# science fiction hour

Mike Hewson
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### RE: RE: So such a

Quote:
Quote:
So such a submarine, even if motionless with respect to the sea floor, will deduce the sum of the presence of the Earth ( gravitational 'attraction' ) and rotation of the Earth ( centrifugal 'repulsion' ) ! :-) :-)

If the Earth 'stopped' rotating (not suddenly) how much more
would we weigh ? without the centrifugal repulsion ?

Bill

Excellent question. ;-)

First point to note is that despite urban myth we won't 'float off' if the Earth stops rotating around it's own axis. So no need to tether yourself any more or less than you might for other reasons. :-)

Second is that the size of the effect depends on the latitude ie. where are we speaking of ? If you stand at either pole precisely - not magnetic, but at 'true' some 90 degrees away from the equator ie. with the Earth's axis of rotation going through you body from tip to toe - then you'll get your weight being that due to gravitational attraction of the Earth alone. And that will be true regardless of Earth's rotation rate, including zero. The radial distance to the centre of the Earth is ( per NASA ) 6356.8 km at either pole. Now if we assume the equatorial radius remains as it is now when during rotation then that is ( again per NASA ) 6378.1 km. Factor in the inverse square law then one's weight will be reduced at the equator c/w pole by

(6356.8/6378.1)^2 ~ 0.99333

Thirdly, and this is what you are really after, what is the magnitude of the centrifugal effect in comparison to gravity ? For this one can use the equation for acceleration of circular motion at a given radius and constant angular velocity :

acceleration = v^2 / r

.... here v is the tangential speed with respect to the offset from the rotation axis. At the equator one does a single Earth equatorial circumference traversal in 24 hours. 24 hours is 24 * 60 * 60 = 86400 seconds. So

v = distance / time = one circumference / day = 6378.1 * 1000 * 2 * PI / 86400

~ 463.82 m/s

... and you may recall in discussions elsewhere we deduced that you could get about a half kilometer per second boost when launching to space ( going Eastward ) from the equator. Hence

a = (463.82)^2/6378.1 * 1000 ~ 0.033 m/s^2

As 'standard' g is given as 9.8 m/s^2 ( give or take ) then the fractional change here is ( 0.033/9.8 ) ~ 0.003 which is about 3 parts in a thousand. So I can lose that weight fraction, with my mass remaining constant, simply by moving from a pole to the tropics. What a good idea ! :-) :-)

Fourthly, A Point Of Mild Order : To be exact 'weight' is a force measured in Newtons ie. a mass in kilograms multiplied by an acceleration in m/s^2 [ F = ma ]. In everyday language for those living on the same planet, if we are not perturbed by mild variances due to location, weights are quoted as that which is due to a certain mass. We scale our scales by g in other words. So I've just 'weighed in' at 91 kg of mass, when to be precise I've just 'forced in' at 91 * 9.8 = 891.8 Newtons. Because 891.8 Newtons is provided by the spring in the gadget to stop me going through the floor. As a Mr Middle Age Spread I prefer to quote the lower number though. Sounds better ! :-)

Fifthly : Let me expand upon 'tangential speed with respect to the offset from the rotation axis'. In particular this is not 'the radius to the centre of the Earth', which would only be true on the equator. Currently I'm at about 38 degrees south of the equator and as time progresses I travel along a circular arc with a centre being the closest point to me on the axis of rotation. That point is not the Earth's centre but rather someway along the axis of rotation b/w the Earth's centre and the south pole. Anyway you want to multiply the above answer - three parts in one thousand - by the cosine* of your latitude. So for me cos(38 degrees ) ~ 0.78, and thus the fractional weight change is about 5 parts per two thousand 'lighter' because the Earth is rotating. Which is a mere 220g on the scales for me. Drats .... :-)

Cheers, Mike.

( edit ) Ooooh ... oooohh ! Follow up query to the audience. So what rotation rate gives the centrifugal/rotation component equal to the attraction of gravity ? Say at the equator. Answers to be forwarded with a fifty dollar note to my address at ...... :-)

* not quite exact but it'll do.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Bill592
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### RE: ( edit ) Ooooh ...

Quote:
( edit ) Ooooh ... oooohh ! Follow up query to the audience. So what rotation rate gives the centrifugal/rotation component equal to the attraction of gravity ? Say at the equator. Answers to be forwarded with a fifty dollar note to my address at ...... :-)

Hmmm .... I won't be able to answer that anytime soon however,

3\1000 surprised me ! Those Subs must have some very precision
instruments !

Bill

Bikeman (Heinz-...
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### Since we are dealing with

Since we are dealing with small effects, maybe we should also be pedantic here and kill one myth, namely that the Earth takes 24 h for one full rotation.

Actually it takes 23 h, 56 min, and ca 4 seconds for one rotation, at least in most contexts (=frames of reference) that are relevant for astronomers, and I think also for the question at hand. The 24 h myth was just invented for people who are not astronomers and have 9to5 jobs ;-).

See https://en.wikipedia.org/wiki/Sidereal_time#Exact_duration_and_its_variation for an explanation. Now of course you could also figure in the centrifugal force due to the Earth's orbit around the sun, which would cause someone at the equator to appear heavier during the day as opposed to at night...but wait...what about the tidal forces of the moon .... .... arrrgggggh!!!

Cheers
HB

Mike Hewson
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### Touche ! Sidereal indeed is

Touche ! Sidereal indeed is the correct measure here. :-)

For me the really mind-numbing aspect of GR is that one can view gravity - yup, good old gravity - as an apparent force by letting go of the fixity of space and time. Intuitively that is a hard ask but that is what physics is all about. Understanding what is actually there rather than merely what you would like to be.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Chris S
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### RE: The Earth rotates once

Quote:
The Earth rotates once in about 24 hours with respect to the sun and once every 23 hours 56 minutes and 4 seconds with respect to the stars (see below). Earth's rotation is slowing slightly with time; thus, a day was shorter in the past. This is due to the tidal effects the Moon has on Earth's rotation. Atomic clocks show that a modern-day is longer by about 1.7 milliseconds than a century ago, slowly increasing the rate at which UTC is adjusted by leap seconds.

[url=https://en.wikipedia.org/wiki/Earth's_rotation#Rotation_period]Rotation[/url]

Waiting for Godot & salvation :-)

Why do doctors have to practice?
You'd think they'd have got it right by now

Stranger7777
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### But even after this I guess

But even after this I guess that the linear velocity of a man sitting on the Equator is still the same as it was a hundred or may be a thousand years ago. Who can dissuade me?

Daniels_Parents
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### RE: ... a man sitting on

Quote:
... a man sitting on the Equator ...

(1) The radius from a arbitrary point on the Equator to the center of the earth differs because of earth is not a perfect ball, so the tangential velocity (what you mean with linear velocity I guess) of such a man differs slightly too.

(2) If the same man is sitting thousand years at the same point on the Equator his tangetial velocity is with high probability not stable because of the earth is slowly changing it's shape over time (distance between this man and the earth's center), so this is another reason why tangential velocity of such a man is slightly changing.

(3) As expalined here the rotation speed of the earth decreases over time (tidal effect), one more reason for changing tangetial speed of this man.

Nothing is perfectly the same as time passes inexorably :-)

BTW: Did you know that your weght (havy bones or so) will increase in case of you are running from the Equator straight to a pole ? We do not speak here of accumulating weight due to your feasting on the way :) If you have decided for the Northpole you also will feel a force drifting you slightly to the East ...

Have fun,
Arthur

I know I am a part of a story that starts long before I can remember and continues long beyond when anyone will remember me [Danny Hillis, Long Now]

Zalster
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### RE: (2) If the same man is

Quote:
(2) If the same man is sitting thousand years at the same point on the Equator his tangetial velocity is with high probability not stable because of the earth is slowly changing it's shape over time (distance between this man and the earth's center), so this is another reason why tangential velocity of such a man is slightly changing.

The one thing people keep forgetting is the moon is slowly moving away from the earth. There for the interaction between the 2 bodies is slowly changing, not only affecting the shape of the earth but the rate at which is rotates. (rate of 1.5 inches a year)

A man sitting at the equator probably wouldn't realize it but, 1000 years ago the moon was 1.578 miles closer to the earth. He won't be able to tell the difference but there would be a very slight change in his velocity.

Mike Hewson
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### The issue is going to be the

The issue is going to be the 'equation of state' of the Earth's material and I expect would be complex due to in-homogeneity. This refers to the mechanical response in various conditions ie. shape change from impressed forces.

I think an hypothesis for pulsar glitches also deals with the 'compliance' of the star's material ie. how much shear/stress it can sustain before a rapid shifting of mass occurs and the moment of inertia jumps to a new value. Or put another way the pulsar timing behaviour is a 'window' into the equation of state. Bear in mind that the massive surface gravity renders any features small and so we may be dealing with very tiny displacements, given the absurd density of the material. I believe a 'mountain' on a pulsar is a several centimeters 'high' ! :-)

Quote:
If you have decided for the Northpole you also will feel a force drifting you slightly to the East ...

Such Coriolis effects can seem a mystery. I find a basic understanding may come from the following simple example : if I was in Northern America and I wanted to shoot a ballistic ( unguided, so it's trajectory is determined from initial conditions only ) missile directly over the North Pole to a target in Asia then, unless otherwise adjusted, I am not going to hit a point some 180 degrees of longitude away. Even assuming the Earth has no atmosphere etc, while the missile is in flight the Earth has rotated underneath it meantime. As the Earth rotates towards the East ( as that is where sunrise comes from ) then it will strike a point to the West of what would otherwise be expected ( in the case of the Earth not rotating ).

The intuitive problem is that in everyday life, with low velocities and short distances, then the time of flight of objects is typically short enough to ignore the 'Earth has rotated underneath me' issue. Of course that in turn depends upon the degree of accuracy desired, so that in the case of artillery laying ( indirect fire especially ) such is a concern.

Cheers, Mike.

NB There are quite a few instances/cases of the Coriolis effect and there is no single pat answer ( to the East/West etc ... ) for all. However I reckon the best viewpoint to consider a problem from is a distant inertial ( un-accelerated ) position. Try the following : why is it that from Chum Creek ( ~ 40 degrees south of the equator ) I can shoot a shell northwards and have it tend to drift westerly from the aiming direction, but if I shoot southwards it will drift easterly ... :-)

{ ASIDE : It may seem surprising but one can readily use the principles of general relativity to resolve such matters. }

( edit ) Or try this one : I'm on the equator with an artillery piece. I shoot a shell precisely to the East. I shoot a shell precisely to the West. Assuming the shells are identical and firing circumstances otherwise match - initial angles and speeds, local topology etc plus no weather effects of interest - will they land the same distance from the gun ?

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Bill592
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### RE: ( edit ) Or try this

Quote:
( edit ) Or try this one : I'm on the equator with an artillery piece. I shoot a shell precisely to the East. I shoot a shell precisely to the West. Assuming the shells are identical and firing circumstances otherwise match - initial angles and speeds, local topology etc plus no weather effects of interest - will they land the same distance from the gun ?

I'm guessing that the gun shooting 'West' would go further.

I know a rocket firing east at the equator gets a boost
however, not sure the two effects are comparable.

Bill