Quote:
Originally Posted by
Hiro
I flipped the words centrifugal and centripetal when responding to a post about centrifugal force reducing the feeling of gravity (
@mstones post). A terminology faux pas, but the math was always and still is still correct when you unflip my booboo term. How nobody else realized that and mentioned it then this is surprising having a look back at the original post.
OK this is better. Let's have one more attempt to see if we can reconcile this. I have no problem forgiving a mistyped word, but I was referring to post 117 where you said:
"
Bad assumptions. You are trying to measure from the wrong frame and wrong coordinate system to make that direct comparison. When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years)."
You used that to dispute the simple derivation of weight at the equator that I posted in post 116. However, the only assumptions that I made were the equations of motion for simple, constant, circular motion, which I then derived from first principles in post 131.
Further, those equations are independent of the frame of reference, provided that it is an inertial frame of reference. You said "
when the frame of reference is the center of rotation of the earth". That would be an inertial frame of reference if it does not rotate with the earth, or a noninertial frame of reference if it does rotate with the earth. Since you ruled out the existence of centrifugal forces (which only exist in noninertial rotating frames of reference), I concluded that you too were referring to the nonrotating inertial frame in which my analysis was conducted.
Quote:
Originally Posted by
Hiro
We can both agree that @mstones quoted virtual centrifugal force absolutely cannot reduce the force of gravity because it doesn't actually exist, not to mention the fact that the virtual centrifugal vector points in a direction that would add to the apparent gravity vector, contradicting a lessening.
I agree that in the nonrotating, inertial frame of reference of the center of the earth, there is no centrifugal force  but two caveats:
 In the rotating, noninertial frame of reference that is the apparent frame of reference to an observer on the surface of the earth, that force does exist, and the analysis in that frame of reference does produce exactly the same result.
 In that noninertial frame of reference, the centrifugal force vector points outwards (+ve r direction), so it is in opposition to the gravitational force vector that points towards the center of the earth (ve r direction) and would reduce it.
However, let's stay in the inertial frame of reference where we don't have a centrifugal force.
Quote:
Originally Posted by
Hiro
As for the rest, it's simple. Gravity is gravity. I don't care about centripetal forces as a component of gravity, because they aren't  that has been my whole point. Yes they are why the earth accreted the way is did, but they are NOT a component of gravity. Period. Ever. (combined forces apparent gravity != gravity)
Well yes, I agree, but that's not what you said earlier. And it is not actually that simple, because we were discussing
g, our measure of gravity on earth. I made that distinction clear in #116, and yet you still dismissed the analysis as flawed for a bunch of other spurious reasons, most notably claiming (see above) that there was no centripetal acceleration on the equator.
So  let me just restate the argument one more time to make sure that we are not just talking about different things. Maybe you can identify precisely where you disagree, if you still do.
The original discussion was about variation in weight,
W, with latitude, where the weight of an object of mass
m is defined by
W =
mg . [1]
From my derivation in #131 that I won't write out again, the centripetal acceleration,
a, of an object on the equator, equatorial radius
R, angular velocity of rotation
ω, is
a = 
ω²
R . [2]
The force, F, exerted on the object due to the gravitational attraction of the earth, mass
M, is given by
F = 
GMm/
R² . [3]
W, the weight of the object, is by definition the normal reaction between the object and the surface of the earth, and since the resultant force on the object must equal the product of its mass and acceleration,
F +
W =
ma . [4]
Substituting for
W,
a and
F from [1], [2] and [3] respectively, we get

GMm/
R² +
mg = 
aω²
R . [5]
Rearranging, and dividing through by m,
g =
GM/
R² 
ω²
R . [6]
Writing this more generally than for just on the equator, for latitude
L, since we need two different
Rs for the two terms on the RHS, one for distance to the center of the earth,
R(
L), in the first term, and one for distance to the axis of rotation,
r(
L), in the second term, and take into account that the acceleration vector is normal to the axis of rotation rather than the surface of the earth
g =
GM/
R² 
ω²
rsin
L , [7]
or explicitly writing r =
Rsin
Lg =
GM/
R² 
ω²
Rsin²
L , [7]
OK  so this is the g that we use as acceleration due to gravity on earth  it is what we multiply mass by to get weight, and it is the acceleration towards the earth's surface of an object when dropped  and it varies with latitude, not only because the oblate shape of the earth means that R varies slightly, but also because L goes from 90˚ at the poles to 0˚ at the equator. It is the quantity that was being discussed.
In our inertial frame of reference, no question that it differs from the gravitational field strength (
GM/
R²) by the centripetal acceleration, as I pointed out in an earlier post when I was trying to reconcile the different views. Alternatively, you can view it as equal to the gravitational field strength in the frame of reference of an observer on the surface of the earth, which as I mentioned earlier, is a rotating, noninertial frame of reference.
And, as a side note related to that observation, remember that although centripetal acceleration and gravity are different in cause, a central result of Newtonian physics is that gravity and acceleration at a point are indistinguishable  in other words an infinitesimally small* observer cannot tell whether he is accelerating in an inertial frame of reference versus moving at constant velocity in an inertial frame of reference in the presence of a gravitational field.
*
Only holds for infinitesimally small, since for an observer of finite extent, gravitational attraction will vary detectably with R, whereas actual acceleration will not.Quote:
Originally Posted by
Hiro
What is so freaking hard about this?
Well  I'd really like to ask you the same question. Are we done?