Galileos idea and the unknown phenomenon.

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Galileo's concept of a fixed gravitational acceleration value on Earth is widely accepted, though variations exist due to Earth's crust. The discussion highlights how the descent of cannonballs is affected by air resistance, particularly when using lighter materials. The calculations provided reference Newton's equations, demonstrating that in a vacuum, all masses fall at the same rate, but air resistance alters this for larger objects. The equations for air resistance are detailed, with specific constants for spherical objects, emphasizing the quadratic nature of the problem.

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Galileos idea of a fixed value for the gravitational acceleration on the Earth iss now accepted for most pratical calulations allthough variations in the Earth's crust mean that this value is not precisely uniform. However there is another phenomenon that will affected the decent of the cannon balls. If the cannon balls were made of a very light material there would be a very noticble difference in the fall time for different sized balls.

Why is this?

Please involve calculations if possible. I would much appreciate any form of help.:smile:
 
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If you think of something as a point mass, in an environment with no fluid resistance, then any mass's, large or small, experiencing the same gravitational pull, will accelerate and thus fall at the same rate.

When you introduce fluid resistance, ie air resistance, you get two things happening. Firstly, obviously, you get the larger (surface area wise) objects being slowed more and secondly you get terminal velocity limiting the speed of the object.

Calculation wise Newton tells us:
[tex]{v^2} = {u^2} + 2as[/tex]
Thus if initial velocity, displacement and acceleration are equal for all objects, regardless of mass, final velocity will be the same as well.

In terms of air resistance, the equations differ for different shapes. For a sphere of diameter D the initial equation is the following:
[tex]F(v) = -c1v - c2v *|v|[/tex] (sorry still gettin used to tex, c1 and c2 are constants )

These constants are experimentally determined and have the following values:
c1 = [tex]1.55 x {10^(-4)} * {D^2}[/tex]
c2 = [tex]0.22 {D^2}[/tex]

Thus usually c2 is used as the c1 value is small enough to drop out. By using c2 we realize we are dealing with a quadratic situation.

hope this makes sense
-G
 

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