I have searched and saw a similar question but i dont know if it's exactly the same. If I had a rope around earth, let's say a few meters above (and supposing a flat surface, and suddenly dropped it at the same time from all of the points, how would it fall? Or what would happen?  I know gravity’s pull is not exactly the same around earth, but if it was the case? If the rope is a few meters above ground, it's circumference is bigger than earth’s. Imagine it's a solid circumference and doesn't bend or whatever, it's pretty solid. How would it fall? It's supposed to fall all around earth at the same time but since its circumference is bigger, it can't touch surface all at the same time. If its just a rope, where does the circumference start deforming if gravity pulls equally? What parts of the rope touch the ground earlier?
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What would happen if we decided to build a giant ring that managed to wrap around the whole world, end to end that was supported with pillars all along the ring and then the supports all suddenly removed?  Would the ring float in place? or if it fell, what direction would it fall?
53153
I am currently stuck on the following question.   A stone is dropped from a stationary helicopter 500m above the ground, at the equator. How far from the point vertically below the helicopter does it land and in what direction? Solve this problem by conserving angular momentum.   My current interpretation is that when the stone falls in order to converse angular momentum the angular velocity must increase. This leads to the stone travelling to the east.   I equated the angular momentum at the start       L0=*(re+h)mVo  where re is radius of earth, h is the height above that, m is mass and Vo is inital velocity. To the angular momentum at any time  L=(re+h-1/2gt^2)m(Ve+Vo)  where the 1/2at^2 term comes from gravitational acceleration and Ve is the additional velocity.   I rearranged to find Ve and then integrated over the time it takes to fall the 500m in order to work out the extra distance the stone traveled. However, my answer is 1/2 what the answer should be. I have checked the maths and it seems to be fine. So I believe my set up is wrong. Any help would be appreciated.
75807
I'm programming an animation of a bouncing ball, and I want it to be as realistic as possible.  I fully understand the physics while the ball is rising and falling: It accelerates downward at 9.8 meters/second/second.  But once it hits the ground, I'm lost. I know that it experiences some compression which translates it's velocity upward again, at which point gravity is again the only force acting on it. But I don't know how to model the compression and deflection of the ball hitting the ground.  (Also, I'm only dealing with strictly vertical linear velocity and acceleration. The ball isn't traveling horizontally or rotating at all. And I'm assuming negligible loss to air resistance.)  I know there are constants to deal with: how much the ball resists compression and how much energy is lost during the bounce, but I don't know how they are related.  Can anybody give me a one- or two-paragraph explanation and/or (a) formula(e) and/or point me to a resource where I can read more? (yes, I've already spent an hour or so googling it)  I'm happy to continue reading about it until I understand, I just don't know where to turn at this point.  Thanks in advance!
18076
So I came across this picture on facebook   My logic says it should show 200N, because you are pulling it with 100N on each side. On the other hand (as one of the facebook comments also stated, also saying is he is wrong he wasted 20k$ on physics degree), if you replace the right side with a wall you get this situation:    In this case I guess everyone agrees that the dynamometer would show 100N force. But if you isolate the dynamometer and "cut" the ropes on either picture, you would get the same thing. Because the system needs to be balanced, there would be 100N on each side.    And I study engineering too but I am confused by this. Why do we get the same distribution of forces, but dynamometer shows a different force?
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In General Relativity, Einstein established that gravity is due to the curvature produced by objects in space.  We all know that gravity is proportional to mass. The picture Einstein painted looks like this: The bigger the object, the bigger the curvature and hence the stronger gravity.  Let's say that the material inside the Earth is taken out and Earth is made less dense. Now the gravitational pull will be decreased according to Newton's formula due to reduction in mass.  But as per Einstein's picture the dimensions of curvature will be same, as the size of Earth is the same still and the gravity of Earth should not be altered.  I've pictured the scenario something like this:   Please clear up my confusion.
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I'm going to provide my argument for why I think the tension in a rope should be twice the force exerted on either side of it.  First, let's consider a different example. Say, there is a person named A and a block in space. A pushes on the block with a force of 100 N. Then, the block will also push A with a force of 100 N by Newton's third law. Now, consider the case where instead of the block, there is a person B who is also pushing on A with a force of 100 N while A is pushing on him. A will experience a force of 100 N because he pushed on B, AND another 100 N because he is being pushed by B. Hence he will experience a force of 200 N. Similarly, B also experiences 200 N of force.  Now, back to the original problem. There are two people A and B in space with a taut rope (no tension currently) in between them. If only A is pulling and B is not, then I agree that the tension is equal to the force A exerts. This situation (in my opinion) becomes analogous to the above if B is also pulling. So, say both of them pull from either side with a force of 100 N. Then the rope at the end of B will pull B with a force of 100 N (this pull is caused by A). By Newton's third law, the rope will experience a pull of 100 N. But B is also pulling his end of the rope with 100 N. Therefore, the tension should be 200 N. Similarly, the end of the rope at A must pull A with 100 N of force (because B is pulling from the other side) and hence experience a force of 100 N itself by Newton's third law plus another 100 N because A is pulling on the rope.  Apparently, the answer is not this (according to my searching on the web). So, could anyone tell me why this reasoning is wrong? Thanks.  EDIT : So apparently people don't agree with my first example, leave alone the second. This is to the downvoters and the upvoters of the highest-rated answer: You all agree that if only A pushes B with a force of 100 N, then A and B both will get pushed by a force of 100 N in opposite directions, right? Then, in the case where B is also pushing with a force of 100 N, it doesn't make sense that the answer would be exactly the same. It doesn't seem right that no matter what B does, B and A will always experience the same force as they would have if B hadn't applied any force.  EDIT 2 : I'm going to provide here a link to a question that I posted:  According to the answer and the comments there, the reason as to why my first example is incorrect is different to the one provided here. So maybe you should all read the answer and the comments provided by the person and reconsider what you think.
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Special mention to the  that kicked this off:  Anyway, the host points out that a tunnel that connects a pair of points on the earth's surface can be thought of as a gravity train - where the force of gravity along the tunnel allows an object to fall through it and emerge on the other side. Any force perpendicular to the tunnel (in case the tunnel doesn't pass through the center of the earth) is ignored.   I worked it out and the equation is $a = -\frac{4}{3}\pi\rho G d$, where $d$ is the distance from the center of the tunnel and $\rho$ is the density of the earth.   Clearly, it is simple harmonic and therefore the period is constant and has no dependence on which two points were used to make the tunnel.   Does anyone have an intuition for why this should be the case? I imagine some Gauss' law type of argument should work here, but I cannot see it.  EDIT: More to the point, why even when the tunnel does not pass through the center, one obtains the same period. Is there any deeper explanation as to why this should be the case?
103546
So I have some confusion here, I am sure I knew this at some point.  Let's say an object of 10 kg is dropped from a height of 10 m.  When it reaches the ground, the work done on the object should be the force ($mg$) x distance or 10 kg x 9.8 m/s/s x 10 m.  That gives 980 joules of work done on the object by gravity.  But the object did not gain 980 joules of .  It lost 980 joules of GPE and gained 980 joules of kinetic energy (up to the point of it reaching ground level).   Using the change in GPE and KE, it looks like no work was done on the object because the loss in GPE equals the gain in KE.  So   a) am I right that no net work was done on the car by sum of all forces?    b) is the work done by gravity equal to force time distance, or is it equal to the change in mechanical energy of the object which is zero?
83340
This question is actually related to my earlier question ("").  The fact that objects move a lot in the universe and that the universe is expanding, can imply that gravity is a repulsive force that increases with distance.. so the farthest objects repel us more.  This can still explain several existing observations, e.g., why does the apple fall?   Motion is the result of such repulsion. Two objects unlucky enough not to be moving relative to each other get squished due to the repulsion of the rest of the universe  around them. The earth repels the apple less than the stars so it is pushed towards the earth.   Furthermore, it can explain the expanding universe without the need for dark energy.  This could be demonstrated in a thought experiment. If we take a lot of same-charge particles (with small mass) such as electrons and lock them in a large box at a low enough temperature. The mutual repulsion of the particles may cause similar motion as if due to gravitational attraction.   Another experiment would be to measure the slight changes in our weight during day and night when the sun and earth align (if their masses are large enough to detect the feeble change in repulsion).   [EDIT: the question in the original form may not have been clear. It is "can we model".. with a yes/no answer and why (not). If downvoting, please justify.
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The earth is not a sphere, because it bulges at the equator.   I tried fiddling with centripetal force equations and gravity, but I couldn't derive why this bulge occurs.    Is there   (a) a mathematical explanation using forces (not energies) and   (b) a simple intuitive explanation to explain to others why the bulge occurs?
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Edit : Added a picture for better understanding of my querry. All the texts I have studied has used an non-inertial frame to explain the phenomenon. But every time I see something explained with pseudo forces, I try to realize in real forces.   But in this case I tried to explain it in a frame which is centered in Earth's center and not rotating. But I couldn't explain what happens to the tangential component (to the point on Earth's surface where $g$ to be measured) of centripetal acceleration in this scenario. It doesn't cancel out.   But then I thought the tangential component is so low that practically it would have no effect.  Is my explanation wrong?
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Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$  The experimentally verified near-Earth gravitational potential is  $$U ~=~ mgh.\tag{2}$$  The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance.  $(1)$ scales as $\frac 1r$.  So the greater the distance from the Earth, the less potential energy an object should have.  But $(2)$ scales proportionally to distance.  So the greater the distance from the Earth, the more potential energy an object should have.  How is this reconcilable?
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I was taught in school that acceleration due to gravity is constant. But recently, when I checked Physics textbook, I noted that  $$F = \dfrac{G  m_1   m_2}{r^2}. $$  So, as the body falls down, $r$ must be changing, so should acceleration due to gravity.
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If I apply a straight upward(perpendicular to ground) force against gravity of $5\ \mathrm{N}$ and lift an object "A" 10 meters, then the work done is:  $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J}$$ But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:  $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J} $$   In cases 1 and 2, the work done is equal, but the height would be different because in case one, height is equal to displacement, but in case 2 it will be less than 10 meters obviously. Then potential energy (i.e mgh) of both objects would be different. Doesn't conflict with the equation work = energy?  Work is force times displacement ($W = F S \cos(\theta)$); where $\theta$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.
112311
I was searching about 'why do not moon crash to earth due to its gravity?'Then by reading physics stake exchange existing questions I came to know that its the sideways motion of the moon that keeps moon in its orbit.   So my question is:- Moon is in circular or ellipthitical orbit and during its orbit moon is always pulled towards the centre of the earth but moon is changing its direction at every point because it is in circular or ellipthitical orbit, so the direction of force pulling moon towards the centre of the earth will also change at every point on the other hand the sideways motion of moon is always in a particular direction then there must be a point in the moon's orbit when sideways motion and gravity will act in same direction and moon should crash to earth.Then why it does not crash?
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I was reading somewhere about a really cheap way of travelling: using balloons to get ourselves away from the surface of the earth. The idea held that because the earth rotates, we should be able to land in a different place after some time. As we all know, this doesn't happen.  Someone said that the reason why this doesn't happen is because the atmosphere (air, clouds etc.) also revolves around the earth (with the same angular velocity as the earth's rotation). Since we are also part of the atmosphere, our position doesn't change relatively.  Well, I'm not convinced with that answer. Why does the atmosphere rotate along with earth?  Gravitational force is towards the centre of the earth, but I don't see how it's making the atmosphere rotate.
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When drag force ($bV$) equals to object's weight (mg) then upward and downward force becomes equal. As a result the object comes to rest. If this is true, how is a body moving with constant velocity?