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Neglecting the effects of rotation on the 'apparent' strength of gravity, the gravitational field strength is similar to the electric field strength around a uniformly charged circular disk (or cylinder). This is because gravitational and electrostatic forces both obey $1\over r^2$ laws. The field strength along the axis of the disk is quite easy to derive (...


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First you should understand why gravity does not change as you move around on Earth's surface. The easiest way to explain this is using symmetry. The Earth is roughly a sphere. On a sphere there are no special points so gravity must act the same everywhere. A flat planet can actually be made to have constant gravity. If you had a flat planet that extended ...


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Although your question isn't posed very clearly I worked on the problem of non-spherical objects sliding/tumbling down a slope and don't mind sharing the following insight with you. Left, a cuboid and right, a sphere of comparable dimensions and mass. The inclination has been chosen deliberately high. Let's look at the forces and torques acting on both ...


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You have two questions. Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?" The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your ...


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Centrifugal force is not considered as a real force in any standard terms, but it may be used to measure resultant forces or reactions in a situation in an easier way. Like what @AJMansfield said, you could use it as a 'correction force', just like how you use Gravity! Theoretically, no, it is not real. Depending on the situation, it may be felt and ...


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If your test mass is being placed in an orbit at same angular speed as the lunar orbit i guess you'd need to super impose its centripetal force with the gravitational force. The new null orbit will probably be much lower because the null point is already near the orbit altitude.


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Just to expand a little bit... The centrifugal force on a particle of mass $m$ is given by $$\vec F_{ce}=-m\vec\omega\times(\vec\omega\times\vec r),$$ where $\vec\omega$ is the Earth's angular acceleration and $\vec r$ is the position of the particle. This force must be taken into account when using Newton's second law in the non inertial frame. Similarly to ...


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That question refers to apparent weight, as opposed to the true weight. Your true weight, $W=mg$, is the force of gravity pulling on you. Your apparent weight is how hard the ground has to push on you to keep you from falling. This is the weight you "feel". When you are accelerating up or down in an elevator while standing on a scale, the scale reads ...


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There are two reasons. One is that the Earth is not perfectly spherically symmetric and so at any point on the Earth's surface, the net gravitational force does not point directly towards the centre of the Earth. Secondly, because the Earth is rotating, there is a pseudo-force, the centrifugal force, that appears to act directly away from the Earth's ...



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