Re: Rotating Earth and its effect on Gravity

On Jan 3, 1:14 am, blackboab <blackb...@xxxxxxxxx> wrote:
Reading a physics textbook to refresh my knowledge (pretty basic) of
the theory of Gravitation and read that the measured gravity is
affected by the rotation of the earth .

the formula is g=a-wR^2

where g is the measured gravitational acceleration , a is the
gravitational acceleration derived from Newton's law of Gravitation ,
w is the earths rotational speed derived from (2*pi) / (24*60*60) and
R is the earth's radius.

It is clear that if the earths rotation were to increase in speed then
the effective gravitational force would decrease .
it would reach zero metres per second per second at a rotation speed
of 84 minutes instead of 24 hours and if the rotational speed
increased beyond that then objects would start flying off the earth's
surface,.

given all that why do textbooks like Halliday and Resnick insist on
maintaining the fiction of a centripetal force directed towards the
centre of the rotating earth when it is a centrifugal force acting
outwards which causes the force of gravity to reach zero and then
become negative ?

Because the second term in the expression is not due to a force. It is
instead just a piece of the real acceleration, moved over
algebraically to the other side of the equation. Centrifugal force is
a fictitious force, something that appears only as a result of this
algebraic manipulation, something functionally equivalent to moving to
an accelerated frame of reference.

Let's start with Newton's second law:
ma = Sum(F)
Now, let's apply that to something sitting on the surface of the
earth. In this case, there are two forces acting on the something:
gravity downwards and the contact force N that the ground pushes up on
the something to keep it from penetrating the solid ground. We have:
ma = [GMm/R^2] - N
The first term on the right is due to gravity.
The term on the left includes an a which is not the same a that you
referred to. Here it is simply the acceleration. Note that it is NOT
zero, because sitting on the Earth still involves going around in a
circle and so there is acceleration. That acceleration is a = w^2R.
(You made a small typo calling this wR^2.)
So we have
mw^2R = [GMm/R^2] - N, for an object sitting on the ground. Here you
can see that N is *not* quite equal to the force of gravity downwards.
It can't be, for if it were, there could be no acceleration, and if
there were no acceleration, the something couldn't move in a circle to
stay on the surface of the Earth.

Now, let's lift the something off the ground and let it go, so that
there can no longer be any contact force N acting on the something.
Now we know N=0. Well, clearly, the acceleration is going to change.
The *additional* acceleration is what we measure as g:
m(g + w^2R) = GMm/R^2.

Be sure to look at the equation above carefully. Note that all the
accelerations are on the left and all the forces are on the right, so
that it still looks like Newton's 2nd law.

But now I do a little algebraic maneuver, dividing through by m and
moving one of the terms on the left to the right:
g = GM/R^2 - w^2R
The first term on the right is what you called a, the "gravitational
acceleration derived from Newton's law of gravitation".

But moving an acceleration from the left side to the right side is
just an algebraic stunt. It doesn't magically convert an acceleration
into a force. All you're doing is taking a force, finding out a total
acceleration, then subtracting off a piece of that acceleration to see
what's left. g is the "what's left", the part of the acceleration over
and above what you have just from sitting on the ground of the
rotating earth.

Moving an acceleration from the acceleration side to the force side
gives you a term that *behaves* like a force, but it isn't a real
force. This is why centrifugal "force" is really just a fictitous
force at best, an artifact of algebra.

PD
.

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