Gravity acceleration and deceleration

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Topic:	Science > Physics
User:	""
Date:	02 Feb 2008 12:13:28 AM
Object:	Gravity acceleration and deceleration

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line. This is instantaneous weightless
acceleration. Also matter does not have to follow a curve.
A rocket or something thrown upward experiences gravitational
deceleration in freefall.
Mitch Raemsch Twice Nobel Laureate 2008
.

User: "Jimbo"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 01:29:16 AM

<mitchgrav@hotmail.com> wrote in message
news:b0a140ce-aa81-404f-9b89-7663f622be76@s8g2000prg.googlegroups.com...

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line. This is instantaneous weightless
acceleration.

So if I step off a chair two feet from the ground I will hit the ground at
32 feet 'persecond'? If I drop out of a window 40 stories up will I also hit
the ground at 32 feet 'persecond'?
Please elaborate on your comment.
Chris
.

User: "Rock Brentwood"
Title: Re: Gravity acceleration and deceleration	05 Feb 2008 12:00:06 AM

On Feb 2, 1:29=A0am, "Jimbo" <chris...@cox.net> wrote:

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line. This is instantaneous weightless
acceleration.

So if I step off a chair two feet from the ground I will hit the ground at=
32 feet 'persecond'?

There is, in fact, no substitute for direct first-hand experience.
I recent times my reaction speed has accelerated significantly. It
used to be fast, now it's several orders of magnitude faster than that
(for example) quoted for someone like the Kung Fu master William
Cheung (he rated at 1/12 second). In more recent times still, this has
start to become accompanied also by a heightened speed of thought that
(subjectively) is somewhat equivalent to time coming to slowing down
to a near dead stop. Objectively, this has been accompanied by the
occurrence of extreme rapid-fire complex decision-making (of the kind
that normally is only associated with conscious decision-making
involving multiple steps or inferences in logic; e.g. how do you react
to 2 or more objects falling at once; or 2 or more accidents occurring
at the same time?)
In the latest incident that occurred, an object started falling to the
ground and I snatched it clean out of the air before it even had time
to move 1 millimeter. During that period, I was perfectly aware and
conscious of the passage of time and of the goings on. It was like the
object was just flat-out hanging there in the middle of the air as if
it were somehow hooked and caught on the fabric of space, itself.
What's different about this incident, compared to previous ones, is
that formerly the reaction did not even BEGIN until after the motion
had gotten to the 1 millimeter stage. Count in the mechanical delay of
actually overcoming the inertia of the arm to move it (which, on these
time scales, is effectively very large, and is really something only a
weight-lifter can overcome), and that comes out to about 1
centimeter's motion before the object is intercepted in the air. This
time, the reaction was COMPLETED, mechanical delay and all, within the
1 millimeter period.
At around 4900mm for 1 second, a 1 mm drop is equivalent to 1/70
second. The quickest reaction time I've actually been able to observe,
up to then, had been around 1/140 second. This had to have been quite
a bit faster in order to be able to count in mechanical delay for a
1/70 second total.
.

User: "CJD - The Cannibal"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 01:02:18 AM

proclaimed:

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line. This is instantaneous weightless
acceleration. Also matter does not have to follow a curve.

A rocket or something thrown upward experiences gravitational
deceleration in freefall.

Mitch Raemsch Twice Nobel Laureate 2008

More evidence of god Mitch?
.

User: ""
Title: Re: Gravity acceleration and deceleration	03 Feb 2008 07:54:30 AM

On Feb 2, 2:02=A0am, "CJD - The Cannibal" <cj_dunnaway-
recov...@yahooY.com> wrote:

mitchg...@hotmail.com proclaimed:

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line. This is instantaneous weightless
acceleration. Also matter does not have to follow a curve.

A rocket or something thrown upward experiences gravitational
deceleration in freefall.

Mitch Raemsch Twice Nobel Laureate 2008

More evidence of god Mitch?

Mitch Every thing fallows the curve of space(Wheeler) Elevator can
curve light beam. Reality is no straight line from A to
B I have a concave(Einstein ) and
convex(Glazier)curving of space. It gives the answer to the
accelerating space of the universe at greater and greater distances.It
has been well received. Bert
.

User: "PD"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 09:13:48 AM

On Feb 2, 12:13=A0am,

wrote:

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line.

No, it doesn't immediately arrive at that speed, and it doesn't stay
at that speed even when it does arrive at that speed. Oh, and speed
isn't acceleration.

This is instantaneous weightless
acceleration.

No, a weightless object wouldn't accelerate. Objects accelerate in
free fall near the earth *because* they have weight. That weight is
otherwise known as the force of gravity acting on that object, and
force produces acceleration, as we've known for over 300 years.

Also matter does not have to follow a curve.

This is true, but I'm not sure of your point.

A rocket or something thrown upward experiences gravitational
deceleration in freefall.

This is also true, but again I'm not sure of your point.

Mitch Raemsch Twice Nobel Laureate 2008

A Nobel Laureate typically has to be able to type four sentences
without making two major mistakes. If you've received these in the
mail, could you please ship them back?
.

User: "Thomas Heger"
Title: Re: Gravity acceleration and deceleration	04 Feb 2008 02:24:52 AM

"PD" <TheDraperFamily@gmail.com> schrieb im Newsbeitrag
news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
On Feb 2, 12:13 am,

wrote:
PD
"No, a weightless object wouldn't accelerate. Objects accelerate in
free fall near the earth *because* they have weight. That weight is
otherwise known as the force of gravity acting on that object, and
force produces acceleration, as we've known for over 300 years."
If you think of objects as free falling, they are moving on their worldline.
They are at rest in their space. If you push them away from free-falling
worldlines you need to apply a force. If they have mass, you need to give
the object some inertia. This 'push' is what we usually call gravity. In
fact, it is inertia, gravity is more the geometry of those free-falling
worldlines. You can't push a weightless object, because massless objects
pace away with c.
Thomas Heger
.

User: "NoEinstein"
Title: Re: Gravity acceleration and deceleration	05 Feb 2008 05:41:49 AM

On Feb 4, 3:24 am, "Thomas Heger" <hba...@hotmail.com> wrote:

"PD" <TheDraperFam...@gmail.com> schrieb im Newsbeitragnews:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
On Feb 2, 12:13 am,

Dear Thomas: You were doing so well till you got to the following
paragraph. Haven't you heard? I've disproved SR and GR. There is no
space-time, and no world lines. -- NoEinstein --

If you think of objects as free falling, they are moving on their worldline.
They are at rest in their space. If you push them away from free-falling
worldlines you need to apply a force. If they have mass, you need to give
the object some inertia. This 'push' is what we usually call gravity. In
fact, it is inertia, gravity is more the geometry of those free-falling
worldlines. You can't push a weightless object, because massless objects
pace away with c.

Thomas Heger

User: "Androcles"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 09:28:57 AM

"PD" <TheDraperFamily@gmail.com> wrote in message
news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
On Feb 2, 12:13 am,

wrote:

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line.

No, it doesn't immediately arrive at that speed, and it doesn't stay
at that speed even when it does arrive at that speed. Oh, and speed
isn't acceleration.

This is instantaneous weightless
acceleration.

Also matter does not have to follow a curve.

This is true, but I'm not sure of your point.

A rocket or something thrown upward experiences gravitational
deceleration in freefall.

This is also true, but again I'm not sure of your point.

Mitch Raemsch Twice Nobel Laureate 2008

A Nobel Laureate typically has to be able to type four sentences
without making two major mistakes. If you've received these in the
mail, could you please ship them back?
Good idea... Maybe Einstein's heirs will do the same for
the same reasons.

.

User: "The Ghost In The Machine"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 12:52:36 PM

In sci.physics.relativity, Androcles
<Headmaster@Hogwarts.physics>
wrote
on Sat, 02 Feb 2008 15:28:57 GMT
<Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:

"PD" <TheDraperFamily@gmail.com> wrote in message
news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
On Feb 2, 12:13 am,

wrote:

Drop something and it immediately arives at a speed of 32 feet
persecond falling in a straight line.

No, it doesn't immediately arrive at that speed, and it doesn't stay
at that speed even when it does arrive at that speed. Oh, and speed
isn't acceleration.

One can also establish how long it takes to get at that speed.
Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
one second instantaneously, 2 seconds if one takes the average.
(In other words, 2 seconds from the drop the stone is moving
at 64 ft/s, and the average speed -- as well as x/t -- is 32 ft/s.)
[rest snipped by TGITM for brevity]
--
#191,

Been there, done that, didn't get the T-shirt.
--
Posted via a free Usenet account from http://www.teranews.com
.

User: "Androcles"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 01:40:32 PM

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:koqe75-98a.ln1@sirius.tg00suus7038.net...
| In sci.physics.relativity, Androcles
| <Headmaster@Hogwarts.physics>
| wrote
| on Sat, 02 Feb 2008 15:28:57 GMT
| <Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:
| >
| > "PD" <TheDraperFamily@gmail.com> wrote in message
| > news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
| > On Feb 2, 12:13 am,

wrote:
| >> Drop something and it immediately arives at a speed of 32 feet
| >> persecond falling in a straight line.
| >
| > No, it doesn't immediately arrive at that speed, and it doesn't stay
| > at that speed even when it does arrive at that speed. Oh, and speed
| > isn't acceleration.
|
| One can also establish how long it takes to get at that speed.
| Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| one second instantaneously, 2 seconds if one takes the average.
| (In other words, 2 seconds from the drop the stone is moving
| at 64 ft/s, and the average speed -- as well as x/t -- is 32 ft/s.)
|
| [rest snipped by TGITM for brevity]
Take it up with Phuckwit Duck, he wrote it.
Here's a question for you though.
If the acceleration due to gravity is g fps/s at the
top of a mineshaft at sea level and also g fps/s at the
top of Everest, how high do you have to be before
an inverse square law kicks in and it is less than g, and why?
(Hint: The Earth is not a point, the local g field is approximately
parallel.)
.

User: "The Ghost In The Machine"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 05:23:08 PM

In sci.physics.relativity, Androcles
<Headmaster@Hogwarts.physics>
wrote
on Sat, 02 Feb 2008 19:40:32 GMT
<QG3pj.28957$801.2573@fe1.news.blueyonder.co.uk>:

wrote:
| >> Drop something and it immediately arives at a speed of 32 feet
| >> persecond falling in a straight line.
| >
| > No, it doesn't immediately arrive at that speed, and it doesn't stay
| > at that speed even when it does arrive at that speed. Oh, and speed
| > isn't acceleration.
|
| One can also establish how long it takes to get at that speed.
| Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| one second instantaneously, 2 seconds if one takes the average.
| (In other words, 2 seconds from the drop the stone is moving
| at 64 ft/s, and the average speed -- as well as x/t -- is 32 ft/s.)
|
| [rest snipped by TGITM for brevity]

Take it up with Phuckwit Duck, he wrote it.

Here's a question for you though.
If the acceleration due to gravity is g fps/s at the
top of a mineshaft at sea level and also g fps/s at the
top of Everest, how high do you have to be before
an inverse square law kicks in and it is less than g, and why?
(Hint: The Earth is not a point, the local g field is approximately
parallel.)

g = GM/d^2 is never constant, as d is never constant in
the usual run of problems (e.g., a ball thrown upward).
SR is therefore applicable without error *nowhere on Earth*,
and indeed nowhere in explorable space, as g_sun near the
orbit of Pluto (5.976 * 10^12 m) is 3.80601 * 10^-6 N/kg.
This is not zero; therefore space is not flat. Not that
the error is all that big, but there will be one.
(As a side issue, there are other anomalies; Earth is
erroneously modeled as a mass point and we're standing on
a massless platform 6.378 * 10^6 meters from that point.
The real accelerations should more properly be calculated
taking the entire oblate spheroid into account, and its
mass is not uniformly distributed. These anomalies will
show up as deviations during satellite orbit, and can be
used to find new mass anomalies, such as undiscovered oil
fields.)
--
#191,

Useless C++ Programming Idea #889123:
std::vector<...> v; for(int i = 0; i < v.size(); i++) v.erase(v.begin() + i);
--
Posted via a free Usenet account from http://www.teranews.com
.

User: "Androcles"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 06:24:27 PM

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:sjaf75-ivc.ln1@sirius.tg00suus7038.net...
| In sci.physics.relativity, Androcles
| <Headmaster@Hogwarts.physics>
| wrote
| on Sat, 02 Feb 2008 19:40:32 GMT
| <QG3pj.28957$801.2573@fe1.news.blueyonder.co.uk>:
| >
| > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
| > news:koqe75-98a.ln1@sirius.tg00suus7038.net...
| > | In sci.physics.relativity, Androcles
| > | <Headmaster@Hogwarts.physics>
| > | wrote
| > | on Sat, 02 Feb 2008 15:28:57 GMT
| > | <Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:
| > | >
| > | > "PD" <TheDraperFamily@gmail.com> wrote in message
| > | >
news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
| > | > On Feb 2, 12:13 am,

wrote:
| > | >> Drop something and it immediately arives at a speed of 32 feet
| > | >> persecond falling in a straight line.
| > | >
| > | > No, it doesn't immediately arrive at that speed, and it doesn't stay
| > | > at that speed even when it does arrive at that speed. Oh, and speed
| > | > isn't acceleration.
| > |
| > | One can also establish how long it takes to get at that speed.
| > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | one second instantaneously, 2 seconds if one takes the average.
| > | (In other words, 2 seconds from the drop the stone is moving
| > | at 64 ft/s, and the average speed -- as well as x/t -- is 32 ft/s.)
| > |
| > | [rest snipped by TGITM for brevity]
| >
| > Take it up with Phuckwit Duck, he wrote it.
| >
| > Here's a question for you though.
| > If the acceleration due to gravity is g fps/s at the
| > top of a mineshaft at sea level and also g fps/s at the
| > top of Everest, how high do you have to be before
| > an inverse square law kicks in and it is less than g, and why?
| > (Hint: The Earth is not a point, the local g field is approximately
| > parallel.)
| >
|
| g = GM/d^2 is never constant, as d is never constant in
| the usual run of problems (e.g., a ball thrown upward).
Of course, but it is a function of something.
| SR is
I wasn't asking about SR drool. Try to concentrate on the question,
this is purely Newtonian.
If the acceleration due to gravity is g fps/s at the
top of a mineshaft at sea level and also g fps/s at the
top of Everest, how high do you have to be before
an inverse square law kicks in and it is less than g, and why?
(Hint: The Earth is not a point, the local g field is approximately
parallel.)
.

User: "Jeckyl"
Title: Re: Gravity acceleration and deceleration	02 Feb 2008 06:33:17 PM

"Androcles" <Headmaster@Hogwarts.physics> wrote in message
news:%Q7pj.29003$801.1233@fe1.news.blueyonder.co.uk...

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
news:sjaf75-ivc.ln1@sirius.tg00suus7038.net...
| In sci.physics.relativity, Androcles
| <Headmaster@Hogwarts.physics>
| wrote
| on Sat, 02 Feb 2008 19:40:32 GMT
| <QG3pj.28957$801.2573@fe1.news.blueyonder.co.uk>:
| >
| > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
| > news:koqe75-98a.ln1@sirius.tg00suus7038.net...
| > | In sci.physics.relativity, Androcles
| > | <Headmaster@Hogwarts.physics>
| > | wrote
| > | on Sat, 02 Feb 2008 15:28:57 GMT
| > | <Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:
| > | >
| > | > "PD" <TheDraperFamily@gmail.com> wrote in message
| > | >
news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
| > | > On Feb 2, 12:13 am,

wrote:
| > | >> Drop something and it immediately arives at a speed of 32 feet
| > | >> persecond falling in a straight line.
| > | >
| > | > No, it doesn't immediately arrive at that speed, and it doesn't
stay
| > | > at that speed even when it does arrive at that speed. Oh, and
speed
| > | > isn't acceleration.
| > |
| > | One can also establish how long it takes to get at that speed.
| > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | one second instantaneously, 2 seconds if one takes the average.
| > | (In other words, 2 seconds from the drop the stone is moving
| > | at 64 ft/s, and the average speed -- as well as x/t -- is 32 ft/s.)
| > |
| > | [rest snipped by TGITM for brevity]
| >
| > Take it up with Phuckwit Duck, he wrote it.
| >
| > Here's a question for you though.
| > If the acceleration due to gravity is g fps/s at the
| > top of a mineshaft at sea level and also g fps/s at the
| > top of Everest, how high do you have to be before
| > an inverse square law kicks in and it is less than g, and why?
| > (Hint: The Earth is not a point, the local g field is approximately
| > parallel.)
| >
|
| g = GM/d^2 is never constant, as d is never constant in
| the usual run of problems (e.g., a ball thrown upward).

Of course, but it is a function of something.

| SR is

I wasn't asking about SR drool. Try to concentrate on the question,
this is purely Newtonian.

If the acceleration due to gravity is g fps/s at the
top of a mineshaft at sea level and also g fps/s at the
top of Everest,

Who said it was .. other than you? Sounds like you're playing the liars
game of strawmen.
.

User: "The Ghost In The Machine"
Title: Re: Gravity acceleration and deceleration	03 Feb 2008 09:46:19 AM

In sci.physics.relativity, Androcles
<Headmaster@Hogwarts.physics>
wrote
on Sun, 03 Feb 2008 00:24:27 GMT
<%Q7pj.29003$801.1233@fe1.news.blueyonder.co.uk>:

wrote:
| > | >> Drop something and it immediately arives at a speed of 32 feet
| > | >> persecond falling in a straight line.
| > | >
| > | > No, it doesn't immediately arrive at that speed, and it doesn't stay
| > | > at that speed even when it does arrive at that speed. Oh, and speed
| > | > isn't acceleration.
| > |
| > | One can also establish how long it takes to get at that speed.
| > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | one second instantaneously, 2 seconds if one takes the average.
| > | (In other words, 2 seconds from the drop the stone is moving
| > | at 64 ft/s, and the average speed -- as well as x/t -- is 32 ft/s.)
| > |
| > | [rest snipped by TGITM for brevity]
| >
| > Take it up with Phuckwit Duck, he wrote it.
| >
| > Here's a question for you though.
| > If the acceleration due to gravity is g fps/s at the
| > top of a mineshaft at sea level and also g fps/s at the
| > top of Everest, how high do you have to be before
| > an inverse square law kicks in and it is less than g, and why?
| > (Hint: The Earth is not a point, the local g field is approximately
| > parallel.)
| >
|
| g = GM/d^2 is never constant, as d is never constant in
| the usual run of problems (e.g., a ball thrown upward).

Of course, but it is a function of something.

| SR is

I wasn't asking about SR drool. Try to concentrate on the question,
this is purely Newtonian.

If the acceleration due to gravity is g fps/s at the
top of a mineshaft at sea level and also g fps/s at the
top of Everest, how high do you have to be before
an inverse square law kicks in and it is less than g, and why?
(Hint: The Earth is not a point, the local g field is approximately
parallel.)

Ah, interesting thought. I'll have to work out,
briefly put, the mass distribution such that
[1] It is in the shape of a mountain somewhere, of height 9.9 km or so,
[2] one has constant g on any point on the surface of that mountain,
[3] work out g in other points of space nearby.
I'll have to get back to you on that. Note that the mass will
have to be of variable density.
Also, the inverse square law will *never* kick in anyway,
in that case. One can of course approximate it to any
nonzero degree of tolerance, to be sure, if one is far
enough away.
--
#191,

Conventional memory has to be one of the most UNconventional
architectures I've seen in a computer system.
--
Posted via a free Usenet account from http://www.teranews.com
.

User: "Androcles"
Title: Re: Gravity acceleration and deceleration	03 Feb 2008 11:35:46 AM

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:b74h75-t3n.ln1@sirius.tg00suus7038.net...
| In sci.physics.relativity, Androcles
| <Headmaster@Hogwarts.physics>
| wrote
| on Sun, 03 Feb 2008 00:24:27 GMT
| <%Q7pj.29003$801.1233@fe1.news.blueyonder.co.uk>:
| >
| > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
| > news:sjaf75-ivc.ln1@sirius.tg00suus7038.net...
| > | In sci.physics.relativity, Androcles
| > | <Headmaster@Hogwarts.physics>
| > | wrote
| > | on Sat, 02 Feb 2008 19:40:32 GMT
| > | <QG3pj.28957$801.2573@fe1.news.blueyonder.co.uk>:
| > | >
| > | > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
| > message
| > | > news:koqe75-98a.ln1@sirius.tg00suus7038.net...
| > | > | In sci.physics.relativity, Androcles
| > | > | <Headmaster@Hogwarts.physics>
| > | > | wrote
| > | > | on Sat, 02 Feb 2008 15:28:57 GMT
| > | > | <Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:
| > | > | >
| > | > | > "PD" <TheDraperFamily@gmail.com> wrote in message
| > | > | >
| > news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
| > | > | > On Feb 2, 12:13 am,

wrote:
| > | > | >> Drop something and it immediately arives at a speed of 32 feet
| > | > | >> persecond falling in a straight line.
| > | > | >
| > | > | > No, it doesn't immediately arrive at that speed, and it doesn't
stay
| > | > | > at that speed even when it does arrive at that speed. Oh, and
speed
| > | > | > isn't acceleration.
| > | > |
| > | > | One can also establish how long it takes to get at that speed.
| > | > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | > | one second instantaneously, 2 seconds if one takes the average.
| > | > | (In other words, 2 seconds from the drop the stone is moving
| > | > | at 64 ft/s, and the average speed -- as well as x/t -- is 32
ft/s.)
| > | > |
| > | > | [rest snipped by TGITM for brevity]
| > | >
| > | > Take it up with Phuckwit Duck, he wrote it.
| > | >
| > | > Here's a question for you though.
| > | > If the acceleration due to gravity is g fps/s at the
| > | > top of a mineshaft at sea level and also g fps/s at the
| > | > top of Everest, how high do you have to be before
| > | > an inverse square law kicks in and it is less than g, and why?
| > | > (Hint: The Earth is not a point, the local g field is approximately
| > | > parallel.)
| > | >
| > |
| > | g = GM/d^2 is never constant, as d is never constant in
| > | the usual run of problems (e.g., a ball thrown upward).
| >
| >
| > Of course, but it is a function of something.
| >
| >
| > | SR is
| >
| > I wasn't asking about SR drool. Try to concentrate on the question,
| > this is purely Newtonian.
| >
| > If the acceleration due to gravity is g fps/s at the
| > top of a mineshaft at sea level and also g fps/s at the
| > top of Everest, how high do you have to be before
| > an inverse square law kicks in and it is less than g, and why?
| > (Hint: The Earth is not a point, the local g field is approximately
| > parallel.)
| >
|
| Ah, interesting thought. I'll have to work out,
| briefly put, the mass distribution such that
|
| [1] It is in the shape of a mountain somewhere, of height 9.9 km or so,
| [2] one has constant g on any point on the surface of that mountain,
| [3] work out g in other points of space nearby.
|
| I'll have to get back to you on that. Note that the mass will
| have to be of variable density.
| Also, the inverse square law will *never* kick in anyway,
| in that case.
Ignore the mountain, assume a perfect sphere, homogeneous density.
You can still be in an airplane at 30,000 feet when the flight attendant
drops your meal in your lap at 32 fps/s. Remarkably, the underbelly
of the plane doesn't shield gravity.
What I'm getting at here is the Earth is NOT a point.
If you flatten the Earth into a thin disc like a CD ROM
and stand near the centre then your vertical weight will
be zero and your horizontal weight slightly greater than zero.
If you are on the rim then you have your full vertical weight,
no horizontal weight and are rotated 90 degrees. If you step
off the edge of the world you fall through the centre, skinning
your butt on the way down and back up the other side.
If you are space-walking a long way from the Earth then it is
effectively a point and the inverse square law (almost) applies,
you still fall toward the centre.
For the spherical Earth F = GMm/r^2 fails below the Earth's
surface, you are weightless at the centre and will not accelerate
at 32 fps/s.
So what I want is an exact function that approximates F = GMm/r^2
when a long way from Earth, is equal to F = m*32 fps/s at or near
the Earth's surface and F = 0 at the centre so that I can calculate
my exact weight at the BOTTOM of a 30,000 ft deep mine shaft
as well as at 30,000 feet in a plane.
Therefore this function will include both r and R where R is the
radius of the Earth, just as we include both M and m where m
is a point. M cannot be a point.
F = f(G, M, m, r, R).
What is the function f() that
F = GMm/r^2 is an approximation to?
| One can of course approximate
Not the question asked, F = GMm/r^2 is already an approximation.
.

User: "The Ghost In The Machine"
Title: Re: Gravity acceleration and deceleration	04 Feb 2008 01:07:07 AM

In sci.physics.relativity, Androcles
<Headmaster@Hogwarts.physics>
wrote
on Sun, 03 Feb 2008 17:35:46 GMT
<SXmpj.40181$3m6.8268@fe2.news.blueyonder.co.uk>:

I'm not too worried about the plane, but I am wondering about what
is essentially a volume integral. Briefly put, the gravitational
acceleration on a point is simply the sum of all mass-elements
around that point (including far away planets, but their effect is
considerably lessened, to the point where we probably can ignore
them in this particular problem):
F(p) = integral(x in V) -Gm(x - p)/||x - p||^3 dV
where p is an arbitrary point somewhere, G is Newton's
universal constant, d_M is a constant (about 5500 kg/m^3,
if memory serves), and V is the Earth.
If one wants to introduce mass anomalies, one can of course
extend the formula:
F(p) = integral(x in V) -Gm(x)(x - p)/||x - p||^3 dV
where d_M(x) is a function of x, describing the density of
the stuff at or around x.
Since you've stipulated a perfect sphere, no rotation,
no mass anomalies, one can replace the volume integral
with a rather ugly triple integral, and hope for the best:
F(p) = integral(-r_0 < x < r_0)
integral(-sqrt(r_0^2 - x^2) < y < sqrt(r_0^2 - x^2)
integral(-sqrt(r_0^2 - x^2 - y^2) < z < sqrt(r_0^2 - x^2 - y^2)
(-G * d_M * [x - p_x, y - p_y, z - p_z]
/ sqrt( (x-p_x)^2 + (y-p_y)^2 + (z-p_z)^2)^3 dz dy dx
where r_0 is V's radius -- about 6378 km.
It is possible without loss of generality to take p_x = p_y
= 0, by appropriate rotation of the coordinate axes, and
convert this into a simpler integral by using cylindrical
coordinates (z, r, theta). Of course F is still a vector,
but now it can depend solely on the distance from the
sphere's center, which I'll call d, and substituting d
for p, one gets:
F(d) = integral(-r_0 < z < r_0)
integral(0 <= r < r_0)
integral(0 <= theta < 2*Pi)
-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2)^3 dV
= integral(-r_0 < z < r_0)
integral(0 <= r < r_0)
integral(0 <= theta < 2*Pi)
(-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r d{theta} dr dz
Since theta appears nowhere in the formula, it drops out pretty easily:
F(d) = 2 * Pi * integral(-r_0 < z < r_0)
integral(0 <= r < r_0)
(-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r dr dz
We can evaluate the inner integral by using a variant of the chain rule:
F(d) = 2 * Pi * integral(-r_0 < z < r_0)
(-G * m * d_M * (d - z) * (2/3)
* (1 / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 + r_0^2) ) dz
= - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
((d - z) / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 + r_0^2) ) dz
If d > r_0, the integral can be split into two parts, one of which
is almost trivial to evaluate:
F(d) = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
(1 - (d - z) / sqrt( (d - z)^2 + r_0^2) ) dz
F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
+ (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
((d - z) / sqrt( (d-z)^2 + r_0)^2) dz
= - (4/3) * G * m * d_M * Pi * (2 * r_0)
- (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
((z - d) / sqrt( (z-d)^2 + r_0)^2) dz
By substituting z-d = w (or w+d = z) and dz = dw, one gets
F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
- (2/3) * G * m * d_M * Pi * integral( - r_0 - d < w < r_0 - d)
(2w / sqrt( w^2 + r_0^2) ) dw
Since w != 0, it is immediately apparent F'(d) is nonzero for any
d > r_0; therefore F(d) is not constant for d > r_0.
One can use the chain rule again, and, since k^2 = (-k)^2, we
can eliminate some redundant stuff, finally getting:
F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
- (2/3) * G * m * d_M * Pi *
( sqrt( (d + r_0)^2 + r_0^2) - sqrt( (d - r_0)^2 + r_0^2))
= - (4/3) * G * m * d_M * Pi * (2 * r_0)
- (2/3) * G * m * d_M * Pi * r_0 *
( sqrt( (d/r_0 + 1)^2 + 1) - sqrt( (d/r_0 - 1)^2 + 1))
= - (4/3) * G * m * d_M * Pi * (2 * r_0)
- (2/3) * G * m * d_M * Pi * d *
( sqrt( (1 + r_0/d)^2 + (r_0/d)^2 )
- sqrt( (1 - r_0/d)^2 + (r_0/d)^2) )
(I'll admit I'm not at all sure this is entirely correct, but I'm
getting a little tired, and it's been a long time since I've
had to fuss with integrals. :-) If anyone can point me to the
correct solution of this problem, that would be helpful.)

What I'm getting at here is the Earth is NOT a point.
If you flatten the Earth into a thin disc like a CD ROM
and stand near the centre then your vertical weight will
be zero and your horizontal weight slightly greater than zero.
If you are on the rim then you have your full vertical weight,
no horizontal weight and are rotated 90 degrees. If you step
off the edge of the world you fall through the centre, skinning
your butt on the way down and back up the other side.

If you are space-walking a long way from the Earth then it is
effectively a point and the inverse square law (almost) applies,
you still fall toward the centre.

For the spherical Earth F = GMm/r^2 fails below the Earth's
surface, you are weightless at the centre and will not accelerate
at 32 fps/s.

So what I want is an exact function that approximates F = GMm/r^2
when a long way from Earth, is equal to F = m*32 fps/s at or near
the Earth's surface and F = 0 at the centre so that I can calculate
my exact weight at the BOTTOM of a 30,000 ft deep mine shaft
as well as at 30,000 feet in a plane.
Therefore this function will include both r and R where R is the
radius of the Earth, just as we include both M and m where m
is a point. M cannot be a point.

F = f(G, M, m, r, R).

What is the function f() that
F = GMm/r^2 is an approximation to?

Good question.

| One can of course approximate

Not the question asked, F = GMm/r^2 is already an approximation.

--
#191,

Linux. Because life's too short for a buggy OS.
--
Posted via a free Usenet account from http://www.teranews.com
.

User: "Androcles"
Title: Re: Gravity acceleration and deceleration	04 Feb 2008 05:44:01 AM

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:r5qi75-131.ln1@sirius.tg00suus7038.net...
| In sci.physics.relativity, Androcles
| <Headmaster@Hogwarts.physics>
| wrote
| on Sun, 03 Feb 2008 17:35:46 GMT
| <SXmpj.40181$3m6.8268@fe2.news.blueyonder.co.uk>:
| >
| > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
| > news:b74h75-t3n.ln1@sirius.tg00suus7038.net...
| > | In sci.physics.relativity, Androcles
| > | <Headmaster@Hogwarts.physics>
| > | wrote
| > | on Sun, 03 Feb 2008 00:24:27 GMT
| > | <%Q7pj.29003$801.1233@fe1.news.blueyonder.co.uk>:
| > | >
| > | > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
| > message
| > | > news:sjaf75-ivc.ln1@sirius.tg00suus7038.net...
| > | > | In sci.physics.relativity, Androcles
| > | > | <Headmaster@Hogwarts.physics>
| > | > | wrote
| > | > | on Sat, 02 Feb 2008 19:40:32 GMT
| > | > | <QG3pj.28957$801.2573@fe1.news.blueyonder.co.uk>:
| > | > | >
| > | > | > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote
in
| > | > message
| > | > | > news:koqe75-98a.ln1@sirius.tg00suus7038.net...
| > | > | > | In sci.physics.relativity, Androcles
| > | > | > | <Headmaster@Hogwarts.physics>
| > | > | > | wrote
| > | > | > | on Sat, 02 Feb 2008 15:28:57 GMT
| > | > | > | <Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:
| > | > | > | >
| > | > | > | > "PD" <TheDraperFamily@gmail.com> wrote in message
| > | > | > | >
| > | >
news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
| > | > | > | > On Feb 2, 12:13 am,

wrote:
| > | > | > | >> Drop something and it immediately arives at a speed of 32
feet
| > | > | > | >> persecond falling in a straight line.
| > | > | > | >
| > | > | > | > No, it doesn't immediately arrive at that speed, and it
doesn't
| > stay
| > | > | > | > at that speed even when it does arrive at that speed. Oh,
and
| > speed
| > | > | > | > isn't acceleration.
| > | > | > |
| > | > | > | One can also establish how long it takes to get at that speed.
| > | > | > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | > | > | one second instantaneously, 2 seconds if one takes the
average.
| > | > | > | (In other words, 2 seconds from the drop the stone is moving
| > | > | > | at 64 ft/s, and the average speed -- as well as x/t -- is 32
| > ft/s.)
| > | > | > |
| > | > | > | [rest snipped by TGITM for brevity]
| > | > | >
| > | > | > Take it up with Phuckwit Duck, he wrote it.
| > | > | >
| > | > | > Here's a question for you though.
| > | > | > If the acceleration due to gravity is g fps/s at the
| > | > | > top of a mineshaft at sea level and also g fps/s at the
| > | > | > top of Everest, how high do you have to be before
| > | > | > an inverse square law kicks in and it is less than g, and why?
| > | > | > (Hint: The Earth is not a point, the local g field is
approximately
| > | > | > parallel.)
| > | > | >
| > | > |
| > | > | g = GM/d^2 is never constant, as d is never constant in
| > | > | the usual run of problems (e.g., a ball thrown upward).
| > | >
| > | >
| > | > Of course, but it is a function of something.
| > | >
| > | >
| > | > | SR is
| > | >
| > | > I wasn't asking about SR drool. Try to concentrate on the question,
| > | > this is purely Newtonian.
| > | >
| > | > If the acceleration due to gravity is g fps/s at the
| > | > top of a mineshaft at sea level and also g fps/s at the
| > | > top of Everest, how high do you have to be before
| > | > an inverse square law kicks in and it is less than g, and why?
| > | > (Hint: The Earth is not a point, the local g field is approximately
| > | > parallel.)
| > | >
| > |
| > | Ah, interesting thought. I'll have to work out,
| > | briefly put, the mass distribution such that
| > |
| > | [1] It is in the shape of a mountain somewhere, of height 9.9 km or
so,
| > | [2] one has constant g on any point on the surface of that mountain,
| > | [3] work out g in other points of space nearby.
| > |
| > | I'll have to get back to you on that. Note that the mass will
| > | have to be of variable density.
| >
| >
| > | Also, the inverse square law will *never* kick in anyway,
| > | in that case.
| >
| > Ignore the mountain, assume a perfect sphere, homogeneous density.
| > You can still be in an airplane at 30,000 feet when the flight attendant
| > drops your meal in your lap at 32 fps/s. Remarkably, the underbelly
| > of the plane doesn't shield gravity.
|
| I'm not too worried about the plane, but I am wondering about what
| is essentially a volume integral. Briefly put, the gravitational
| acceleration on a point is simply the sum of all mass-elements
| around that point (including far away planets, but their effect is
| considerably lessened, to the point where we probably can ignore
| them in this particular problem):
|
| F(p) = integral(x in V) -Gm(x - p)/||x - p||^3 dV
The force between p_i and p_j is still proportional to 1/||p_i - p_j||^2,
where p_n = (x_n, y_n, z_n), which hat did you pluck a cube from?
|
| where p is an arbitrary point somewhere, G is Newton's
| universal constant, d_M is a constant (about 5500 kg/m^3,
| if memory serves), and V is the Earth.
|
| If one wants to introduce mass anomalies, one can of course
| extend the formula:
|
| F(p) = integral(x in V) -Gm(x)(x - p)/||x - p||^3 dV
|
| where d_M(x) is a function of x, describing the density of
| the stuff at or around x.
|
| Since you've stipulated a perfect sphere, no rotation,
| no mass anomalies, one can replace the volume integral
| with a rather ugly triple integral, and hope for the best:
|
| F(p) = integral(-r_0 < x < r_0)
| integral(-sqrt(r_0^2 - x^2) < y < sqrt(r_0^2 - x^2)
| integral(-sqrt(r_0^2 - x^2 - y^2) < z < sqrt(r_0^2 - x^2 - y^2)
| (-G * d_M * [x - p_x, y - p_y, z - p_z]
| / sqrt( (x-p_x)^2 + (y-p_y)^2 + (z-p_z)^2)^3 dz dy dx
|
| where r_0 is V's radius -- about 6378 km.
|
| It is possible without loss of generality to take p_x = p_y
| = 0, by appropriate rotation of the coordinate axes, and
| convert this into a simpler integral by using cylindrical
| coordinates (z, r, theta). Of course F is still a vector,
| but now it can depend solely on the distance from the
| sphere's center, which I'll call d, and substituting d
| for p, one gets:
|
| F(d) = integral(-r_0 < z < r_0)
| integral(0 <= r < r_0)
| integral(0 <= theta < 2*Pi)
| -G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2)^3 dV
|
| = integral(-r_0 < z < r_0)
| integral(0 <= r < r_0)
| integral(0 <= theta < 2*Pi)
| (-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r d{theta} dr
dz
|
| Since theta appears nowhere in the formula, it drops out pretty easily:
|
| F(d) = 2 * Pi * integral(-r_0 < z < r_0)
| integral(0 <= r < r_0)
| (-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r dr dz
|
| We can evaluate the inner integral by using a variant of the chain rule:
|
| F(d) = 2 * Pi * integral(-r_0 < z < r_0)
| (-G * m * d_M * (d - z) * (2/3)
| * (1 / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 + r_0^2) ) dz
|
| = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
| ((d - z) / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 +
r_0^2) ) dz
|
|
| If d > r_0, the integral can be split into two parts, one of which
| is almost trivial to evaluate:
|
| F(d) = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
| (1 - (d - z) / sqrt( (d - z)^2 + r_0^2) ) dz
|
| F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| + (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
| ((d - z) / sqrt( (d-z)^2 + r_0)^2) dz
|
| = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
| ((z - d) / sqrt( (z-d)^2 + r_0)^2) dz
|
| By substituting z-d = w (or w+d = z) and dz = dw, one gets
|
| F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi * integral( - r_0 - d < w < r_0 - d)
| (2w / sqrt( w^2 + r_0^2) ) dw
|
| Since w != 0, it is immediately apparent F'(d) is nonzero for any
| d > r_0; therefore F(d) is not constant for d > r_0.
|
| One can use the chain rule again, and, since k^2 = (-k)^2, we
| can eliminate some redundant stuff, finally getting:
|
| F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi *
| ( sqrt( (d + r_0)^2 + r_0^2) - sqrt( (d - r_0)^2 + r_0^2))
|
| = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi * r_0 *
| ( sqrt( (d/r_0 + 1)^2 + 1) - sqrt( (d/r_0 - 1)^2 + 1))
|
| = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi * d *
| ( sqrt( (1 + r_0/d)^2 + (r_0/d)^2 )
| - sqrt( (1 - r_0/d)^2 + (r_0/d)^2) )
|
| (I'll admit I'm not at all sure this is entirely correct, but I'm
| getting a little tired, and it's been a long time since I've
| had to fuss with integrals. :-) If anyone can point me to the
| correct solution of this problem, that would be helpful.)
|
| >
| > What I'm getting at here is the Earth is NOT a point.
| > If you flatten the Earth into a thin disc like a CD ROM
| > and stand near the centre then your vertical weight will
| > be zero and your horizontal weight slightly greater than zero.
| > If you are on the rim then you have your full vertical weight,
| > no horizontal weight and are rotated 90 degrees. If you step
| > off the edge of the world you fall through the centre, skinning
| > your butt on the way down and back up the other side.
| >
| > If you are space-walking a long way from the Earth then it is
| > effectively a point and the inverse square law (almost) applies,
| > you still fall toward the centre.
| >
| > For the spherical Earth F = GMm/r^2 fails below the Earth's
| > surface, you are weightless at the centre and will not accelerate
| > at 32 fps/s.
| >
| > So what I want is an exact function that approximates F = GMm/r^2
| > when a long way from Earth, is equal to F = m*32 fps/s at or near
| > the Earth's surface and F = 0 at the centre so that I can calculate
| > my exact weight at the BOTTOM of a 30,000 ft deep mine shaft
| > as well as at 30,000 feet in a plane.
| > Therefore this function will include both r and R where R is the
| > radius of the Earth, just as we include both M and m where m
| > is a point. M cannot be a point.
| >
| > F = f(G, M, m, r, R).
| >
| > What is the function f() that
| > F = GMm/r^2 is an approximation to?
|
| Good question.
Yes, and what I want is a good answer, you erred right at the start
with a cubic power and then got carried away trying to solve an
error.
Speaking of cubes, if you take the eight vertices at a cube's
corners as fixed points of unit mass and calculate the forces acting
on a unit point mass P at the cube's centre, the resultant is zero.
For any other P the non-zero resultant is the sum (for i = 0 to 7)
of 1/(P - p_i)^2.
Briefly put, the gravitational acceleration on a point is simply the sum
of all mass-elements around that point.
A point mass that is free to move will move toward the vertex
it is closest to, and not the centre of the cube. Thus the arrangement
is unstable. Now if we replace the centre of the faces of the cube
with unit point masses, with P near the centre the movement will
be toward the centre of a face and not a corner vertex because that
will be nearer to P.
Now the problem extends to a spherical shell, the resultant
takes the free mass to its surface, whether from inside or outside
the sphere. A pea inside a blown egg will attach to the shell,
not the centre of the egg.
.

User: "The Ghost In The Machine"
Title: Re: Gravity acceleration and deceleration	04 Feb 2008 09:14:14 PM

In sci.physics.relativity, Androcles
<Headmaster@Hogwarts.physics>
wrote
on Mon, 04 Feb 2008 11:44:01 GMT
<5UCpj.80994$Kt4.73638@fe3.news.blueyonder.co.uk>:

wrote:
| > | > | > | >> Drop something and it immediately arives at a speed of 32
feet
| > | > | > | >> persecond falling in a straight line.
| > | > | > | >
| > | > | > | > No, it doesn't immediately arrive at that speed, and it
doesn't
| > stay
| > | > | > | > at that speed even when it does arrive at that speed. Oh,
and
| > speed
| > | > | > | > isn't acceleration.
| > | > | > |
| > | > | > | One can also establish how long it takes to get at that speed.
| > | > | > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | > | > | one second instantaneously, 2 seconds if one takes the
average.
| > | > | > | (In other words, 2 seconds from the drop the stone is moving
| > | > | > | at 64 ft/s, and the average speed -- as well as x/t -- is 32
| > ft/s.)
| > | > | > |
| > | > | > | [rest snipped by TGITM for brevity]
| > | > | >
| > | > | > Take it up with Phuckwit Duck, he wrote it.
| > | > | >
| > | > | > Here's a question for you though.
| > | > | > If the acceleration due to gravity is g fps/s at the
| > | > | > top of a mineshaft at sea level and also g fps/s at the
| > | > | > top of Everest, how high do you have to be before
| > | > | > an inverse square law kicks in and it is less than g, and why?
| > | > | > (Hint: The Earth is not a point, the local g field is
approximately
| > | > | > parallel.)
| > | > | >
| > | > |
| > | > | g = GM/d^2 is never constant, as d is never constant in
| > | > | the usual run of problems (e.g., a ball thrown upward).
| > | >
| > | >
| > | > Of course, but it is a function of something.
| > | >
| > | >
| > | > | SR is
| > | >
| > | > I wasn't asking about SR drool. Try to concentrate on the question,
| > | > this is purely Newtonian.
| > | >
| > | > If the acceleration due to gravity is g fps/s at the
| > | > top of a mineshaft at sea level and also g fps/s at the
| > | > top of Everest, how high do you have to be before
| > | > an inverse square law kicks in and it is less than g, and why?
| > | > (Hint: The Earth is not a point, the local g field is approximately
| > | > parallel.)
| > | >
| > |
| > | Ah, interesting thought. I'll have to work out,
| > | briefly put, the mass distribution such that
| > |
| > | [1] It is in the shape of a mountain somewhere, of height 9.9 km or
so,
| > | [2] one has constant g on any point on the surface of that mountain,
| > | [3] work out g in other points of space nearby.
| > |
| > | I'll have to get back to you on that. Note that the mass will
| > | have to be of variable density.
| >
| >
| > | Also, the inverse square law will *never* kick in anyway,
| > | in that case.
| >
| > Ignore the mountain, assume a perfect sphere, homogeneous density.
| > You can still be in an airplane at 30,000 feet when the flight attendant
| > drops your meal in your lap at 32 fps/s. Remarkably, the underbelly
| > of the plane doesn't shield gravity.
|
| I'm not too worried about the plane, but I am wondering about what
| is essentially a volume integral. Briefly put, the gravitational
| acceleration on a point is simply the sum of all mass-elements
| around that point (including far away planets, but their effect is
| considerably lessened, to the point where we probably can ignore
| them in this particular problem):
|
| F(p) = integral(x in V) -Gm(x - p)/||x - p||^3 dV

The force between p_i and p_j is still proportional to 1/||p_i - p_j||^2,
where p_n = (x_n, y_n, z_n), which hat did you pluck a cube from?

|
| where p is an arbitrary point somewhere, G is Newton's
| universal constant, d_M is a constant (about 5500 kg/m^3,
| if memory serves), and V is the Earth.
|
| If one wants to introduce mass anomalies, one can of course
| extend the formula:
|
| F(p) = integral(x in V) -Gm(x)(x - p)/||x - p||^3 dV
|
| where d_M(x) is a function of x, describing the density of
| the stuff at or around x.
|
| Since you've stipulated a perfect sphere, no rotation,
| no mass anomalies, one can replace the volume integral
| with a rather ugly triple integral, and hope for the best:
|
| F(p) = integral(-r_0 < x < r_0)
| integral(-sqrt(r_0^2 - x^2) < y < sqrt(r_0^2 - x^2)
| integral(-sqrt(r_0^2 - x^2 - y^2) < z < sqrt(r_0^2 - x^2 - y^2)
| (-G * d_M * [x - p_x, y - p_y, z - p_z]
| / sqrt( (x-p_x)^2 + (y-p_y)^2 + (z-p_z)^2)^3 dz dy dx
|
| where r_0 is V's radius -- about 6378 km.
|
| It is possible without loss of generality to take p_x = p_y
| = 0, by appropriate rotation of the coordinate axes, and
| convert this into a simpler integral by using cylindrical
| coordinates (z, r, theta). Of course F is still a vector,
| but now it can depend solely on the distance from the
| sphere's center, which I'll call d, and substituting d
| for p, one gets:
|
| F(d) = integral(-r_0 < z < r_0)
| integral(0 <= r < r_0)
| integral(0 <= theta < 2*Pi)
| -G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2)^3 dV
|
| = integral(-r_0 < z < r_0)
| integral(0 <= r < r_0)
| integral(0 <= theta < 2*Pi)
| (-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r d{theta} dr
dz
|
| Since theta appears nowhere in the formula, it drops out pretty easily:
|
| F(d) = 2 * Pi * integral(-r_0 < z < r_0)
| integral(0 <= r < r_0)
| (-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r dr dz
|
| We can evaluate the inner integral by using a variant of the chain rule:
|
| F(d) = 2 * Pi * integral(-r_0 < z < r_0)
| (-G * m * d_M * (d - z) * (2/3)
| * (1 / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 + r_0^2) ) dz
|
| = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
| ((d - z) / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 +
r_0^2) ) dz
|
|
| If d > r_0, the integral can be split into two parts, one of which
| is almost trivial to evaluate:
|
| F(d) = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
| (1 - (d - z) / sqrt( (d - z)^2 + r_0^2) ) dz
|
| F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| + (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
| ((d - z) / sqrt( (d-z)^2 + r_0)^2) dz
|
| = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
| ((z - d) / sqrt( (z-d)^2 + r_0)^2) dz
|
| By substituting z-d = w (or w+d = z) and dz = dw, one gets
|
| F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi * integral( - r_0 - d < w < r_0 - d)
| (2w / sqrt( w^2 + r_0^2) ) dw
|
| Since w != 0, it is immediately apparent F'(d) is nonzero for any
| d > r_0; therefore F(d) is not constant for d > r_0.
|
| One can use the chain rule again, and, since k^2 = (-k)^2, we
| can eliminate some redundant stuff, finally getting:
|
| F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi *
| ( sqrt( (d + r_0)^2 + r_0^2) - sqrt( (d - r_0)^2 + r_0^2))
|
| = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi * r_0 *
| ( sqrt( (d/r_0 + 1)^2 + 1) - sqrt( (d/r_0 - 1)^2 + 1))
|
| = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| - (2/3) * G * m * d_M * Pi * d *
| ( sqrt( (1 + r_0/d)^2 + (r_0/d)^2 )
| - sqrt( (1 - r_0/d)^2 + (r_0/d)^2) )
|
| (I'll admit I'm not at all sure this is entirely correct, but I'm
| getting a little tired, and it's been a long time since I've
| had to fuss with integrals. :-) If anyone can point me to the
| correct solution of this problem, that would be helpful.)
|
| >
| > What I'm getting at here is the Earth is NOT a point.
| > If you flatten the Earth into a thin disc like a CD ROM
| > and stand near the centre then your vertical weight will
| > be zero and your horizontal weight slightly greater than zero.
| > If you are on the rim then you have your full vertical weight,
| > no horizontal weight and are rotated 90 degrees. If you step
| > off the edge of the world you fall through the centre, skinning
| > your butt on the way down and back up the other side.
| >
| > If you are space-walking a long way from the Earth then it is
| > effectively a point and the inverse square law (almost) applies,
| > you still fall toward the centre.
| >
| > For the spherical Earth F = GMm/r^2 fails below the Earth's
| > surface, you are weightless at the centre and will not accelerate
| > at 32 fps/s.
| >
| > So what I want is an exact function that approximates F = GMm/r^2
| > when a long way from Earth, is equal to F = m*32 fps/s at or near
| > the Earth's surface and F = 0 at the centre so that I can calculate
| > my exact weight at the BOTTOM of a 30,000 ft deep mine shaft
| > as well as at 30,000 feet in a plane.
| > Therefore this function will include both r and R where R is the
| > radius of the Earth, just as we include both M and m where m
| > is a point. M cannot be a point.
| >
| > F = f(G, M, m, r, R).
| >
| > What is the function f() that
| > F = GMm/r^2 is an approximation to?
|
| Good question.

Yes, and what I want is a good answer, you erred right at the start
with a cubic power and then got carried away trying to solve an
error.

There was an error, yes; it wasn't there. But the reason I said
-d/||d||^3
instead of
1/||d||^2
is quite simple; F is a *vector* field, not a scalar one.
In other words, if I'm, say, over San Francisco and you're
over Sydney, our force vectors aren't pointing in the same
direction at all.
Magnitude doesn't quite cut it.

Speaking of cubes, if you take the eight vertices at a cube's
corners as fixed points of unit mass and calculate the forces acting
on a unit point mass P at the cube's centre, the resultant is zero.

Correct.

For any other P the non-zero resultant is the sum (for i = 0 to 7)
of 1/(P - p_i)^2.

Actually, sum(i=0,7) GMm/(P-p_i)^2, where M is the point mass, m the
mass of each cube vertex, G the universal constant of gravitation,
P the position of the point mass, and p_i the cube vertex.
But you're there up to a constant.

Briefly put, the gravitational acceleration on a point is simply the sum
of all mass-elements around that point.

And an integral is ... ?

A point mass that is free to move will move toward the vertex
it is closest to, and not the centre of the cube. Thus the arrangement
is unstable. Now if we replace the centre of the faces of the cube
with unit point masses, with P near the centre the movement will
be toward the centre of a face and not a corner vertex because that
will be nearer to P.

Now the problem extends to a spherical shell, the resultant
takes the free mass to its surface, whether from inside or outside
the sphere. A pea inside a blown egg will attach to the shell,
not the centre of the egg.

A pea inside of a blown spherical shell, absent other
forces, gets no force at all.
--
#191,

/dev/signature: Not a text file
--
Posted via a free Usenet account from http://www.teranews.com
.

User: "Androcles"
Title: Re: Gravity acceleration and deceleration	04 Feb 2008 10:32:08 PM

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:6t0l75-pie.ln1@sirius.tg00suus7038.net...
| In sci.physics.relativity, Androcles
| <Headmaster@Hogwarts.physics>
| wrote
| on Mon, 04 Feb 2008 11:44:01 GMT
| <5UCpj.80994$Kt4.73638@fe3.news.blueyonder.co.uk>:
| >
| > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
| > news:r5qi75-131.ln1@sirius.tg00suus7038.net...
| > | In sci.physics.relativity, Androcles
| > | <Headmaster@Hogwarts.physics>
| > | wrote
| > | on Sun, 03 Feb 2008 17:35:46 GMT
| > | <SXmpj.40181$3m6.8268@fe2.news.blueyonder.co.uk>:
| > | >
| > | > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
| > message
| > | > news:b74h75-t3n.ln1@sirius.tg00suus7038.net...
| > | > | In sci.physics.relativity, Androcles
| > | > | <Headmaster@Hogwarts.physics>
| > | > | wrote
| > | > | on Sun, 03 Feb 2008 00:24:27 GMT
| > | > | <%Q7pj.29003$801.1233@fe1.news.blueyonder.co.uk>:
| > | > | >
| > | > | > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote
in
| > | > message
| > | > | > news:sjaf75-ivc.ln1@sirius.tg00suus7038.net...
| > | > | > | In sci.physics.relativity, Androcles
| > | > | > | <Headmaster@Hogwarts.physics>
| > | > | > | wrote
| > | > | > | on Sat, 02 Feb 2008 19:40:32 GMT
| > | > | > | <QG3pj.28957$801.2573@fe1.news.blueyonder.co.uk>:
| > | > | > | >
| > | > | > | > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net>
wrote
| > in
| > | > | > message
| > | > | > | > news:koqe75-98a.ln1@sirius.tg00suus7038.net...
| > | > | > | > | In sci.physics.relativity, Androcles
| > | > | > | > | <Headmaster@Hogwarts.physics>
| > | > | > | > | wrote
| > | > | > | > | on Sat, 02 Feb 2008 15:28:57 GMT
| > | > | > | > | <Z_%oj.65742$Kt4.9586@fe3.news.blueyonder.co.uk>:
| > | > | > | > | >
| > | > | > | > | > "PD" <TheDraperFamily@gmail.com> wrote in message
| > | > | > | > | >
| > | > | >
| > news:fab838ca-32e0-4504-b90c-bb631ba3f949@s8g2000prg.googlegroups.com...
| > | > | > | > | > On Feb 2, 12:13 am,

wrote:
| > | > | > | > | >> Drop something and it immediately arives at a speed of
32
| > feet
| > | > | > | > | >> persecond falling in a straight line.
| > | > | > | > | >
| > | > | > | > | > No, it doesn't immediately arrive at that speed, and it
| > doesn't
| > | > stay
| > | > | > | > | > at that speed even when it does arrive at that speed.
Oh,
| > and
| > | > speed
| > | > | > | > | > isn't acceleration.
| > | > | > | > |
| > | > | > | > | One can also establish how long it takes to get at that
speed.
| > | > | > | > | Since g = 32 ft/s/s, the time it takes to reach 32 f/s is
| > | > | > | > | one second instantaneously, 2 seconds if one takes the
| > average.
| > | > | > | > | (In other words, 2 seconds from the drop the stone is
moving
| > | > | > | > | at 64 ft/s, and the average speed -- as well as x/t -- is
32
| > | > ft/s.)
| > | > | > | > |
| > | > | > | > | [rest snipped by TGITM for brevity]
| > | > | > | >
| > | > | > | > Take it up with Phuckwit Duck, he wrote it.
| > | > | > | >
| > | > | > | > Here's a question for you though.
| > | > | > | > If the acceleration due to gravity is g fps/s at the
| > | > | > | > top of a mineshaft at sea level and also g fps/s at the
| > | > | > | > top of Everest, how high do you have to be before
| > | > | > | > an inverse square law kicks in and it is less than g, and
why?
| > | > | > | > (Hint: The Earth is not a point, the local g field is
| > approximately
| > | > | > | > parallel.)
| > | > | > | >
| > | > | > |
| > | > | > | g = GM/d^2 is never constant, as d is never constant in
| > | > | > | the usual run of problems (e.g., a ball thrown upward).
| > | > | >
| > | > | >
| > | > | > Of course, but it is a function of something.
| > | > | >
| > | > | >
| > | > | > | SR is
| > | > | >
| > | > | > I wasn't asking about SR drool. Try to concentrate on the
question,
| > | > | > this is purely Newtonian.
| > | > | >
| > | > | > If the acceleration due to gravity is g fps/s at the
| > | > | > top of a mineshaft at sea level and also g fps/s at the
| > | > | > top of Everest, how high do you have to be before
| > | > | > an inverse square law kicks in and it is less than g, and why?
| > | > | > (Hint: The Earth is not a point, the local g field is
approximately
| > | > | > parallel.)
| > | > | >
| > | > |
| > | > | Ah, interesting thought. I'll have to work out,
| > | > | briefly put, the mass distribution such that
| > | > |
| > | > | [1] It is in the shape of a mountain somewhere, of height 9.9 km
or
| > so,
| > | > | [2] one has constant g on any point on the surface of that
mountain,
| > | > | [3] work out g in other points of space nearby.
| > | > |
| > | > | I'll have to get back to you on that. Note that the mass will
| > | > | have to be of variable density.
| > | >
| > | >
| > | > | Also, the inverse square law will *never* kick in anyway,
| > | > | in that case.
| > | >
| > | > Ignore the mountain, assume a perfect sphere, homogeneous density.
| > | > You can still be in an airplane at 30,000 feet when the flight
attendant
| > | > drops your meal in your lap at 32 fps/s. Remarkably, the underbelly
| > | > of the plane doesn't shield gravity.
| > |
| > | I'm not too worried about the plane, but I am wondering about what
| > | is essentially a volume integral. Briefly put, the gravitational
| > | acceleration on a point is simply the sum of all mass-elements
| > | around that point (including far away planets, but their effect is
| > | considerably lessened, to the point where we probably can ignore
| > | them in this particular problem):
| > |
| > | F(p) = integral(x in V) -Gm(x - p)/||x - p||^3 dV
| >
| > The force between p_i and p_j is still proportional to 1/||p_i -
p_j||^2,
| > where p_n = (x_n, y_n, z_n), which hat did you pluck a cube from?
| >
| >
| >
| > |
| > | where p is an arbitrary point somewhere, G is Newton's
| > | universal constant, d_M is a constant (about 5500 kg/m^3,
| > | if memory serves), and V is the Earth.
| > |
| > | If one wants to introduce mass anomalies, one can of course
| > | extend the formula:
| > |
| > | F(p) = integral(x in V) -Gm(x)(x - p)/||x - p||^3 dV
| > |
| > | where d_M(x) is a function of x, describing the density of
| > | the stuff at or around x.
| > |
| > | Since you've stipulated a perfect sphere, no rotation,
| > | no mass anomalies, one can replace the volume integral
| > | with a rather ugly triple integral, and hope for the best:
| > |
| > | F(p) = integral(-r_0 < x < r_0)
| > | integral(-sqrt(r_0^2 - x^2) < y < sqrt(r_0^2 - x^2)
| > | integral(-sqrt(r_0^2 - x^2 - y^2) < z < sqrt(r_0^2 - x^2 - y^2)
| > | (-G * d_M * [x - p_x, y - p_y, z - p_z]
| > | / sqrt( (x-p_x)^2 + (y-p_y)^2 + (z-p_z)^2)^3 dz dy dx
| > |
| > | where r_0 is V's radius -- about 6378 km.
| > |
| > | It is possible without loss of generality to take p_x = p_y
| > | = 0, by appropriate rotation of the coordinate axes, and
| > | convert this into a simpler integral by using cylindrical
| > | coordinates (z, r, theta). Of course F is still a vector,
| > | but now it can depend solely on the distance from the
| > | sphere's center, which I'll call d, and substituting d
| > | for p, one gets:
| > |
| > | F(d) = integral(-r_0 < z < r_0)
| > | integral(0 <= r < r_0)
| > | integral(0 <= theta < 2*Pi)
| > | -G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2)^3 dV
| > |
| > | = integral(-r_0 < z < r_0)
| > | integral(0 <= r < r_0)
| > | integral(0 <= theta < 2*Pi)
| > | (-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r
d{theta} dr
| > dz
| > |
| > | Since theta appears nowhere in the formula, it drops out pretty
easily:
| > |
| > | F(d) = 2 * Pi * integral(-r_0 < z < r_0)
| > | integral(0 <= r < r_0)
| > | (-G * m * d_M * (d - z) / sqrt( (d - z)^2 + r^2))^3 r dr dz
| > |
| > | We can evaluate the inner integral by using a variant of the chain
rule:
| > |
| > | F(d) = 2 * Pi * integral(-r_0 < z < r_0)
| > | (-G * m * d_M * (d - z) * (2/3)
| > | * (1 / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 + r_0^2) ) dz
| > |
| > | = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
| > | ((d - z) / sqrt( (d - z)^2 ) - 1 / sqrt( (d - z)^2 +
| > r_0^2) ) dz
| > |
| > |
| > | If d > r_0, the integral can be split into two parts, one of which
| > | is almost trivial to evaluate:
| > |
| > | F(d) = - (4/3) * G * m * d_M * Pi * integral(-r_0 < z < r_0)
| > | (1 - (d - z) / sqrt( (d - z)^2 + r_0^2) ) dz
| > |
| > | F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| > | + (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
| > | ((d - z) / sqrt( (d-z)^2 + r_0)^2) dz
| > |
| > | = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| > | - (4/3) * G * m * d_M * Pi * integral( - r_0 < z < r_0)
| > | ((z - d) / sqrt( (z-d)^2 + r_0)^2) dz
| > |
| > | By substituting z-d = w (or w+d = z) and dz = dw, one gets
| > |
| > | F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| > | - (2/3) * G * m * d_M * Pi * integral( - r_0 - d < w < r_0 - d)
| > | (2w / sqrt( w^2 + r_0^2) ) dw
| > |
| > | Since w != 0, it is immediately apparent F'(d) is nonzero for any
| > | d > r_0; therefore F(d) is not constant for d > r_0.
| > |
| > | One can use the chain rule again, and, since k^2 = (-k)^2, we
| > | can eliminate some redundant stuff, finally getting:
| > |
| > | F(d) = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| > | - (2/3) * G * m * d_M * Pi *
| > | ( sqrt( (d + r_0)^2 + r_0^2) - sqrt( (d - r_0)^2 +
r_0^2))
| > |
| > | = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| > | - (2/3) * G * m * d_M * Pi * r_0 *
| > | ( sqrt( (d/r_0 + 1)^2 + 1) - sqrt( (d/r_0 - 1)^2 + 1))
| > |
| > | = - (4/3) * G * m * d_M * Pi * (2 * r_0)
| > | - (2/3) * G * m * d_M * Pi * d *
| > | ( sqrt( (1 + r_0/d)^2 + (r_0/d)^2 )
| > | - sqrt( (1 - r_0/d)^2 + (r_0/d)^2) )
| > |
| > | (I'll admit I'm not at all sure this is entirely correct, but I'm
| > | getting a little tired, and it's been a long time since I've
| > | had to fuss with integrals. :-) If anyone can point me to the
| > | correct solution of this problem, that would be helpful.)
| > |
| > | >
| > | > What I'm getting at here is the Earth is NOT a point.
| > | > If you flatten the Earth into a thin disc like a CD ROM
| > | > and stand near the centre then your vertical weight will
| > | > be zero and your horizontal weight slightly greater than zero.
| > | > If you are on the rim then you have your full vertical weight,
| > | > no horizontal weight and are rotated 90 degrees. If you step
| > | > off the edge of the world you fall through the centre, skinning
| > | > your butt on the way down and back up the other side.
| > | >
| > | > If you are space-walking a long way from the Earth then it is
| > | > effectively a point and the inverse square law (almost) applies,
| > | > you still fall toward the centre.
| > | >
| > | > For the spherical Earth F = GMm/r^2 fails below the Earth's
| > | > surface, you are weightless at the centre and will not accelerate
| > | > at 32 fps/s.
| > | >
| > | > So what I want is an exact function that approximates F = GMm/r^2
| > | > when a long way from Earth, is equal to F = m*32 fps/s at or near
| > | > the Earth's surface and F = 0 at the centre so that I can calculate
| > | > my exact weight at the BOTTOM of a 30,000 ft deep mine shaft
| > | > as well as at 30,000 feet in a plane.
| > | > Therefore this function will include both r and R where R is the
| > | > radius of the Earth, just as we include both M and m where m
| > | > is a point. M cannot be a point.
| > | >
| > | > F = f(G, M, m, r, R).
| > | >
| > | > What is the function f() that
| > | > F = GMm/r^2 is an approximation to?
| > |
| > | Good question.
| >
| > Yes, and what I want is a good answer, you erred right at the start
| > with a cubic power and then got carried away trying to solve an
| > error.
|
| There was an error, yes; it wasn't there. But the reason I said
|
| -d/||d||^3
|
| instead of
|
| 1/||d||^2
|
| is quite simple; F is a *vector* field, not a scalar one.
| In other words, if I'm, say, over San Francisco and you're
| over Sydney, our force vectors aren't pointing in the same
| direction at all.
|
| Magnitude doesn't quite cut it.
|
| >
| > Speaking of cubes, if you take the eight vertices at a cube's
| > corners as fixed points of unit mass and calculate the forces acting
| > on a unit point mass P at the cube's centre, the resultant is zero.
|
| Correct.
|
| > For any other P the non-zero resultant is the sum (for i = 0 to 7)
| > of 1/(P - p_i)^2.
|
| Actually, sum(i=0,7) GMm/(P-p_i)^2, where M is the point mass, m the
| mass of each cube vertex, G the universal constant of gravitation,
| P the position of the point mass, and p_i the cube vertex.
|
| But you're there up to a constant.
|
| > Briefly put, the gravitational acceleration on a point is simply the sum
| > of all mass-elements around that point.
|
| And an integral is ... ?
|
| > A point mass that is free to move will move toward the vertex
| > it is closest to, and not the centre of the cube. Thus the arrangement
| > is unstable. Now if we replace the centre of the faces of the cube
| > with unit point masses, with P near the centre the movement will
| > be toward the centre of a face and not a corner vertex because that
| > will be nearer to P.
| >
| > Now the problem extends to a spherical shell, the resultant
| > takes the free mass to its surface, whether from inside or outside
| > the sphere. A pea inside a blown egg will attach to the shell,
| > not the centre of the egg.
| >
|
| A pea inside of a blown spherical shell, absent other
| forces, gets no force at all.
|
Let's cut to the chase with a picture:
http://www.androcles01.pwp.blueyonder.co.uk/InverseSquare.jpg
On the right hand side you weigh a 1/4 of your normal weight when
on the red shell and your normal weight when on the gold shell
(Earth's surface). What Jim Black calls the flux is distributed over
4 times the area at 2r, hence the inverse square law.
On the left hand side:
The Earth is an oblate spheroid (M&M shaped) because it is rotating.
(It also has a couple of lobes like a yankee football
http://z.about.com/d/esl/1/0/S/2/football.gif
because the Moon makes tides, but ignore that)
Are you heavier at the pole than the equator because of the inverse
square law, or are you lighter because you are closer to the centre
of the Earth where you would weigh nothing?
.

User: "Jeckyl"
Title: Re: Gravity acceleration and deceleration	04 Feb 2008 10:43:13 PM

"Androcles" <Headmaster@Hogwarts.physics> wrote in message
news:cFRpj.88863$Kt4.48117@fe3.news.blueyonder.co.uk...

Are you heavier at the pole than the equator because of the inverse
square law, or are you lighter because you are closer to the centre
of the Earth where you would weigh nothing?

Do you even know *why* you would have no weight at the centre of the earth?
.

User: "The Ghost In The Machine"
Title: Re: Gravity acceleration and deceleration	05 Feb 2008 10:19:52 PM

In sci.physics.relativity, Jeckyl
<noone@nowhere.com>
wrote
on Tue, 5 Feb 2008 15:43:13 +1100
<13qfqb46678j81e@corp.supernews.com>:

"Androcles" <Headmaster@Hogwarts.physics> wrote in message
news:cFRpj.88863$Kt4.48117@fe3.news.blueyonder.co.uk...

Are you heavier at the pole than the equator because of the inverse
square law, or are you lighter because you are closer to the centre
of the Earth where you would weigh nothing?

Do you even know *why* you would have no weight at the centre of the earth?

I'd have to look, but can calculate the centrifugal
acceleration readily enough. The Earth is an oblate
spheroid, equatorial radius 6378100 m, polar radius
6356800 m, mean radius 6371000 m. "G-force", or the
acceleration because of gravity, is given as 9.80665
N/kg in http://en.wikipedia.org/wiki/G-force, and,
interestingly, equatorial acceleration is given as 0.99732
g in http://en.wikipedia.org/wiki/Earth (the sidebar).
However, since it is far from clear precisely where 'g'
is measured, the best I can do is stipulate that on the
equator one sheds a little weight; one cannot conclude
that at the poles one is at exactly g.
If one assumes that g at the poles is exactly thus, one gets:
Pole 6356800 m 9.80665 N/kg = 1 g
Mean 6371000 m 9.76298 N/kg = 0.9955469 g
Equator 6378100 m 9.74126 N/kg = 0.9933321 g
If one uses the mean instead, one gets:
Pole 6356800 m 9.85051 N/kg = 1.004472 g
Mean 6371000 m 9.80665 N/kg = 1 g
Equator 6378100 m 9.78483 N/kg = 0.997775 g
Bear in mind that this is without the Earth's spin.
With that spin, one can calculate that, with angular
velocity {omega}, the edge speed is r * {omega}, and the
accceleration is the second derivative of the point's
motion, which is simply
(-r * {omega}^2 cos {omega} * t, -r * {omega}^2 sin {omega} * t)
or length r * {omega} ^ 2, or a lessening in weight of
0.03373056 N/kg. This gives a computed measurable value
at the equator of about 9.751099 N/kg -- not quite matching
the given value of 9.780327 N/kg.
All this is showing is that the model isn't quite working,
admittedly (the model being the more or less standard
"mass-at-the-center' one). However, it's clear that one
is heavier at the poles, regardless.
--
#191,

New Technology? Not There. No Thanks.
--
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.