IFERS - Exposing the 'Global' Conspiracy From Atlantis to Zion
Would you like to react to this message? Create an account in a few clicks or log in to continue.

Debunking Star and planet distances using physics

2 posters

Go down

Debunking Star and planet distances using physics Empty Debunking Star and planet distances using physics

Post by Spake Face Sun Nov 27, 2022 9:15 am

For this analysis I will be using:


Germany (Stuttgart) When October 1972 The resolving power of the human eye is 0.0003 of a radian or an arc of one minute (1/60th of a degree), which corresponds to 100 microns at 25 cm (10 in). A micron is a thousandth of a millimetre, hence 100 microns is 0.003937, or less than four thousandths of an inch or 2.5 cm.

Smallest visible object | Guinness World Records
The resolving power of the human eye is 0.0003 of a radian or an arc of one minute (1/60th of a degree), which corresponds to 100 microns at 25 cm (10 in).

1/60 of a degree = 0.016666667 = smallest resolution of the naked eye in degrees

Diameter of a truck 16 feet
Distance from observer 10.41667 miles
Angular size in degrees 0.016667863013531032 = just resolvable

Diameter of the sun in miles 864938
Average distance to the sun in miles 91400000
Angular size in degrees 0.54219832728588

Diameter of the moon in miles 0.5178905434225652
Average distance to the moon in miles 238855
Angular size in degrees 0.5178905434225652

Diameter of mars in miles 4 222
Distance from Earth in miles 3 900 000
Angular size in degrees 0.007135775244588674

First problem is here. By dividing the smallest resolvable size of the eye by the angular size of mars we can establish that Mars should be 2.335649049 times smaller than anything visible from Earth. By using NASA’s figures, we should never be able to see Mars without a telescope.

Diameter of Jupiter in miles 88846.13
Average distance from Earth in miles 444280400.85
Angular size in degrees 0.011457872659624184

1.454603965 times smaller than the smallest resolvable size.

When dealing with stars they are measured in solar radius: 1 R☉ = 432,288 miles

1 Lightyear = 5 880 000 000 000

It is hard to find the size of Barnards star. I’m using Wikipedia's info here

Radius of Barnards star = 84728.448 miles *2 = Diameter of 169456.896
Distance from Earth (thoughtco.com) = 5.96 ly in miles = 35 044 800 000 000 or 35.0448 trillion
Angular size in degrees 0.00000027705009528861455

Barnards star should be 60157.59345846 times smaller than the smallest resolvable size but the Wikipedia says it is visible by the naked eye.
Remember the smallest visible size is comparable to 100 microns(100 000 nano meters) at 25 cm, so 100 000 / 60157.59345846 will give a comparable size at 25 cm distance, which is 1.662300538 nm
I think good telescopes can magnify 200 – 300 times. Even if 1000 times can be achieved, it is far from 60 158 times, which is impossible.
Barnards star is the next closest star aside from Alpha Centauri, which is a group of three stars and makes it more complicated to analyse here.

Wolf 359 has a solar radius of 0.144 which means it’s diameter in miles is 124498.944
Distance of 7.78 Light years or in miles is 4.57464×10¹² or 45 746 400 000 000
Angular size in degrees = 1.559306047219555e-7 or 0.0000001559306047219555

This means Wolf 359 would have an apparent size 106885.155930222 times smaller than the smallest visible object. Thoughtco.com says it is too dim to be visible without a telescope. Maybe their telescope has 107 000 times magnification. Not likely. It would be like seeing an object 0.93558361 nm from 25 cm distance with the naked eye, or less than a billionth of a meter.

Lalande 21185 has a solar radius of 0.392 which makes its diameter in miles 338913.79
Distance of 8.29 LY or in miles it’s 4.87452×10¹³ = 48 745 200 000 000
Angular size in degrees = 3.9836393708063734e-7 or 0.00000039836393708063734

This one is only 41837.790644755 times smaller than the smallest visible size. By multiplying this number by the size, we get the size that would make it visible which is a diameter of 14 179 404 192.64 miles. They claim you need a telescope to see it so it would have to be around 14 trillion miles wide to be visible by telescope.

Epsilon Eridani has a solar radius of 0.375 making the diameter in miles 324216
Distance is 10.52 LY or 6.18576×10¹³ miles or 61 857 600 000 000
Angular size in degrees = 3.003060004043723e-7 or 0.0000003003060004043723

The angular size of Epsilon Eridani should be 55498.947665241 times smaller than the smallest resolvable size with the naked eye but thoughtco.com says it is visible with the naked eye. It would have to have a minimum diameter of 17 993 646 816.23 miles to become barely visible. If we count the solar system as ending at Neptune, then it would have a radius of 4.545 billion km and a diameter of 9.09 billion which works out to 5 648 264 117.1 miles. This means we should be able to easily fit 3 of our solar systems side by side inside of this star to make it visible.

All these can be found at thoughtco.com Ten closest stars.

Let’s try this:


Farthest Star You Can See With The Unaided Eye - Cosmoknowledge
Are you one of those wondering what is the farthest star visible with the naked eye? Well, this is the right page for you!

V762 Cassiopeiae has a diameter of 166 679 600.53 miles according to universeguide.com
Distance 2764.1 LY= 1.6252908×10¹⁶ = 16 252 908 000 000 000 or 16.252908 quadrillion miles
Angular diameter is 5.87589472683628e-7 or 0.000000587589472683628

V762 Cassiopeiae should be 28 364.475156235 times smaller than the smallest visible size. To actually be visible with the naked eye it would have to be 4.727779388×10¹² or 4 727 779 388 000 miles wide. That’s nearly 5 quadrillion miles. 837.032270454 of our solar systems side by side would equal the diameter the star would have to be to be visible. The diameter of this star should be 0.804044114 of a light year wide to become barely visible.

This is only looking at apparent size and does not take into account the inverse square law.


Inverse-square law - Wikipedia
The divergence of a vector field which is the resultant of radial inverse-square law fields with respect to one or more sources is proportional to the strength of the local sources, and hence zero outside sources. Newton's law of universal gravitation follows an inverse-square law, as do the effects of electric, light, sound, and radiation phenomena. ...

Doubling the distance reduces illumination to 1/4 intensity.

For example, the intensity of radiation from the Sun is 9126 watts per square meter at the distance of Mercury (0.387 AU); but only 1367 watts per square meter at the distance of Earth (1 AU)—an approximate threefold increase in distance results in an approximate ninefold decrease in intensity of radiation.

AU = astronomical unit is a unit of length, roughly the distance from Earth to the Sun and approximately equal to 150 million kilometers or 8.3 light-minutes.

Let P be the total power radiated from a point source (for example, an omnidirectional isotropic radiator). At large distances from the source (compared to the size of the source), this power is distributed over larger and larger spherical surfaces as the distance from the source increases. Since the surface area of a sphere of radius ris A= 4πr2, the intensity I (power per unit area) of radiation at distance ris

It’s kinda complex but you can see that the intensity at Mercury is almost 9 times greater than at Earth.

At Earth it’s 1367 watts. At 2 AU it would be 341.75 W. At 4 AU it would be at 85.4375 W. At 8 AU it would be at 21.359375 W. At 16 AU it would be at 5.33984375 W. At 32 AU it would be at 1.334960938 W.

At 64 AU it would be at 0.333740234 W.

At 128 AU Intensity of the sun would be at 0.083435059 W, which is really dim.

128 AU is 1062.4 light minutes or 17.706666667 light hours or about 19.2 billion km.

There are 8760 hours in a year, so 17 or 20 light hours for the light from the sun to become dim to see, is not much compared to 5.6 light years, or 10.52 light years or 2764.1 light years.

I'm not a mathematician, so if I've made some mistakes or if you can show it more clearly, please reply and let me know. Thanks.

By considering these laws of physics alone(each one separately debunks the whole paradigm) they utterly destroy the mainstream figures but together they make the impossibility of it astronomical. There is no possible way anything of these sizes could be visible at those distances and there is no possible way light could propagate across those distances either. To believe that both of these laws are broken and you see stars hundreds or thousands of light years, or even 30 light hours away is ludicrous.

The stars orbit around the north pole and are all similar distances in elevation, I think less than 3000 miles but could be much less like 100 to 200 miles high. Actually I'm just guessing about this, but it makes more sense than the official narrative. Mainstream propaganda.

Check out these short videos:


Nikon P900 Stars- Real Stars and Planets vs NASA images. Stars are not what they tell us!
Why do some stars twinkle differently than others? Why do Sirius and Capella flash through all the colors while other stars do not?


Shooting Morning Star with Nikon p900 zooming x83. Best after 0:48
Shooting Morning Star with Nikon p900 zooming x83

And Eric's video is very good: How the Southern Stars Work on Flat Earth

Thanks for reading and spread the truth.

Spake Face

Posts : 6
Points : 637
Reputation : 0
Join date : 2022-09-06
Location : Surrey, BC, Canada

Admin and nowhereelsetogo like this post

Back to top Go down

Debunking Star and planet distances using physics Empty Fake Planets and Stars

Post by DR Huddleston Mon Nov 28, 2022 4:21 pm

Hi Eric,
Longtime "follower", previously in South Africa last we talked on twitter, David Huddleston, here.

I saw the other night, 3 planets/stars in a row.
I got the feeling that I should watch the one in the middle.
It was stationary for a few minutes then began to slowly move down and away from the 2 flanking sky objects. I checked myself to make sure I wasn't "seeing things", briefly, so as not to lose the moving sky object, then continued to watch what appeared to be a star or planet type sky object move further down and away from its flanking objects. After approx 60 seconds of slow downward movement it changed direction and moved up and to the right at a slightly increased pace.
Then the object took off .... in about half a second it was gone from view.
Cloudless night.
Not an aeroplane.
I often see objects that glow white from all angles of view as they pass overhead.
Rockets only glow from behind.....and usually an orangish glow.
"Satellites" are much smaller and dimmer.
This, shall I say, craft, really looked like a star sitting between 2 other stars. Then it moved slowly, then quickly away.
I think more people would see things like this if they just looked up at night (and day) more often.....instead of down (at their phones or whatever).

DR Huddleston

Posts : 3
Points : 545
Reputation : 0
Join date : 2022-11-28

Barber78 and kellydavid7 like this post

Back to top Go down

Back to top

- Similar topics

Permissions in this forum:
You cannot reply to topics in this forum