From Yahoo Answers

**Question:**

**Answers:**If air density were constant with altitude (which is not) AND acceleration of gravity were constant with altitude (which is not) then you would need a column almost 8 kilometers tall to have 101,000 N/m^2 or 1 Atmosphere at sea level. P = density x acceleration of gravity x altitude 101,000 = 1.29 x 9.81 x altitude altitude = around 7,981 Kilometers. The issue is that air density changes with altitude and air pressure too. Same thing happens to acceleration due to gravity. Considering that: 50% of the atmosphere (mass) is below an altitude of 5.6 km. 90% of the atmosphere (mass) is below an altitude of 16 km. 99.99997% of the atmosphere (mass) is below 100 km. i would say that MOST of the atmospheric pressure we feel at sea level is MOSTLY due to a column of air (with decreasing density as altitude increases) about 16 kilometers tall. The entire atmosphere contributes to the atmospheric pressure at sea level but there is no clear boundary between our atmosphere and the outer space. For this reason we could assume a reasonable and practical figure like 16 kilometers instead of mentioning the upper part of the Exosphere, which is at about 10,000 kilometers from sea level. Nonetheless, even the small molecules found in the Exosphere exert pressure on us, but their contribution is too small to be considered.

**Question:**my example is if your leaving earth or approaching earth going inside the earth toward its core. than would the acceleration formula change in these two examples?

**Answers:**The strength (or apparent strength) of Earth's gravity varies with latitude, altitude, and local topography and geology. For most purposes the gravitational force is assumed to act in a line directly towards a point at the centre of the Earth, but for very precise work the direction can also vary slightly because the Earth is not a perfectly uniform sphere. Gravity is weaker at lower latitudes (nearer the equator), for two reasons. The first is that in a rotating non-inertial or accelerated reference frame, as is the case on the surface of the Earth, there appears a 'fictitious' centrifugal force acting in a direction perpendicular to the axis of rotation. The gravitational force on a body is partially offset by this centrifugal force, reducing its weight. This effect is smallest at the poles, where the gravitational force and the centrifugal force are orthogonal, and largest at the equator. This effect on its own would result in a range of values of g from 9.789 m s 2 at the equator to 9.832 m s 2 at the poles. The second reason is that the Earth's equatorial bulge (itself also caused by centrifugal force), causes objects at the equator to be farther from the planet's centre than objects at the poles. Because the force due to gravitational attraction between two bodies (the Earth and the object being weighed) varies inversely with the square of the distance between them, objects at the equator experience a weaker gravitational pull than objects at the poles. In combination, the equatorial bulge and the effects of centrifugal force mean that sea-level gravitational acceleration increases from about 9.780 m/s at the equator to about 9.832 m/s at the poles, so an object will weigh about 0.5% more at the poles than at the equator. Gravity decreases with altitude, since greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to the top of Mount Everest (8,850 metres) causes a weight decrease of about 0.28%. (An additional factor affecting apparent weight is the decrease in air density at altitude, which lessens an object's buoyancy.) It is a common misconception that astronauts in orbit are weightless because they have flown high enough to "escape" the Earth's gravity. In fact, at an altitude of 250 miles (roughly the height that the Space Shuttle flies) gravity is still nearly 90% as strong as at the Earth's surface, and weightlessness actually occurs because orbiting objects are in free-fall. If the Earth was of perfectly uniform composition then, during a descent to the centre of the Earth, gravity would decrease linearly with distance, reaching zero at the centre. In reality, the gravitational field peaks within the Earth at the core-mantle boundary where it has a value of 10.7 m/s , because of the marked increase in density at that boundary. ..

**Question:**Hi I needed help. Size, shape and mass of objects. I am not sure whether any of these variables have any effect on acceleration of an object. If so could please tell me what effect it has for instance: does it cause the acceleration to increase, decrease or stay the same?

**Answers:**shape has no effect on rate of acceleration per se, although depending on shape and where the force is applied the object will accelerate differently--either in a different direction, or rotationally as opposed to just directionally. size doesn't directly affect acceleration either, except in two ways: 1. generally, the larger an object is, the more mass it will typically have, and mass affects acceleration--but size isn't a good indicator because the density of the object will also determine how massive it is. 2. air resistance; a large object will encounter more air and so, the faster it goes, will experience less acceleration even with the same force because the force of air resistance will be pushing in the opposite direction. mass is really where it's at in terms of acceleration. the equation is F=ma, where F=force, m=mass and a=acceleration. so the greater the mass, the less the acceleration with a constant force.

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