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Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons: The centrifugal force cancels out the gravity minimally, more so at the equator than at the poles. The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field. TL;DR version: There are ...


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Here's a simple argument that doesn't require any knowledge of fancy stuff like equipotentials or rotating frames of reference. Imagine that we could gradually spin the earth faster and faster. Eventually it would fly apart. At the moment when it started to fly apart, what would be happening would be that the portions of the earth at the equator would at ...


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The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post. Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator. Or equivalently, the field strength$^2$ $g$ ...


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Yes, the gradient in spherical coordinates contains angular terms. It has to, because the force need not point to the center. The relevant equation is $$\Delta U=\frac {\partial U}{\partial r}e_r+\frac {\partial U}{r\partial \theta}e_\theta+\frac {\partial U}{ r \sin \theta \partial\phi}e_\phi$$ You have no $\phi$ variation, so can ignore the last term.


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The potential energy is $U\left(r\right) = k \left(r - r_0\right)^2 / 2$, where $r_0$ is the equilibrium position of the spring. Since this is a central potential (depends only on $r$), angular momentum $L$ is conserved, and we can use the concept of an effective potential energy (see equation 335 here) $$ \begin{eqnarray} U_{eff}\left(r\right) &=& ...


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At 1:02 in that video, you see the "people" at the end of the swing do something like a half revolution with a radius of 15 meter in 0.3 seconds (rough estimate of dimensions and time - feel free to come up with your own estimates by counting frames etc - then substitute those numbers in the equations below). The centrifugal acceleration is given by $$a = ...


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would the steering wheel reach the end of travel at a smaller angle? NO. the range of travel of the wheel is set by mechanical parts in the chain. They don't know how fast you are going. There has been talk of "steer by wire" where a computer reads the angle of the steering wheel and decides how much to move the front wheels. As opposed to the throttle, ...


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Regardless of whether you are driving at 65mph or 1mph, as long as your steering angle is the same, you will travel the same path. (Of course in real life understeer or oversteer has to be considered) However, due to the much greater centripetal acceleration experienced, where it is proportional to the square of the velocity, $$a_c = \frac{v^2}{r}$$ You ...


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Diavolo does exert a force on the loop by Newton's 3rd law, just as you said. The reason this force isn't shown on the free body diagram is that only forces on the object of interest are shown, not forces by the object. Diavolo is the object of interest, so we don't include forces that he exerts on other things. This is all justified because only forces ...


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Option 4, none of the above. Your option 1 is wrong because points don't rotate. Your option 2 is closer to correct, but ultimately still wrong. You're overly hung up on points (the origin). It might help to get a handle on what "rotation" is. Points don't rotate. Better said, a rotated point is indistinguishable from the original. What about one ...


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An other way of seeing this is to imagine that since the object is "climbing up", a part of the cinetic energy is along the $r$ axis, and the rest is in angular speed. The angular speed is thus necessarely inferior than the tower.


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Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower? Slower. Provided that the space elevator is geo-stationary, the higher you go above geosynchronous orbit, the greater your velocity ...


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If you are asking the question "How high does a person have to be in order to be "weightless" because gravity is canceled from the rotation of the Earth?" as paraphrased by CoilKid, then what you are talking about is a Geostationary Orbit. You can only achieve this orbit in a equatorial plane for the orbitting object to seem truly stationary with respect to ...


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Think about this. If gravity is of equal magnitude as the centrifugal force, a satellite there will have an angular velocity same as the earth rotating angular velocity. However, this is just a geosynchronous satellite! The equation is $$\frac{GMm}{r^2}=mr\frac{4\pi^2}{T^2}$$ where $T=24h$. Hope it helps!



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