Standing on the North Pole, look at any point on the horizon that you choose. That
line of sight, extended indefinitely on the
surface of the earth, will pass through the South Pole. Therefore, that direction, by
definition is South.
One of two singularities on the Earth's surface. The other, of course, is the South
Pole where every direction you look is North.
There is an interesting little "trick" question that takes advantage of that.
Where on Earth can you go 1 mile South, 1 mile East, and 1 mile North and be back
exactly where you started?
Wichita, KS USA
The others are good references, but this might be simpler and to the
point: Look at the period being proportional to the root of (l/g).
This means that it is proportional to the length [gets longer as the
root of the length increases, not important here since that won't
change] and inversely proportional to the root of gravitational
acceleration, g, as measured on that lump of dirt.
So the period will increase [slower clock] as the gravity decreases,
and is measured by the root of that value [about 1/6th Earth's
Look at it this way: It takes longer [for the pendulum bob] to fall
the same distance on the moon.
So, was this your son's weekend homework assignment? :-)
Hoyt Weathers thought it a good use of my time to say:
Wow! I wish there was a newsgroup where this was on-topic! More "space
math" than I have seen in a long time. The way I see it, though,(like you
really want to know) is that the issue has been "under simplified". The
clock works through the use of (a) falling weight(s). This weight has its
"weight" based on the total force required to overcome the
friction/inertia/gearing needed to move Mickey's hands around the clockface.
If the moon's gravity is .XXX times the earth's gravity, then that clock
will move .XXX times slower on the moon. AFAIK, there is no other
weight-save the pendulum(which is "driven" by the rate of the falling
weight), and therefore there is no additional 'gravitational' requirement,
only the torque needed to spin the gears.
But, I could be full of it, and not know!
Nope. The ratio is sqrt(x).
Doug Miller (alphageek-at-milmac-dot-com)
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Are your eyes brown? <grin>
The pendulum regulates how fast the clock 'ticks'.
The energy of the 'falling' weight is just there to make up for the
friction losses in the gear chain, and the entropy loss in the pendulum
itself. This is why the weight falls _so_much_ slower than it would
if it were 'free falling'. It is only _allowed_ to fall enough to provide
the 'make-up' energy.
The period of oscillation of a pendulum is (a) proportional to the square-root
of the length of the pendulum and (b) inversely proportional to the square-root
of the local gravitational constant.
Thus, if all else remains the same, and the gravitational constant is
lowered to .xxx, the clock will run slower, by a factor of 1/sqrt(.xxx).
I'm surprised with over 40 posts, that no one has mentioned the effect
gravity has on time. While the pendulum will have the largest effect for
this experiment, EVERY clock, regardless of type be it atomic, spring,
quartz, etc, will run faster on the moon due to the lower gravity.
Einstein's general theory of relativity.
firstname.lastname@example.org (Robert Bonomi) wrote:
Granted, the velocity of the moon has to be considered but it is an
additional factor and all relatavistic effects have to be taken into
account, as it is in the GPS satellite system.
General relativity in the global positioning system
decided to post "Waaay OT- Lunar physics question" to
In deciding on using a division of the physical properties of the earth to
help create a universal (almost) system of measurement, of about
1/10,000,000 of 1/4 of half of a complete meridian passing through France
-- which they messed up anyway because of several factors, not the least of
which was failing to recognize completely the true oddity of the shape of
this ball we live on -- the French Academy rejected a measurement based on
moment of pendulums of given arc at a certain latitude mostly because they
determined that local distortions of gravity, such as provided by
mountains, would fail to provide the exactitude which these astronomers
wished for their standard.
Such a measurement taken over a fixed period of time, at one given
location, would not necessarily match that of another at another location.
Other factors, such as measuring the length of the pendulum, determining
the timing of the swing, atmospheric density and the quality and
repeatability of the observations, to name but a few, all proved that a
pendulum (eg clock) was not really that accurate. Given all of this, I
think that if you wish to measure time on the moon, you would do better
with a Swatch or Timex, at the very least.
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