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Earth's orbit is a slight ellipse, so to conserve momentum its speed increases when it is closest to the Sun. If the speed changes there is an acceleration. If there is an acceleration there is a force. Even if the change is small and gradual, wouldn't we experience a force because the Earth is so massive?

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    $\begingroup$ The Earth is moving at a rough speed of 19 miles/second at a distance of roughly 93 000 000 miles from the Sun, this gives the centripetal acceleration $v^2/R \approx 4 \times 10^{-6}$ miles/s^2 or about $0.02$ ft/s^2. Contrast this with $g \approx 32$ ft/s^2. $\endgroup$ – jim Jul 14 '16 at 17:12
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    $\begingroup$ @jim -- It's much smaller than that. What we can feel is the tidal acceleration, the difference between the acceleration of us toward the Sun versus the Earth as a whole toward the Sun. This is very, very small. $\endgroup$ – David Hammen Jul 14 '16 at 19:12
  • $\begingroup$ @jim That's the best answer, as far as I'm concerned. No need for GR (but where's the fun in that?) $\endgroup$ – wedstrom Jul 15 '16 at 15:35
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    $\begingroup$ We don't directly feel the tides either... $\endgroup$ – keshlam Jul 17 '16 at 8:02
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We don't feel any acceleration because the Earth and all of us humans on it is in free fall around the Sun. We don't feel the centripetal acceleration any more than the astronauts on the ISS feel the acceleration of the ISS towards the Earth.

This happens because of the way general relativity describes motion in gravitational field. The motion of a freely falling object is along a line called a geodesic, which is basically the equivalent of a straight line in curved spacetime. And because the freely falling object is moving in a straight line it experiences no force.

To be a bit more precise about this, the trajectory followed by a freely falling object is given by the geodesic equation:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

Explaining what this means is a bit involved, but actually we don't need the details. All we need to know is that the four-acceleration of a body $\mathbf A$ is given by another equation:

$$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{2} $$

But if use equation (1) to substitute for $d^2x^\alpha/d\tau^2$ in equation (2) we get:

$$ A^\alpha = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu = 0 $$

So for any freely falling body the four acceleration is automatically zero. The acceleration you feel, the "g force", is the size of the four-acceleration - technically the norm of the four-acceleration or the proper acceleration.

Nothing in this argument has referred to the shape of the orbit. Whether the orbit is hyperbolic, parabolic, elliptical or circular the same conclusion applies. The orbitting observer experiences no acceleration.

You might be interested to read my answer to How can you accelerate without moving?, where I discuss this in a bit more detail. For an even more technical approach see How does "curved space" explain gravitational attraction?.

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    $\begingroup$ Since you're at it, include the GR equations of motion for a rigid body of finite size and an expression for the tidal forces, please :) . $\endgroup$ – Count Iblis Jul 14 '16 at 17:13
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    $\begingroup$ I am not sure about this answer. The Equivalence Principle says that there is no local experiment capable of distinguishing a uniforme gravitational field to an accelerated frame. But the Earth rotating around the Sun is definitely a non local experiment. A gravitational field can definitely be detect. Tides do this. $\endgroup$ – Diracology Jul 14 '16 at 17:18
  • $\begingroup$ That's a good point, and the solar effect on the tides is lower at aphelion than perihelion. If you could measure the rate at which this changes then I guess you'd be measuring the effect of the change in the distance to the Sun. However I don't think this is what the question was asking. $\endgroup$ – John Rennie Jul 14 '16 at 17:23
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    $\begingroup$ ...does this question really require GR to answer? $\endgroup$ – BlueRaja - Danny Pflughoeft Jul 14 '16 at 19:35
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    $\begingroup$ Newtonian gravity obeys the equivalence principle. At this amount of mass and acceleration, and tidal effects, they give the same result to the extent anybody can measure it (for now). Do not need GR for this $\endgroup$ – Bob Bee Jul 15 '16 at 5:23
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John Rennie's answer is right from the viewpoint of General Relativity -- but since the question is tagged with Newtonian mechanics, it deserves a Newtonian answer too.

In the Newtonian framework, I think the best answer to "why don't we experience this force" is that we can't feel forces that apply to our body at all. What we actually experience with our senses are only forces between different parts of our body.

When you stand on the surface of Earth, you feel neither the gravitational pull of the Sun nor the gravitational pull of the Earth. Strictly speaking you don't even feel the contact force between your footsoles and the ground -- but you do feel the compressive force between the skin of your feet and the bones inside the foot. And to a lesser extent you feel your bones being compressed, and your flesh being stretched by hanging from your skeleton. All these internal forces balance out the gravitational pull on your body, such that there's zero net force applying to each part of it (ignoring the pull from the sun and moon), and you stay in place compared to the earth.

This is what produces the sensation of being pulled towards the earth: the internal forces in your body that resist that pull.

However, for the pull from the sun, there's nothing that balances it out. Every particle in your body simply falls towards the sun, with the acceleration given by the strength of the sun's gravitational field -- and every particle in the earth and in the air around you is doing the same, so no internal forces are needed anywhere to keep the various parts of your body in the same relative position. Therefore there is nothing to feel.

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    $\begingroup$ The question is about feeling the difference in speed when the earth rounds one of the ends of it's elliptical orbit, not feeling gravity from the earth and sun pulling on us. Maybe I'm not putting it together but I don't get how your answer addresses that? $\endgroup$ – Zack Jul 15 '16 at 13:26
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    $\begingroup$ @Zack: The differences in the speed of Earth at different points in its orbits are caused by the pull of the sun. Because the orbit is not exactly at right angles to the direction to the sun, at various points in the orbit the gravitational pull from the sun will have a component in the direction of travel, or against it. The point of the answer is that no part of the sun's pull is something you can feel -- in particular its component in the direction of the earth's momentary velocity isn't. $\endgroup$ – Henning Makholm Jul 15 '16 at 13:50
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    $\begingroup$ @DavidSchwartz: No, you can't. Your box can only detect how its acceleration differs from what is imposed by it by gravitation (and any other force the applies directly to everything in the box in exact proportion to its mass). If you could seal your box so completely as to keep gravitation out, then you might be on to something, but there is no way to do that. $\endgroup$ – Henning Makholm Jul 15 '16 at 17:34
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    $\begingroup$ @DavidSchwartz: Yes, because acceleration from a rocket would need to be enforced on the content of the box through forces transmitted by the material in its walls -- and those forces are ultimately what accelerometers detect. A box sitting on the surface of earth will detects the same 1G acceleration, now from the earth pushing on the underside of the box. In both cases the difference between the actual motion of the box and how it would move if gravity were the only force applying to it is 1G, so the readings will be the same. $\endgroup$ – Henning Makholm Jul 15 '16 at 17:40
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    $\begingroup$ @DavidSchwartz: Gravitation is special because it applies directly and uniformly to every particle in the body it applies to, without being distributed by mechanical stresses from a point (or region) of the body where the entire force applies. $\endgroup$ – Henning Makholm Jul 15 '16 at 17:42
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According to the Equivalence Principle a free falling system cannot locally detect a gravitational field. However Earth is a large enough system such that non-local effects turn out to be appreciable. Solar tides are - although small - detectable. So in principle one can experience the Sun's gravitational field even though we are in free fall. What I claim is that the acceleration change through the elliptical orbit is too small.

The Earth's angular momentum, with respect to the focus of the ellipse, is $L=mr^2\dot\phi$, where $m$, $r$ and $\dot\phi$ are the mass, the distance to the center of the Sun and the angular velocity, respectively. Therefore $$mr_p^2\dot\phi_p=mr_a^2\dot\phi_a,$$ where the indices $p$ and $a$ denotes "perihelion" and "aphelion". For an ellipse, $$r_p=\frac{r_0}{1+\epsilon},\quad r_a=\frac{r_0}{1-\epsilon}.$$ Hence $$\frac{\dot\phi_p}{\dot\phi_a}=\left(\frac{1+\epsilon}{1-\epsilon}\right)\approx 1,0340,$$ since the Earth's orbit eccentricity is, $\epsilon\approx 0,0167$. We have a change of about three per cent in six months. The mean angular acceleration is $$\bar\alpha=\frac{0.0340\cdot\phi_a}{180\cdot 24\cdot 60\cdot 60}\sim 10^{-9}\phi_a\, \mathrm{rad/s^2}.$$ Notice that $\phi_a$ is of order $$\frac{2\pi}{365\cdot 24\cdot 60\cdot 60}\sim 10^{-7}\, \mathrm{rad/s}.$$ So the angular acceleration is of order $10^{-16}\mathrm{rad/s^2}$. If you multiply this value by the mean distance to the Sun, $r\sim 10^{11}\, \mathrm{m}$, we get an acceleration of order $10^{-6}\, \mathrm{m/s^2}$. That is negligible compared to the acceleration due to the Earth's gravity, $9.8\, \mathrm{m/s^2}$.

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    $\begingroup$ Yes, but even if the acceleration change was enormous, we still would not feel it because of what John Rennie said in his first sentence. The Earth's orbit around the Sun (like any orbit of any body around any other body) is a "free-fall" trajectory. $\endgroup$ – Solomon Slow Jul 14 '16 at 18:17
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    $\begingroup$ @jameslarge Earth is a large enough system such that non-local effects turn out to be appreciable. Solar tides are - altough small - detectable. If the Equivalence Principle was valid here we could not sense Sun's gravitational field. We do though. $\endgroup$ – Diracology Jul 14 '16 at 19:19
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    $\begingroup$ @Diracology Well, the equivalence principle is valid here. It's just that it applies to point particles, not planets. As you break the planet into pieces, you'll find that while the center of mass is on a geodesic, the individual parts aren't (or at least, they're not on the same geodesic). This results in tidal acceleration since both gravity and electromagnetism prevent the planet from breaking into pieces. If you keep the whole system perfectly local (and we have no reason to believe it is not local, that's the core of GR), you still have tidal forces. $\endgroup$ – Luaan Jul 15 '16 at 8:49
  • $\begingroup$ @Luaan Exactly, that was what I meant. $\endgroup$ – Diracology Jul 15 '16 at 12:43
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John Rennie has answered the question in terms on general relativity, but it can also be answered with Newtonian physics. Your question is very similar to this one:

Why does the moon stay with the Earth?

and I can refer you to my answer there. In short, the Sun isn't only pulling on the Earth itself, it's pulling on everything on it as well, including us, with the same gravitational force. Therefore we experience the same gravitational acceleration due to the Sun as the rest of the Earth does. From Galileo's and Newton's laws of motion, it follows that we move on the same free-fall path as the Earth around the Sun, so we remain stationary relative to the Earth.

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Even if the orbit were a perfect circle, there's some acceleration towards the sun. If there weren't acceleration then the earth would move in a straight line (instead of a circle); but it doesn't move in a straight line therefore there's acceleration.

In a sense, the earth doesn't feel the acceleration because it doesn't try to resist it: if you stand on something then you resist gravity (resist falling) and feel a force on your feet; if you don't stand on something and fall then (ignoring air resistance) you feel nothing (except maybe nausea because you're accustomed to feeling gravity).

An orbit can be described as a situation where instead of "falling onto" something you perpetually "fall around" it. Because you're in an infinite "free fall" (whether circular or elliptical) you feel no force -- there is a force (of gravity) but you don't resist it (you don't push against it) and so you don't feel it.

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You definitely don't need to use General Relativity to answer this question.

It depends upon what you mean by "feel". If "feel" means "detectable by sophisticated instruments" then, yes, it can be "felt". But your body is not a very sophisticated detection instrument.

According to what I've read elsewhere, the Earth speeds up by $1000$ $m/s$ as it moves from its greatest distance from the Sun to its closest distance to the Sun. That takes six months or, roughly, 15,768,000 seconds. Use the following equation to roughly figure the acceleration:

$$V_{f} - V{o} = a*t$$

The acceleration of the Earth comes out to about $0.0000634$ $m/s^2$.

The whole Earth and everything on it is roughly accelerating at that rate and your body can't detect that very minor acceleration.

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My answer is more metaphysics than physics.

The reason we do not "feel" the acceleration is that the change is within the tolerances of our bodies. That said, I am sure there have been people born who are more attuned to these forces. But for the most part, for most of use, there are so many forces acting on our senses that the acceleration of the earth is one we have learned to ignore, or just cannot feel.

An example is a seismograph. A simple one can be made from a pencil paper, and a flexible piece of plastic. With greater tremors this would work fine to mark size of earthquakes. However, the more rigid the plastic, the less movement you will see, and the greater the force applied to it required to make it move. Professional seismographs are made of much more sensitive material.

We are like that rigid plastic. We feel changes, but we are not made for detecting subtle changes such as the earth's acceleration.

That is not to say we cannot develop the skill to be more attuned to the earth's speed. In my martial arts studies, I have seen amazing "superhuman" feats that most anyone can do, if only one would spend the required time to develop such skills.

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    $\begingroup$ Please provide constructive comments rather than just downvoting the post. $\endgroup$ – Chiel ten Brinke Jul 15 '16 at 11:46
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    $\begingroup$ There's nothing to feel in the first place, except for second-order effects like stronger or weaker tides. The actual centripetal force due to gravity that curves the Earth's path into an orbit affects things on the surface of the Earth as well as the Earth itself. The "small gradual changes" in the Earth's speed that the OP talks can't be felt directly, regardless of sensitivity. (With enough sensitivity, you can detect tidal forces and infer the big picture, but there's no detectable simple force in the same direction as the change in earth's velocity, not even buried under other signals) $\endgroup$ – Peter Cordes Jul 16 '16 at 8:40
  • $\begingroup$ I disagree. Just because the effect is infinitesimal does not mean it cannot be felt. If it can be measured, it can be felt because it has an effect. Also, if it cannot be measured but can be calculated, it eventually can be measured. Science is all about the discovery of things once ignored. $\endgroup$ – Sensii Miller Jul 18 '16 at 16:00

protected by Qmechanic Jul 14 '16 at 19:30

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