Introduction

Using a and b positions on a length as base with density condensation power as exponentiation along with division to determine number of times a length can be proportionally divided into smaller parts toward position a (i.e. 0) from position b. This kind of creates a case of uniform density condensation.

I connected this formula to gravity because of the way gravity of this Earth gravitates things toward it - meaning mass and energy surely could be dense deep into the core or something like that if so. The formula visualizes that in a simple way, 1-dimension only.

Formula

1. L / 2^x = PL

2. PL * 2^x = L (to double-check #1, I called it density inflation)

Formula Definition

L = length

/ = division

* = multiplication

2^x = a and b position on length (beginning and ending) as objects (i.e. 2) with density condensation power as exponentiation.

PL = Part of length

Equation Examples

I have attached a picture I developed to this post, illustrating what I'm talking about.

Question

How to make formula more useful in understanding areas of gravity if it is applicable, that is?

What is density condensation?

4 hours ago, tylers100 said:

I have attached a picture I developed to this post, illustrating what I'm talking about.

That’s not helping me understand it.

We already have an equation for gravity. Why not just use that?

1 hour ago, swansont said:

What is density condensation?

That’s not helping me understand it.

We already have an equation for gravity. Why not just use that?

"What is density condensation?"

More density; denser.

Gravity Equation

Okay. I'll try to spend some time with already existing equations to understand better.

Edited December 31, 2025Dec 31 by tylers100
Clarification.

9 minutes ago, tylers100 said:

More density; denser.

We usually just say density increase

17 hours ago, swansont said:

We usually just say density increase

Gravity Equations

I'll make some time to re-look into existing gravity equations on wiki and etc.

Meanwhile

The 1-dimensional geometry with increased density is something I think may be important as it may could be responsible for driving the gravity acceleration here on Earth. I mean, yeah total mass as baseline for gravity strength since proportionality and all but that maybe only gives a pulling constant rate of falling or constant rate of gravitating toward whereas gravity acceleration which happen to be around 9.8 metres per second ^ 2 meaning objects falling with increased rate over time because of: increased density might does that - effective strength of gravity, I think.

But honestly, I'm still learning about distance, time, and speed equations and how the geometry with increased density can factor into all of that.

The geometry with increased density as I'm currently talking about, could be what "uneven distribution of mass" that Einstein referred to. Maybe not.

"Gravity is described by the general theory of relativity, proposed by Albert Einstein in 1915, which describes gravity in terms of the curvature of spacetime, caused by the uneven distribution of mass."

https://en.wikipedia.org/wiki/Gravity

Attach

I have attached another picture I developed that may help you understand what I'm talking about.

Also, I realized on my previous post along with 1st attached picture titled, "1-Dimensional Geometry with Density Condensation and Inflation", didn't include P variable referring to density increase so I have included P this time.

Edited December 31, 2025Dec 31 by tylers100
Added reference link for the uneven distribution of mass quote.

2 hours ago, tylers100 said:

The 1-dimensional geometry with increased density is something I think may be important as it may could be responsible for driving the gravity acceleration here on Earth. I mean, yeah total mass as baseline for gravity strength since proportionality and all but that maybe only gives a pulling constant rate of falling or constant rate of gravitating toward whereas gravity acceleration which happen to be around 9.8 metres per second ^ 2 meaning objects falling with increased rate over time because of: increased density might does that - effective strength of gravity, I think.

Gravity is also the reason for orbits of celestial bodies, and closed orbits require it follow the 1/r^2 pattern of Newtonian gravity, and depend on mass, not density. There’s not all that much room leftover for any other effect, after what Newton or GR cover.

The objects feeling gravity are generally outside of each other. You can’t distinguish densities from just the gravitation.

Gravity ( Newtonian ), is a 1/r² interaction between two centers of mass.
IOW, You would feel the same gravitational 'tug' from 1000 tons of feathers as you would from 1000 tons of lead, at a distance of 1000 km. frm their center of mass. The density of the feathers and lead don't warrant consideration at all.

Is you are using density in some other non-standard way, you are completely baffling us as to what you mean, and, most likely also confusing yourself. Try explaining in the simplest terms possible, without the pictures and pseudo-technical wording.
What does "base with density condensation power as exponentiation along with division" or "uniform density condensation" even mean ???

1 hour ago, MigL said:

Gravity ( Newtonian ), is a 1/r² interaction between two centers of mass.
IOW, You would feel the same gravitational 'tug' from 1000 tons of feathers as you would from 1000 tons of lead, at a distance of 1000 km. frm their center of mass. The density of the feathers and lead don't warrant consideration at all.

Is you are using density in some other non-standard way, you are completely baffling us as to what you mean, and, most likely also confusing yourself. Try explaining in the simplest terms possible, without the pictures and pseudo-technical wording.
What does "base with density condensation power as exponentiation along with division" or "uniform density condensation" even mean ???

Simplest Terms

The formula tells how many times to divide a length by 2, into smaller parts toward 0.

These smaller parts toward 0, are increased density.

Confusion

Now, I'm a bit not sure because maybe I'm confused. For some reason, these smaller parts look like increased density to me - because they are so packed closely to each other toward 0. Maybe I confused myself between visual and mathematics.

2 minutes ago, tylers100 said:

The formula tells how many times to divide a length by 2, into smaller parts toward 0.

To what end? Why is this necessary?

2 minutes ago, swansont said:

To what end? Why is this necessary?

Maybe I think to potentially see if can understand and see if can conceptually create an artificial gravity - a plating on ground via density method, for onboard spaceflight and exploration. Kind of alternative, maybe more better than rotating version.

But you guys pointed out density doesn't matter.

12 minutes ago, tylers100 said:

Maybe I think to potentially see if can understand and see if can conceptually create an artificial gravity - a plating on ground via density method, for onboard spaceflight and exploration. Kind of alternative, maybe more better than rotating version.

If it’s from mass, there’s nothing artificial about it, but you need a tremendous amount of it. The only impact from increased density is making the mass occupy a smaller volume, but there a limit to how much we can do this; the densest metal (osmium) is less than 10x the density of aluminum.

The only 'density' that I would consider would be in the case of GR, where you could conceptualize a 4dimensional grid, and where higher Gravity would be manifested by curvature of the grid lines, and closer packing, or higher density, of the grid linesnear gravitating sources.
Alternatively, thinking along the lines of Faraday's field lines as applied to Newtonian Gravity, one notes that the 3dimensional density of field lines increases nearing a source of Gravity, but this leads to discounting the large effects of time on Gravity, and only gives (reasonably ) valid results outside the gravitating body ( field lines will still originate at the CoM ).

This comes down to a ( non-mathematical ) visualization of Gravity, which can sometimes be useful for understanding the Physics of a situation, if not the Mathematics, so I use it myself, and encourage it.
Your examples only deal with one dimension, but, at some point, you need to scale this to 3 or 4 dimensions, so you need to clarify 'density' ( of what exactly ? )

Several of the answers above have provided excellent clues into gravity vs density but let's refine that with mass density.

Lets do a couple thought experiments and for simplicity we will keep the total mass constant in each case. Lets set at 1 solar mass ( mass of our sun).

Case 1) spread that mass out evenly everywhere where no coordinate has greater mass than any other coordinate. No matter which location you choose you can state it's the effective center of mass.

Gravity in the above case is zero everywhere. It does not matter what density of mass each coordinate has it could be as dense as one can fathom. As long as the mass density is uniform everywhere Newtons Shell theorem applies.

Case 2) you have one region with higher mass density than other regions (anistropic distribution) Now you have a clear cut center of mass as the center of that region is clearly a higher density than the surrounding regions. Now you have gravity where the difference follows Newtons laws of gravity.

Now Case 3 is rather special take that one solar mass above and let's assume it has the same volume as our sun.

The strength of gravity one measures depends on the radius from the center of the sun. If however you collapse the radius of the sun below its Schwartzchild radius it becomes a blackhole. However the mass does not change.

The radius where you can measure gravity has decreased so at the event horizon the strength is such that nothing can escape.

Yet the force of gravity is still the same if you were to measure gravity from Earth.

Hope that helps remember at no point did of the 3 scenarios change the total mass. It is the distribution of mass that leads to gravity and the radius from the center of mass.

Edited January 2Jan 2 by Mordred

In the first part of this lecture Alan Guth explains problems with applying Newton's law of gravity, including the shell theorem, to an infinite distribution of mass.

Lecture 6: The Dynamics of Homogeneous Expansion, Part II | The Early Universe | Physics | MIT OpenCourseWare

In particular, the gravity in this case is not necessarily zero.

I recall that video always enjoyed Guths lectures as well as articles. Static vs inertial in terms of different observers can often give surprising results. Guth does an excellent job demonstrating some of the effects in that video

More Density = More Mass Fill in

I think the formula; l / 2^xP in geometry terms specify how increased density can be created thus allowing more mass to fill in then maybe a bit more gravity?

Questioning and Self-Doubt

But I'm a bit confused now because the formula includes division which divides length into half by 2 number of times until 0 is reached. It divides whereas increased density should be multiplicated? I'm not sure.

Limitation

But there is a catch; once the formula if it such one at that; l / 2^xP - specifically P (increased density) make length reach 0, that is when a limitation is reached for increased density.

Attach

Visual version titled, "Increased Density Transition Across Dimensions"

2 hours ago, tylers100 said:

I think the formula; l / 2^xP in geometry terms specify how increased density can be created thus allowing more mass to fill in then maybe a bit more gravity?

1/2^xP doesn’t look to be a valid formula. xP needs to be dimensionless, which matters even if there was some physical basis for it making any sense.

19 hours ago, tylers100 said:

increased density can be created thus allowing more mass to fill in then maybe a bit more gravity?

More mass always = more gravity. Unless you are adding mass from somewhere, that quantity (and its gravitational effects) will remain constant regardless of the density, which only relates to the volume that mass occupies.

1 hour ago, npts2020 said:

More mass always = more gravity. Unless you are adding mass from somewhere, that quantity (and its gravitational effects) will remain constant regardless of the density, which only relates to the volume that mass occupies.

Under that condition, because you are assuming the same r. But not in general.

Re-assessment of the Formula

Formula Variables:

L = length

2 = proportion

P = density

PL = part of length

Formula:

L / 2^xP = PL

or

L / 2^P = PL

The formulas is seemingly a mostly geometry formula that divides a length into smaller parts until 0. That is when a limitation is reached. Maybe the formula has some uses for understanding a potential limitation when dividing in proportion way because of the 2.

Equation Example

1 L / 2^31 P = 0

Once reached 0, that is when a limitation is realized. Then a decrement of density power by 1:

1 L / 2^30 P = 0.000000001 (this had a nearly maximum number of density power to determine a minimum number of available length)

That could be a good thing, because without limitation - how does one begin and end at what point?

The 2 is for proportion reason, a part of symmetry I later considered.

Useful or Not

I'm not sure whether if the formula is useful or not, just the realization of the limitation in length due to 2 proportion in 1-dimensional geometry of what I was doing.

Previously, I thought the formula divides length into smaller parts with decrementing lengths toward 0 thus appearance of density of that. But now, it might be possible that these divided parts in 1-dimensional length according to the formula could be still of uniform case of decrementing division because of the 2 proportion thing.

I think the formula determines from top down to bottom, not bottom to up.

Premise

Maybe the premise was that if a 1-dimensional diameter or radius (e.g. singular focus on a planet or moon) factored into the formula then could determine proportional limitation and gravity yield in that. And after all, the gravity acceleration 9.8 m per second ^ 2 is approximated to be at every location on this planet Earth thus a possible 1 dimensional understanding could achieve that. Maybe not. Maybe this is just another one of my struggles to understand gravity in a different way.

1 hour ago, tylers100 said:

The formulas is seemingly a mostly geometry formula that divides a length into smaller parts until 0.

No matter how many times you divide by 2, you will never reach 0

Do you have any numerical example of applying this to an actual, physical situation, such that it can be tested?

1 hour ago, swansont said:

No matter how many times you divide by 2, you will never reach 0

Do you have any numerical example of applying this to an actual, physical situation, such that it can be tested?

Earth's Radius

I tried the following:

Earth's 12742 km diameter converted to 12742000 metres then divided into half (6371000 metres) to obtain radius:

6371000 metres / 2^53 density = 0.000000001

So the 53 density is nearly maximum number. But now, I don't know how useful that 53 density could give answer or something useful about gravity. And the result, I'm not sure but possible time? Similar to the time formula:

time = distance / speed

Maybe Invalid Formula After All

Maybe the formula I made is similar to the time formula? Somehow I express some doubts about the formula I made and its usability. Maybe it is invalid after all.

What’s the physical significance of 53?

1 hour ago, swansont said:

What’s the physical significance of 53?

Density of Itself-Pulling Speed Maybe

Density of radius's itself-pulling speed, maybe. I'm still a bit unsure, tbh. The density power was specified based on my guesswork in order to obtain the most minimum length number ever possible before hitting nearly 0 without really thought about what the density is of what exactly.

I think back to what @MigL said about density of what exactly. So, maybe I'm nearing on the speed?

3 hours ago, tylers100 said:

6371000 metres / 2^53 density = 0.000000001

So the 53 density is nearly maximum number

... for your calculator?