how would a ship move if u wanted to stimulate gravity around earth?
436378
i guess the question is how can the force of gravity be simulated on a spaceship. the answer is like the previous person said, by making it spin. this creates something called centripital force, which is an acceleration toward the center of the ship. the percived effect, however, would be that the people would be stuck to the outside of the ship ( i think).
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...
182304
great question - I wish I knew more but tides would be different. Also - the day would be shorter. The equatorial regions would be hotter, and the regions further away would be colder. The earth's rotation would be more stable with less wobble.
799088
Assuming:\n1. that the ships area does not tapper as it goes down (which of course they do in real life)\n2. that the density of water is 1x10^3 kg/m^3\n\nwe know that the new arrangement is a ballanced force arrangement.\ntherefore the change in weight must equal the change in upthrust.\n  delta W = delta upthrust\n\nupthrust before:\nF upb = densityxgxVolume b\nFupb = densityxgxAreaxh    (h=depth of ship)\n\nupthrust end:\nFupe = densityxgxVolume e\nFupe = densityxgxAreax(h-10.5) \n\nchange in upthrust:\ndelta Fup = denistyxg(Ah -A(h-10.5))\ndelta Fup = densityxgx(10.5A)\n\nchange in weight:\ndelta W = mg\n\nsubs in:\nmg = densityxgx(10.5A)\nm   = densityx(10.5A)\nm/density = 10.5A\nA = m/(10.5xdensity)\n\nm = 1507480kg,  density = 1000kg/m^3\n\nA = 1507480/(10500)\nA = 143.57 m^2\n\nsmall in comparisson to flight deck, however as i said this is assuming a uniform area of hull ie floating cuboid, aship would be much more like a triangular prism with one point sunk into the water. i'll be dammed if i'm going through the calculas to work out that one!!!!!
925295
Since anti-gravitation is still SF, the only way I can think of is to take a steep dive with an aeroplane. You won't be weightless in relation to the earth, but in relation to the plane it seems like so.\nAnother situation may appear when you actually could get to the center of the Earth. There are still gravitational forces, but you are pulled equally in every direction by them, so you would be floating.
844498
You have uncovered one of the secrets of the universe. Simple Harmonic Motion (SHM) surrounds us in every aspect of the universe. It is the lowest common denominator. Even Newton got it slightly wrong in his First Law where he said that a body will continue in uniform motion until an external force acts upon it. There is nowhere in the known universe that gravity does not exist.\n\nConsider an air table which is flat and which extends in all directions by a million miles, and it rests on the earth. place a puck on it and give it a gentle nudge. It will move off in the direction you push it at an apparently constant velocity, however it will eventually stop and return to the point where it started because of the earth's gravitational field. It will shoot past and head off in the opposite direction until it stops again, being pulled back by the earth's gravitational field. It will continue this way forever (ignoring all friction), oscillating in simple harmonic motion. \n\nNewton's problem was that he was restricted to observing a very small portion of the SHM path, so it seemed linear.\n\nAnyway, congratulations, you are on the path to unlocking the secrets of the universe.
966128
use the universal law of gravitation to find the force of gravity at the inner radius.  Use 1kg as your second mass (mass is irrelevent in this next part anyways), and use this force to find the accel of gravity.  Using the force and accel of gravity, you can find the kinetic energy needed in order to overcome the potential gravity of the planet (m cancels).  with kinetic energy you can find velocity, and with velocity and radius, period.  Hope that helps.\n\nfor the second part, repeat, except with the outer radius as D.
93072
As we are not, the Earth is spinning at a pretty constant speed (1 rotation / day).  Based on the position we are on the Earth, we are traveling around at different speeds, the closer one is to the equater, the faster one is traveling.\nAt the same time, the force of gavity is pulling us toward the center of the Earth, this force acts as the centripetal force (NOT to be confused with the ficticious centrifugal force) that keeps us in circular motion.\n\nIf the Earth's rotation we to increase, the force needed to keep up down on the ground would need to increase too.  However, the force (gravity) is only so strong, it can only pull up down by some much.  So it one were to increase the spin of the Earth enough, it is possible for the centripetal acceleration to equal the gravitational acceleration....therefore we would be weightless.\n\nBut now the question is, what happens is we further increase the Earth's rotation past this point of weightlessness?\nIf you spin the Earth faster, our weight would not be enough to keep us in the circular motion we are currently in, so it would have the effect of seeminly lifting up off the ground.  HOWEVER, we would not fly up into space.  The moment we leave the ground, the centripetal acceleration needed is reduced (a = v^2 / r, when r increases, a decreases).  If we could maintain a constant speed once we leave the ground, we would eventually lift up enough to that we get to a height so that the centripetal acceleration equals the gravitational acceleration again and we would hover at that height.
1014901
Very impractical for at least two reasons:\n\n1) It would have to be an unbelievably powerful magnetic field to generate any significant lift. The Earth's magnetic field is very weak.\n\n2) As soon as it left the ground the vehicle would be trying to flip over to an attitude where the Earth's magnetic field attracted it. You can't have a south magnetic pole without a north one and vice versa. As soon as this flip happened, the vehicle would crash back to Earth hard.
109780
As you move "downward" towards the center of the Earth, gravity will decrease. As you pass through the exact, physical center of Earth, gravity will be zero. As you move away from the center, gravity will begin to increase.\n\nYou can roughly compute the force of gravity with this formula --\n\ng = G*m/r^2\n\ng - gravitational force in Newtons\nG - gravitational constant (6.673^minus 11)\nm - mass of Earth\nr - distance from center of mass
1175221
I do not understand why you specified standing on the "Galapadous Island [Galapagos Islands?]" as opposed to anywhere else on Earth.\nNever-the-less,\n\nSince you are on Earth, you would experience an acceleration due to gravity.  Similarly, on Earth, one would experience a centripetal acceleration caused by the force of gravity.  Gravity acts as the centripetal acceleration, these two accelerations are not really separate.\n\nAt the same time, the Earth (with you on it) is orbiting the sun, which is intern, orbiting the galaxy, which is intern orbiting the center of a galactic cluster, and so on.  All this orbiting implies a force of gravity and a centripetal acceletation (again, the same acceleration).\n\n\nEDIT:\nYour question was about acceleration.  Acceleration requires a net force to be present.  In most of the examples you listed, those force acting on a body are very weak and are easily canceled out by forces of friction.  If the force is canceled out and no net force remains, there is no acceleration.
596583
Mass. I think.Depending on how strong the gravity is your mass changes.
514578
Yes, Gravity works everywhere, but there are some kinds of gas that result lighter than gravity so they float or even goes up.
247359
Mass interacting with gravity
110744
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
374997
The sun is the larger object. So, it has more gravity. The earth moves around the sun and centrigfugal force keeps it going around rather than crashing into it. It continues to move because there is no gravity or atmosphere in space to slow it down. Once a spaceship or rocket leaves the earth's atmosphere and pull, it keeps going even without its rockets on. It just keeps going in the direction it was pointed. \n\nTo learn more, see http://spaceplace.nasa.gov/en/kids/phonedrmarc/2002_july.shtml
852481
I think you pose a very insightful and valid question. I've wondered this same thing myself and dreamed about the day we can created a structure like the one you describe.\n\nHowever, to create a rotating torus large enough to mimic earth's gravity would be unfeasable with the enormous cost right now to build in space. Plus, the way the space station is built puts almost no stress on the station. If the station were rotating it would have to constantly withstand the forces of "gravity".