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Re: [Phys-l] method for measuring the size of the earth



No. I haven't tried it yet. If anyone is going to the beach over
spring break maybe they can report back as to how well the method
actually works in practice.

David Bowman

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Vern
Lindberg
Sent: Thursday, February 26, 2009 9:25 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] method for measuring the size of the earth

Do you have sample numbers?

Vern

On Feb 26, 2009, at 9:14 AM, David Bowman wrote:

I've come up with a fairly simple means of measuring/calculating
the size of the earth using only local measurements (not requiring
multiple sightings at far away locations like Eratothenes' method
needs).

The idea is to observe and time the motion of the terminator at
sunset/sunrise ascending or descending the face of a building,
pole, or other tall structure with an exposed vertical face. It
is a fairly simple exercize in trigonometry to realize that if one
is situated on the equator during an equinox that the terminator
ascends vertically at sunset and descends vertically at sunrise
with a constant acceleration that is related to the rotation rate
of the earth and the radius of the earth. In this situation it is
a high school-level calculation to see that if a is the vertical
acceleration of the terminator then the radius R of the earth is

R = a*(T/(2*[pi]))^2

where T is the duration of one solar day.

If the observing location is not on the equator then there is a
correction for the local latitude of the observer. And if the
time of observation is not during an equinox then there is another
correction involving the declination of the sun due to the
tilted path of the sunrise/sunset at the horizon. Figuring out
these corrections is *not* at the high school level, but they
don't involve anything more than a lot of complicated
trigonometry. The date/place corrected formula involving these
corrections is

R*F = a*(T/(2*[pi]))^2

where F is a fudge factor that depends on the complications.
According to my calculations, the explicit value of F is

F = cos^2(D) + sin^2(L)*(cos^2(D) - tan^2(D))

where D is the declination of the sun on the date of observation
and L is the local latitude of the observation.

The actual value of a, being only a few cm/s^2, is plenty slow
enough for easy observation of motion of the terminator up or down
the observing wall using a wristwatch. To get the acceleration only
a few timings are needed between some fiducial marks on the wall
whose separation distance is measured.

BTW, in order for the method to work properly the horizon needs
to be an unobstructed 'true' horizon that accurately represents the
effectively smoothed surface of the earth. This means that the
observation ought to be done along a seacoast or the coast of a
great lake that can't be seen across from the top of the observing
wall. For those whose budgets are unlimited this means that a field
trip to Miami Beach for the purpose of observing the sunrise on the
beachfront hotels would make a very nice educational experience.

David Bowman
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

Vern Lindberg
vern.lindberg@rit.edu
585-475-2546
http://www.rit.edu/~vwlsps/




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Forum for Physics Educators
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