# Tag Info

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Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the "background force" we don't notice. Earthquakes are not at all smooth and the accelerations involved are large and change ...

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Dan's answer is essentially good, but miss one effect : the Coriolis effect. You can imagine a planet spinning much more rapidly than the earth, but at a constant angular speed. On that quickly rotating planet, the explanation of Dan would still stand, but as soon as on moves, we would feel a lateral Coriolis force. The Coriolis acceleration is ...

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By definition an orbit occurs when gravity balances with the "centrifugal" force. It is essentially a free fall situation. So the answer is the same reason why you don't get stuck to the ceiling of a free falling elevator. Both the spacecraft and the occupants are moving in-sync.

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There was some doubt about Lubos' answer (which I've accepted), so this is just a verification. I copied the method Lubos described and found the potential difference for an ellipsoid with different eccentricities. Sure enough, for an oblate spheroid, if you make the center-equator distance a fraction $e$ larger than the center-pole distance, the ...

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Ok, here is my (hopefully rigorous) demonstration of the origin of these forces here, from first principles. I've tried to be pretty clear what's happening with the maths. Bear with me, it's a bit lengthy! Angular velocity vector Let us start with the principal equation defining angular velocity in three dimensions, $$\dot{\vec{r}} = \vec{\omega} \times ... 9 The real force at work is centripetal force, or a force pushing inwards. Imagine you have a bucket on a string, and you swing that around in a circle: As you swing the bucket, it travels in a circle. The red line shows the path the bucket takes. In order to make it swing like this, you have to apply a constant force on the rope -- this is the green arrow ... 8 Yes, the ball would land in front of you. If you watch from outside the space station, the ball moves in a straight line at constant speed while you move in a circle at constant speed. That means the distance the ball takes to get from point A (where you release it) to point B (where it hits the floor) is shorter than the distance you take. Further, ... 8 Centrifugal force is a particular example of a fictitious force. It is introduced so that Newton's second law holds in a rotating reference frame. Newton's second law says$$F = ma$$This means that whenever we find an object accelerating (speeding up, slowing down, turning, or some combination), we can look around and find a physical reason why this ... 7 Actually, the astronaut would only float completely free in the middle of the space station. Elsewhere, he will stick slightly to whatever side is closer to the Earth than is the middle, or farther from the Earth than is the middle. The reason is the tidal force from the Earth, which will be very small but probably detectable. If the acceleration from ... 7 Because it's effect is smaller than the variation in g due to earth's bulge (caused by the same centrifugal force) or the local geology - when you use 9.8m/s^2 that's just an approximation. The effect of the bulge and centrifugal force mean that 'g' at the equator is about 0.5% lower than 'g' at the poles edit: velocity at equator 40,000 km / 24 h = ... 7 It sounds like you don't want the normal rotating spaceship like in "2001" because you get motion sickness. No one really gets "motion sickness" just from moving, though. That's impossible because moving with constant velocity is physically the same as being stationary. What you get is "acceleration sickness". You feel the bumpiness of a car ride. Even ... 6 Well, it depends... If you just made the sun much heavier, so the earth would have to move faster in it's orbit, you wouldn't feel any different. It's just that the year would be shorter and the tides higher. If you just put a rocket behind the earth and somehow put it on rails so it couldn't go to a different orbit, then you'd feel it. You'd be heavier in ... 6 It is an interesting question and I know exactly what you mean. I often look at the likes of Stoner performing what you describe and think "wow!". Right, for the answer... Although the front wheel of any bicycle plays a key role in providing stability, it is not required for cornering. The ability to change direction is provided by a centripetal force, ... 5 There are two equivalent descriptions^1 of the reduced two-body problem with a central potential V(r): In an inertial frame with no fictitious forces: Here \frac{1}{2}\mu r^{2}\dot{\theta}^{2} is the angular part of the kinetic energy. In a rotating frame following the reduced particle with fictitious forces and only 1D radial kinematics: Here ... 5 They should feel the same. You only feel forces in orbit if there is something causing sensations, and nothing does in either case. Even on earth, you don't feel the "force" of gravity; you feel the force of the floor pushing you up so that you don't start falling under gravity's influence. In orbit, there is no floor, so you don't feel gravity. You ... 4 I have often thought of the possibility of using a liquid inside the capsule (with the astronaut in an appropriate breathing apparatus of course). In addition to the increased drag induced on movement, one might be able to cleverly design circulation cells that would induce a force pushing the astronaut towards a grated floor. I think arm motion would be ... 4 Neglecting friction, the force experienced is the Centrifugal Force F=\frac{mv^2}{r} (it would be less if you included friction since the car actually slips) vectorially added to the orthogonal gravitational force F_g=mg, i.e. F = m\sqrt{\left(\frac{v^2}r\right)^2 + g^2} where g = 9.81 \frac{m}{s^2}1. Divide this by F_g to obtain a result in Gs. ... 4 I think that all the right physics is contained in Martin Beckett's answer and the comments on it, but I'd like to restate it in a way that may bring out what I think the key point is. In practice, when we do experiments in a lab near Earth's surface, we use a value of g that's been determined empirically at that location. For instance, we might determine ... 4 Your analysis is correct. At small speeds, the mass hangs straight down. You might start with your picture; it is quite misleading. (Edit: queueoverflow has now fixed up the picture!) The text of your problem assumes that your arm is held exactly at the same angle as the string, making effectively one long string. Your picture shows a different story. ... 4 In this case, the gravitational force is the centripetal force, i.e. the force which keeps the satellite moving in orbit. As you have correctly surmised, the net force is towards the Earth, and the satellite will accelerate in that direction. What makes you think there is another force "holding the satellite in orbit"? 4 No one else has taken a crack at it, so I'll just point you in the direction of the answer. He won't notice unless the pseudo-forces due to rotation (centrafugal, Coriolis and Euler) are large enough to notice. So take a human being to have a height of 2 meter. Using the conventions in the wikipedia pages and assuming that the angular velocity of the can ... 4 1) You surely feel the pressure when you accelerate. Whether you attribute it to fictitious forces or other forces depends on your choice of the "reference frame" (vantage point). From the viewpoint of your body's reference frame, which is not an inertial frame, there exist fictitious forces (inertia and/or centrifugal and/or Coriolis' force) that are ... 4 The acceleration of uniform circular motion is a very basic computation that we do for first year students.$$ a = \frac{v^2}{r} $$which for someone standing on the Earth's equator comes to$$ a_\text{equator} \approx \frac{\left(465\text{ m/s}\right)^2}{6400\text{ km}} = 0.03\text{ m/s}^2$$or less than 1% of g. That is a measurable quantity, but not very ... 3 I look at two models of a "fat earth": a spherically symmetric interior with an aspherical surface layer in hydrostatic equilibrium. This analysis generalizes from the constant density assumed in other answers and thereby exhibits the sensitivity of the flattening to the surface density. I compare the result to those of various other answers. To estimate ... 3 The force that you feel in a Tilt-A-Whirl is not possible in a galaxy and does not correspond to anything in our solar system. In a Tilt-A-Whirl, you are subject to two additional constraints that are not present in the solar system galaxy interaction: The base of the Tilt-A-Whirl platform is not flat so there are times when the Earth's gravity slows the ... 3 The g force is a unit of acceleration. 1 g is equal to 9.80665 m s-2. So the correct formula is$$ \text{G force} = \frac{\text{Acceleration in m s}^{-2}}{9.8}.  However, when describing uniform circular motion (i.e. $\boldsymbol\omega$ is constant) in free space, the only acceleration felt by the person rotating (in their frame of reference) is the ...

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