Re: [Phys-l] NYT article: Centrifugal force

[John Denker <jsd@av8n.com>]

On 07/03/2009 06:43 AM, alex brown wrote:
> How would one go about calculating this one third of a degree?... I
> would just like to know the starting point so I can have a go.

Actually it's a tenth of a degree.

Suppose our observer is in a boat in the middle of a calm
secluded pond.  The surface of the pond defines our operational 
definition of horizontal.

I do the calculation in the frame of the boat.  This frame
partakes of the rotation of the earth, and therefore includes
centrifugal fields.

There can be no net horizontal acceleration acting on the
water;  otherwise it would not stay where it is.  Therefore
the acceleration must be purely vertical.  This acceleration
defines what we mean by weight, and defines what we mean by
"down".  Things accelerate "down" when we drop them.  The
dominant contributions to the acceleration are:
 -- the "universal" term, GM/r^2 directed in the r direction, and 
 -- the centrifugal term, ρω^2 directed in the ρ direction.

Here r denotes the radius vector from the center of the earth
to the boat, and ρ denotes the projection of r in the direction
perpendicular to the axis of rotation, i.e. the projection onto
a plane parallel to the equator.

Calculate the centrifugal term from the definition.  Estimate
the other term as nearly equal to "standard" gravity.
  http://physics.nist.gov/cgi-bin/cuu/Value?gn

Add these two acceleration terms in the usual way, tip to tail,
with due regard to angles.  The resultant is the total local 
acceleration field, and defines what we mean by vertical.

Calculate the angle of the resultant relative to r.  It's about
0.1 degree at temperate latitudes.


On 07/03/2009 07:04 AM, Dennis Erickson wrote:

> > The Moon is 1/2 degree wide.

Yeah.  Apparently I can't divide fractions this morning.

In any case, 20% of the diameter of the full moon still seems 
like a lot.