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First in answer to your sign question.
The formula will work as long as the results are taken in context with the original directions selected as positive... which was not consistent.
The forces and acceleration should have opposite signs and the formula for total acceleration should be AT = A1 - A2 but it was more confusing at first glance that way .
You said: ( from the other post )
Please show me how the equation AT = (G / R(^2) ) x ( M2 + M1 + ...) could do that for just two dissimilar mass's that we are trying to determine difference or lack of difference of their rate of fall in the neighborhood of a massive object.
I wont try the to write the math for this, but by inspection it is easy enough to compile an equation.
Let say we have masses 1, 2, 3, ... ,n.
The total acceleration for mass 1 would be equal to the sum of acceleration vectors caused by each of (n-1) masses. ( all masses but itself )
By multiplying by a correction factor C1, C2, ... Cn to reduce the total acceleration vector to the component of the vector which is in the direction we are inspecting, we can say that the total acceleration of mass 1 in a specific direction is:
A = F / M ...and I know you can quote the equation of force due to gravity.
j is tag identifying any specific mass we choose.
n represents a set of all object tags excluding j.
Rj3 is the radius from mass j to mass 3.
you can check the other post for the rest.
Which gives us
Aj = [ Cj x G x (Mj x M1)/(Rj2)^2 + C3 x G x (Mj x M3)/(Rj3)^2 + .... + Cj(n-1) x G x (Mj x Mn))/(Rjn) ^ 2 ] / Mj
Which reduces to
Aj = G x Summation of [ Cn x Mn / Rnj^2 ]
This is interesting....
The acceleration of n does not depend on the quantity mass of mass in n.
( this part is weird .... uh I mean .. Definitely counterintuitive.)
However....
If you have two masses x and y accelerating towards each other....
then the component of the acceleration of x would be affected by the quantity of mass in y.
and the component of the acceleration of y would be affected by the quantity of mass in x.
making the total acceleration of both affected by both masses.
When you say the mass of the basketball does not affect its falling acceleration you are saying the basketball falls toward the earth but the earth does not fall toward the basketball.
A very close approximation.
Here is the part I think you will like.
If you drop a basket ball and a bowling ball ( at the same time ) then the earth will fall toward both balls with an acceleration component due to the combined mass of both balls.
The balls will fall toward the earth with an acceleration independent of their own mass.
Therefore if the bowling ball and the basketball are dropped AT THE SAME TIME ( in a vacuum with the same center of gravity ) they will fall toward the earth at the same total acceleration.
Im going to play some basketball now, my brain hurts. |
