From Newsgroup: sci.physics

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out) then why is it that the deeper you go into a plant,
like Jupiter for example, the pressure increases and the center of a
star, when it forms, ignites into nuclear fusion?
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From Newsgroup: sci.physics

Popping Mad <rainbow@colition.gov> wrote:

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out) then why is it that the deeper you go into a plant,
like Jupiter for example, the pressure increases and the center of a
star, when it forms, ignites into nuclear fusion?

Pressure and gravity are not the same force.

Pressure is the result of mass that is above you.

Gravity is the result of mass that is around you.

On the surface of the Earth all the mass above you is air and the
pressure because of the weight of that air is about 14 psi. As you
descend into the Earth there is more and more massive stuff above you
pushing you down and the pressure goes up.

As you descend into the Earth the vector sum of gravity changes. You
now have mass above you pulling you up. The gravity thus decreases.

The simplist way to put it is that as you go down, the gravity of the
stuff above you cancels the gravity of the stuff below you but the
weight of stuff above you is always increasing.
--
penninojim@yahoo.com
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From Newsgroup: sci.physics

Popping Mad wrote:
^^^^^^^^^^^
This is not a chat group. Politeness suggests that your *real* name belongs there.

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out)

First of all, it is not good to think of gravitation as a property of an
object as in "Earth *has* gravity". That is NOT how it works. [This is frequently taught wrong in schools.] Instead, it is an *interaction*
_between_ objects. According to Newton's theory, objects _attract each
other_ because they have non-zero mass. So (according to Newton) it is not
so that "earth does gravity" but that Terra (_Earth_) has non-zero mass, and
so do other objects (including people like you and me), and so everything is attracted to everything else. (As the story goes, he realized that an apple and Earth attract each other in the same as Earth and the Moon attract each other, and Earth and the Sun are attracted to each other: Gravitation was *universal*, not limited to Earth. Thus he could also explain how Kepler's planetary orbits arose, and predict them.)

One way to understand how the magnitude of the gravitational acceleration at the center of Terra is approximately zero is to consider that a test object with a negligible non-zero mass (a "test mass") located there will be
attracted gravitationally by all the matter that surrounds it (which also
has non-zero mass) in all directions of space approximately in the same way (*exactly* in the same way if the planet were spherically-symmetric and had
a uniform mass density; we know that this is not so for any planet, but it
is still a good approximation), so the net gravitational force on it and its net gravitational acceleration is approximately (would be *exactly*) zero.

Another (*mathematically* much more simple) way (to *calculate* this) is to consider Gauss' Law for gravitation. In differential form it is

rec ria g(R) = -4-C G -U,

where by g(R) I mean the vector field of gravitational acceleration that depends on the radius vector R; G is the Newtonian gravitational constant,
and -U is the mass density.

[(Here) we imagine the gravitational field to be like a field describing
the flow of a fluid. rec ria g(R) is its divergence, a measure how much
the flow is spreading out. Since the field points inwards, the flow is
negative. The source of the fluid, so to speak, is represented by the
density on the right-hand side which in general is also a field (it
depends on position). See the URIs at the bottom for details.]

If we take the integral over a spherical volume centered at R = 0 which has uniform mass density (that is what we assumed in the previous
interpretation) and radius r, we find

re#_V(r) dV [rec ria g(R)] = 4-C G re#_V(r) dV -U = -4-C G M(r).

[The right-hand side results because by definition of the mass density,
the mass density multiplied by (or, if not uniform, integrated over)
the volume gives the mass contained in that volume.]

According to Gauss' Theorem, the left-hand side can be written

re#_V(r) dV [rec ria g(R)] = re>_S^2(r) [dS ria g(R)],

where S^2(r) is the (2-)sphere (a _surface) with radius r. Since the gravitational acceleration is spherically symmetric and directed towards
the center (R = 0) (here), this becomes

re>_{S^2(r)} [dS ria g(R)] = re>_{S^2(r)} ds ||g(R) cos[dS, g(R)]
= re>_{S^2(r)} ds g(r) cos(180-#)
= -g(r) re2_0^-C d++ sin(++) re2_0^2-C d-a
= -g(r) re2_{-1}^1 du re2_0^2-C d-a
= -4-C r^2 g(r).

So finally we have

-4-C r^2 g(r) = -4-C G M(r)
<==> g(r) = G M(r)/r^2.

But notice that this, the magnitude of the gravitational acceleration,
depends on the *enclosed* mass up to the radius r, NOT just the total mass
of the body.

One might think that because of the division by r^2, g(r) --> +inf as
r --> 0. But because it is M(r) and not just M, that is actually not so.
If the mass density is constant, then for all radii r we have

-U = M(r)/V(r).

<==> M(r) = -U V(r).

So for a spherical volume with uniform mass density, we have

M(r) = 4/3 -C r^3 -U.

So the magnitude of the gravitational acceleration at the radius r if r is smaller than or equal to the radius of the body, is actually *proportional
to the radius* if the body is spherically-symmetric and its mass density is
the same everywhere:

g(r) = G M(r)/r^2 = 4/3 -C G r^3 -U/r^2 = 4/3 -C G -U r. (!)

So for the center of that body, we find:

g(r = 0) = 0. reA

See also:

<https://www.youtube.com/watch?v=hbmf0bB38h0&list=PL41EYJuJ5YuDn3d13ryZwpzGBXewXa9AH&index=24>

Or, if you do not need the introduction, you can just start watching here:

<https://youtu.be/hbmf0bB38h0?list=PL41EYJuJ5YuDn3d13ryZwpzGBXewXa9AH&t=3688>

[It is a common exercise in the courses of Mathematical Methods of Physics
and (Classical) Electrodynamics to use Gauss' Law and Gauss' Theorem to determine the gravitational/electric field in the interior, at the surface,
and outside an object (for the electric field, Gauss' law is just one of the Maxwell's equations). To make the calculation easier, the mass/charge densities of those objects usually exhibit a simple symmetry, e.g. spherical symmetry. One is supposed to realize that symmetry, and choose the
so-called "Gaussian surface" over which to integrate, and the coordinates to use, accordingly.]

then why is it that the deeper you go into a plant, like Jupiter for
example, the pressure increases and the center of a star, when it
forms, ignites into nuclear fusion?

Consider the terrestrial atmosphere and hydrosphere first. There is atmospheric pressure on the terrestrial surface because we are living at the bottom of an "ocean" of air: The air above you and the planet attract each other gravitationally (in Newton's theory of gravitation), and each layer of air pushes down on the layer below it and compresses it (thus the
atmospheric pressure and density decreases with increasing altitude).

[In fact, you can arrive very closely at the standard atmospheric pressure
by estimating the mass of the terrestrial atmosphere from the density of
air, and approximating it as a thin uniform spherical shell (so that you
can calculate as if it were a cuboid) out of nitrogen and oxygen in the
known proportions (say 79 % nitrogen and 21 % oxygen, ignoring trace
gases), with a thickness of ca. 100 km, and calculating the pressure that
the atmosphere therefore exerts as the gravitational force divided by the
surface area.]

Then when you go underwater, like in a lake and the actual (water) ocean,
the water and the rest of the planet also attract each other
gravitationally. So now the underwater pressure is produced not only by the atmosphere, but also by the water column above your head. The deeper you
go, the larger that column, So the underwater pressure increases with depth.
But water as a liquid (with particles bound to each other) is barely compressible, so the water temperature does not increase (much) with depth
(see below).

[This is in fact how buoyancy arises: The pressure by a fluid at the
bottom of a immersed object or substance is greater than at the top, and
equal on the sides. So if the object's/substance's mass density is less
than that of the fluid that it displaces, it will be pushed up, and it
can even float on that fluid (like a boat or ship where the submerged
part consists mostly of air).

If the substance is also a fluid, layers with less mass density will
arrange themselves to be above layers with greater mass density. That is
also how, for example, water ice is found at the top and not the bottom
of a frozen lake or river, allowing the animals requiring the lake or
river to survive the winter: the lake or river never completely freezes
because water below 4 -#C moves to the top and freezes there, and the ice
at the top prevents further heat loss of the water below it.]

It is the same with gas giants, only their atmospheres are much deeper (it
is assumed, now also from gravitational measurements, that they have a core
out of metallic hydrogen, but they are still huge), and the substances are
all *gaseous* (despite the low temperature), thus *compressible*, so the atmospheric pressure and temperature towards the core can increase even
further (when a gaseous substance is compressed, momentum is imparted on its freely moving particles, so they move faster which we understand as a higher temperature; cf. the equation of state of an ideal gas).

Now you can understand how stars form. In fact, if the mass of Jupiter
would be approximately 82 times its current mass, the pressure and
temperature in its interior would be high enough that nuclear fusion becomes possible and it would become a star. (Don't worry, there is not enough
matter in the Sol System or its vicinity left for that to happen. The mass
of Sol alone constitutes ca. 99.8 % of the mass of the Sol System :))

HTH.
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

On 12/6/25 3:06 PM, Thomas 'PointedEars' Lahn wrote:

is not
so that "earth does gravity" but that Terra (_Earth_) has non-zero m

I wish I could read this and understnd it. It stops being
understandable right here.
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From Newsgroup: sci.physics

Popping Mad wrote:
^^^^^^^^^^^
AISB.

On 12/6/25 3:06 PM, Thomas 'PointedEars' Lahn wrote:

is not so that "earth does gravity" but that Terra (_Earth_) has non-zero m

I wish I could read this and understnd it. It stops being
understandable right here.

It helps to read texts carefully; to read whole words, sentences and paragraphs. The entire sentence was:

So (according to Newton) it is not so that "earth does gravity" but that
Terra (_Earth_) has non-zero mass, and so do other objects (including
people like you and me), and so everything is attracted to everything
else.

Which part of that do you not understand?
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

Thomas 'PointedEars' Lahn wrote:

First of all, it is not good to think of gravitation as a property of an object as in "Earth *has* gravity". That is NOT how it works. [This is frequently taught wrong in schools.] Instead, it is an *interaction* _between_ objects. According to Newton's theory, objects _attract each other_ because they have non-zero mass. So (according to Newton) it is not so that "earth does gravity" but that Terra (_Earth_) has non-zero mass,

["non-zero X" simply means "X is not equal to zero"]

and so do other objects (including people like you and me), and so everything is attracted to everything else. (As the story goes, he realized that an apple
and Earth attract each other in the same as Earth and the Moon attract each

^^^^^^^^^^^
in the same _way_

other, and Earth and the Sun are attracted to each other: Gravitation was *universal*, not limited to Earth. Thus he could also explain how Kepler's planetary orbits arose, and predict them.)

One way to understand how the magnitude of the gravitational acceleration at the center of Terra is approximately zero is to consider that a test object with a negligible non-zero mass (a "test mass") located there will be attracted gravitationally by all the matter that surrounds it (which also
has non-zero mass) in all directions of space approximately in the same way (*exactly* in the same way if the planet were spherically-symmetric and had
a uniform mass density; we know that this is not so for any planet, but it
is still a good approximation), so the net gravitational force on it and its net gravitational acceleration is approximately (would be *exactly*) zero.

Granted, the "incredibly unbroken" one-paragraph sentence above is far too long, and reminds me of a certain Dr. Fassbinder }:-)

An attempt to rewrite it:

One way to understand how the magnitude of the gravitational acceleration
at the center of Terra is approximately zero is to consider a test object
with a negligible non-zero mass (a "test mass") there. It will be attracted gravitationally by all the matter that surrounds it (which also has non-zero mass) in all directions of space approximately in the same way. So the net gravitational force on it and its net gravitational acceleration are approximately zero.

[The test object would be attracted *exactly* in the same way if the planet were spherically-symmetric and had a uniform mass density; we know that this
is not so for any planet, but it is still a good approximation.]
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

Thomas 'PointedEars' Lahn wrote:

Popping Mad wrote:

then why is it that the deeper you go into a plant, like Jupiter for
example, the pressure increases and the center of a star, when it
forms, ignites into nuclear fusion?

Consider the terrestrial atmosphere and hydrosphere first. There is atmospheric pressure on the terrestrial surface because we are living at the bottom of an "ocean" of air: The air above you and the planet attract each other gravitationally (in Newton's theory of gravitation), and each layer of air pushes down on the layer below it and compresses it (thus the
atmospheric pressure and density decreases with increasing altitude).

[In fact, you can arrive very closely at the standard atmospheric pressure
by estimating the mass of the terrestrial atmosphere from the density of
air, and approximating it as a thin uniform spherical shell (so that you
can calculate as if it were a cuboid) out of nitrogen and oxygen in the
known proportions (say 79 % nitrogen and 21 % oxygen, ignoring trace
gases), with a thickness of ca. 100 km, and calculating the pressure that
the atmosphere therefore exerts as the gravitational force divided by the
surface area.]

This is another paragraph that can be improved ':-)

[In fact, you can arrive very closely at the standard atmospheric pressure
by approximating the atmosphere as a thin uniform spherical shell out of
nitrogen and oxygen in the known proportions (say 79 % nitrogen and 21 %
oxygen, ignoring trace gases), with a thickness of ca. 10 km. This
approximation allows you to calculate its volume as if it were a
rectangular cuboid. You can estimate the mass of the terrestrial
atmosphere from the density of air and this estimate of its volume. Then
you can calculate the pressure that the atmosphere therefore exerts as
the gravitational force divided by the surface area.]

[...]
It is the same with gas giants, only their atmospheres are much deeper (it
is assumed, now also from gravitational measurements, that they have a core out of metallic hydrogen, but they are still huge), and the substances are all *gaseous* (despite the low temperature), thus *compressible*, so the atmospheric pressure and temperature towards the core can increase even further (when a gaseous substance is compressed, momentum is imparted on its freely moving particles, so they move faster which we understand as a higher temperature; cf. the equation of state of an ideal gas).

And this one ':-)

It is the same with gas giants as with ocean water, only the substances out
of which their atmospheres consist are all *gaseous* (despite the low temperature), thus *compressible*. When a gaseous substance is compressed, momentum is imparted on its freely moving particles, so those move faster.
We understand that as the gas having a higher temperature; cf. the equation
of state of an ideal gas.

It is assumed, now also from gravitational measurements with space probes,
that the gas giants have a core out of metallic hydrogen. So they do not
only consist of their atmospheres. But they are still huge, and so their atmospheres are much deeper than the terrestrial oceans, and the atmospheric pressure and temperature towards the core can increase much further. In
fact, it is the high pressure that allows exotic states of matter like
metallic hydrogen to exist.

HTH.
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

On 12/7/25 4:27 AM, Thomas 'PointedEars' Lahn wrote:

Which part of that do you not understand?

The part where you mixed up the words, punctuated trains of thought with
other trains of thought, and made it hard to read. I appreciate the
effort, but I can't make sense of that message as it was written. You
also assume my mathmatics is on the level of a well practiced graduate
student, which it is no longer. The last time I did integration was in
my masters degree 40+ years ago. And your use of mathmatical symbols in
ascii wasn't easy to understand.

I assume that since you wrote so much, you wanted to be understood. If
not, my appologies. I didn't realize I was being trolled at such a high
level.
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From Newsgroup: sci.physics

On 12/7/25 11:35 AM, Thomas 'PointedEars' Lahn wrote:

Thomas 'PointedEars' Lahn wrote:

First of all, it is not good to think of gravitation as a property of an
object as in "Earth *has* gravity". That is NOT how it works. [This is
frequently taught wrong in schools.] Instead, it is an *interaction*
_between_ objects. According to Newton's theory, objects _attract each
other_ because they have non-zero mass. So (according to Newton) it is not >> so that "earth does gravity" but that Terra (_Earth_) has non-zero mass,

["non-zero X" simply means "X is not equal to zero"]

and so do other objects (including people like you and me), and so everything
is attracted to everything else. (As the story goes, he realized that an apple
and Earth attract each other in the same as Earth and the Moon attract each

^^^^^^^^^^^
in the same _way_

other, and Earth and the Sun are attracted to each other: Gravitation was
*universal*, not limited to Earth. Thus he could also explain how Kepler's >> planetary orbits arose, and predict them.)

One way to understand how the magnitude of the gravitational acceleration at >> the center of Terra is approximately zero is to consider that a test object >> with a negligible non-zero mass (a "test mass") located there will be
attracted gravitationally by all the matter that surrounds it (which also
has non-zero mass) in all directions of space approximately in the same way >> (*exactly* in the same way if the planet were spherically-symmetric and had >> a uniform mass density; we know that this is not so for any planet, but it >> is still a good approximation), so the net gravitational force on it and its >> net gravitational acceleration is approximately (would be *exactly*) zero.

I understand this in theory. My problem is that if an apple falls from
a tree and hits you in the head, it hurts, despite that it falls from
the tree because of its gravetational attraction to the Earth, not to you.

So, if you drive in the ocean, as you go deeper the preasure
increases... not because of the gravity between you and water, but
because of the force of gravity between the water and the earth. You
will still get the bends and be crushed by the pressure. I am pretty
sure a similar result would happen in the center of a perfectly
shperical planet. I suppose I am having trouble conceptually with the
idea that in the center of the planet you would have effectly null force
of gravity, but there is still a huge amount of pressure from all
directions If I was in the center of the earth, my effective weight
might be zero, but the whole weight of the plaent still surrounds me.

Granted, the "incredibly unbroken" one-paragraph sentence above is far too long, and reminds me of a certain Dr. Fassbinder }:-)

An attempt to rewrite it:

One way to understand how the magnitude of the gravitational acceleration
at the center of Terra is approximately zero is to consider a test object with a negligible non-zero mass (a "test mass") there. It will be attracted gravitationally by all the matter that surrounds it (which also has non-zero mass) in all directions of space approximately in the same way. So the net gravitational force on it and its net gravitational acceleration are approximately zero.

check

[The test object would be attracted *exactly* in the same way if the planet were spherically-symmetric and had a uniform mass density; we know that this is not so for any planet, but it is still a good approximation.]

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From Newsgroup: sci.physics

Popping Mad wrote:
^^^^^^^^^^^
Please modify your newsreader's settings so that you post under your real
name. It is considered polite here. (As you can see, you can *also*
include your nickname if you want.)

On 12/7/25 11:35 AM, Thomas 'PointedEars' Lahn wrote:

Thomas 'PointedEars' Lahn wrote:

First of all, it is not good to think of gravitation as a property of an >>> object as in "Earth *has* gravity". That is NOT how it works. [This is >>> frequently taught wrong in schools.] Instead, it is an *interaction*
_between_ objects. According to Newton's theory, objects _attract each
other_ because they have non-zero mass. So (according to Newton) it is not >>> so that "earth does gravity" but that Terra (_Earth_) has non-zero mass,

["non-zero X" simply means "X is not equal to zero"]

and so do other objects (including people like you and me), and so everything
is attracted to everything else. (As the story goes, he realized that an apple
and Earth attract each other in the same as Earth and the Moon attract each >> ^^^^^^^^^^^

in the same _way_

other, and Earth and the Sun are attracted to each other: Gravitation was >>> *universal*, not limited to Earth. Thus he could also explain how Kepler's >>> planetary orbits arose, and predict them.)

[One way to understand how the magnitude of the gravitational acceleration >>> at the center of Terra is approximately zero is to consider a test object >>> with a negligible non-zero mass (a "test mass") there. It will be attracted
gravitationally by all the matter that surrounds it (which also has non-zero
mass) in all directions of space approximately in the same way. So the net >>> gravitational force on it and its net gravitational acceleration are
approximately zero.]

I understand this in theory. My problem is that if an apple falls from
a tree and hits you in the head, it hurts, despite that it falls from
the tree because of its gravetational attraction to the Earth, not to you.

That is not quite correct. There is also a small gravitational force
between the person and the apple. And the apple attracts Terra (_Earth_) in the same way that Terra attracts the apple. However, as the person is at
rest relative to Terra, they consider the apple to be the moving object
(this was one of Newton's great insights, building on Galilei's principle of relativity).

However, the strength of the force between the person and the apple is negligibly small compared to that between Earth and the apple because the
mass of Terra is so much larger than that of the person:

The strength of the gravitational force between the person (M =~ 70 kg) and
the apple (m =~ 10 g) at a distance of 1 m is approximately

F = G M m/r^2
=~ 6.674 |u 10^-11 m^3/(kg s^2) |u 70 kg |u 0.01 kg/(1 m)^2
=~ 4.67 |u 10^-11 N.

The gravitational force between Terra (M =~ 5.97 |u 10^24 kg, R =~ 6371 km)
and the apple 1 m away from the former's surface is approximately

F = G M m/r^2
=~ 6.674 |u 10^-11 m^3/(kg s^2) |u 5.97 |u 10^24 kg |u 0.01 kg/
(6371 km + 1 m)^2
=~ 0.0982 N,

i.e. 9 orders of magnitude larger. [This is based on Earth modeled as a
point mass in its center. One can see that if one divides that by the mass
of the apple, one obtains approximately the average surface gravitational acceleration which therefore is (without air resistance) the same for all objects near the terrestrial surface (Galilei's discovery): ca. 9.8 m/s^2. <https://www.youtube.com/watch?v=oYEgdZ3iEKA>]

Whatever the total force on the apple while it is in free fall, when the
apple hits the person, the person's atoms also exert an oppositely directed force on the apple, and vice-versa, that prevents the apple from continuing
to fall freely towards the center of gravitation of Terra. When the apple
hits the head of the person, due to the sudden increase in pressure on the latter's skin/body and potential damage to the body from that, the nervous system of the person sends electric signals to the person's brain which interprets this as pain :'-)

[If there is no person and no other object, the atoms of the surface of the planet, exerting the same opposite contact forces, prevent the apple from continuing to fall freely. It turns out that is how the apple has a weight (not: mass which it has regardless of gravitation) in the first place.]

So, if you drive in the ocean, as you go deeper the preasure
increases... not because of the gravity between you and water, but

_gravitation_

because of the force of gravity between the water and the earth.

Both; but the gravitational force between the water and the person is,
again, negligibly small compared to the gravitational force between the
water and the rest of the planet. It is, however, the latter force that
the person experiences *partially* as it acts on the surface of the
person's body.

You will still get the bends and be crushed by the pressure.

Yes, if you go to deep, because of what I explain above.

I am pretty sure a similar result would happen in the center of a perfectly shperical planet.

It would.

I suppose I am having trouble conceptually with the idea that in the center of the planet you would have effectly null force of gravity,

_gravitation_

but there is still a huge amount of pressure from all directions

We are only talking about the *gravitational* force on the object, that
would make *it* move, and *its* acceleration due to gravitation. Consider
a person to be in a person-sized hole filled with air, and the surrounding
rock to be prevented by other forces from falling in. That person would _float_ due to the mass of the rock that it is *surrounded* by.

If I was in the center of the earth, my effective weight might be zero,

It would be, approximately.

but the whole weight of the plaent still surrounds me.

That is *how* the above can be explained.
--
PointedEars

Twitter: @PointedEars2
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From Newsgroup: sci.physics

On 12/8/25 12:51 AM, Thomas 'PointedEars' Lahn wrote:

Please modify your newsreader

umm - No - now you are a troll and will be /dev/nulled. You have enough trouble to control yourself, let alone to control anyone else. I may
not be polite, but I am more polite than you.


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From Newsgroup: sci.physics

Popping Mad wrote:

On 12/8/25 12:51 AM, Thomas 'PointedEars' Lahn wrote:

Please modify your newsreader

The whole paragraph reads:

Please modify your newsreader's settings so that you post under your real
name. It is considered polite here. (As you can see, you can *also*
include your nickname if you want.)

Your insinuation that I suggested that you modify your *newsreader* (which
may not be possible) is quite disingenuous of you and, together with your "funny" "name", and ignoring all my other detailed explanations regarding
your question, ^strong indication, that *you* are just trolling (me) instead.

umm - No - now you are a troll and will be /dev/nulled. You have enough

^^^^^^^^^^^^^^^^^^^

trouble to control yourself, let alone to control anyone else. I may
not be polite, but I am more polite than you.

No, you are not. Your "name" appears to be an accurate description of your state of mind instead.

Score adjusted, F'up2 poster
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

Popping Mad wrote:

On 12/8/25 12:51 AM, Thomas 'PointedEars' Lahn wrote:

Please modify your newsreader

The whole paragraph reads:

Please modify your newsreader's settings so that you post under your real
name. It is considered polite here. (As you can see, you can *also*
include your nickname if you want.)

Your insinuation that I suggested that you modify your *newsreader* (which
may not be possible) is quite disingenuous of you and, together with your "funny" "name", and ignoring all my other detailed explanations regarding
your question, *strong* indication, that *you* are just trolling (me) instead.

I already regret having wasted my precious free time on you. But,
hopefully, my explanations will help someone else who is actually
interested in physics.

umm - No - now you are a troll and will be /dev/nulled. You have enough

^^^^^^^^^^^^^^^^^^^

trouble to control yourself, let alone to control anyone else. I may
not be polite, but I am more polite than you.

No, you are not. Your "name" appears to be an accurate description of your state of mind instead.

Score adjusted, F'up2 poster
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics

Thomas 'PointedEars' Lahn <PointedEars@web.de> wrote:

Popping Mad wrote:
^^^^^^^^^^^
Please modify your newsreader's settings so that you post under your real name. It is considered polite here. (As you can see, you can *also*
include your nickname if you want.)

On 12/7/25 11:35 AM, Thomas 'PointedEars' Lahn wrote:

Thomas 'PointedEars' Lahn wrote:

First of all, it is not good to think of gravitation as a property of an >>>> object as in "Earth *has* gravity". That is NOT how it works. [This is >>>> frequently taught wrong in schools.] Instead, it is an *interaction*
_between_ objects. According to Newton's theory, objects _attract each >>>> other_ because they have non-zero mass. So (according to Newton) it is not
so that "earth does gravity" but that Terra (_Earth_) has non-zero mass, >>>

["non-zero X" simply means "X is not equal to zero"]

and so do other objects (including people like you and me), and so everything
is attracted to everything else. (As the story goes, he realized that an apple
and Earth attract each other in the same as Earth and the Moon attract each

^^^^^^^^^^^
in the same _way_

other, and Earth and the Sun are attracted to each other: Gravitation was >>>> *universal*, not limited to Earth. Thus he could also explain how Kepler's
planetary orbits arose, and predict them.)

You are rambling and you are off topic and not being helpful.

[One way to understand how the magnitude of the gravitational acceleration >>>> at the center of Terra is approximately zero is to consider a test object >>>> with a negligible non-zero mass (a "test mass") there. It will be attracted
gravitationally by all the matter that surrounds it (which also has non-zero
mass) in all directions of space approximately in the same way. So the net
gravitational force on it and its net gravitational acceleration are
approximately zero.]

I understand this in theory. My problem is that if an apple falls from
a tree and hits you in the head, it hurts, despite that it falls from
the tree because of its gravetational attraction to the Earth, not to you.

That is not quite correct. There is also a small gravitational force
between the person and the apple.

That is irrelevant to the point. I think you are writing to hear
yourself.

I didn't ask or address a question about degrees of freedom of all the graviational forces that one can measure in this situation. I was just
trying to understand why the pressure at the center, or as one
approaches the center of a planet increases to incredable levels which
the actual pravatational force once experiences reaches zero.

You are not helping at all in addressing this. You are all over the
place. And your language is difficult to parse. And now you have
detoured to be the first person is 40 years to ask me to change my
news reader's settings.

I'd make an arm chair diagnosis, but it is not worth it. No help is
coming from this post on usenet. I'll need to find a different venue to
see if I can clarify this a little.

And the apple attracts Terra (_Earth_) in
the same way that Terra attracts the apple. However, as the person is at rest relative to Terra, they consider the apple to be the moving object
(this was one of Newton's great insights, building on Galilei's principle of relativity).

However, the strength of the force between the person and the apple is negligibly small compared to that between Earth and the apple because the mass of Terra is so much larger than that of the person:

The strength of the gravitational force between the person (M =~ 70 kg) and the apple (m =~ 10 g) at a distance of 1 m is approximately

F = G M m/r^2
=~ 6.674 ? 10^-11 m^3/(kg s^2) ? 70 kg ? 0.01 kg/(1 m)^2
=~ 4.67 ? 10^-11 N.

The gravitational force between Terra (M =~ 5.97 ? 10^24 kg, R =~ 6371 km) and the apple 1 m away from the former's surface is approximately

F = G M m/r^2
=~ 6.674 ? 10^-11 m^3/(kg s^2) ? 5.97 ? 10^24 kg ? 0.01 kg/
(6371 km + 1 m)^2
=~ 0.0982 N,

i.e. 9 orders of magnitude larger. [This is based on Earth modeled as a point mass in its center. One can see that if one divides that by the mass of the apple, one obtains approximately the average surface gravitational acceleration which therefore is (without air resistance) the same for all objects near the terrestrial surface (Galilei's discovery): ca. 9.8 m/s^2. <https://www.youtube.com/watch?v=oYEgdZ3iEKA>]

Whatever the total force on the apple while it is in free fall, when the apple hits the person, the person's atoms also exert an oppositely directed force on the apple, and vice-versa, that prevents the apple from continuing to fall freely towards the center of gravitation of Terra. When the apple hits the head of the person, due to the sudden increase in pressure on the latter's skin/body and potential damage to the body from that, the nervous system of the person sends electric signals to the person's brain which interprets this as pain :'-)

[If there is no person and no other object, the atoms of the surface of the planet, exerting the same opposite contact forces, prevent the apple from continuing to fall freely. It turns out that is how the apple has a weight (not: mass which it has regardless of gravitation) in the first place.]

So, if you drive in the ocean, as you go deeper the preasure
increases... not because of the gravity between you and water, but

_gravitation_

because of the force of gravity between the water and the earth.

Both; but the gravitational force between the water and the person is,
again, negligibly small compared to the gravitational force between the
water and the rest of the planet. It is, however, the latter force that
the person experiences *partially* as it acts on the surface of the
person's body.

You will still get the bends and be crushed by the pressure.

Yes, if you go to deep, because of what I explain above.

I am pretty sure a similar result would happen in the center of a perfectly >> shperical planet.

It would.

I suppose I am having trouble conceptually with the idea that in the center >> of the planet you would have effectly null force of gravity,

_gravitation_

but there is still a huge amount of pressure from all directions

We are only talking about the *gravitational* force on the object, that
would make *it* move, and *its* acceleration due to gravitation. Consider
a person to be in a person-sized hole filled with air, and the surrounding rock to be prevented by other forces from falling in. That person would _float_ due to the mass of the rock that it is *surrounded* by.

If I was in the center of the earth, my effective weight might be zero,

It would be, approximately.

but the whole weight of the plaent still surrounds me.

That is *how* the above can be explained.

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics

Jim Pennino <jimp@gonzo.specsol.net> wrote:

Popping Mad <rainbow@colition.gov> wrote:

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out) then why is it that the deeper you go into a plant,
like Jupiter for example, the pressure increases and the center of a
star, when it forms, ignites into nuclear fusion?

Pressure and gravity are not the same force.

Pressure is the result of mass that is above you.

Gravity is the result of mass that is around you.

On the surface of the Earth all the mass above you is air and the
pressure because of the weight of that air is about 14 psi. As you
descend into the Earth there is more and more massive stuff above you
pushing you down and the pressure goes up.

As you descend into the Earth the vector sum of gravity changes. You
now have mass above you pulling you up. The gravity thus decreases.

The simplist way to put it is that as you go down, the gravity of the
stuff above you cancels the gravity of the stuff below you but the
weight of stuff above you is always increasing.

Yeah that is the part that is hard to reconcile. It might not be
pulling on YOU but it is pulling on the mass above (and in the center)
around you.

To clarify this a little bit, lets assume you enter a black hole in a
vacume. You might not feel any presure on your body.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics

Ruben Safir wrote:

Thomas 'PointedEars' Lahn <PointedEars@web.de> wrote:

Popping Mad wrote:

On 12/7/25 11:35 AM, Thomas 'PointedEars' Lahn wrote:

Thomas 'PointedEars' Lahn wrote:

First of all, it is not good to think of gravitation as a property of an >>>>> object as in "Earth *has* gravity". That is NOT how it works. [This is >>>>> frequently taught wrong in schools.] Instead, it is an *interaction* >>>>> _between_ objects. According to Newton's theory, objects _attract each >>>>> other_ because they have non-zero mass. So (according to Newton) it is not
so that "earth does gravity" but that Terra (_Earth_) has non-zero mass, >>>>

["non-zero X" simply means "X is not equal to zero"]

and so do other objects (including people like you and me), and so everything
is attracted to everything else. (As the story goes, he realized that an apple
and Earth attract each other in the same as Earth and the Moon attract each

^^^^^^^^^^^
in the same _way_

other, and Earth and the Sun are attracted to each other: Gravitation was >>>>> *universal*, not limited to Earth. Thus he could also explain how Kepler's
planetary orbits arose, and predict them.)

You are rambling and you are off topic and not being helpful.

I don't care what you think because you have no clue what you are talking about.

[One way to understand how the magnitude of the gravitational acceleration
at the center of Terra is approximately zero is to consider a test object >>>>> with a negligible non-zero mass (a "test mass") there. It will be attracted
gravitationally by all the matter that surrounds it (which also has non-zero
mass) in all directions of space approximately in the same way. So the net
gravitational force on it and its net gravitational acceleration are >>>>> approximately zero.]

I understand this in theory. My problem is that if an apple falls from
a tree and hits you in the head, it hurts, despite that it falls from
the tree because of its gravetational attraction to the Earth, not to you. >>

That is not quite correct. There is also a small gravitational force
between the person and the apple.

That is irrelevant to the point.

No, it is not. That gravitation is an interaction between *all* objects is
the key thing to understand here.

I think you are writing to hear yourself.

I think that you are a luser who is having delusions of entitlement.

I didn't ask or address a question about degrees of freedom of all the graviational forces that one can measure in this situation.

You have no clue what "degrees of freedom" are either, obviously.

I was just trying to understand why the pressure at the center, or as one approaches the center of a planet increases to incredable levels which
the actual pravatational force once experiences reaches zero.

"incradable"? "pravational"? Learn to read, then write. The sentence
above is incoherent babble, even after correcting typos.

You are not helping at all in addressing this.

I was, you are just too self-absorbed to recognize it.

You are all over the place.

It is a complex problem for which there is no one-line explanation. I have answered your question correctly and broke the problem down for you as best
as I could. It is not my fault if you are not smart enough to grasp it
because you are too lazy to read.

And your language is difficult to parse.

Not my problem. My answer is correct, and my language is exact. I even
made considerable efforts to make it easier to understand after the fact.
But you come here with a consumer attitude, and there is no pleasing you.

No help is coming from this post on usenet.

Usenet is not a right, stupid.

And trim your quotes to the relevant minimum next time.

Score adjusted.

<http://www.catb.org/~esr/faqs/smart-questions.html>
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics

Ruben Safir <mrbrklyn@panix.com> wrote:

Jim Pennino <jimp@gonzo.specsol.net> wrote:

Popping Mad <rainbow@colition.gov> wrote:

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out) then why is it that the deeper you go into a plant, >>> like Jupiter for example, the pressure increases and the center of a
star, when it forms, ignites into nuclear fusion?

Pressure and gravity are not the same force.

Pressure is the result of mass that is above you.

Gravity is the result of mass that is around you.

On the surface of the Earth all the mass above you is air and the
pressure because of the weight of that air is about 14 psi. As you
descend into the Earth there is more and more massive stuff above you
pushing you down and the pressure goes up.

As you descend into the Earth the vector sum of gravity changes. You
now have mass above you pulling you up. The gravity thus decreases.

The simplist way to put it is that as you go down, the gravity of the
stuff above you cancels the gravity of the stuff below you but the
weight of stuff above you is always increasing.

Yeah that is the part that is hard to reconcile. It might not be
pulling on YOU but it is pulling on the mass above (and in the center)
around you.

The gravity felt is the vector sum of the total of all the Earth around
you, which is zero.

The pressure felt is the simple sum of all the Earth that is above you,
which is a big number.

I'm assuming you know the difference between a vector and simple sum.

To clarify this a little bit, lets assume you enter a black hole in a
vacume. You might not feel any presure on your body.

--
penninojim@yahoo.com
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics

Jim Pennino wrote:

Popping Mad <rainbow@colition.gov> wrote:

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out) then why is it that the deeper you go into a plant,
like Jupiter for example, the pressure increases and the center of a
star, when it forms, ignites into nuclear fusion?

Pressure and gravity are not the same force.

I appreciate your wanting to put it in simple language, but let us not oversimplify.

Pressure is NOT a force. It is the result of a force acting on a surface:

P = F/A.

"Gravity" is NOT a force. _Gravitation_ is an interaction between objects.
It can be described by a force acting between those objects, and that
(Newton's theory) is the simplest correct explanation here that one can come
up with.

Pressure is the result of mass that is above you.

This pressure is *produced* by gravitation (and is therefore fully called "gravitational pressure" in astrophysics); in the theory that we chose to
use (see above), it is the *result* of the gravitational force acting
between the parts of Earth.

If we would imagine a person within Earth, this person just happens to be *between* those parts, and so the gravitational force between the opposite parts is pushing "down" on them (towards the center of gravitation to which
all the parts are mutually attracted because they are attracted to each
other). This produces the pressure e.g. on the skin of that person (which
is their surface, with the area A_person):

P(r) = F_G(r)/A_person =~ G M(r)/(r^2 A_person).

[As we are approaching the center of gravitation, M(r) increases and r decreases, so P(r) increases.]

Gravity is the result of mass that is around you.

Therefore, this is wrong. However, because gravitation is an interaction between *all* objects (as I pointed out several times now), it is also an interaction between Earth and the person; and, as you say correctly, in our thought experiment the matter of Earth is all around the person. So in the center of Earth, the gravitational force in one direction cancels the one in the opposite direction (that is how that point is *defined*).

On the surface of the Earth all the mass above you is air

Correct.

and the pressure because of the weight of that air is about 14 psi.

I am not well-versed in this non-SI unit, but IIRC the standard atmospheric pressure is ca. 1023.25 hPa.

As you descend into the Earth there is more and more massive stuff above you pushing you down and the pressure goes up.

Correct, in a sense.

As you descend into the Earth the vector sum of gravity changes. You
now have mass above you pulling you up. The gravity thus decreases.

Correct, in a sense. As I point out above, there is no quantity called "gravity". What decreases here is the magnitude/norm of the net
gravitational force/acceleration, as calculated. (It does not make sense to speak of the increase or decrease of a vector.)

The simplist way to put it is that as you go down, the gravity of the
stuff above you cancels the gravity of the stuff below you but the
weight of stuff above you is always increasing.

This wording can be confusing because the person is weightless in the center
of gravitation.
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics

Jim Pennino wrote:

Popping Mad <rainbow@colition.gov> wrote:

If the center of the earth would have 0 force do to gravity (all the
Gravity cancels out) then why is it that the deeper you go into a plant,
like Jupiter for example, the pressure increases and the center of a
star, when it forms, ignites into nuclear fusion?

Pressure and gravity are not the same force.

I appreciate your wanting to put it in simple language, but let us not oversimplify.

Pressure is NOT a force. It is the result of a force acting on a surface:

P = F/A.

"Gravity" is NOT a force. _Gravitation_ is an interaction between objects.
It can be described by a force acting between those objects, and that
(Newton's theory) is the simplest correct explanation here that one can come
up with.

Pressure is the result of mass that is above you.

This pressure is *produced* by gravitation (and is therefore fully called "gravitational pressure" in astrophysics); in the theory that we chose to
use (see above), it is the *result* of the gravitational force acting
between the parts of Earth.

If we would imagine a person within Earth, this person just happens to be *between* those parts, and so the gravitational force between the opposite parts is pushing "down" on them (towards the center of gravitation to which
all the parts are mutually attracted because they are attracted to each
other). This produces the pressure e.g. on the skin of that person (which
is their surface, with the area A_person):

P(r) = F_G(r)/A_person =~ G M(r)^2/(r^2 A_person).

[As we are approaching the center of gravitation, M(r) increases and r decreases, so P(r) increases.]

Gravity is the result of mass that is around you.

Therefore, this is wrong. However, because gravitation is an interaction between *all* objects (as I pointed out several times now), it is also an interaction between Earth and the person; and, as you say correctly, in our thought experiment the matter of Earth is all around the person. So in the center of Earth, the gravitational force in one direction cancels the one in the opposite direction (that is how that point is *defined*).

On the surface of the Earth all the mass above you is air

Correct.

and the pressure because of the weight of that air is about 14 psi.

I am not well-versed in this non-SI unit, but IIRC the standard atmospheric pressure is ca. 1023.25 hPa.

As you descend into the Earth there is more and more massive stuff above you pushing you down and the pressure goes up.

Correct, in a sense.

As you descend into the Earth the vector sum of gravity changes. You
now have mass above you pulling you up. The gravity thus decreases.

Correct, in a sense. As I point out above, there is no quantity called "gravity". What decreases here is the magnitude/norm of the net
gravitational force/acceleration, as calculated. (It does not make sense to speak of the increase or decrease of a vector.)

The simplist way to put it is that as you go down, the gravity of the
stuff above you cancels the gravity of the stuff below you but the
weight of stuff above you is always increasing.

This wording can be confusing because the person is weightless in the center
of gravitation.
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

On 12/13/25 11:48 PM, Jim Pennino wrote:

I'm assuming you know the difference between a vector and simple sum.

thank you for clarifying this
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