6 "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.  
> **username_9**: @username_5 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.  
> **username_5**: @username_9 True. But if I were in freefall around the Sun, with no interference from the Earth, I wouldn't feel anything. But I'm not exactly orbiting the Sun, because my position & velocity is slightly modified by the Earth. For that matter, I can't feel the Earth's gravity either, but I can feel the normal force exerted by the ground or floor that's stopping me from accelerating towards the centre of the Earth.  
> **username_10**: @username_5 The acceleration you feel towards the center of the earth is ~2000 times greater than the acceleration towards the sun. Oh, and you are in free-fall around the sun: the Earth is in free-fall around the sun and you are part of "the Earth."  
> **username_5**: @username_10 Sure, the Earth's gravitational force on me is much greater than the gravitational force on me from the Sun, I'm not disputing that. I'm not *quite* in freefall around the Sun for the reasons I mentioned earlier. Please see the leapsecond.com link I posted above.  
> **username_10**: Proof of freefall: 1) (Presumed fact) Earth as a whole is in freefall around the sun. 2) Therfore a random chunk of soil is in freefall around the sun. 3) Therfore a dead body which will decay to a random chunk of soil is in freefall. 4) Therefore a living body (excluding those with sufficient means of aided acceleration, eg a rocket) are in freefall around the sun. QED. In order to not be in freefall around the sun you need [at least 18.25 km/s of delta-v](https://space.stackexchange.com/a/3613).  
> **username_5**: @username_10 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.  
> **username_10**: @username_5 Yes, that is true. A whopping 0.0059 m/s² worth. If the entire earth vanished into nothingness instantly leaving you floating in space with your current velocity with respect to the sun intact you would....continue to orbit the sun. That's close enough to "freefall" as to be indistinguishable to your nervous system.  
> **username_11**: `or when the airplane circles the airport before landing, are much less likely to spill your drink.`  No, this is entirely different.  Aircraft make [coordinated turns](https://en.wikipedia.org/wiki/Coordinated_flight), and the fact that you don't spill your drink is because the pilot is carefully managing the acceleration vector to keep it pointing at the floor of the aircraft.  If the turn were flat, like in a car, everyone would be flying around the cabin like a barrel of monkeys the drinks would be pasted to the windows.  
> **username_5**: @username_10 No, it's even smaller than that. See John Rennie's calculation in the answer linked by username_16. Yes, of course I'd continue to orbit the Sun if the Earth magically disappeared. I never disputed that. Nor did I claim that my nervous system could detect the solar (or lunar) tidal force. It's hard enough to do that with a precision pendulum and a caesium atomic clock, as described in the leapsecond.com articles by Tom van Baak in my 2nd comment above.  
> **username_4**: @J Can you elaborate on that?  If everything's weight was suddenly pointed in an unexpected direction, that would certainly cause chaos.  The intended point, though, is that the turn itself is very slow and gradual, and thus the resulting acceleration is small enough to be virtually unnoticable - regardless of what's necessary to keep your weight pointed towards the floor of the aircraft.  https://en.wikipedia.org/wiki/Standard_rate_turn  
> **username_11**: @username_4 The point is that the acceleration induced by a standard turn in an aircraft is not negligible, it is just because that aircraft is kept oriented such that the total acceleration vector acting on the aircraft (gravity plus the acceleration of the turn) is maintained normal to the floor of the aircraft.  Look out the window next time you're in an aircraft in a standard turn - the bank angle is huge.  You see the ground out one window and the sky out the other. The g-force into the seat will increase to perhaps 1.15g in a turn. If that 0.15g was sideways instead you would really notice it.  
> **username_4**: @J That number is indeed concerningly higher than I expected (and indeed much higher than the numbers for the question's topic), although I'm still not quite convinced it's non-negligible in terms of human perception.  It's comparable to the moon's gravity, for instance.  I can't find any numbers for roundabouts to crunch at the moment, but I'm wondering how they would measure up.  
> **username_12**: Both objects in freefall and objects on the ground are subject to the ‘centrifugal’ force simply because we live and measure things in a noninertial reference frame. At the equator, this is enough to change the measured acceleration to at least 9.79 m/s^2 from centrifugal factors alone, which isn’t enough to reasonably feel but noticeable when dealing when dealing with things where that precision matters. Furthermore, the Coriolis effect is an artifact from the fact that we live on a spinning planet, and while that may not be felt day to day, you certainly see it’s effects in hurricanes.  

---

### Answer by username_16 on 2019-06-12. Score: 11

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](https://physics.stackexchange.com/a/121785/11645), 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.

> **Comments:**
> **username_5**: 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.  

---

### Answer by username_13 on 2019-06-13. Score: 9

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.

> **Comments:**
> **username_14**: It could be used for navigation, like gyrocompasses do. But human senses are not sensitive enough.  
> **username_15**: 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)?  

---

