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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?

Tom Veatch Wichita, KS USA

wrote:

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.

http://physics.about.com/cs/dimensional/a/230603b.htm

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 gravity.].

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? :-)

Bill.

alt.astronomy alt.physics sci.physics

[snip]

Nope. The ratio is sqrt(x).

-- Regards, Doug Miller (alphageek-at-milmac-dot-com)

For a copy of my TrollFilter for NewsProxy/Nfilter, send email to autoresponder at filterinfo-at-milmac-dot-com You must use your REAL email address to get a response.

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).

But, but,... The moon is moving -faster- than the earth, whenever the moon is more than 1/2 full, time is -slower- on the Moon, due to velocity effects. To an observer outside the Earth-Moon, that is.

#### Site Timeline

- posted on May 18, 2004, 3:04 am

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?

Tom Veatch Wichita, KS USA

- posted on May 17, 2004, 12:38 am

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.

http://physics.about.com/cs/dimensional/a/230603b.htm

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 gravity.].

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? :-)

Bill.

- posted on May 18, 2004, 6:01 am

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!

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!

- posted on May 18, 2004, 12:38 pm

alt.astronomy alt.physics sci.physics

[snip]

Nope. The ratio is sqrt(x).

-- Regards, Doug Miller (alphageek-at-milmac-dot-com)

For a copy of my TrollFilter for NewsProxy/Nfilter, send email to autoresponder at filterinfo-at-milmac-dot-com You must use your REAL email address to get a response.

- posted on May 19, 2004, 12:51 am

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

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).

- posted on May 19, 2004, 10:45 pm

In rec.woodworking

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.

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.

- posted on May 20, 2004, 3:13 am

But, but,... The moon is moving -faster- than the earth, whenever the moon is more than 1/2 full, time is -slower- on the Moon, due to velocity effects. To an observer outside the Earth-Moon, that is.

- posted on May 20, 2004, 6:12 am

In rec.woodworking
snipped-for-privacy@host122.r-bonomi.com (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 http://www.phys.lsu.edu/mog/mog9/node9.html

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 http://www.phys.lsu.edu/mog/mog9/node9.html

- posted on May 22, 2004, 3:54 am

decided to post "Waaay OT- Lunar physics question" to
rec.woodworking:

Hoyt, 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.

/ts

Hoyt, 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.

/ts

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