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 Acceleration of free fall, DYK Fact #358
algebraic topology
Posted: 02:35 Thursday 14 February 2008


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Note by Athene_noctua: The posts in this thread were originally made in this thread. I have split them into a thread of their own as I really feel the discussion is too interesting to be part of the other thread. Thanks to algebraic topology for such a vibrant topic! Worship.gif

Did you know?

The terrestrial acceleration of free fall, g, varies at different points on the Earth, being less at the Equator than at the Poles. From Newton’s law of gravitation, g is given by GMR2, where G is the gravitational constant, M the mass of the Earth and R the radius of the Earth (which varies). Now plug in these values:

G = 6.674 28 × 10−11 m3 kg−1 s−2
 
M = 5.973 6 × 1024 kg
 
R = 6.378 137 × 106 m (equatorial)
  6.356 752 × 106 m (polar)

These should give g as approximately 9.8006 m s−2 at the Equator and 9.8667 m s−2 at the Poles.



This post has been edited by Athene_noctua on 01:12 Saturday 01 March 2008
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Athene_noctua
Posted: 07:50 Thursday 14 February 2008


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Did you know? The Moon has a mass of 7.347673×1022 kg and radius 1.7373×106 m – hence the lunar acceleration of free fall is 6.674 28×10−11 × 7.347673×1022 ⁄ (1.7373×106)2 ≈ 1.62 m s−2.




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algebraic topology
Posted: 17:47 Tuesday 26 February 2008


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Did you know?

The variation of g on the surface of the Earth is due to the fact that the Earth is not a perfect sphere. Let us assume that it is an oblate ellipsoid – in other words, its cross section in a longitudinal plane is an ellipse. Let Re be the equatorial radius and Rp the polar radius. Then the equation of the cross-sectional ellipse is given by x2Re2 + y2Rp2 = 1. If Rθ is the distance from the Earth’s surface to the Earth’s centre at latitude θ, we have x = Recosθ, y = Rpsinθ. Thus:

Rθ2cos2Re2 + Rθ2cos2Rp2 = 1 ⇒ Rθ2 = Re2Rp2Re2sin2θ + Rp2cos2θ

The acceleration of free fall at latitude θ, gθ, is therefore given by

gθ = GM(cos2θ⁄Re2 + sin2θ⁄Rp2)

So the acceleration of free fall in London (latitude 51.5°N) is approximately 9.841 m s−2 while the acceleration of free fall in New York (latitude 40.8°N) is approximately 9.829 m s−2.

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Athene_noctua
Posted: 18:33 Tuesday 26 February 2008


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Did you know? The increase of gθ with θ may explain why people tend to be apparently more overweight the further from the Equator they live. ROFL.gif



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Jane Bennet
Posted: 03:41 Thursday 28 February 2008


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QUOTE (algebraic topology @ 17:47 GMT Tuesday 26 February 2008)

gθ = GM(cos2θ⁄Re2 + sin2θ⁄Rp2)


Did you know? You can also write the equation as

gθ = gecos2θ + gpsin2θ

where ge and gp are the values of the acceleration of free fall at the Equator and the Poles respectively. biggrin.gif

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Athene_noctua
Posted: 12:00 Thursday 28 February 2008


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Did you know? The graph of gθ against θ is almost a straight line! blink.gif



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algebraic topology
Posted: 00:44 Saturday 01 March 2008


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Did you know?

Differentiating gθ with respect to θ gives

gθ′ = −2gecosθsinθ + 2gpsinθcosθ = (gpge)sin2θ

This is positive for θ between 0° and 90° – hence, though the variation may be slight, the acceleration of free fall is still a strictly increasing function of latitude between the Equator and either Pole.

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William McCormick
Posted: 00:20 Wednesday 16 April 2008


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QUOTE (Jane Bennet @ 03:41 GMT Thursday 28 February 2008)
QUOTE (algebraic topology @ 17:47 GMT Tuesday 26 February 2008)

gθ = GM(cos2θ⁄Re2 + sin2θ⁄Rp2)


Did you know? You can also write the equation as

gθ = gecos2θ + gpsin2θ

where ge and gp are the values of the acceleration of free fall at the Equator and the Poles respectively. biggrin.gif


Did you know that standing on the equator you are moving West to East at a rate of 1567 feet per second?
Standing on the North or South pole you are just revolving once in 24 hours.

I believe your equations need this data. Although I cannot make heads or tails out of them.


Sincerely,


William McCormick
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Jane Bennet
Posted: 09:34 Wednesday 16 April 2008


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QUOTE (William McCormick @ 00:20 BST Wednesday 16 April 2008)
Did you know that standing on the equator you are moving West to East at a rate of 1567 feet per second?

That’s only 478 metres per second. It’s actually much faster than that by my calculations.

When you’re rotating with the Earth on the Equator, your centripetal acceleration is provided by the Earth’s gravitation – so your tangential speed ve is given by ve2Re = g = GMRe2 (using Algebraic topology’s formula for the value of the acceleration of free fall). Hence ve = √(GMRe). Plugging in actual values gives ve ≈ 7 906 m s−1 (nearly 26 000 feet per second).

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William McCormick
Posted: 23:00 Wednesday 16 April 2008


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QUOTE

That’s only 478 metres per second. It’s actually much faster than that by my calculations.

When you’re rotating with the Earth on the Equator, your centripetal acceleration is provided by the Earth’s gravitation – so your tangential speed ve is given by ve2Re = g = GMRe2 (using Algebraic topology’s formula for the value of the acceleration of free fall). Hence ve = √(GMRe). Plugging in actual values gives ve ≈ 7 906 m s−1 (nearly 26 000 feet per second).



This is actually how I come to my conclusion.


The Earth is stated to be 12,800,000 meters in diameter by Wiki answers.
That would make it 41,994,754 feet in diameter.
Times 3.14159..... equals 131,930,292.93568
Divided by 24 leaves you with 5,497,095 feet an hour
Divided by 60 leaves you with 91,618.258983111111..... feet a minute
Divided by 60 leaves you with 1,526.97098305....... Or 1,527 feet of west to east travel a second.



Sincerely,


William McCormick
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Jane Bennet
Posted: 23:27 Wednesday 16 April 2008


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I believe you’re right. My calculation was for the tangential speed of an object orbiting the Earth under the Earth’s gravity – a totally different problem. Blush.gif

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William McCormick
Posted: 00:33 Thursday 17 April 2008


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QUOTE (Jane Bennet @ 23:27 GMT Wednesday 16 April 2008)

I believe you’re right. My calculation was for the tangential speed of an object orbiting the Earth under the Earth’s gravity – a totally different problem. Blush.gif


I guess my question is, if you are going that fast on the equator, wouldn't you weigh less, due to centrifuge force at the equator?

And then my next question how much less?

I would go down to the merry go round at the park and experiment with different ratios of diameter and speed.

We used to have one at our park that was oiled to perfection.

It would spin and spin. I was always amazed to the pressure in my head, laying down on the merry go round, like a spoke from the hub. With my head at the outer circumference. Even when it was barely turning the pressure in my head was noticeable.

Maybe I took one turn to many? Ha-ha.



Sincerely,


William McCormick
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Jane Bennet
Posted: 07:04 Thursday 17 April 2008


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What “centrifuge force”? I don’t see any such force anywhere. NoGood.gif

And I don’t think being rotated along with the Earth has any effect on our weight compared with not being rotated.

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algebraic topology
Posted: 22:00 Thursday 17 April 2008


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Centrifugal foree is a very unhelpful concept and is best avoided in analysis of problems. It crops up in two scenarios:

  1. In a frame of reference that moving in circular motion, centrifugal force appears as an inertial force that tends to pull objects away from the centre of rotation or revolution. Since the frame of reference is accelerating (its direction is constantly changing), a “fictitious” force is present.

  2. When an object is moving in a circle, it is kept in its circular orbit by a centripetal force. By Newton’s third law, the object itself exerts an equal and opposite force on the source of the centripetal force – this is a centrifugal force. For example, when an object is being whirled on a string, the tension in the string provides the centripetal force that keeps the object whirling in a circle; at the same time, the object exerts a centrifugal force on the string keeping it taut.

When we analyse an object in circular motion, we normally view the object in a stationary frame of reference – we do not, as in (i), choose to place ourselves in the object’s frame of reference. And in the case of (ii), we are usually much more interested in forces exerted on the object by surroundings than in forces exerted by the object on its surroundings. For these two reasons, centrifugal force is normally just a red herring in analyses of circular motion.

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William McCormick
Posted: 01:39 Friday 18 April 2008


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QUOTE (algebraic topology @ 22:00 GMT Thursday 17 April 2008)

Centrifugal foree is a very unhelpful concept and is best avoided in analysis of problems. It crops up in two scenarios:

  1. In a frame of reference that moving in circular motion, centrifugal force appears as an inertial force that tends to pull objects away from the centre of rotation or revolution. Since the frame of reference is accelerating (its direction is constantly changing), a “fictitious” force is present.

  2. When an object is moving in a circle, it is kept in its circular orbit by a centripetal force. By Newton’s third law, the object itself exerts an equal and opposite force on the source of the centripetal force – this is a centrifugal force. For example, when an object is being whirled on a string, the tension in the string provides the centripetal force that keeps the object whirling in a circle; at the same time, the object exerts a centrifugal force on the string keeping it taut.

When we analyse an object in circular motion, we normally view the object in a stationary frame of reference – we do not, as in (i), choose to place ourselves in the object’s frame of reference. And in the case of (ii), we are usually much more interested in forces exerted on the object by surroundings than in forces exerted by the object on its surroundings. For these two reasons, centrifugal force is normally just a red herring in analyses of circular motion.


Centrifuge force; pumps many, or most of the fluids in our lives. It is a real calculation and ratio.

Many fans, vacuums, and exhaust systems use centrifuge force to pump or suck air as well.

Many also use a friction wheel, to pump air. Some use a vein system to pump, that is not a centrifuge pump. But most high volume or waste pumps that may move solids use centrifuge action to pump.

I do not know how much centrifuge force reduces the weight of someone on the equator. However it is real, and certainly measurable.

I used to work with a machinist that also did work for William Casey, the ex-CIA director, and he could calculate exactly how much a pump would pump or suck, based on centrifuge force. I could guess or build a pump and then create a ratio. However I bet there are formulas. I am sure I could create one.

He would grab a calculator, and go bang, bang, bang, bang, and tell you exactly how much it would pump. He taught me how the silencer on a gun works.


Sincerely,


William McCormick
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William McCormick
Posted: 01:44 Friday 18 April 2008


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QUOTE (Jane Bennet @ 07:04 GMT Thursday 17 April 2008)

What “centrifuge force”? I don’t see any such force anywhere. NoGood.gif

And I don’t think being rotated along with the Earth has any effect on our weight compared with not being rotated.


When they launch a satellite for dish TV for instance. You aim your dish at the satellite hopefully once. And it stays tuned in to that satellite.

The reason it does, is because the satellite is actually above a point on the earth and is matching the rotation angle as the earth turns.

The reason it stays there is centrifuge force.


Sincerely,


William McCormick
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William McCormick
Posted: 01:59 Friday 18 April 2008


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I forget what the speed to obtain orbit, a few feet above the earths surface is. However at a certain speed. An object with no aerodynamics can in fact travel around the world, and not hit the earth.

The speed is just a relation to centrifuge action. I believe it is around, 32,000 feet per second near sea level. But that is just from memory. We did the calculation once in school. But I was probably clowning around.

If you take the speed of a satellite that maintains its orbit at a certain elevation, and also does not move from a spot on earth while directly overhead.
That objects speed is probably the speed of an object stably orbiting at sea level on earth.


Sincerely,


William McCormick
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algebraic topology
Posted: 03:11 Friday 18 April 2008


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QUOTE (William McCormick)
The reason it stays there is centrifuge force.

No, no, no! The reason it stays in geostationary orbit is centripetal force! The gravitational pull of the Earth provides the centripetal force that keeps the satellite in orbit!



QUOTE (William McCormick)
I forget what the speed to obtain orbit, a few feet above the earths surface is. However at a certain speed. An object with no aerodynamics can in fact travel around the world, and not hit the earth.

The speed is just a relation to centrifuge action. I believe it is around, 32,000 feet per second near sea level. But that is just from memory. We did the calculation once in school. But I was probably clowning around.

If you take the speed of a satellite that maintains its orbit at a certain elevation, and also does not move from a spot on earth while directly overhead.
That objects speed is probably the speed of an object stably orbiting at sea level on earth.

Jane Bennet did some calculations earlier; she got v = (GMR)1⁄2. Now v = RT, where T is the period of revolution. Hence RT = (GMR)1⁄2 and rearranging gives R = (T ⁄ 2π)2⁄3(GM)1⁄3. Therefore

v = RT = (2πGMT)1⁄3

Now substitute T = 8.6164×104 s and other values as appropriate. You should get v ≈ 3 075 m s−1. So a satellite in geostationary orbit has a tangential speed of approximately 3 kilometres per second. In your primitive language, that’s just under 10 000 feet per second – so either your calculations are wrong or your memory is faulty.

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William McCormick
Posted: 01:29 Saturday 19 April 2008


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QUOTE (algebraic topology @ 03:11 GMT Friday 18 April 2008)
QUOTE (William McCormick)
The reason it stays there is centrifuge force.

No, no, no! The reason it stays in geostationary orbit is centripetal force! The gravitational pull of the Earth provides the centripetal force that keeps the satellite in orbit!



QUOTE (William McCormick)
I forget what the speed to obtain orbit, a few feet above the earths surface is. However at a certain speed. An object with no aerodynamics can in fact travel around the world, and not hit the earth.

The speed is just a relation to centrifuge action. I believe it is around, 32,000 feet per second near sea level. But that is just from memory. We did the calculation once in school. But I was probably clowning around.

If you take the speed of a satellite that maintains its orbit at a certain elevation, and also does not move from a spot on earth while directly overhead.
That objects speed is probably the speed of an object stably orbiting at sea level on earth.

Jane Bennet did some calculations earlier; she got v = (GMR)1⁄2. Now v = RT, where T is the period of revolution. Hence RT = (GMR)1⁄2 and rearranging gives R = (T ⁄ 2π)2⁄3(GM)1⁄3. Therefore

v = RT = (2πGMT)1⁄3

Now substitute T = 8.6164×104 s and other values as appropriate. You should get v ≈ 3 075 m s−1. So a satellite in geostationary orbit has a tangential speed of approximately 3 kilometres per second. In your primitive language, that’s just under 10 000 feet per second – so either your calculations are wrong or your memory is faulty.


You can note that taking a fairly heavy object on a string, and turning your body around and around allowing the object to move into a created orbit for lack of a better term.

That the object in fact will achieve an orbit around you. Defying gravity. Little David's sling shot worked like this.

The reason that the object pulls the string that you are holding, is that an object in motion likes to travel in a straight line.

It requires energy to take an object off its straight path. So at some speed orbiting the earth, the straight line that the object would like to travel in, is enough to keep it from falling.

When we were kids we would do this kind of stuff all day long.


Sincerely,


William McCormick
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algebraic topology
Posted: 23:53 Sunday 20 April 2008


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QUOTE (William McCormick)
Defying gravity.

Well, it may look like defying gravity – but, believe it or not, gravity may actually be causing the object to move in a circle! Take, for instance, the case of a stone being whirled in a horizontal circle on a string, with the pivot above the plane of the stone’s circular orbit. Gravity acts on the stone, causing it to tug on the string; this produces a tension in the string, and the horizontal component of the tension then provides the centripetal force keeping the stone in its circular orbit. Hence, gravity is indirectly responsible for the stone’s circular motion. The stone not really “defying” gravity after all, but is rather being helped by it to move around in a circle.

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William McCormick
Posted: 02:17 Monday 21 April 2008


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QUOTE (algebraic topology @ 23:53 GMT Sunday 20 April 2008)
QUOTE (William McCormick)
Defying gravity.

Well, it may look like defying gravity – but, believe it or not, gravity may actually be causing the object to move in a circle! Take, for instance, the case of a stone being whirled in a horizontal circle on a string, with the pivot above the plane of the stone’s circular orbit. Gravity acts on the stone, causing it to tug on the string; this produces a tension in the string, and the horizontal component of the tension then provides the centripetal force keeping the stone in its circular orbit. Hence, gravity is indirectly responsible for the stone’s circular motion. The stone not really “defying” gravity after all, but is rather being helped by it to move around in a circle.


You can do the same thing in space as well.

Gravity is gravity. It is actually a force pressing down on you. The planets surface is abundant with electrons compared to space. When ambient radiation from space comes close to the planet it is slowed down to gravity velocity electrons.

The planet itself accelerates a good portion of the electrons passing through the planet, to a velocity faster then gravity. That is why you are not weightless. Near the earth.

Sincerely,


William McCormick
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algebraic topology
Posted: 12:17 Monday 21 April 2008


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QUOTE (William McCormick)
Gravity is gravity. It is actually a force pressing down on you.

Gravity is an attractive force. It is due to mass. Or, if you look at it at a subnuclear level, it is a force mediated by the exchange of virtual particles called gravitons. In any case, it’s always attractive. “Repulsive” gravitational force has never been discovered.

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William McCormick
Posted: 00:56 Thursday 24 April 2008


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QUOTE (algebraic topology @ 12:17 GMT Monday 21 April 2008)
QUOTE (William McCormick)
Gravity is gravity. It is actually a force pressing down on you.

Gravity is an attractive force. It is due to mass. Or, if you look at it at a subnuclear level, it is a force mediated by the exchange of virtual particles called gravitons. In any case, it’s always attractive. “Repulsive” gravitational force has never been discovered.


Gravity being a pressing force was in fact the norm for many years on earth. I believe that during World War Two it fell out of favor. Because it was a very useful thing, to understanding the Universe.

The governments of earth openly stated that counterintelligence would be practiced. And it was. Now we believe things somehow attract, when it has never been demonstrated as something real, possible or even plausible.

There is zero scientific evidence that attraction is possible. Yet all evidence shows clearly that things repel and can repel each other.

It would be much easier to prove a flat world then to prove attraction. Mass is actually an illusion created by time and structures of sub-atomic particles. This can also be demonstrated.


Sincerely,


William McCormick
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algebraic topology
Posted: 08:09 Thursday 24 April 2008


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QUOTE (William McCormick)
There is zero scientific evidence that attraction is possible. Yet all evidence shows clearly that things repel and can repel each other.

Dude, the graviton theory of gravitation in quantum field theory is a “attractive” theory of gravity (no pun intended). Read about gravitons here: http://en.wikipedia.org/wiki/Graviton.

Granted, Newton’s theory of gravitation analyses gravity as an attractive force but does not explain why it is attractive. The graviton theory explains why: two masses attract each other gravitationally because they exchange gravitons. The question is not lack of any explanation but your not being aware of one, that’s all.


QUOTE (William McCormick)
It would be much easier to prove a flat world then to prove attraction.

What?! Are you claiming that just because something is extremely difficult to prove, it must be false? If that were the case, Fermat’s last theorem would have been dismissed as false a long time ago!

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