<p>While the question is rather hard to follow, the formula simply doesn't work.</p>

<p>For Mercury the formula using a naive application of the numbers predicts:
four layers</p>

<pre><code>Orbital Velocity (km/s)        47.4
Density (kg/m3)              5427
Orbital Inclination (degrees)   5.1
n                            -328.65368
Aphelion (10**6 km)            69.8
Layers                          4
</code></pre>

<p>That is clearly not true, since Mercury has no atmosphere. You can vary the formula, by choosing different units, or taking the inclination relative to a different plane, but no matter what I try, can't get the formula to give a sensible value for all the planets. </p>

<p>The formula is inconsistent, since you generally can't add velocity (km/s) to sqrt(density)*inclination (kg^0.5 degrees/m^(1.5)) as these quantities have different units. It rarely makes sense to add quantities with different units.</p>

<p>There is, moreover, no reason to suppose that the orbital characteristics of a planet, and its density would have any effect on the number of layers in the atmosphere, which is not a very well defined quantity anyway.</p>

<p>Sorry but this formula is entirely wrong. </p>

CC BY-SA 4.0
2018-11-13T23:38:04.213Z
I would love to see a list of the actual atmospheres. The one at the site does not give an accurate answers. Earth says 1. Also, you'll ave to notice that the degrees are calculated by earth as well.
thexiv
CC BY-SA 4.0
2018-11-13T23:43:15.983Z
Okay, this is the thing: I think that the surface bars are wrong on that sheet. I think Mercury has 4.
thexiv
CC BY-SA 4.0
2018-11-13T23:48:32.977Z
So you're saying that there is no way that the gravity of the planet keeps the gases there. That's inane. No offense. The solar winds during orbit would be difficult not to append with, as well. As for the differing metric units, you can assign different weights to different areas. Altitude and pressure on liquid. So therefore I cannot fully find your answer comprehensive. Thus i voted it down.
thexiv
CC BY-SA 4.0
2018-11-14T14:26:44.550Z
@thexiv: That's not what JamesK is saying. He is saying that your formula is a bad predictor for the number of atmospheric layers, based on an example that has no atmosphere.
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-11-14T14:29:37.923Z
I asked an astronaut. I'm waiting on a reply for Venus and Mars and mercury. I highly doubt Venus has 92 layers and Earth has 1. I was taught we had 4. I'm saying we don't have a good example except for earth and the moon (which has 4 and 0 respectively)
thexiv
CC BY-SA 4.0
2018-11-14T19:00:51.703Z
@AtmosphericPrisonEscape Here is a table of my work.
thexiv
CC BY-SA 4.0
2018-11-14T21:46:19.490Z
I have absolutely no idea what you are talking about.  You seem to mix pressure (in bars) with "layers".  I maintain the last line of my answer. Your formula is entirely wrong.
James K
CC BY-SA 4.0
2018-11-14T23:28:16.853Z
I've edited to use your definition of inclination. However it still doesn't give reasonable values for Mercury or the other planets.
James K
CC BY-SA 4.0
2018-11-13T23:28:18.307Z
2018-11-14T23:26:07.313Z
2018-11-14T23:26:07.313Z
James K
<p>This math states, that by multiplying the gravitational pull (we assume square of density of the planet), by its inclination (being how high and out it goes) and subtracting that loop from the velocity in a slope formula that measures to the borderline of being horizontal on the sun's x-axis, most parallel with the solar system's other planets, relegating the offset in order as to consign the preferred balance of the planet with the Sun. We can take then from the furthest point out once the Sun can lesser not take from the product and sum and difference of the clouds on the planet so that we may therefore continue on and calculate the priori aft dividing by 100 to delineate in a reciprocity to the true count, the atmospheric makeup of which the count of surface bars is.</p>

<p>This site is were to get the constants for the planetary facts you will need to use the formula.</p>

<p><a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/" rel="nofollow noreferrer">https://nssdc.gsfc.nasa.gov/planetary/factsheet/</a></p>

<p>n = Orbital Velocity - root of density * inclination to sun<br>
round((Apogee - n)/100) = Planet's Atmospheric layer count;</p>

<p>This is an easy formula to figure out how many layers of atmosphere your favorite planets have on them. It's preliminary because I don't have the satellite data from each planet to synthesize with my data from this formula. </p>

<p>If this is wrong, please tell me. If it is not, I suppose tell me. But I am only in search criticism and laud.</p>

<pre><code>NASA Deg.   Real Deg.   Planet + Rank   Density Orbital Vel.    Apogee  Bars (Atmospheres)

17.2        15.3        Pluto+4         2095    4.7             7375.9  6.6 Bars

7           5.1         Mercury+3       5427    47.4            69.8    2.5 Bars

1.8         3.7         Neptune+2       1638    5.4             4545.7  44.01 Bars

1.3         3.2         Jupiter+1       1326    13.1            816.6   34.13 Bars

1.9         -           Mars -0         3933    24.1            2492.2  2.1 Bars

2.5         -0.6        Saturn -1       687     9.7             1514.5  15.07 Bars

3.4         -1.5        Venus -2        5243    35.0            108.9   2.52 Bars

0           -1.9        Earth -3        5514    29.8            152.1   3.22 Bars

0.8         -2.7        Uranus -4       1271    6.8             3003.6  31.06 bars
</code></pre>

CC BY-SA 4.0
2018-11-13T19:42:13.607Z
This looks weird, where is this from? Atmospheres are always continuous, any equidistant layer numbers we assign to them must be arbitrary and probably the result of some computer simulation technique.
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-11-13T19:46:54.477Z
@AtmosphericPrisonEscape  That's not true, even on Earth.  We often see sharp shear layers between upper and lower wind patterns, for example, with different densities and humidities.
Carl Witthoft
CC BY-SA 4.0
2018-11-13T19:48:03.530Z
But.. where did you get that formula from?   It looks highly unlikely to be valid
Carl Witthoft
CC BY-SA 4.0
2018-11-13T19:48:33.720Z
@CarlWitthoft: I said "equidistant". Look at the formula provided by OP. This very much looks like the discretization of atmospheric structure based on oribital velocities, that's why it's weird.
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-11-13T19:50:39.663Z
I actually created it myself. I was just tinkering with numbers and came up with that. It seemed to get most of them so I assume there's something wrong with other bases.
thexiv
CC BY-SA 4.0
2018-11-13T19:51:30.543Z
@AtmosphericPrisonEscape apologies - I did in fact miss that word
Carl Witthoft
CC BY-SA 4.0
2018-11-13T19:52:02.460Z
@AtmosphericPrisonEscape I am inviting you to read this again. I answered your question.
thexiv
CC BY-SA 4.0
2018-11-13T20:22:04.910Z
@thexiv: You created it through tinkering with numbers? If you didn't derive it from physical principes, then it is wrong. I don't see what those quantities in your formula have to do with each other. You could post your 'tinkering' of course, then we'll be able to have a more thorough look at this.
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-11-13T20:44:57.787Z
Okay, calc is to derive a wake or actuality of a linear structure that can bring about certain derivation. i believe that's the point. I was using the certainties in the NASA planetary fact sheet that most stated balance in the equation. To justify its mass at a square to perhaps hold the gravitation, and to derive the orbit's structure, which holds almost a blowing wind from the planet. Thus I came to the construct above and was quite taken with the numbers ordinance. @AtmosphericPrisonEscape
thexiv
CC BY-SA 4.0
2018-11-13T22:07:26.567Z
@thexiv: A derivation by words is very hard to follow. Please back it up by the math you've done/
AtmosphericPrisonEscape
CC BY-SA 4.0
2018-11-13T22:16:16.360Z
I've taken 2 college math courses. I had a great teacher. In this math, is states, that by multiplying the gravitational pull (we assume square of density of the planet), by its inclination (being how high and out it goes and subtracting that loop from the velocity in a slope formula that measures to the borderline of being flat, we can take from the furthest point out what the sun can less take from the product and sum and difference of the cloud around the planet so that we may therefore calculate the prior notation of how many atmospheres there are on each planet @AtmosphericPrisonEscape
thexiv
CC BY-SA 4.0
2018-11-13T23:01:22.620Z
@AtmosphericPrisonEscape There we go, full explanation in the question
thexiv
CC BY-SA 4.0
2018-11-13T23:24:53.107Z
@AtmosphericPrisonEscape Do I have this logically? I'm a schizophrenic.
thexiv
<amateur-observing><planetary-atmosphere>
Solution for counting atmospheric layers of planets according to relational data
thexiv
# Solution for counting atmospheric layers of planets according to relational data

(Asked by: username_0 on 2018-11-13)

This math states, that by multiplying the gravitational pull (we assume square of density of the planet), by its inclination (being how high and out it goes) and subtracting that loop from the velocity in a slope formula that measures to the borderline of being horizontal on the sun's x-axis, most parallel with the solar system's other planets, relegating the offset in order as to consign the preferred balance of the planet with the Sun. We can take then from the furthest point out once the Sun can lesser not take from the product and sum and difference of the clouds on the planet so that we may therefore continue on and calculate the priori aft dividing by 100 to delineate in a reciprocity to the true count, the atmospheric makeup of which the count of surface bars is.
This site is were to get the constants for the planetary facts you will need to use the formula.
https://nssdc.gsfc.nasa.gov/planetary/factsheet/
n = Orbital Velocity - root of density * inclination to sun
round((Apogee - n)/100) = Planet's Atmospheric layer count;
This is an easy formula to figure out how many layers of atmosphere your favorite planets have on them. It's preliminary because I don't have the satellite data from each planet to synthesize with my data from this formula. 
If this is wrong, please tell me. If it is not, I suppose tell me. But I am only in search criticism and laud.
NASA Deg.   Real Deg.   Planet + Rank   Density Orbital Vel.    Apogee  Bars (Atmospheres)

17.2        15.3        Pluto+4         2095    4.7             7375.9  6.6 Bars

7           5.1         Mercury+3       5427    47.4            69.8    2.5 Bars

1.8         3.7         Neptune+2       1638    5.4             4545.7  44.01 Bars

1.3         3.2         Jupiter+1       1326    13.1            816.6   34.13 Bars

1.9         -           Mars -0         3933    24.1            2492.2  2.1 Bars

2.5         -0.6        Saturn -1       687     9.7             1514.5  15.07 Bars

3.4         -1.5        Venus -2        5243    35.0            108.9   2.52 Bars

0           -1.9        Earth -3        5514    29.8            152.1   3.22 Bars

0.8         -2.7        Uranus -4       1271    6.8             3003.6  31.06 bars

**Tags:** amateur-observing, planetary-atmosphere
> **Comments:**
> **username_1**: This looks weird, where is this from? Atmospheres are always continuous, any equidistant layer numbers we assign to them must be arbitrary and probably the result of some computer simulation technique.  
> **username_2**: @AtmosphericPrisonEscape  That's not true, even on Earth.  We often see sharp shear layers between upper and lower wind patterns, for example, with different densities and humidities.  
> **username_2**: But.. where did you get that formula from?   It looks highly unlikely to be valid  
> **username_1**: @CarlWitthoft: I said "equidistant". Look at the formula provided by OP. This very much looks like the discretization of atmospheric structure based on oribital velocities, that's why it's weird.  
> **username_0**: I actually created it myself. I was just tinkering with numbers and came up with that. It seemed to get most of them so I assume there's something wrong with other bases.  
> **username_2**: @AtmosphericPrisonEscape apologies - I did in fact miss that word  
> **username_0**: @AtmosphericPrisonEscape I am inviting you to read this again. I answered your question.  
> **username_1**: @thexiv: You created it through tinkering with numbers? If you didn't derive it from physical principes, then it is wrong. I don't see what those quantities in your formula have to do with each other. You could post your 'tinkering' of course, then we'll be able to have a more thorough look at this.  
> **username_0**: Okay, calc is to derive a wake or actuality of a linear structure that can bring about certain derivation. i believe that's the point. I was using the certainties in the NASA planetary fact sheet that most stated balance in the equation. To justify its mass at a square to perhaps hold the gravitation, and to derive the orbit's structure, which holds almost a blowing wind from the planet. Thus I came to the construct above and was quite taken with the numbers ordinance. @AtmosphericPrisonEscape  
> **username_1**: @thexiv: A derivation by words is very hard to follow. Please back it up by the math you've done/  
> **username_0**: I've taken 2 college math courses. I had a great teacher. In this math, is states, that by multiplying the gravitational pull (we assume square of density of the planet), by its inclination (being how high and out it goes and subtracting that loop from the velocity in a slope formula that measures to the borderline of being flat, we can take from the furthest point out what the sun can less take from the product and sum and difference of the cloud around the planet so that we may therefore calculate the prior notation of how many atmospheres there are on each planet @AtmosphericPrisonEscape  
> **username_0**: @AtmosphericPrisonEscape There we go, full explanation in the question  
> **username_0**: @AtmosphericPrisonEscape Do I have this logically? I'm a schizophrenic.  

---

## 1 answers

### Answer by username_3 on 2018-11-13. Score: 3

While the question is rather hard to follow, the formula simply doesn't work.
For Mercury the formula using a naive application of the numbers predicts:
four layers
Orbital Velocity (km/s)        47.4
Density (kg/m3)              5427
Orbital Inclination (degrees)   5.1
n                            -328.65368
Aphelion (10**6 km)            69.8
Layers                          4

That is clearly not true, since Mercury has no atmosphere. You can vary the formula, by choosing different units, or taking the inclination relative to a different plane, but no matter what I try, can't get the formula to give a sensible value for all the planets. 
The formula is inconsistent, since you generally can't add velocity (km/s) to sqrt(density)*inclination (kg^0.5 degrees/m^(1.5)) as these quantities have different units. It rarely makes sense to add quantities with different units.
There is, moreover, no reason to suppose that the orbital characteristics of a planet, and its density would have any effect on the number of layers in the atmosphere, which is not a very well defined quantity anyway.
Sorry but this formula is entirely wrong.

> **Comments:**
> **username_0**: I would love to see a list of the actual atmospheres. The one at the site does not give an accurate answers. Earth says 1. Also, you'll ave to notice that the degrees are calculated by earth as well.  
> **username_0**: Okay, this is the thing: I think that the surface bars are wrong on that sheet. I think Mercury has 4.  
> **username_0**: So you're saying that there is no way that the gravity of the planet keeps the gases there. That's inane. No offense. The solar winds during orbit would be difficult not to append with, as well. As for the differing metric units, you can assign different weights to different areas. Altitude and pressure on liquid. So therefore I cannot fully find your answer comprehensive. Thus i voted it down.  
> **username_1**: @thexiv: That's not what JamesK is saying. He is saying that your formula is a bad predictor for the number of atmospheric layers, based on an example that has no atmosphere.  
> **username_0**: I asked an astronaut. I'm waiting on a reply for Venus and Mars and mercury. I highly doubt Venus has 92 layers and Earth has 1. I was taught we had 4. I'm saying we don't have a good example except for earth and the moon (which has 4 and 0 respectively)  
> **username_0**: @AtmosphericPrisonEscape Here is a table of my work.  
> **username_3**: I have absolutely no idea what you are talking about.  You seem to mix pressure (in bars) with "layers".  I maintain the last line of my answer. Your formula is entirely wrong.  
> **username_3**: I've edited to use your definition of inclination. However it still doesn't give reasonable values for Mercury or the other planets.  

---

