Re: galileo's gravity experiment


PD wrote:
gubernacullum wrote:
PD wrote:

In this case the size of friction/drag is the same, but the force of
gravity is different. This results in a different acceleration. To see
why requires a couple of algebra steps. Are you willing to work through
that?

PD

thanks pd for your helpful reply. am i to understand from this that the
famous galileo experiment was wrong and that the moon experiment was
wrong and that masses of different size should fall at different rates
wether on the earth or on the moon?

No, not at all. The force of gravity is different on a feather than it
is on a golf ball. The forces acting on the lead ball and the wooden
ball dropped from a tower are also different. So now the interesting
question is, why then are the accelerations the same if the forces are
different? And here is where some illumination occurs.

The force law is F = ma, or put another way, a = F/m
Now, if the force is different on two different objects, shouldn't a be
different? Not necessarily.
In particular, we know that the force of gravity acting on an object is
proportional to it's mass. The more mass it has, the heavier it is (or,
the stronger the force of gravity). That is, the F that is in the
numerator is some constant times mass: km.
If we stick that information in the numerator, we find an interesting
thing:
a = km / m, and right away we see the mass m cancels out. This is what
Galileo figured out: That even though the force of gravity acting on
lead and wooden balls is very different, the acceleration is NOT
different.

Now, suppose we have a balloon and we no longer neglect air friction.
a = F/m, and once again the force of gravity is km, but we need to also
include friction f.
The force F acting on the balloon is not just km but (km - f)
a = (km - f) / m
and if we do the algebra, we see
a = k - f/m
Now the number for the acceleration doesn't have the mass cancel out of
it completely. It's there in that f/m part. So we no longer expect
identical size balloons of two different masses to fall at the same
rate.

So why didn't this apply to Galileo as well? It did, but f/m is a small
number with his balls, compared to the number for balloons, so it just
didn't matter much.

PD

pd you appear to be a clever man but i am not prepared to forgive your
slight-of-hand at this time. it is very easy to misdirect through
equations to the uninformed. let me explain:

you say f=ma, ok.
you say a=f/m, ok.
you say a=km/m, ok.

but lets not proceed in a hurry from this 'conclusion'.

assume a wooden ball 1 and a lead ball 2:

ball 1: a1=k1m1/m1 implies: a1=k1
ball 2: a2=k2m2/m2 implies: a2=k2

i.e. ball 1 and ball 2 have different forces AND different
accelerations acting on them.

further, if we do a unitary analysis, a has units m/s2 and not m/s2/kg.
this would not be acceleration by definition. unless frictional force
was mass dependent, in which case the masses would cancel out and you
would have the same result as in a frictionless scenario.

could you review your analysis please and respond.

.

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