It is so, and what of that? what did you proove?

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a=G*M/r^2 so?
da=-2*G*M*(dr/r^3), so da~dr/r^3 that means, for r big enough, da~0.
Is it so?

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Yep, if you''re far enough away (r large enough) the effect on acceleration is negligible. Also, da~M, so light objects (small enough M) have very little effect...

Remember that accleration of an object when dropped from a certain height is 9.8m/s x s . That means an object dropped from a 1000 ft building or 100 ft building the acceleration will be 9.8m /s x s. That means 9.8 meters per second, per second. Acceleration is always squared that does not mean that it is area.

This is the problem with teaching physics without calculus...

Acceleration is not "squared". It is the rate of change of the rate of change in position. In other words, it is the second derivative of position with respect to time. This is very different from "squared".

This also shows the danger of ignoring the concept of "force" and thinking only about acceleration. On the descent from a 100m or a 1000m building, the amount of gravitational force acting on the body will remain more or less constant. However, the amount of force due to friction, etc. will increase as the velocity increases. At some point, the force due to gravity will be cancelled out by the force due to friction. The acceleration will be zero and you will have terminal velocity. Therefore, when you start to think about the net *force* acting on a body, you can say that the *average* acceleration over the descent from a 1000m building is less than the *average* acceleration over the descent from a 100m building. Having said all that, you can safely ignore the difference unless the descent distances are very different.

Author, "Real Time Rendering Tricks and Techniques in DirectX", "Focus on Curves and Surfaces"

Don''t mess with Micromania. He is good at math.

my mistake...

Author, "Real Time Rendering Tricks and Techniques in DirectX", "Focus on Curves and Surfaces"