| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
20 Jan 2008 01:07:17 PM |
| Object: |
Is the force of gravity 1/r^2 or 1/r????? |
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
http://www.space-electronics.com/Literature/Precise_Measurement_of_Mass.PDF
On pg 15, they claim to have a formula to calculate the force of
gravity to .002%
g = 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
L represents the lattitude and H represents the height.
The strange thing to notice is that the relationship between g and
H(height or radius from the center of the Earth) is linear. If gravity
varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
in the correction, but there isn't. The formula implies that gravity
decreases linearly rather than exponentially as height increases.
I quick search of the internet failed to reveal experiments which
directly measure the 1/r^2 relationship from non-orbiting platforms (1/
r^2 is often justified by looking at the orbits of satellites). But I
was looking for a direct measurement of this force by comparing how
the force of gravity varies by height on an Earth based experiment.
Such an experiment might involve sending a weather balloon from the
ground to the high stratosphere while measuring the force of gravity
to directly verify the 1/r^2 relationship.
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
fhugravity
.
|
|
| User: "Matthew Lybanon" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 09:50:27 AM |
|
|
In article
<496e47e8-c265-4b39-b699-cf3993472833@s12g2000prg.googlegroups.com>,
wrote:
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
http://www.space-electronics.com/Literature/Precise_Measurement_of_Mass.PDF
On pg 15, they claim to have a formula to calculate the force of
gravity to .002%
g = 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
L represents the lattitude and H represents the height.
The strange thing to notice is that the relationship between g and
H(height or radius from the center of the Earth) is linear. If gravity
varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
in the correction, but there isn't. The formula implies that gravity
decreases linearly rather than exponentially as height increases.
I quick search of the internet failed to reveal experiments which
directly measure the 1/r^2 relationship from non-orbiting platforms (1/
r^2 is often justified by looking at the orbits of satellites). But I
was looking for a direct measurement of this force by comparing how
the force of gravity varies by height on an Earth based experiment.
Such an experiment might involve sending a weather balloon from the
ground to the high stratosphere while measuring the force of gravity
to directly verify the 1/r^2 relationship.
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
fhugravity
Try looking for books(!) and (possibly) websites on geodesy. The
description of the gravitational field of a body whose mass is not
distributed exactly with spherical symmetry has a lot of terms. It will
take you some time to work through the details, but it will be be time
well spent.
.
|
|
|
|
| User: "tadchem" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 08:32:58 PM |
|
|
On Jan 20, 2:07 pm, wrote:
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
http://www.space-electronics.com/Literature/Precise_Measurement_of_Ma...
On pg 15, they claim to have a formula to calculate the force of
gravity to .002%
g = 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
L represents the lattitude and H represents the height.
The strange thing to notice is that the relationship between g and
H(height or radius from the center of the Earth) is linear. If gravity
varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
in the correction, but there isn't. The formula implies that gravity
decreases linearly rather than exponentially as height increases.
I quick search of the internet failed to reveal experiments which
directly measure the 1/r^2 relationship from non-orbiting platforms (1/
r^2 is often justified by looking at the orbits of satellites). But I
was looking for a direct measurement of this force by comparing how
the force of gravity varies by height on an Earth based experiment.
Such an experiment might involve sending a weather balloon from the
ground to the high stratosphere while measuring the force of gravity
to directly verify the 1/r^2 relationship.
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
fhugravity
Did you not read the article at all?
Your g formula was preceded IMMEDIATELY with this paragraph:
"By knowing the altitude and latitude of the location where the
scale was calibrated and where the final measurement will take place.
In some
instances you will have no means of calibrating a scale. It may have
been calibrated at the
factory only. In this case you will need to know the acceleration of
gravity at the location
where the scale was calibrated and also at the location where you will
be making your
measurements.
Although anomalies in the earth's crust will affect local gravity,
gravity is mainly function
of latitude and altitude. Gravity at a particular location can be
calculated with a precision
of about 0.002 % using the following formula:"
In the context, g is the acceleration due to gravity at a particular
location on the surface of the earth. The statement that the formula
has .002% pre3cision should tell you that it is only an
approximation. The preceding portions of the document discuss several
other factors in g that cannot be estimated a priori - see the
Abstract (page 2):
"Space vehicle weight is very critical. It is particularly important
to have accurate weight if the
vehicle weight is near the maximum allowable. Scales with accuracies
as good as 0.002% are
now available from companies such as Space Electronics. However,
unless precaution is taken to
correct for local gravity, air buoyancy, and a number of other
factors, the high accuracy of the
scale is not realized. In fact, considerations which are often
overlooked can result in a total error
of several percent. Some of the concerns are:
1. Latitude The acceleration of gravity varies from 9.780 m/s2 at the
equator to 9.832
m/s2 at the poles (a difference of 0.53%). This is in part due to the
centrifugal force
resulting from the rotation of the earth, which varies from zero at
the north pole to a
maximum value at the equator, and in part due to the bulge of the
earth at the equator,
resulting in a greater distance to the mass center of the earth.
2. Altitude The gravitational mass attraction to the earth at a
particular location varies as
the square of the distance to the mass center of the earth, resulting
in a variation by as
much as 0.26% over the surface of the earth.
3. Tidal effects The gravitational mass attraction to the sun and moon
at a particular
location may exhibit a variation as large as 0.003% of the
acceleration of earth gravity at
certain dates during the year when the sun and moon align (this means
that the scale could
indicate a larger weight in the morning than at night).
4. Gravity anomaly Variations in the density of the earth's crust in
the vicinity of an
object result in small variations in gravity at different locations.
Nearby mountain ranges
reduce the force of gravity on the object.
5. Buoyancy The object being measured displaces a volume of air, whose
density can
vary due to the weather. The resulting upward force is a function of
object size. For
large low density objects, failure to correct for buoyancy may result
in errors of 0.5% or
larger.
In addition, errors can be introduced by
6. Moisture condensation or absorption of moisture (evaporates in
vacuum of space)
7. Electrostatic attraction to the draft shield surrounding the scale
8. Magnetic attraction to nearby objects and to the earth's magnetic
north
9. Downdrafts or updrafts (due to a temperature difference between
object and ambient air)"
Do you get it yet?
RTFM!!!
Tom Davidson
Richmond, VA
.
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| User: "Androcles" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 10:15:37 PM |
|
|
"tadchem" <tadchem@comcast.net> wrote in message
news:3db3685d-20c5-46d8-bf86-93698241d288@l1g2000hsa.googlegroups.com...
| 1. Latitude The acceleration of gravity varies from 9.780 m/s2 at the
| equator to 9.832
| m/s2 at the poles (a difference of 0.53%). This is in part due to the
| centrifugal force
| resulting from the rotation of the earth, which varies from zero at
| the north pole to a
| maximum value at the equator, and in part due to the bulge of the
| earth at the equator,
| resulting in a greater distance to the mass center of the earth.
Ugh... The explanation was written by a journalist.
Extrapolate the oblate spheroid into a cylinder with rounded edge.
A(_________________C__________________)B
Now clearly gravity is greater at the equator than at the pole, you
are pulled equally toward A and B, whereas at A or B you pulled
only toward C. However, the REASON for the oblateness is
the centrifugal force. (This does of course assume a uniform density)
| 2. Altitude The gravitational mass attraction to the earth at a
| particular location varies as
| the square of the distance to the mass center of the earth, resulting
| in a variation by as
| much as 0.26% over the surface of the earth.
The omnidirectional inverse square law applies only to
points. If you are standing at the pole you are pulled toward
the plane of the equator, not toward the centre.
/A
/
P / C
Q \
\
\B
In other words there is a cone with base (AB) and apex P;
the inverse square law cannot apply since that is a reversed cone
with apex C. Someone at Q will fall toward the centre C, yes, but
a point mass at P and a point mass at Q results in P falling toward
Q (and Q toward P). The Earth is not a point until you are a long way
from it. The path P and Q take can actually diverge from parallel
if there is a mountain at B and none at A.
| 3. Tidal effects The gravitational mass attraction to the sun and moon
| at a particular
| location may exhibit a variation as large as 0.003% of the
| acceleration of earth gravity at
| certain dates during the year when the sun and moon align (this means
| that the scale could
| indicate a larger weight in the morning than at night).
This is getting even more quirky. The tidal bulge is balanced on
opposite sides of the Earth. Since the Earth is flying out from
the Sun as it falls toward it we have a centrifugal force away
from the Sun, but the same force applies to the spacecraft.
| 4. Gravity anomaly Variations in the density of the earth's crust in
| the vicinity of an
| object result in small variations in gravity at different locations.
| Nearby mountain ranges
| reduce the force of gravity on the object.
That one is acceptable. Now we are getting to the truth and
away from the theoretical inverse square law with a point-like
model.
| 5. Buoyancy The object being measured displaces a volume of air, whose
| density can
| vary due to the weather. The resulting upward force is a function of
| object size. For
| large low density objects, failure to correct for buoyancy may result
| in errors of 0.5% or
| larger.
Will not apply to spacecraft, only its initial weight measurement.
| In addition, errors can be introduced by
<trivia snipped>
Someone is trying to sell something.
.
|
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 01:43:25 PM |
|
|
On Jan 20, 2:07 pm, wrote:
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
http://www.space-electronics.com/Literature/Precise_Measurement_of_Ma...
On pg 15, they claim to have a formula to calculate the force of
gravity to .002%
g = 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
L represents the lattitude and H represents the height.
The strange thing to notice is that the relationship between g and
H(height or radius from the center of the Earth) is linear. If gravity
varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
in the correction, but there isn't. The formula implies that gravity
decreases linearly rather than exponentially as height increases.
I quick search of the internet failed to reveal experiments which
directly measure the 1/r^2 relationship from non-orbiting platforms (1/
r^2 is often justified by looking at the orbits of satellites). But I
was looking for a direct measurement of this force by comparing how
the force of gravity varies by height on an Earth based experiment.
Such an experiment might involve sending a weather balloon from the
ground to the high stratosphere while measuring the force of gravity
to directly verify the 1/r^2 relationship.
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
Both corrections come from doing a small-number approximation
of 1/(r + epsilon)^2 relative to 1/r^2.
The correction for altitude is derived in section 1.4, page 7,
of the document.
It's quite simple. The ratio of g on the ground compared to
the acceleration g_h at height h is
g/g_h = (r+h)^2/r^2
= 1 + 2rh/r^2 + h^2/r^2
= 1 + 2h/r + (h/r)^2
Under the assumption that h/r is "small", then
(h/r)^2 is "negligible". What "small" and "negligible"
mean numerically depends on how much precision
you need in your estimate. For the top of Mount
Everest, h/r ~ 0.001 so the error in dropping the
(h/r)^2 term is down around the 6th significant digit.
So g/g_h ~ 1 + 2h/r, and g_h/g ~ 1 - 2h/r.
Again this comes from a small-number approximation,
that 1/(1+x) = 1 - x + x^2 - ... ~ 1 - x
However, I do have an issue with his coefficient. For
h in meters, I get 2/r = 2/6.378E6 = 3.14 x 10^-7,
not 3.92 x 10^-7.
- Randy
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 02:46:32 PM |
|
|
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:0bef8ef7-3f5e-48aa-a434-4ade56b1afab@v4g2000hsf.googlegroups.com...
| On Jan 20, 2:07 pm, wrote:
| > It is well understood that the force of gravity is a 1/r^2 force.
| > However, I was looking at a document describing how to precisely
| > meausure mass.
| >
| > http://www.space-electronics.com/Literature/Precise_Measurement_of_Ma...
| >
| > On pg 15, they claim to have a formula to calculate the force of
| > gravity to .002%
| >
| > g = 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
| >
| > L represents the lattitude and H represents the height.
| >
| > The strange thing to notice is that the relationship between g and
| > H(height or radius from the center of the Earth) is linear. If gravity
| > varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
| > in the correction, but there isn't. The formula implies that gravity
| > decreases linearly rather than exponentially as height increases.
| >
| > I quick search of the internet failed to reveal experiments which
| > directly measure the 1/r^2 relationship from non-orbiting platforms (1/
| > r^2 is often justified by looking at the orbits of satellites). But I
| > was looking for a direct measurement of this force by comparing how
| > the force of gravity varies by height on an Earth based experiment.
| > Such an experiment might involve sending a weather balloon from the
| > ground to the high stratosphere while measuring the force of gravity
| > to directly verify the 1/r^2 relationship.
| >
| > If anyone could point me to a reference on the web experimentally
| > proving the 1/r^2 relationship or can explain why the gravity
| > correction formula is linear, I would appreciate it.
|
| Both corrections come from doing a small-number approximation
| of 1/(r + epsilon)^2 relative to 1/r^2.
|
| The correction for altitude is derived in section 1.4, page 7,
| of the document.
|
| It's quite simple. The ratio of g on the ground compared to
| the acceleration g_h at height h is
| g/g_h = (r+h)^2/r^2
| = 1 + 2rh/r^2 + h^2/r^2
| = 1 + 2h/r + (h/r)^2
|
| Under the assumption that h/r is "small", then
| (h/r)^2 is "negligible". What "small" and "negligible"
| mean numerically depends on how much precision
| you need in your estimate. For the top of Mount
| Everest, h/r ~ 0.001 so the error in dropping the
| (h/r)^2 term is down around the 6th significant digit.
|
| So g/g_h ~ 1 + 2h/r, and g_h/g ~ 1 - 2h/r.
|
Everest is denser than air.
From an altitude of 5 miles flying in a plane where one might take
bathroom scales is not the same as from the top of Everest where
there is mass directly below, but ok, the difference is negligible.
One other point. There is no acceleration g_h due to gravity at
the top of Everest. Once again you confuse motion with force.
The difference is not negligible, you are babbling.
| Again this comes from a small-number approximation,
| that 1/(1+x) = 1 - x + x^2 - ... ~ 1 - x
|
| However, I do have an issue with his coefficient. For
| h in meters, I get 2/r = 2/6.378E6 = 3.14 x 10^-7,
| not 3.92 x 10^-7.
|
| - Randy
.
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 03:13:39 PM |
|
|
On Jan 20, 3:46 pm, "Androcles" <Engin...@hogwarts.physics_d> wrote:
Everest is denser than air.
From an altitude of 5 miles flying in a plane where one might take
bathroom scales is not the same as from the top of Everest where
there is mass directly below, but ok, the difference is negligible.
Actually not. You're quite correct, detailed gravitational
surveys of the earth taken from satellites show measurable
effects over the Himalayas, and similarly over deep-sea trenches.
These effects are actually significant when you're trying to launch
ballistic missiles and predict where they'll go (which is the main
reason the US government cares about high-accuracy gravitational
surveys).
One other point. There is no acceleration g_h due to gravity at
the top of Everest.
If you drop things when you're standing on top of Mt.
Everest, they don't fall? You may want to check your data
on that one. People are actually able to walk around on
Everest, they don't float.
Once again you confuse motion with force.
You do know what the "g" represents in "weight = mg",
right?
How many whiskeys did you take before you wrote this
post?
The difference is not negligible, you are babbling.
The difference between (h/r) + (h/r)^2 and
(h/r) is negligible if (h/r) < 0.001 and you only
care about 2 or 3 significant digits of precision.
- Randy
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 04:27:30 PM |
|
|
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:463fefe4-b21d-4337-895d-c822f3639310@p69g2000hsa.googlegroups.com...
| On Jan 20, 3:46 pm, "Androcles" <Engin...@hogwarts.physics_d> wrote:
| > Everest is denser than air.
| > From an altitude of 5 miles flying in a plane where one might take
| > bathroom scales is not the same as from the top of Everest where
| > there is mass directly below, but ok, the difference is negligible.
|
| Actually not. You're quite correct, detailed gravitational
| surveys of the earth taken from satellites show measurable
| effects over the Himalayas, and similarly over deep-sea trenches.
| These effects are actually significant when you're trying to launch
| ballistic missiles and predict where they'll go (which is the main
| reason the US government cares about high-accuracy gravitational
| surveys).
Well ok, but given the tolerance you stated it would be negligible.
I was agreeing with you, I'm not going to pick the nit's nit. As to
ballistic missiles, wind is of far greater significance at higher altitudes
making the flight from London to New York an hour longer than
the flight from New York to London (jet stream), so that's a
bigger nit to pick than mascons.
08:20 21 Jan
11:00 21 Jan
Heathrow (London)
John F Kennedy (New York)
11:00-8:20 = 2:40
+ 5:00 = 7:40
07:45 22 Jan
19:40 22 Jan
John F Kennedy (New York)
Heathrow (London)
19:40 - 07:45 = 11:55
- 5:00 = 6:55
ref: http://tinyurl.com/2tudlk
| > One other point. There is no acceleration g_h due to gravity at
| > the top of Everest.
|
| If you drop things when you're standing on top of Mt.
| Everest, they don't fall?
I didn't say anything about dropping things. If you drop things
then you remove the upward force and acceleration happens.
I'm not falling out of my chair right now, but there is force on my arse.
| You may want to check your data
| on that one. People are actually able to walk around on
| Everest, they don't float.
Not at all, you want to find out about Newton's third law.
| > Once again you confuse motion with force.
|
| You do know what the "g" represents in "weight = mg",
| right?
What does that have to do with d^2z/dt^2?
| How many whiskeys did you take before you wrote this
| post?
None. How many did you have when you confused
force with acceleration?
| > The difference is not negligible, you are babbling.
|
| The difference between (h/r) + (h/r)^2 and
| (h/r) is negligible if (h/r) < 0.001 and you only
| care about 2 or 3 significant digits of precision.
Weighing something doesn't mean dropping it as any mother
with a newborn child will tell you, and they always get weighed.
Nurse: "We'll just find the infant's acceleration, doctor..."
You are an incoherent raving lunatic.
.
|
|
|
| User: "Kenneth Doyle" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 07:13:14 PM |
|
|
"Androcles" <Engineer@hogwarts.physics_d> wrote in news:mVPkj.232997
$S37.114443@fe3.news.blueyonder.co.uk:
| > One other point. There is no acceleration g_h due to gravity at the
| > top of Everest.
|
| If you drop things when you're standing on top of Mt.
| Everest, they don't fall?
I didn't say anything about dropping things. If you drop things
then you remove the upward force and acceleration happens.
You said that there is no acceleration due to gravity. By what is the
acceleration you subsequently referr to caused?
I'm not falling out of my chair right now, but there is force on my arse.
You're posting from the top of Mt. Everest?
.
|
|
|
|
|
|
|
| User: "" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 11:31:02 PM |
|
|
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Another interesting factoid to consider is that according to orbital
mechanics:
http://en.wikipedia.org/wiki/Orbital_mechanics
The specific potential energy associated with a planet of mass M is
given by:
-GM/r
This is the potential energy an object would have it it were hovering
over a gravitational object. Once again, we see that this energy is
not 1/r^2, it is 1/r. So we have some factors that show up as 1/r with
gravity. Could the strength of gravity be directly related to the
potential energy which is linear?
Unless the experiment has been done on Earth (versus orbital
experiments) to distinguish between the 1/r^2 and 1/r case, I think it
still might be possible for gravity not to follow 1/r^2 near the Earth
surface, so I challenge someone to show me the evidence.
fhugravity
On Jan 20, 11:43=A0am, Randy Poe <poespam-t...@yahoo.com> wrote:
It's quite simple. The ratio of g on the ground compared to
the acceleration g_h at height h is
=A0 =A0 =A0 =A0g/g_h =3D (r+h)^2/r^2
=A0 =A0 =A0 =A0 =A0=3D 1 + 2rh/r^2 + h^2/r^2
=A0 =A0 =A0 =A0 =A0=3D 1 + 2h/r + (h/r)^2
Under the assumption that h/r is "small", then
(h/r)^2 is "negligible". What "small" and "negligible"
mean numerically depends on how much precision
you need in your estimate. For the top of Mount
Everest, h/r ~ 0.001 so the error in dropping the
(h/r)^2 term is down around the 6th significant digit.
So g/g_h ~ 1 + 2h/r, and g_h/g ~ 1 - 2h/r.
Again this comes from a small-number approximation,
that 1/(1+x) =3D 1 - x + x^2 - ... ~ 1 - x
However, I do have an issue with his coefficient. For
h in meters, I get 2/r =3D 2/6.378E6 =3D 3.14 x 10^-7,
not 3.92 x 10^-7.
=A0 =A0 =A0 =A0 =A0 =A0- Randy- Hide quoted text -
- Show quoted text -
.
|
|
|
| User: "N:dlzc D:aol T:com \dlzc" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 06:59:25 AM |
|
|
Dear franklinhu:
<franklinhu@yahoo.com> wrote in message
news:2c1eb0df-30ce-4185-9781-43d23d7a5ee7@l1g2000hsa.googlegroups.com...
....
I would really like to know if it has been
experimentally verified that gravity falls
as 1/r^2 and not 1/r.
Yes. The Moon's orbit does not work, for example, unless it is
1/r^2.
David A. Smith
.
|
|
|
|
| User: "Paul Schlyter" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
22 Jan 2008 01:46:31 AM |
|
|
In article <2c1eb0df-30ce-4185-9781-43d23d7a5ee7
@l1g2000hsa.googlegroups.com>, says...
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
It's been verified, to high precision, that gravity does indeed fall off as=
=20
1/r^2. If gravity did obey a different law, the orbits of the planets woul=
d=20
behave differently.
Another interesting factoid to consider is that according to orbital
mechanics:
http://en.wikipedia.org/wiki/Orbital_mechanics
The specific potential energy associated with a planet of mass M is
given by:
-GM/r
This is the potential energy an object would have it it were hovering
over a gravitational object. Once again, we see that this energy is
not 1/r^2, it is 1/r. So we have some factors that show up as 1/r with
gravity. Could the strength of gravity be directly related to the
potential energy which is linear?
They are indeed "directly related": take the derivative of the potential
energy and you'll get the force of gravity. If you remember your high=20
school math you'll recall that the derivative of 1/r is -1/r^2 and the=
=20
minus sign here means that the force of gravity is directed towards lower,=
=20
not higher, potential energy.
So the fact that the potential energy varies as 1/r merely confirms that th=
e
force of gravity varies as 1/r^2.
Unless the experiment has been done on Earth (versus orbital
experiments) to distinguish between the 1/r^2 and 1/r case, I think it
still might be possible for gravity not to follow 1/r^2 near the Earth
surface, so I challenge someone to show me the evidence.
Why would the force of gravity follow a different law near the Earth's=20
surface than far out into space?
Of course, if you take a sufficiently small part of the 1/r^2 curve, you ca=
n=20
always make a linear approximation to that small part of the curve.
Perhaps that's what you're looking for when you're talking about "near the
Earth's surface", and yes, you can find a linear approximation there which=
=20
works if your demand for accuracy isn't too high. But that doesn't mean
gravity follows a different law near the Earth's surface - it merely means
that over such a restricted altitude interval, a linear approximation can
be useful. But remember that your approximation will break down if you mov=
e=20
sufficiently far outside your restricted altitude interval.
fhugravity
On Jan 20, 11:43=A0am, Randy Poe <poespam-t...@yahoo.com> wrote:
It's quite simple. The ratio of g on the ground compared to
the acceleration g_h at height h is
=A0 =A0 =A0 =A0g/g_h =3D (r+h)^2/r^2
=A0 =A0 =A0 =A0 =A0=3D 1 + 2rh/r^2 + h^2/r^2
=A0 =A0 =A0 =A0 =A0=3D 1 + 2h/r + (h/r)^2
Under the assumption that h/r is "small", then
(h/r)^2 is "negligible". What "small" and "negligible"
mean numerically depends on how much precision
you need in your estimate. For the top of Mount
Everest, h/r ~ 0.001 so the error in dropping the
(h/r)^2 term is down around the 6th significant digit.
So g/g_h ~ 1 + 2h/r, and g_h/g ~ 1 - 2h/r.
Again this comes from a small-number approximation,
--=20
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stjarnhimlen dot se
WWW: http://stjarnhimlen.se/
.
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| User: "PD" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 08:48:59 AM |
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On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
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| User: "" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
28 Jan 2008 12:03:29 PM |
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On Jan 21, 6:48=A0am, PD <TheDraperFam...@gmail.com> wrote:
On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
=46rom this discussion, I can only conclude that the 1/r^2 force
relationship has NEVER been experimentally verified on Earth using
earthbound instruments. You would need an instrument accurate up to 6
digits of precision to directly distinguish between a 1/r and 1/r^2
force, so this is perhaps why it has never been done. Based on the
math, we are extremely confident of the 1/r^2 force, but without ever
directly measuring it, can we really be sure?
.
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| User: "Timo A. Nieminen" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
28 Jan 2008 02:36:02 PM |
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This message is in MIME format. The first part should be readable text,
while the remaining parts are likely unreadable without MIME-aware tools.
--38239155-19110-1201552562=:1540
Content-Type: TEXT/PLAIN; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: QUOTED-PRINTABLE
On Mon, 28 Jan 2008, wrote:
On Jan 21, 6:48=A0am, PD <TheDraperFam...@gmail.com> wrote:
On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
From this discussion, I can only conclude that the 1/r^2 force
relationship has NEVER been experimentally verified on Earth using
earthbound instruments. You would need an instrument accurate up to 6
digits of precision to directly distinguish between a 1/r and 1/r^2
force, so this is perhaps why it has never been done. Based on the
math, we are extremely confident of the 1/r^2 force, but without ever
directly measuring it, can we really be sure?
Usenet is hardly the place to conclude "NEVER been experimentally=20
verified". Akin to concluding that just because you've never seen a moth=20
emerging from a silkworm cocoon, that they NEVER do. Why not _try_ looking=
=20
to see what's in the research literature? A few seconds finds:
Non-Newtonian Gravity and New Weak Forces: an Index of Measurements and=20
Theory
E Fischbach et al 1992 Metrologia 29 213-260=20
doi:10.1088/0026-1394/29/3/001
--=20
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
--38239155-19110-1201552562=:1540--
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| User: "Paul Schlyter" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
29 Jan 2008 03:42:21 AM |
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In article <eaeb480f-d61c-4d90-aa53-7ff09a9cd449@s13g2000prd.googlegroups.com>,
<franklinhu@yahoo.com> wrote:
On Jan 21, 6:48=A0am, PD <TheDraperFam...@gmail.com> wrote:
On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
=46rom this discussion, I can only conclude that the 1/r^2 force
relationship has NEVER been experimentally verified on Earth using
earthbound instruments. You would need an instrument accurate up to 6
digits of precision to directly distinguish between a 1/r and 1/r^2
force, so this is perhaps why it has never been done.
If you want to measure this near the Earth's surface, then there are
a lot of factors which makes this very complex. The Earth isn't a
homogenouos sphere, instead it's quite "lumpy" - somewhat like the
lunar "mascons" but to a lesser extent. So if you measure the gravity
here at the Earth's surface, you'll find many local deviations due
to local mass concentrations here and there. These deviations are
quite visible when you measure the force of gravity to 6 digits of
precision.
Based on the math, we are extremely confident of the 1/r^2 force,
but without ever directly measuring it, can we really be sure?
If you want to pull it that far, you cannot be sure of anything,
really. Even if you do measure something, how can you be absolutely
sure there was no significant errors in your measurements????
Perhaps you should apply Occam's razor here - it works quite well
in the majority of circumstances, and it says you should adopt the
simplest model which explains all our experimental data. We know
that the 1/r^2 law is obeyed to a very high precision wherever we
have been able to measure it - which is throughout the solar system,
and in a number of other stellar systems (all double stars which have
had their orbits measured follow orbits which imply an 1/r^2 law of
gravity).
It's too complex to measure the gravity very near the Earth's surface
though to determine whether the 1/r^2 law is followed there or not,
however this means there are no data which contradicts an 1/r^2 law
of gravity, also near the Earth's surface.
So you have two choices:
1. Use Occam's razor, and assume that gravity follows the 1/r^2 law
everywhere, also near the Earth's surface. Change your mind only
when experimental evidence of the contrary appears.
or:
2. Discard Occam's razor, and assume that the conditions near the
Earth's surface are so special that gravity follows a law there which
is different from the rest of the universe - after all, no-one can
present you with direct measurements which disproves you, right?
But then you might as well assume that e.g. green elephant-sized
ants live on the far side of Sirius - can anyone by direct measurements
disprove THAT? <evil grin>
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stjarnhimlen dot se
WWW: http://stjarnhimlen.se/
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| User: "PD" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
29 Jan 2008 11:17:43 AM |
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On Jan 28, 12:03=A0pm, wrote:
On Jan 21, 6:48=A0am, PD <TheDraperFam...@gmail.com> wrote:
On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
From this discussion, I can only conclude that the 1/r^2 force
relationship has NEVER been experimentally verified on Earth using
earthbound instruments. You would need an instrument accurate up to 6
digits of precision to directly distinguish between a 1/r and 1/r^2
force, so this is perhaps why it has never been done. Based on the
math, we are extremely confident of the 1/r^2 force, but without ever
directly measuring it, can we really be sure?
Well, first of all, it has been directly measured. However even on
principle, I see no value in *requiring* that the measurement be done
with earth-bound instruments by climbing, say, a tower. As I've
explained to Seto many times, just because an experimental measurement
is direct and seems straightforward to you is NOT necessarily
justification for doing it. It would be justified if it could be
demonstrated that such a measurement is inherently more precise or
avoids sources of systematic error that would be problematic for other
methods. Unless an experimental setup produces an experimental result
of higher quality or makes the measurement in a physical domain where
it hasn't been established to hold, then it's not worth doing. Very
often, an *indirect* measurement produces an experimental result of
higher quality than a measurement from a dead-simple experimental
concept. This is part of the art of experimental design.
So in answer to your question, there are a bunch of *indirect*
measurements that span a whole range of h where r =3D h + R and R is the
Earth's radius. Examples of this are ballistic launches to high enough
altitude where the difference between r and R makes a predictable
difference in the landing spot; in the predictable orbital periods of
low-orbit, medium-orbit, and geosynchronous satellites, including the
Space Shuttle; and others. Those results yield in fact better
measurements of the 1/r^2 dependence of gravity than would a
measurement using purely earthbound instruments.
PD
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| User: "Paul Schlyter" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
29 Jan 2008 04:13:11 PM |
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In article <7d441b08-622e-4975-b987-8f7682ca19a3@v67g2000hse.googlegroups.com>,
PD <TheDraperFamily@gmail.com> wrote:
On Jan 28, 12:03=A0pm, wrote:
On Jan 21, 6:48=A0am, PD <TheDraperFam...@gmail.com> wrote:
On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
From this discussion, I can only conclude that the 1/r^2 force
relationship has NEVER been experimentally verified on Earth using
earthbound instruments. You would need an instrument accurate up to 6
digits of precision to directly distinguish between a 1/r and 1/r^2
force, so this is perhaps why it has never been done. Based on the
math, we are extremely confident of the 1/r^2 force, but without ever
directly measuring it, can we really be sure?
Well, first of all, it has been directly measured. However even on
principle, I see no value in *requiring* that the measurement be done
with earth-bound instruments by climbing, say, a tower. As I've
explained to Seto many times, just because an experimental measurement
is direct and seems straightforward to you is NOT necessarily
justification for doing it. It would be justified if it could be
demonstrated that such a measurement is inherently more precise or
avoids sources of systematic error that would be problematic for other
methods. Unless an experimental setup produces an experimental result
of higher quality or makes the measurement in a physical domain where
it hasn't been established to hold, then it's not worth doing. Very
often, an *indirect* measurement produces an experimental result of
higher quality than a measurement from a dead-simple experimental
concept. This is part of the art of experimental design.
So in answer to your question, there are a bunch of *indirect*
measurements that span a whole range of h where r =3D h + R and R is the
Earth's radius. Examples of this are ballistic launches to high enough
altitude where the difference between r and R makes a predictable
difference in the landing spot; in the predictable orbital periods of
low-orbit, medium-orbit, and geosynchronous satellites, including the
Space Shuttle; and others. Those results yield in fact better
measurements of the 1/r^2 dependence of gravity than would a
measurement using purely earthbound instruments.
PD
Actually, the 1/r^2 law of gravity has been verified by earth-bound
experiments. Not gravity from the Earth though, but gravity between
masses in a laboratory. And, yes, the 1/r^2 law does hold there too.
Check out: http://en.wikipedia.org/wiki/Cavendish_experiment
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stjarnhimlen dot se
WWW: http://stjarnhimlen.se/
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| User: "Randy Poe" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
28 Jan 2008 12:24:07 PM |
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|
On Jan 28, 1:03 pm, wrote:
On Jan 21, 6:48 am, PD <TheDraperFam...@gmail.com> wrote:
On Jan 20, 11:31 pm, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r. The difference between the
two is apparently very little for altitudes you could reach on Earth,
but wouldn't it be something if the experiment did not show 1/r^2??
Read Douglas Giancoli's textbook on introductory physics (either the
algebra-based one, or the calculus-based one will do), which is no
doubt on the shelves at a nearby university library.
Newton figured out the acceleration of the moon, based on v^2/r, where
r is 250,000 miles. He asked the question whether this acceleration
was really the acceleration of an object falling near the earth (the
famous "apple" anecdote arises here), scaled down by distance. He knew
the radius of the earth was about 4000 miles, so that the Moon's orbit
is about 60 times further out. But the acceleration of the moon is
about 1/3600 that of something near the surface of the Earth. This
observation leads one to think that the acceleration (and hence the
force producing it) scales like 1/r^2.
PD
From this discussion, I can only conclude that the 1/r^2 force
relationship has NEVER been experimentally verified on Earth using
earthbound instruments.
You may have decided to conclude that, but not from
this discussion since your conclusion is directly opposite
to statements made in this discussion.
You would need an instrument accurate up to 6
digits of precision to directly distinguish between a 1/r and 1/r^2
force, so this is perhaps why it has never been done.
Many more digits of precision than that are available for
many experiments, including gravitational ones.
Based on the
math, we are extremely confident of the 1/r^2 force, but without ever
directly measuring it, can we really be sure?
If 1/r says the moon should behave like X, and
1/r^2 says it should behave like Y, and what we see
is Y and nothing like X, then yes we can really be sure
that X is wrong.
What's your alternative, that the force really falls off
as 1/r but somehow the planets manage to behave
like it's 1/r^2? How do they manage to operate as if
the force is something that it's not?
- Randy
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| User: "Ockham" |
|
| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
28 Jan 2008 12:59:18 PM |
|
|
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:167e6989-3425-4472-8f02-147a83d29853@j20g2000hsi.googlegroups.com...
| On Jan 28, 1:03 pm, wrote:
| > On Jan 21, 6:48 am, PD <TheDraperFam...@gmail.com> wrote:
| >
| >
| >
| > > On Jan 20, 11:31 pm, wrote:
| >
| > > > If h/r is small for any reasonable h you could obtain on the Earth,
| > > > then I could see why you could drop the r^2 term. It would seem that
| > > > the effect of (h/r)^2 term would be very small and difficult to
| > > > measure, but that we could measure it.
| >
| > > > I would really like to know if it has been experimentally verified
| > > > that gravity falls as 1/r^2 and not 1/r. The difference between the
| > > > two is apparently very little for altitudes you could reach on
Earth,
| > > > but wouldn't it be something if the experiment did not show 1/r^2??
| >
| > > Read Douglas Giancoli's textbook on introductory physics (either the
| > > algebra-based one, or the calculus-based one will do), which is no
| > > doubt on the shelves at a nearby university library.
| >
| > > Newton figured out the acceleration of the moon, based on v^2/r, where
| > > r is 250,000 miles. He asked the question whether this acceleration
| > > was really the acceleration of an object falling near the earth (the
| > > famous "apple" anecdote arises here), scaled down by distance. He knew
| > > the radius of the earth was about 4000 miles, so that the Moon's orbit
| > > is about 60 times further out. But the acceleration of the moon is
| > > about 1/3600 that of something near the surface of the Earth. This
| > > observation leads one to think that the acceleration (and hence the
| > > force producing it) scales like 1/r^2.
| >
| > > PD
| >
| > From this discussion, I can only conclude that the 1/r^2 force
| > relationship has NEVER been experimentally verified on Earth using
| > earthbound instruments.
|
| You may have decided to conclude that, but not from
| this discussion since your conclusion is directly opposite
| to statements made in this discussion.
|
| > You would need an instrument accurate up to 6
| > digits of precision to directly distinguish between a 1/r and 1/r^2
| > force, so this is perhaps why it has never been done.
|
| Many more digits of precision than that are available for
| many experiments, including gravitational ones.
|
| > Based on the
| > math, we are extremely confident of the 1/r^2 force, but without ever
| > directly measuring it, can we really be sure?
|
| If 1/r says the moon should behave like X, and
| 1/r^2 says it should behave like Y, and what we see
| is Y and nothing like X, then yes we can really be sure
| that X is wrong.
|
| What's your alternative, that the force really falls off
| as 1/r but somehow the planets manage to behave
| like it's 1/r^2? How do they manage to operate as if
| the force is something that it's not?
|
| - Randy
What's his hang-up with 1/r? Why not 1/2r or r/1 or just plain 1?
If the Earth (or the Sun) were flattened into a disc then that
would distort the g-field and 1/r^2 would not apply but reality
says that is not the case. Sheesh, all it says is the force is
distributed over an area that is four times as large when twice
as far away. Obviously that doesn't apply to an apple twice
as high up a tree, the law fails when very close to the Earth
because the Earth is not a point to the apple, it is a plane.
With 6 digits of precision the apple is accelerated at 32 fps^2
whether from the top of a skyscraper or from the second floor.
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| User: "Randy Poe" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 08:32:10 AM |
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On Jan 21, 12:31 am, wrote:
If h/r is small for any reasonable h you could obtain on the Earth,
then I could see why you could drop the r^2 term. It would seem that
the effect of (h/r)^2 term would be very small and difficult to
measure, but that we could measure it.
I would really like to know if it has been experimentally verified
that gravity falls as 1/r^2 and not 1/r.
Yes. I have read that the exponent is known to be 2.000...0
to extremely high precision, one of the most precisely-known
physical constants.
Now let me see if I can Google for what experiments
are behind that claim...
Well, this isn't what I was thinking of, but here's an
experimental paper testing the inverse square law:
http://flux.aps.org/meetings/YR01/APR01/abs/S1290.html#SJ9.002
(abstract only)
and some more
http://www.physorg.com/news88939401.html
http://physicsworld.com/cws/article/print/21822
These experiments are looking for tiny deviations
from 1/r^2. The difference between 1/r^2 and 1/r
would be much more noticeable. None of Kepler's
orbital laws would hold, for instance, not even
approximately. Planets wouldn't have closed
elliptical orbits.
- Randy
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| User: "" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 11:12:14 AM |
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Thanks for trying to come up with some references, I have seen
references like these before, but, I am not interested in measuring 1/
r^2 under submillimeter conditions where electostatic conditions could
dominate, nor am I interested in indirect measurements using orbiting
bodies like the moon, planets, etc. I am looking for a simple direct
measurement which can only be done by taking a gravity sensing
instrument and raising it straight up through the gravity field over a
static position. This is a very simple experiment and I would be
surprised if nobody has done this basic experiment with a high degree
of accuracy. I find it puzzling that I can find no references to such
a basic experiment. I think there may have been some experiments
involving towers, but I can't seem to locate those refrences now.
The problem with relying on orbiting bodies is that you are relying on
a mathematical assumption that certain equations (centrifugal force)
and (gravitational force) are equal. There is a cancellation of terms
and it does produce the correct value for velocity for a given radius.
But this already a-prioi presumes that your assumptions about the
forces are correct, so while being very convincing in the orbital
case, cannot be used as a subsitute for direct measurements. You can
validate your assmptions with direct measurements, but you cannot
presume the results of direct measurements based on only your
assumptions.
Well, this isn't what I was thinking of, but here's an
experimental paper testing the inverse square law:http://flux.aps.org/meet=
ings/YR01/APR01/abs/S1290.html#SJ9.002
(abstract only)
and some morehttp://www.physorg.com/news88939401.htmlhttp://physicsworld.c=
om/cws/article/print/21822
These experiments are looking for tiny deviations
from 1/r^2. The difference between 1/r^2 and 1/r
would be much more noticeable. None of Kepler's
orbital laws would hold, for instance, not even
approximately. Planets wouldn't have closed
elliptical orbits.
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0- Randy
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| User: "Mike Dworetsky" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 11:34:34 AM |
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)
<franklinhu@yahoo.com> wrote in message
news:25af42b8-ebe8-44c1-8950-0b3d31e0688f@v4g2000hsf.googlegroups.com...
Thanks for trying to come up with some references, I have seen
references like these before, but, I am not interested in measuring 1/
r^2 under submillimeter conditions where electostatic conditions could
dominate, nor am I interested in indirect measurements using orbiting
bodies like the moon, planets, etc. I am looking for a simple direct
measurement which can only be done by taking a gravity sensing
instrument and raising it straight up through the gravity field over a
static position. This is a very simple experiment and I would be
surprised if nobody has done this basic experiment with a high degree
of accuracy. I find it puzzling that I can find no references to such
a basic experiment. I think there may have been some experiments
involving towers, but I can't seem to locate those refrences now.
The problem with relying on orbiting bodies is that you are relying on
a mathematical assumption that certain equations (centrifugal force)
and (gravitational force) are equal. There is a cancellation of terms
and it does produce the correct value for velocity for a given radius.
But this already a-prioi presumes that your assumptions about the
forces are correct, so while being very convincing in the orbital
case, cannot be used as a subsitute for direct measurements. You can
validate your assmptions with direct measurements, but you cannot
presume the results of direct measurements based on only your
assumptions.
Reply:
Sorry, my indentation character seems to be on the fritz at the moment. It
seems you are looking for experiments that measure the change in
gravitational acceleration with altitude?
So what you are looking for amounts to asking for an experimental
verification that the Earth isn't flat? I'm not exactly sure what your
question is about.
Why wouldn't the successful orbiting of a satellite with the correct period
for its altitude be a verification for you?
Well, this isn't what I was thinking of, but here's an
experimental paper testing the inverse square
law:http://flux.aps.org/meetings/YR01/APR01/abs/S1290.html#SJ9.002
(abstract only)
and some
morehttp://www.physorg.com/news88939401.htmlhttp://physicsworld.com/cws/article/print/21822
These experiments are looking for tiny deviations
from 1/r^2. The difference between 1/r^2 and 1/r
would be much more noticeable. None of Kepler's
orbital laws would hold, for instance, not even
approximately. Planets wouldn't have closed
elliptical orbits.
- Randy
--
Mike Dworetsky
(Remove pants sp*mbl*ck to reply
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| User: "Paul Schlyter" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
22 Jan 2008 03:12:40 AM |
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In article <HqWdnbw4i-8wSgnanZ2dnUVZ8rCdnZ2d@bt.com>,
Mike Dworetsky <platinum198@pants.btinternet.com> wrote:
)
Why wouldn't the successful orbiting of a satellite with the correct period
for its altitude be a verification for you?
A satellite in circular orbit only shows how strong the gravity is at
that altitude, not how the force of gravity varies with altitude.
However, few (if any) orbits are perfect circles or nearly so. Most
orbits are ellipses, i.e. their altitude vary with the position in
the orbit. And these elliptical orbits require a 1/r^2 gravitational
law. If gravity obeyed a different law, the shapes of these orbits
would be different. A fundamental textbook in celestial mechanics
will show you why - the better ttextbooks usually have a charpter
about what orbits would be like if gravity obeyed a different law.
Read for instance the subcharpter "To find the law of force, given the
orbit" in the charpter "Central orbits" in Danby's classical textbook
"Fundamentals of Celestial Mechanics". And an elliptical orbit with
the central body at one focus of the ellipse (as observed) *implies*
that gravity must follow a 1/r^2 law.
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stjarnhimlen dot se
WWW: http://stjarnhimlen.se/
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| User: "Sjouke Burry" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
22 Jan 2008 03:24:49 AM |
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Paul Schlyter wrote:
In article <HqWdnbw4i-8wSgnanZ2dnUVZ8rCdnZ2d@bt.com>,
Mike Dworetsky <platinum198@pants.btinternet.com> wrote:
)
Why wouldn't the successful orbiting of a satellite with the correct period
for its altitude be a verification for you?
A satellite in circular orbit only shows how strong the gravity is at
that altitude, not how the force of gravity varies with altitude.
Any deviation from predicted gravity, will make a huge difference
in fuel consumption when launching a vehicle.
And the margins in fuel reserves are so small, that the vehicle will
end up nowhere near the intended location/orbit.
The fact that almost all craft DO end up in the right place, tells
you that there is nothing much wrong with the expected gravity profile.
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| User: "Paul Schlyter" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
22 Jan 2008 03:12:40 AM |
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In article <25af42b8-ebe8-44c1-8950-0b3d31e0688f@v4g2000hsf.googlegroups.com>,
<franklinhu@yahoo.com> wrote:
The problem with relying on orbiting bodies is that you are relying on
a mathematical assumption that certain equations (centrifugal force)
and (gravitational force) are equal. There is a cancellation of terms
and it does produce the correct value for velocity for a given radius.
But this already a-prioi presumes that your assumptions about the
forces are correct, so while being very convincing in the orbital
case, cannot be used as a subsitute for direct measurements. You can
validate your assmptions with direct measurements, but you cannot
presume the results of direct measurements based on only your
assumptions.
If the force of gravity obeyed a law different from the 1/r^2 law, the
*shapes* of the orbits would be different than observed! Few orbits
are a circle or almost a circle. Most orbits are ellipes, and some
are very eccentric ellipses.
You need to read a fundamental textbook about celestial mechanics to
understand why. DO that, and then come back if you want to know more.
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stjarnhimlen dot se
WWW: http://stjarnhimlen.se/
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| User: "Randy Poe" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 11:27:00 AM |
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On Jan 21, 12:12 pm, wrote:
Thanks for trying to come up with some references, I have seen
references like these before, but, I am not interested in measuring 1/
r^2 under submillimeter conditions where electostatic conditions could
dominate, nor am I interested in indirect measurements using orbiting
bodies like the moon, planets, etc. I am looking for a simple direct
measurement which can only be done by taking a gravity sensing
instrument and raising it straight up through the gravity field over a
static position. This is a very simple experiment and I would be
surprised if nobody has done this basic experiment with a high degree
of accuracy. I find it puzzling that I can find no references to such
a basic experiment. I think there may have been some experiments
involving towers, but I can't seem to locate those refrences now.
The problem with relying on orbiting bodies is that you are relying on
a mathematical assumption that certain equations (centrifugal force)
and (gravitational force) are equal.
I think you're talking about the equation of motion which
says m*a = m*d^2x/dt^2 = gravitational force.
Or for a circular orbit, mv^2/r = gravitational force.
These equations say "the force which this body is moving
under is the gravitational force". If they are unequal, then
there is another force present. What force do you suggest?
There is a cancellation of terms
and it does produce the correct value for velocity for a given radius.
But this already a-prioi presumes that your assumptions about the
forces are correct,
That assumption being that the force acting on the body
is the gravitational force. How would you modify that?
Where is there room to change that assumption?
- Randy
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| User: "Randy Poe" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 09:47:33 AM |
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On Jan 21, 12:31 am, wrote:
I didn't see this part of your post when responding before:
Another interesting factoid to consider is that according to orbital
mechanics:
http://en.wikipedia.org/wiki/Orbital_mechanics
The specific potential energy associated with a planet of mass M is
given by:
-GM/r
This is the potential energy an object would have it it were hovering
over a gravitational object.
This is the potential an object has everywhere
in the gravitational field of an object of mass M, whether it
is hovering or not. (Nit: The potential energy is actually
-GMm/r. The term "potential" is used for the part -GM/r
which doesn't depend on the test mass m).
Once again, we see that this energy is
not 1/r^2, it is 1/r. So we have some factors that show up as 1/r with
gravity.
That's correct. We all know this. A 1/r potential is the
same thing as a 1/r^2 force, effectively (see below).
Could the strength of gravity be directly related to the
potential energy which is linear?
The relationship is that the force is the derivative of the
potential energy. That means that if the potential is 1/r,
the force is 1/r^2, and if the force is 1/r^2, the potential is 1/r.
Those two facts are not contrasting, they go together.
- Randy
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| User: "tadchem" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
21 Jan 2008 05:42:33 AM |
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On Jan 20, 2:07 pm, wrote:
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
http://www.space-electronics.com/Literature/Precise_Measurement_of_Ma...
On pg 15, they claim to have a formula to calculate the force of
gravity to .002%
g =3D 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
L represents the lattitude and H represents the height.
The strange thing to notice is that the relationship between g and
H(height or radius from the center of the Earth) is linear. If gravity
varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
in the correction, but there isn't. The formula implies that gravity
decreases linearly rather than exponentially as height increases.
I quick search of the internet failed to reveal experiments which
directly measure the 1/r^2 relationship from non-orbiting platforms (1/
r^2 is often justified by looking at the orbits of satellites). But I
was looking for a direct measurement of this force by comparing how
the force of gravity varies by height on an Earth based experiment.
Such an experiment might involve sending a weather balloon from the
ground to the high stratosphere while measuring the force of gravity
to directly verify the 1/r^2 relationship.
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
fhugravity
The pedagogue in me has drawn me back to offer the explanation you
solicited for why the gravity correction for altitude is linear.
Yes, g varies as 1/r^2 from the center of the mass.
What if you can't measure from the center of a large mass?
The you can measure H from the surface of the mass with known radius
r.
Your distance from the center is then r+H.
You can compare that to the gravity AT the surface g=B0 with a
proportion:
g =3D g=B0 * [ r^2 / (r+H)^2 ]
Cute, but a little dodgy, especially if H is much smaller than r.
Mountains and ocean trenches on the earth are generally less than a
few km high/deep (-11km <=3D H <=3D 9 km), but the earth itself is over
6300 km radius (r).
We can safely expand r^2/(r+H)^2 =3D 1/(1 + H/r)^2 as a binomial series
(look it up if you don't understand):
1/(1 + H/r)^2 =3D 1 - 2(H/r) + 3(H/r)^2 - 4(H/r)^3 + ...(etc)
Now we know that the magnitude of H/r is at most 11/6300 =3D 0.0017, so
the third term is about 0.000009, and further terms in the series are
even smaller.
It should be apparent that, if you are interested in calculating to
within .002 precision, the third and higher order terms are not going
to be important.
HTH
Tom Davidson
Richmond, VA
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| User: "Igor" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
20 Jan 2008 01:14:12 PM |
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On Jan 20, 2:07=A0pm, wrote:
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
http://www.space-electronics.com/Literature/Precise_Measurement_of_Ma...
On pg 15, they claim to have a formula to calculate the force of
gravity to .002%
g =3D 9.80613 ( 1 - 0.0026325 cos 2 L ) ( 1 - 3.92 10-7 H )
L represents the lattitude and H represents the height.
The strange thing to notice is that the relationship between g and
H(height or radius from the center of the Earth) is linear. If gravity
varied as 1/r^2, then I would suspect that the would be a 1/H^2 term
in the correction, but there isn't. The formula implies that gravity
decreases linearly rather than exponentially as height increases.
I quick search of the internet failed to reveal experiments which
directly measure the 1/r^2 relationship from non-orbiting platforms (1/
r^2 is often justified by looking at the orbits of satellites). But I
was looking for a direct measurement of this force by comparing how
the force of gravity varies by height on an Earth based experiment.
Such an experiment might involve sending a weather balloon from the
ground to the high stratosphere while measuring the force of gravity
to directly verify the 1/r^2 relationship.
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
fhugravity
This is a formula that tells you how to calculate the effective
gravitational acceleration at any place on the earth. The 1/r^2 has
already been taken into account in the factor of 9.80613, by factoring
in the average radius of the earth. The additional terms involve
centrifugal corrections due to the rotation of the earth and have
absolutely nothing to do with gravity at all.
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| User: "John C. Polasek" |
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| Title: Re: Is the force of gravity 1/r^2 or 1/r????? |
22 Jan 2008 10:15:06 AM |
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On Sun, 20 Jan 2008 11:07:17 -0800 (PST), wrote:
It is well understood that the force of gravity is a 1/r^2 force.
However, I was looking at a document describing how to precisely
meausure mass.
snip
If anyone could point me to a reference on the web experimentally
proving the 1/r^2 relationship or can explain why the gravity
correction formula is linear, I would appreciate it.
fhugravity
Newton fell short a bit by not including the factor 4pi. Clearly his
formula applies to perfect spherical symmetry, and in fact the
influence of M is spread over 4 pi steradians. He should use G'
instead of G:
a = MG'/4pir^2
Once this is underst | |
