How would a simple pendulum change if it were done in deep space rather than earth?
90577
With no gravity, if I pushed the pendulum bob, it would travel around the fulcrum forever until some other force stopped it or until the friction stopped the motion, or the string wound itself up. \n\nIn the simple pendulum, there gravity provides a restoring force. \n\nIf I did not push the plumb bob, then if I just pulled the plumb bob back, it would stay where I left it, and not oscillate. However, as said above, some debris might strike it (a very low probability event) and cause it to move.
720016
It is not logical to think that the earth is hollow.  But the enormous pressure at the earth's center may cause the rock to be in a state that we don't understand.  It may be able to move and flow as a super dense liquid, or semi-solid gas of some sort.  \n\nGravity from an object in space is directly proportional to it's mass.  The gravity on the moon is much less than the gravity on earth because the moon is smaller, and weighs less than the earth.  \n\nThe center of gravity between the earth and moon, would be closer to the moon because the moon has less gravitational pull the farther away from the surface you go while the strength of the earth's gravity is relatively strong.\n\nThe "center" or region of space between the earth/moon system where the gravitational pull from both planets is equal, would be a far distance from the surface of the earth, and a closer distance to the surface of the moon.  \n\nAt this "center of gravity" the pull from the earth would equal the pull from the moon.  If you pass the "center" in space, the pull from the moon becomes stronger on a spaceship while the spaceship feels less effect from the weaker pull of the earth.
364195
None of us would have to worry about what we going to do the next day because we would all be dead.\n\nWhat would happen if the Earth stopped spinning?\nThe probability for such an event is practically zero in the next few billion years. If the Earth stopped spinning suddenly, the atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. All of the land masses would be scoured clean of anything not attached to bedrock. This means rocks, topsoil, trees, buildings, your pet dog, and so on, would be swept away into the atmosphere. \n\nIf the process happened gradually over billions of years, the situation would be very different, and it is this possibility which is the most likely as the constant torquing of the Sun and Moon upon the Earth finally reaches it's conclusion. If the rotation period slowed to 1 rotation every 365 days a condition called 'sun synchronous', every spot in the Earth would have permanent daytime or nighttime all year long. This is similar to the situation on the Moon where for 2 weeks the front-side is illuminated by the Sun, and for 2 weeks the back side is illuminated. This situation for the Earth is not the condition of 'stopped' rotation, but it is as close as the laws of physics will let the Earth get. \n\n \n\nIf it stopped spinning completely...not even once every 365 days, you would get 1/2 year daylight and 1/2 year nightime. During daytime for 6 months, the surface temperature would depend on your latitude, being far hotter that it is now at the equator than at the poles where the light rays are more slanted and heating efficiency is lower. This long-term temperature gradient would alter the atmospheric wind circulation pattern so that the air would move from the equator to the poles rather than in wind systems parallel to the equator like they are now. The yearly change in the Sun's position in the sky would now be just its seasonal motion up and down the sky towards the south due to the orbit of the Earth and its axial tilt. As you moved along constant lines of Earth latitude, you would see the elevation of the Sun increase or decrease in the sky just as we now see the elevation of the Sun change from a single point on the Earth due to the Earth's daily rotation. \n\nFor example, if you were at a latitude of +24 degrees North in the Summer and at a longitude where the Sun was exactly overhead, it would slide gradually to the horizon as Fall approached, but since the Sun has moved 90 degrees in its orbit, it would now be due west. Then as we approach Winter, you would now be located on the dark side of the Earth, and would have to travel in longitude to a location 180 degrees around the Earth to see the Sun 1/2 way up the sky because in the Winter, the Sun is 48 degrees south of its summer location in the sky. It's a little confusing, but if you use a globe of the Earth and orient it the right way, you can see how all this works out. \n\nAs for other effects, presumably the magnetic field of the Earth is generated by a dynamo effect that involves its rotation. If the Earth stopped rotating, it's magnetic field would no longer be regenerated and it would decay away to some low, residual value due to the very small component which is 'fossilized' in its iron-rich rocks. There would be no more 'northern lights' and the Van Allen radiation belts would probably vanish, as would our protection from cosmic rays and other high-energy particles. This is a significant biohazard.
932837
forces are opposite and equal, otherwise you would change speed/direction.
830124
Nothin would happen.\nIf you consider two bodies the earth (name it E) and a large mass (m). \ninitially M rests on E. and you lift mass M. the force used to lift it will also act on E.\nAlso if you consider the mass M and E. as a system force exerted on dropping M would be internal and since internal forces do not cause acc. the Earth will not get unbalanced.\n\nThats good for us isn't it...
123056
The Earth would immediately leave it’s orbit in a direction tangential to the current orbital plane at a velocity equal to it’s current orbital velocity, roughly 11 km/s.  The final path will be modified by the gravitaitonal force of the other solar system bodies, but since the Sun’s gravitational attraction is much greater than the others,  the solar system would just fly apart.  \n\nThere’s no delay in gravity force propagation, so the effects will be instantaneous the moment the gravity is “turned off”.\n\nEDIT:  \n\nmrjeffrey 321:  It is correct.  The information of the sun's gravitational field really does travel faster than the speed of light.  I don't have any vested interest in any particular theory, I prefer to look at the facts and the experimental data.  Ask any astronomer who performs astrometric or planetary position calculations using celestial mechanics.  The positions of the planets are always calculated using instantaneous gravity.  The positions are then corrected for aberration, the visual position error caused by the fact that light is speed limited.  So, we really calculate the actual positions of the planets, and then we correct for where we actually see the light coming from these planets, delayed by the speed of light.\n\nIn fact, if you model the solar system by adding in a delay for "gravity propagation" at the speed of light, the orbits of the planets in the solar system begin to slowly deteriorate.  This is most obvious with observations of certain binary double stars, where any delay in gravity would cause the star pairs to fly apart.  According to actual ephemeris measurements, the Earth is being accelerated to point about 20 arc-seconds ahead of where the sun appears in the sky.  This is because gravity is pulling us instantaneously towards the true position of the sun, not the apparent position we see in the sky.\n\nI'm not an expert on relativity, so I'm not going to argue fine points about what general  or special relativity either does or doesn't say or is interpreted to say.  What matters is the actual evidence, and the evidence says that the effects of gravity are instantaneous.\n\nNote that there was a paper published back in 2003 regarding the measurement of the speed of gravity, which resulted in it being very close to the speed of light.  There are several errors in this paper that were pointed out after publication.  The calculations used were shown to be incorrect.\n\nAn interesting question on this subject is black holes.  If we agree that gravity is a particle or wave, and we agree that nothing, no information, can escape a black hole, then how is it that black holes exhibit gravity? If they are truly so massive that they warp the very fabric of space-time, then how is it that gravity waves can escape when light cannot  If we accept the fact that black holes are "information holes" in the sky, how can they reveal themselves through gravity?
443155
First off, the rotation around it's axis speeds up. \nThe faster the speed, the more "rugby like" a sphere apears . \nThe moon would be orbiting it in a more eleptical orbit\nIts own orbit perhaps would be altered and become more eliptical. A more eliptical orbit results in us getting out of the "life zone" thus ending life on the planet throughout the course of several years... \n\nhmm... HOWEVER, if it was a Vertical Rugby ball then its the total opposite, for it would be rotating around the two pointy sides rather than the flat ones.  Interesting, I'm going to start making models !
435954
i'm assuming that you are ralking bout a hole being drilled right thru the earth and somebody falling into it...............you would accelerate and move towards the center and reach the opposite end of earth, at which point your velocity will reduce to zero and you will again move back to the cetre of earth and then onto the starting place.............in simple words you would execute Simple Harmonic Motion(SHM)
730814
assuming that you could do it and you didnt die and all that, you would go through to the other side.  if you include air resistance, you wouldnt make it to the other side, you would accelarate until you reached the center of the earth or until you reached terminal velocity, then you would start slowing down after you passed the center.  you would then start falling back toward the center in the opposite direction, passing it again, you would oscillate back and forth like this, losing some distance from center on each pass, eventually you would come to rest at the center
1089694
The simple answer is no.  It would actually be shorter.\n\nIn simple newtonian mechanics an object on the top of everest and an object at ground level will take the same amount of time to circle the earth (ie one day) and return to their initial positions.  However the object on top of everest will be moving faster.  Take a large revolving door for example.  If two people enter the door and one goes to the middle of the door (by the pivot) and the other goes next to the edge of the door, the person at the edge will have to move faster than the guy at the middle to keep up.\n\nHowever, in reality newtonian mechanics is only an estimate.  It is Einstein's theory of relativity that brings about the most accurate timings.  Using Time dilation methods it can be shown that the object on top of everest will have an ever so slightly shorter day because it is moving faster.
879542
Ok, this is kind of a long answer, but bear with me.  First off, the night would be no darker than it currently is during a new moon.  Remember, the moon is only up at night half of the time.\n\nHowever, the consequences of Earth having no moon would be much more severe.  Life as we know it would probably die out.\n\nOur moon is actually fairly large compared to our planet...this is what causes the strong tides people experience living on the ocean.  This tidal force also causes a much more subtle effect known as "precession".\n\nEarth's seasons are entirely determined by it's "axial tilt", the amount that our axis of rotation is tipped to the axis of our orbit (it has absolutely nothing to do with Earth being closer or further form the Sun).  During summer in the northern hemisphere, that hemisphere is tipped towards the Sun, while the southern hemisphere is tipped away from the Sun...that's why it's winter there.\n\nNow, throughout its history, Earth has had a relatively steady 23.5 degree tilt, meaning that the seasons have always been roughly the same intensity.  As mentioned, the moon's tidal forces causes precession, meaning the rotation axis carves out a wobble 23.5 degrees in radius every 26,000 years.  It's kind of like a top spinning...not only does it spin quickly, but as it also slowly wobbles from the Earth gravity.  The direction of the rotation axis changes, but the magnitude of the angle doesn't change.\n\nSo, the moon has a steadying effect on our axial tilt...so what?  Well, without it, our axial tilt could vary a lot, being caused by changing tidal forces from Jupiter and the Sun.  There's good evidence to show that this is exactly what happened to Mars...it has no large moons, just a couple tiny captured asteroids which exert very little tidal force.   As a result, it seems that Mars has varied anywhere from 0 degrees axial tilt (no seasons) up to a 60 degree axial tilt (crazy intense seasons) caused by the tidal forces of the Sun and Jupiter.\n\nWith an axial tilt of 60 degrees, that means all locations northward of 30 degrees get 24 hours of sunshine in summer, and absolutely no sunshine in winter.  With such strong seasons, it's likely the oceans would evaporate off...this seems to be the case on Mars, where there is mounting evidence that there was once a large northern ocean that mysteriously disappeared.\n\nSo, Short Answer: The seasons will become incredibly intense, the oceans would evaporate off, and without liquid water, all life would die.
949347
The period of  simple pendulum is\n\nT = 2π√(l / g)      if on such planet g is 1/3 of the Earth then \n\ng = 1/3 (9.81) m/s² = 3.27 m/s²\nl = 1.53 m\nThen\n\nT = 2π√(1.53 / 3.27) = 4.3 s
1044221
I would think that you would spiral toward Australia and then back toward us with the mid point at the center of earth's mass, each time not going as far as before .  \n\nThe reason that you would basically oscillate like that, is that the force of gravity increases the closer the centers of mass are to each other (the closer you got to the center of the earth) and so you would be accelerating the entire way till you got there and you would continue going because there would be no force acting on you to decelerate you.  As you start going away from the center toward the land down under, the force of gravity would act against your current velocity, till they cancel each other out and you start to fall toward the center of the earth this time the other way.  The reason that you would not quite make it out the other side of the hole is wind resistance and other very small forces would be affecting you. \n\nAssuming I'm right, you would eventually end up rotating in 3-dimensions around your stomach (or where ever your center of mass is) at the center of the earth.
151602
it would have no affect on a pendulum. \n\nstudies at the university of chico found that the up and down motion does not conflict in any way with the side to side motion of a pendulum. \n\nthe study also found that most of the research from the university of chico is conducted by a group of approximately 20 people ranging in age from 12 to 25.
1028524
The force would be 1/4 what it is now (since gravity follows an inverse-square of distance law). As it is, moon's gravity has next to no effect on our weight on earth, so weight change would be unnoticable. The main effect would be on the tides. Rather than being the primary determinant of tide height, as it is now, the moon would be a secondary influence and the sun would be the primary. This would reduce tidal heights by (very roughly) one half.
364601
If that direction were opposite the direction of the \nearth's rotation, I think the answer is theoretically yes.\nIn reality there's not enough mass of people relative to the mass of the earth to have any effect.
583414
There is an influence, but it is difficult to detect.  First of all, I'll point out that Foucault's pendulum does not relate to this question.\n\nSo, what is the force?  The Earth moves under both the Sun's and Moon's gravitational influences, but it's the centre of mass of the Earth which falls freely through the gravity fields.  An object on the surface of the Earth experiences a tidal force because it's, generally, not at exactly the same distance from the Sun or Moon as is the centre of the Earth.\n\nLet's consider the Sun, the Moon's case is exactly analogous.  If the pendulum is directly between the Sun and the centre of the Earth, it feels its maximum tidal force, 5.1e-7 Newtons per kilogram.  Unfortunately, that force is vertical, along the pendulum's suspension string/rod, and so is unmeasurable by deflection.  Well, what if the pendulum were 90 degrees from the Earth-Sun line.  Then the force is transverse, and that's best for measuring the force, but because the pendulum is now basically at the same distance from the Sun as the Earth, the tidal force is essentially zero, which doesn't help at all.\n\nSo, we want to calculate the transverse force on the pendulum due to the Sun, as a function of the angle between the pendulum-Earth and Sun-Earth lines.  This is\n\n2 * G * M_sun / R^2 * (r sin T cos T ) / R\nWhere R is the distance from the Earth to the Sun and r is the radius of the Earth.\n\nMaximize this function with respect to the angle T, and we find that the optimal angle is at 45 degrees.  Then, the transverse force is 2.5e-7 N/kg.  Your apparatus has to be capable of detecting this transverse force in the context of a vertical force of 9.81 N/kg.  Your deflection angle will be 2.6e-8 radians.  If you have to measure a deflection of 1 millimetre to see the effect, you will need a pendulum arm some 39 kilometres long.\n\nThe force of the Moon is roughly double this, but you'd still need a tower 20 kilometres high from which to suspend your pendulum.\n\nWith sufficiently sensitive instruments, the tidal force of the Moon can be measured.  In fact, the Moon was "discovered" by the folks at CERN, because of Earth tides.  See http://bulletin.cern.ch/9809/art1/Text_E.html