On gravity’s smallest unit

Sun, Jul 16, 2006 at 11:33 AM

That is a thought I harbor for a while already(released to my
friends and co-inventor)
When doing calculation of Free Fall under Gravity, we use calculus
which assume that the time interval between each acceleration is
infinite small. Is it really the case in reality?
I had a thought that either gravity by itself or the interaction of
gravity with the matter in an object require some minimal time to take
effect. Therefore, instead of assuming the time interval is
1/infinite, we could assume some other value. For instance, 0.0001
second. Now, if we can do an very accurate experiment on Free Fall or
other gravity-only experiment. We can check if we are using the right
time interval. i.e. We can deduce from this experiment of the REAL
smallest interval of time which gravity take effect.
For instance, if we assume the time interval is 1/10^8 of an object
free fall for 3 second. We can do calculation base on this time
interval and compare that to the calculus result.
At t=0, v=0
t=1*10^8, v=g/10^8….
d=summation of v*t until t=3
We could then compare that with the official formula d=4.5g
The general formula of total distance traveled for divide time
interval into 1/p second over a period of n second is:
the official is d=1/2*g*n^2
Their difference is -ng/p

Therefore if we have a very very accurate measure of d of free fall
with very very accurate measurement of gravity at the site. And take
place in vacuum. We can calculate the difference between the result
and the theoretical value from Physics textbook. And substitute into
this formula, presumably we already know p(time it took) and
g(gravity), it is possible to get n. i.e. We are able to deduce the
smallest unit of time that gravity interact with a mass.
If you are still not satisfy with the precision as gravity vary
with distance, we could go back to the beginning and calculate the v
for each interval, summation of all v*t and get the most accurate result




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