Is Space Zero Gravity (0 G)

[Darky]

Hi all,

 

In school, we've been taught that while in space; we can float around and have no weight at all. You can float around, yes. Having no weight at all is an entirely different story.

 

The formula for weight is what? : Mass * Gravity.

 

Now if I'm to prove my point, I need to first prove that space isn't Zero-G (Zero Gravity). Let's for instance take the sun; all planets are in a orbit around it, due to? Yes, Gravity! So if for some reason you jumped out of your rocket, you won't leave the Galaxy. You'll just obtain an orbit around the sun unless you're accelerating at high levels. So this is confirmed that space has gravity. Gravity keeping the planets in orbit. However, that's only the big picture, as stars aren't the only thing giving gravity. It has been proven that anything with mass has a gravitational pull. For example, Earth. If space had Zero-Gravity, why the nincapoops (my polite way of saying '****') is the moon in a fixed orbit around Earth. Yes the gravitational pull is low, but there is always gravity in space.

 

Not only does this correct our learning on gravity and tell us the truth about space, it also proves that even in space, you have a weight, just extremely low. Zero-Gravity may not be the right word to describe space - so let's use Micro-Gravity instead.

 

Regards and hopes that you take this positively and use this to carry on your projects,

 

Darky.

[sanctus]

Noone says that in space there is no gravity. The force of gravity is F=G*m1*m2/r^2 therefore gravitational forces just decreases reaching zero at infinity--> hence never.

Astronauts in orbit as well as satellites are just in free fall...

[Darky]

Noone says that in space there is no gravity. The force of gravity is F=G*m1*m2/r^2 therefore gravitational forces just decreases reaching zero at infinity--> hence never.

Astronauts in orbit as well as satellites are just in free fall...

Hi,

 

Thanks for adding in the maths.

 

Check most sources/books and they prefer to gravity in space as Zero-G (which is clearly wrong).

[A-wal]

Anything in free-fall experiences (almost) 0G.

 

What you feel as your 'weight' is actually the ground pushing you 'up'. The gravity pulling you down is more or less uniform so you don't feel it but the electro-magnetic force pushing you up is concentrated on your points of contact with the ground. That's why it's more comfortable to spread the points of contact by sitting or laying down. ;)

[Darky]

Anything in free-fall experiences (almost) 0G.

 

What you feel as your 'weight' is actually the ground pushing you 'up'. The gravity pulling you down is more or less uniform so you don't feel it but the electro-magnetic force pushing you up is concentrated on your points of contact with the ground. That's why it's more comfortable to spread the points of contact by sitting or laying down. ;)

Hi,

 

You're correct on the most part. However, there is nothing known as 0G. It's 0.0001G at certain times - never 0G. 0G is theoretically only possible when two objects of the exact same mass having the same gravitational pull and something between them (lets call it T1). In such case, T1 would be experiencing 0G due to forces cancelling each other out. However, this is just a theory, and I suppose there may be a tiny gravitational pull of something near to those two objects (near as in the grand-scale of space) making it 1E-2000 G. That is : 10^(-2000) but never, ever, 0G.

[A-wal]

Correct, nothing ever truly experience 0G but it can very easily be so low as to be negligible.

[JMJones0424]

I think the misconception here is that the "g" in "zero g" is not a measure of the force of gravity.  A less confusing terminology would be zero g-force, where "g" is equal to the average acceleration of gravity at sea-level.  "G" is a physical constant.  Zero g does not mean the absence of gravity, but rather the absence of acceleration.  When you are in free-fall in the absence of any resistance to falling and the gravitational field can be approximated as uniform (i.e. you aren't near a black hole), then you experience zero g.  It is incorrect to call this zero G or zero gravity, as neither can exist in the universe we inhabit.

Edited April 27, 2017 by JMJones0424

[JMJones0424]

It is likewise inaccurate to refer to this as "micro G" or "micro gravity".  G doesn't change, so "micro G" is meaningless, and "micro gravity" only exists very much further away than any human construct has every reached.  It is more accurate to refer to this as "micro g", and even more accurate to refer to this as "micro g-force".

[Super Polymath]

Where there's space, there's time. However, where there is no mass, there can be no gravity.

 

Most of the universe is without mass, the vacuum (a term meaning that space would disperse and therefore suck all of the oxygen out of a spacecraft that generates artificial oxygen to sustain its crew in the event of a haul breach) emptiness, there's nothing there other than the continuity of time and emptiness, that emptiness being unoccupied space. 

Edited April 27, 2017 by Super Polymath

[sanctus]

Polymath, again if you create a completely empty universe with only 1 atom somewhere and you are the only other collection of atoms in that universe, independently of where you are you will will eventually be pulled to that atom. At least as long we do not consider the expansion of the universe

[Super Polymath]

Polymath, again if you create a completely empty universe with only 1 atom somewhere and you are the only other collection of atoms in that universe, independently of where you are you will will eventually be pulled to that atom. At least as long we do not consider the expansion of the universe

I'm pretty sure spatial distortions have a gravitational range, or else there'd be no particle horizon limiting the observable universe, no event horizons inhibiting our ability to observe singularities, everything would just be a singularity, which is what inflation is all about. You're right though, even photons may create infinitesimal imprints in the space-time, at least according to Rainbow_gravity_theory - also, special relativity would say so, though it wouldn't say wavelength varies the mass of a photon like RGT. 

Edited April 27, 2017 by Super Polymath

[JMJones0424]

I'm pretty sure spatial distortions have a gravitational range

What is that range?  Imagine a universe consisting of two atoms.  No matter the distance that separates these atoms, gravitational attraction between these two atoms exist.  It may very well be negligible, but it is never zero.

 

 

 

or else there'd be no particle horizon limiting the observable universe

 I understand the limit of the observable universe to be a function of the age of the universe, not a function of the limitation of gravitation.

 

 

 

no event horizons inhibiting our ability to observe singularities

 The event horizon prevents our observation of a singularity because there exists no path for photons to take inside the event horizon that reach areas outside the event horizon.  This doesn't seem to me to provide any evidence whatsoever that "spatial distortions have a gravitational range"

 

 

 

everything would just be a singularity, which is what inflation is all about

  I disagree on both accounts. Edited April 28, 2017 by JMJones0424

[Super Polymath]

 

What is that range?  Imagine a universe consisting of two atoms.  No matter the distance that separates these atoms, gravitational attraction between these two atoms exist.  It may very well be negligible, but it is never zero.

 

 

 

 I understand the limit of the observable universe to be a function of the age of the universe, not a function of the limitation of gravitation.

 

 

 

 

 The event horizon prevents our observation of a singularity because there exists no path for photons to take inside the event horizon that reach areas outside the event horizon.  This doesn't seem to me to provide any evidence whatsoever that "spatial distortions have a gravitational range"

 

 

 

 

  I disagree on both accounts.

 

Consider the fact that the surface area of the event horizon depends upon the mass of the interior singularity. 

 

Space folds only so much. The curvature stops at set point.

 

If we have two atoms, that's a whole other cosmological constant. 

 

How else would the Voyager craft's ability to coast to Alpha Centauri off of the acceleration it had achieved when the rocket fuel had expired? Leaving earth orbit has a demonstrably substantial negating gravitational impact. Only thing that could slow it down would be impacting pebbles of space debris. 

 

Unless you would like to straw-hat non-apparent infinitesimals. In which case, subjectivity a science does not make. The leading experts predicted that Voyager would reach Alpha Centauri within a given & set duration, so argue that point with them, not me. It does not seem obvious that the Voyager's course would be effected by the earth's spatial imprint in the slightest. Thus, the earths gravitational pull must have a finite range. Even if the earth's supposedly infinite gravitational range were being neutralized by every atom in the universe, you're aiding my point in that this neutralization of gravitational forces still equates to zero gravity. Moreover that reality does, in fact, have a mathematical constant that asserts empty space can be outside of the range of any gravitational influence given inflation causes gaps that even quark-gluon plasma cannot fill entirely. Because the totality of all matter & energy in the cosmos, even when broken down into its smallest masses (such as neutrinos, gluons & quarks), is finite whereas flat space is exponentially infinite. 

 

Even when given imperfect flatness or, in the case of rainbow gravity, infinite matter & energy still being dispersed by a dynamically or temporarily inflating infinitude of space, neutralization of gravitational forces still equates to zero G. This presents a logical loophole one cannot escape. The reality is, the vacuum of space has no gravity. 

Edited April 28, 2017 by Super Polymath

[JMJones0424]

Consider the fact that the surface area of the event horizon depends upon the mass of the interior singularity. 

 

Space folds only so much. The curvature stops at set point.

 

If we have two atoms, that's a whole other cosmological constant. 

 

How else would the Voyager craft's ability to coast to Alpha Centauri off of the acceleration it had achieved when the rocket fuel had expired? Leaving earth orbit has a demonstrably substantial negating gravitational impact. Only thing that could slow it down would be impacting pebbles of space debris. 

Neither Voyager is coasting to Alpha Centauri as is clearly described by the source you cite.  However, both of these crafts achieved a velocity necessary to escape the Sun's gravitational field.  What do you mean when you say that "Leaving earth orbit has a demonstrably substantial negating gravitational impact"?  I know of no reason to believe your claims are accurate.

 

I did not at all reference the cosmological constant.  What I did reference was the fact that two atoms, even if they are the only existing things in a hypothetical universe, can measure gravitational attraction between them, regardless of the distance, according to our understanding of gravity.  You have failed to provide the distance at which "spatial distortions have a gravitational range" and I'm pretty sure that you are making this up.  You could provide evidence to support your claim, but you have failed to do so.

Edited April 29, 2017 by JMJones0424

[Super Polymath]

Neither Voyager is coasting to Alpha Centauri

That article says they're both going to be flying for another 90,000 years. These are probes detached from their rockets, there's nothing needed nor available to continuously defy outside mass. The further one gets from an object with mass, the weaker that object's gravitational effects appears. And if there is no gravitational effects, acceleration is no longer needed. By that very reasoning, the only way to "coast" (maintain velocity without propulsion aka rocket fuel in the case of the Voyager) is to be liberated from any gravitational effects. You've failed to define if there even is an infinitesimal influence and in fact have said that the craft has escaped the sun's gravitational field, so by your very own reasoning the sun's gravitational field has a finite range. Velocity achieved can still be decreased by any gravitational field, so you see you've yet to address distance's relation. Where did you read that our understanding of gravity states that distance isn't even a factor?? Space is, by very definition, the only numerical measure of distance, is it not??

 

Think of spatial imprintation as curvature, any imprint such as clay when you make a hole by sticking a needle in it, has a finite area, right? Given our cosmological constant, the finite spatial imprint of two quarks means that a universe with two quarks only will not attract those quarks if the distance exceeds the range of their spatial imprint. If spatial imprintation has no range, than the voyager could not move as space would not exist. 

Edited April 29, 2017 by Super Polymath

[JMJones0424]

The article does not support your claim.  Why do you suppose that after having been propelled into the void, they must undergo acceleration in order to continue upon their path?  

 

You have failed to answer my question.  You claimed, "I'm pretty sure spatial distortions have a gravitational range"  What is this range and why are you pretty sure that it exists?  I am not interested in addressing fabrications.  You made a claim, support it.

[Super Polymath]

You have failed to provide the distance at which "spatial distortions have a gravitational range"

Any point of liberation. An accelerated craft can be decelerated by gravitational attraction by, say, the mass of a large asteroid, can it not?