variation of gravity with altitude?
823207
gravity is a special case  of gravitation.It is the force by which earth pulls a one kg object towards it.\n\nie g=GM/R^2\n M=mass of earth\nR=Radius of earth\n\nnow force acting on a object  of mass 1 kg at a height h(from earths surface) will be \nf=GM/(R+h)^2\nwe can take out R out of bracket so that (R+h)^2 can be R^2{1+h/R}^2\nNow\nGM/R^2=g\nSo we can easily WRITE it as \nf= g/{1+h/R}^2\n\nThis f is that something we usually write as g'\n\nNow g' can be written as g{1+h/R}^(-2)\n\nFOR NOT TOO SOPHISTICATED CALCULATION\nIT CAN BE REDUCED USING BINOMIAL THEORM TO \ng'=g{1-2h/R}
1146161
The barometric formula, sometimes called the exponential atmosphere, is a formula used to model how the pressure (or density) of the air changes with altitude. It is based on the simplifying (not very realistic) assumption that the temperature does not depend on altitude. However, this formula agrees reasonably well with the actual pressure and density variations above the earth's surface up to a height of about 450,000 ft (140 km).\n\n \np=p0exp(-Mg0z/RT)\n\n\n\n \nwhere h is the scale height, ρ (rho) is density, P is pressure, P0 is pressure at ground level (mean sea level pressure is 1013.25 hPa), M = 0.029 kg mol-1 (the mass of 1 mole of air), R = 8.314 J K-1 mol-1 is the gas constant, T is temperature, g0 is the acceleration due to gravity (about 9.8 m s-2 depending on your location) and z is the vertical height above the earth's surface.\n\nUsing the same principles, the above equation can be solved for altitude as a function of pressure. This formulation is known as the hypsometric equation.\n\nAs a rule of thumb, the pressure decreases by about 1% for every 80 metres increase in altitude.
1082685
Simply equate gravity function with centripetal acceleration.\n\nGMe/(r+h)^2 = V^2/(r+h)....where r is earth radius and h is height\n\nsolve for v
849468
gravity at the earth's surface = 9.80 m/s^2
418803
Theoretically, the gravity at the north pole should be higher.  The reason being: The earth is not a perfect sphere, even if you filled in all the oceans and leveled all the mountains.  This is because the earth's spin creates centripital force, which forces the earth out at the equator.  So, at the north pole, you would be closer the the earth's core, and therefore the center of it's gravity well.  The closer you are to the center of a gravity well, the more you weigh.  \n\nIF you are trying to find the exact strength, you could determine your distance from the north pole and amount of centripital force and gravatational force acting on your body at your lattitude, then subtract it from the gravatational force on you at the north pole, but I don't know how you'd do that.  FInd someone with a really fancy calculator.
28151
Acceleration due to gravity (NOT change in acceleration), which could also be described as change in velocity.  That's assuming no air resistance.  If there is air resistance, they will eventually reach terminal velocity after which their velocity will be constant, but this is not implied by "free fall," and given that the question sounds like it's from a basic middle school or high school physics class I'm guessing you're supposed to ignore air resistance.
634175
It´s not quite clear to me what you are looking for, but here goes....\n\nThe air pressure goes down with the height in the atmosphere. In a gas mixture the molecules are free to roam where they can but under the influence of gravity they stick mainly close to the earth. The pressure drops by about 10% over the first kilometer and is about a third of that at groundlevel at the top of the Mount Everest. The decrease is exponential and the gradient depends on the average velocity of the molecules (a few hundred meters per second).\n\nIn the ocean pressure increases linearly with depth approx. 1 atmosphere each 10 meters. The reason is that the weight of the upper layers presses on the lower layers of a liquid.   \n\nJess knows somewhat more.
1054239
no it doesn't affect gravity.  it's a double reversal, so everything is perceived right side up.  hold something in front of your face.  then turn it upside down, that's how your eyes see it, flip it around again (notice how it's right side up now?) this is how your brain sees it.\n\nit's all cool, you won't fall off the face of the earth...  ;-)
534088
gravitational pull is different in different places on earth. but it is fixed. we cannot change it locally or globally. bcoz it depends on earth's mass, which we cant change.
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
474726
You have gravity holding down air at lower elevations due to the mass of the Earth. The higher you rise, the less gas there is contained in the atmosphere (but you probably already know this). As you have less mass and more vacuums in higher altitudes, the need to fill the vacuum becomes much greater. The water starts to fill here and other water tends to coagulate around these base droplets. You eventually find groups of water together that tend to vary based on temperature.
301536
Weight = Acceleration*Mass\n\nAcceleration of the free fall is the earth gravity that is 9.797 645 m/s²\n\nWeight = 9.797645*Mass\n\nfree-falling object is an object which is falling under the sole influence of gravity; such an object has an acceleration on Earth of 9.8 m/s/s, downward. This numerical value for the acceleration of a free-falling object is such an important value that it has been given a special name. It is known as the acceleration of gravity – the acceleration for any object moving under the sole influence of gravity. As a matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it – the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. \n\nThere are slight variations in this numerical value (to the second decimal place) which are dependent primarily upon altitude. The Physics Classroom will use the approximated value of 10 m/s/s in order to reduce the complexity of the many mathematical tasks performed with this number. By so doing, you will be able to better focus on the conceptual nature of physics without sacrificing too much in the way of numerical accuracy. When the moment arises that you need to be accurate (such as in lab work), use the more accurate value of 9.8 m/s/s. \n\ng = 10 m/s/s, downward\n\nthe acceleration is the rate at which an object changes its velocity. Between any two points in an object's path, acceleration is the ratio of velocity change to the time taken to make that change. To accelerate at 10 m/s/s means to change your velocity by 10 m/s each second.\na = dv/t = 10/1 = 10\nthen if a body falls in the first second his speed is 10m/s and in the 2nd second is 20m/s and so on
61784
That's an interesting qts. I assume that you wish to invlove pure physics in this qts. Forget all the heat at the core of the earth and also air resistance.\nConsider pure gravitation theory.\n\nIf say the fellow who fell in the hole has mass m\n\nThen force of gravity exerted by the earth is :\n\n  F = GMm/(r*r) ..... Where G - universal gravitation constant\n\n                            M - Mass of the earth\n                            r - distance between earth and body of mass m\n\n          acceleration due to gravity\n            = (1-h/R)g   .... where h is distance of body beneath the earth's surface and g is acceleration due to gravity at the surface of the earth.\n\nSo as the body comes closer to the core of the earth, h tends to R.\nSo the acceleration of the body goes on decreasing very much. So, body may not actually hover, since his velocity will be constant the instant it is at the core.
824167
The radius is actually 2/3 of the altitude.  Try that
99997
Increased lift at low altitude due to ground effect, yah? \n\nI'm curious.  You apparently know the answer, so why ask the question?
614736
There is no point in space in which you are not being gravitationally attracted to something, therefore you always have a "weight".\n\nNewtons Law of Universal Gravitation,\nF = (G * M * m) / r^2\nwhere F is the force of gravitational attraction and r is the distance between the two masses, M and m.\nThe force of gravity only goes to zero when the distance between the two objects goes to infinity.  You can never be an infinite distance away from anything.\n\nHowever, just because your always within a gravitational field does not mean that you can not experience a sense of weightlessness.  During a free fall, one experiences weighlessness, and this could (theoretically) be experienced at any altitude.\nThere is no set height which one must obtain in order to magically become weightless.  Just because you are X miles above the Earth does not mean you are weightless.
912921
altitude = height = 552km\nAngle of dep = 39.7\n\nTangent of Angle = Alt/ Radius\n\nRadius = Altitude/ Tan(39.7) = 552/tan(39.7) I think... p.s not really sure