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Words, words. "Error" means bad numbers.
That usually happens when you are too goddamned thick to get the factAgain, a heap of words, arguing about definitions.
Most every time
you've actually done a simulation, or presented real numbers (like the
recent newton/kg thing) you've been wrong.
It displays in pounds. I doesn't say if it's pounds-mass orOn Thu, 17 Jun 2010 10:33:10 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 22:08:32 -0500, John Fields
jfields@austininstruments.com> wrote:
On Tue, 15 Jun 2010 14:36:50 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 15 Jun 2010 13:48:18 -0700 (PDT), Rich Grise on Google groups
richardgrise@yahoo.com> wrote:
On Jun 14, 10:09 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 14 Jun 2010 09:38:18 -0700, Archimedes' Lever
It is an applied force, regardless of how you attack my grammatical
error in describing it.
Torque is not force. The units are different.
Well, mass isn't weight either, but people use them interchangeably.
"Weight" isn't clear.
---
Nonsense.
Weight is the product of mass and the force exerted on it by
gravitational acceleration.
---
Often it means mass.
---
More nonsense.
It never means mass, since mass is simply the measure of inertial
resistance.
---
Mass (kg) and force
(newtons) are formal SI things; "weight" isn't.
---
Even more nonsense.
Weight is derived from the relationship between mass and force, both
of which are prime, so your disingenuous attempt at bastardizing
weight by denying it other than prime status, for your own ends is, at
the very least, despicable.
But then, that's how you work, isn't it?
I work in SI units. Mass is kilograms, force is newtons, and I don't
have to worry about what "weight" might mean to various people.
John
You do when you declare your spring tension based bathroom scale to be
measuring mass simply because of the unit utilized on the scale.
"andOn Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Jun 16, 7:11 pm, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Jun 16, 2:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.
No you cannot. What makes you think that G decreases (or
increases)linearly?
I doesn't!
John
for small enough changes it is linear!
---
Not if you take at least 3 samples.- Hide quoted text -
- Show quoted text -
What? if the change is a part in 1,000, then the second order
correction is at the one out of 10^6 level. John L's going to stop 1/2
between SF and Truckee and weigh himself with six figure acurracy?
---
Probably not, but that's not what I had in mind.
Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.
Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.
Now, lets subtract g2 from g1 and call the difference d1.
Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.
Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.
Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.
With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 gal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.
The fly in the ointment is that the gradient decreases from 3.083
gal/m at sea level to 2.638 gal/m at 340 km, which is a change in
gradient of 345 gal/340km, or 1.014 gal/km.
That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.
Tethered balloon?
No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, "
---On Thu, 17 Jun 2010 09:26:30 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 16, 7:11 pm, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Jun 16, 2:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.
No you cannot. What makes you think that G decreases (or
increases)linearly?
I doesn't!
John
for small enough changes it is linear!
---
Not if you take at least 3 samples.- Hide quoted text -
- Show quoted text -
What? if the change is a part in 1,000, then the second order
correction is at the one out of 10^6 level. John L's going to stop 1/2
between SF and Truckee and weigh himself with six figure acurracy?
---
Probably not, but that's not what I had in mind.
Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.
Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.
Now, lets subtract g2 from g1 and call the difference d1.
Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.
Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.
Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.
With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 ľgal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.
The fly in the ointment is that the gradient decreases from 3.083
ľgal/m at sea level to 2.638 ľgal/m at 340 km, which is a change in
gradient of 345 ľgal/340km, or 1.014 ľgal/km.
That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.
Tethered balloon?
No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, and
therefore proved it's not linear.
Isaac Newton demonstrated that gravitational attraction goes as 1/d^2.
It explains the orbits of the planets.
Yesterday, you were squawking that it displays in kg. Make up your mind,You do when you declare your spring tension based bathroom scale to be
measuring mass simply because of the unit utilized on the scale.
It displays in pounds.
No, what you really mean is that "you do not know". IF you did know,I doesn't say if it's pounds-mass or
pounds-force, and I don't care.
It matters when you declare some reading as fact, and ignore any errorAs was discussed and calculated
elsewhere, g doesn't change enough to matter.
But there is also no indication that the reading is actually correctIt certainly doesn't
change much in my bathroom.
Yes, John. You do belong in a toilet bowl, and depending on whichHey, that's your opening for a bathroom/toilet comment.
Please do the world a favor and flush yourself.Go for it!
If you are done, then YOU go fuck off in some other group, asswipe.On Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.
No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.
I'm done with you.
Go pester some other news group.
George H.
I suppose at some point it becomes obvious that some posters aren'tOn Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.
No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.
I'm done with you.
Go pester some other news group.
George H.
You finally put your "I Am Stupid" T-shirt on! Wow! Congratulations,On Thu, 17 Jun 2010 11:26:07 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.
No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.
I'm done with you.
Go pester some other news group.
George H.
I suppose at some point it becomes obvious that some posters aren't
just nasty ignorant SOBs,
Funny, that is what we have been telling you throughout the thread.they are in fact mentally deficient for one
reason or another.
No, John. Slippery little SOB bastards like you are usually pretty slickIf we met them in person, it would be obvious.
You were quite easy to spot, Johnny. It was glaringly apparent yearsOn
usenet, it's harder to tell.
In two spheres mutually gravitating each towards the other if theOn Thu, 17 Jun 2010 08:45:55 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 17 Jun 2010 09:26:30 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 16, 7:11 pm, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Jun 16, 2:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.
No you cannot. What makes you think that G decreases (or
increases)linearly?
I doesn't!
John
for small enough changes it is linear!
---
Not if you take at least 3 samples.- Hide quoted text -
- Show quoted text -
What? if the change is a part in 1,000, then the second order
correction is at the one out of 10^6 level. John L's going to stop 1/2
between SF and Truckee and weigh himself with six figure acurracy?
---
Probably not, but that's not what I had in mind.
Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.
Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.
Now, lets subtract g2 from g1 and call the difference d1.
Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.
Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.
Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.
With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 ľgal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.
The fly in the ointment is that the gradient decreases from 3.083
ľgal/m at sea level to 2.638 ľgal/m at 340 km, which is a change in
gradient of 345 ľgal/340km, or 1.014 ľgal/km.
That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.
Tethered balloon?
No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, and
therefore proved it's not linear.
Isaac Newton demonstrated that gravitational attraction goes as 1/d^2.
It explains the orbits of the planets.
---
Yes, but I think he said 1/r^2.
Don't wrestle a pig. You'll get covered in mud and the pig likes it.On Wed, 16 Jun 2010 19:15:04 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 21:08:31 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 16 Jun 2010 18:50:06 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.
You are so goddamned retarded that you cannot even follow a thread. It
was posted three times already, dumbass.
http://en.wikipedia.org/wiki/File:Earth-G-force.png
More irrelevance from DimBulb.
It is absolutely relevant.
You're full of shit. ...but you like it that way.
if the
matter in places on all sides round about and equi-distant from the
centres is similar,
On 06/17/2010 06:12 AM, krw@att.bizzzzzzzzzzzz wrote:
On Wed, 16 Jun 2010 19:15:04 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 21:08:31 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 16 Jun 2010 18:50:06 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.
You are so goddamned retarded that you cannot even follow a thread. It
was posted three times already, dumbass.
http://en.wikipedia.org/wiki/File:Earth-G-force.png
More irrelevance from DimBulb.
It is absolutely relevant.
You're full of shit. ...but you like it that way.
Don't wrestle a pig. You'll get covered in mud and the pig likes it.
Jeroen Belleman
---On Wed, 16 Jun 2010 21:58:44 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 16 Jun 2010 16:22:13 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 17:58:51 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 16 Jun 2010 14:06:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 16:44:17 -0400, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:
On Wed, 16 Jun 2010 13:01:06 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 16, 2:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.
No you cannot. What makes you think that G decreases (or
increases)linearly?
I doesn't!
John
for small enough changes it is linear!
Well, you can say that about most anything.
But for g, sure.
2 * 2 = 4
1.1 * 1.1 = 1.21
1.01 * 1.01 = 1.0201
1.001 * 1.001 = 1.002001
and like that.
John
Vertical gravity gradient is around 3000 Eotvos (3E-6 sec^-2) at sea
level, and it's not very sensitive to altitude. 6.4 k feet is about
2km, so around 0.06% less acceleration. Less than a slice of
cheesecake. That's assuming a non-rotating uniform spherical earth
with no atmosphere.
Do we need to take the delta of your buoyancy into account at this
level of precision? Assume, say, 2.5 or 3 cubic feet for your volume.
;-) I'm guessing yes.
About 2.4, by my calcs.
---
Seems small for someone with such a fat head.
You dislike me, so you want to prove me wrong at every opportunity.
---
No, I don't.
I'm not interested in _proving_ you wrong, all I'm interested in is
showing you your errors and letting you prove to yourself that you
did, in fact, make an error.
Words, words. "Error" means bad numbers.
The reason I dislike you is because you're disingenuous and because
you won't admit to error when it's pointed out to you even if it's as
plain as the nose on your face that you made an error.
Again, a heap of words, arguing about definitions. Most every time
you've actually done a simulation, or presented real numbers (like the
recent newton/kg thing) you've been wrong.
---Why don't you check your work?
---On Thu, 17 Jun 2010 10:46:09 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
You do when you declare your spring tension based bathroom scale to be
measuring mass simply because of the unit utilized on the scale.
It displays in pounds. I doesn't say if it's pounds-mass or
pounds-force, and I don't care. As was discussed and calculated
elsewhere, g doesn't change enough to matter.
But time wounds all heels, so eventually your true colors (or lack ofOn Thu, 17 Jun 2010 11:26:07 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.
No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.
I'm done with you.
Go pester some other news group.
George H.
I suppose at some point it becomes obvious that some posters aren't
just nasty ignorant SOBs, they are in fact mentally deficient for one
reason or another. If we met them in person, it would be obvious. On
usenet, it's harder to tell.
I'm done with you.On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.
No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.
I know what mass means, bacause it's an SI unit. Weight means whateverOn Thu, 17 Jun 2010 11:00:50 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 17 Jun 2010 10:46:09 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
You do when you declare your spring tension based bathroom scale to be
measuring mass simply because of the unit utilized on the scale.
It displays in pounds. I doesn't say if it's pounds-mass or
pounds-force, and I don't care. As was discussed and calculated
elsewhere, g doesn't change enough to matter.
---
It's not whether it matters or not, it's whether you know the
difference between mass and weight.
Don't be silly. You're just playing with words. Physics doesn't care.From one of your posts early-on it was apparent you didn't, and now
you're trying to play the "it doesn't matter" card in order to make it
seem like your error didn't matter, cheater.
On Thu, 17 Jun 2010 13:20:41 -0500, John Fields
jfields@austininstruments.com> wrote:
On Thu, 17 Jun 2010 08:45:55 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 17 Jun 2010 09:26:30 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 16, 7:11 pm, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Jun 16, 2:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.
No you cannot. What makes you think that G decreases (or
increases)linearly?
I doesn't!
John
for small enough changes it is linear!
---
Not if you take at least 3 samples.- Hide quoted text -
- Show quoted text -
What? if the change is a part in 1,000, then the second order
correction is at the one out of 10^6 level. John L's going to stop 1/2
between SF and Truckee and weigh himself with six figure acurracy?
---
Probably not, but that's not what I had in mind.
Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.
Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.
Now, lets subtract g2 from g1 and call the difference d1.
Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.
Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.
Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.
With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 ľgal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.
The fly in the ointment is that the gradient decreases from 3.083
ľgal/m at sea level to 2.638 ľgal/m at 340 km, which is a change in
gradient of 345 ľgal/340km, or 1.014 ľgal/km.
That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.
Tethered balloon?
No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, and
therefore proved it's not linear.
Isaac Newton demonstrated that gravitational attraction goes as 1/d^2.
It explains the orbits of the planets.
---
Yes, but I think he said 1/r^2.
In two spheres mutually gravitating each towards the other if the
matter in places on all sides round about and equi-distant from the
centres is similar, the weight of either sphere towards the other will
be reciprocally as the square of the distance between their centres.
-- Principia Book 1 "De motu corporum" by Newton
(translated into English by Motte, 1848)