The bright night could be too bright…

If the sun has to follow spherical attenuation to transmit energy to earth, and there is a universal setting for solar systems that every particle shot out from the stars that is supposed to hit a Star or planet, then all energy comes out from every surrounding star will be received by the surrounding solar systems. In this situation, we could get some simpler way in calculation. In this setting, our night sky should be all covered with stars and planets.

`For simplifying our calculation, I set up these assumptions below:`
` `

Assumption:

Now using spherical attenuation, and particles are randomly shooting out at all directions from the suns.

Every particle shoot from the sun going out of its solar system will hit another star or earthly planet. In other word, everywhere we looks at the night sky, we should be only seeing stars in every spot in space, and the night sky is fully 100% covered by shining stars.

Every solar system has only one earth like planet.

Every star has the same size and give out the same amount of light and assuming stars never glow old and die.

Every earth like planet has the same size.

All stars are evenly spaced.

The ratio of the sun's radius to the Earth's radius is 109.

The ratio of the sun’s surface area to the earth’s surface area = 11881.

Radius of earth orbit, R, around the sun = 149,600,000 km, and the orbit is a perfect circle.

Radius of earth = r = 6378km. The earth is a perfect circle.

There is an effective receiving surface area for the Earth (and our sun too) to receive particles from the stars and the sun which is actually equal to  x

The total spherical area  (4π) at 149,600,000 km from the sun, at earth orbit:

= 4×.

The total effective area of earth receiving particles from our sun:

= π x ( = π x 40,678, 884

The energy density W (per second) comes out from our sun to earth are following the spherical attenuation:

Total energy per sec from the sun = S = Power from the sun

= W =

S = W

W = Power density from the sun at earth’s Orbits distance R

Power received on earth =  W

Energy from the sun will evenly disperse in the area of cover the orbit of earth in spherical shape around the sun, and the Earth also occupy a small part in this area of.

=  =  =

= = k = 4.54407877×

The earth is only receiving k of the total sun power at day light, the majority remaining sun power is going outside our solar system.

The sun is shooting over 99.999999% of total output energy out to the whole universe out of our solar system.  Energy from our Sun reaching earth is relatively very small if compare to total energy gives out by the sun due to the distance causing huge spherical attenuation from the sun.

From the assumptions, the ratio of the sun’s surface area to the earth’s surface area = 11881.

The ratio of effective receiving area of the sun to the effective receiving area of the earth for the particles come from stars outside our solar system is also 11881.

However it will have a very small portion of solar systems in the universe could be hiding their earth behind their sun or the planet blocking a little bit of star light from us because of a small percentage of solar system has planet orbits are in the planar position with us. Also the earth orbits are many times larger than the radius of the sun and therefore the hiding time and the blocking time are very short relative to the time cycle once around the sun. The sun is 11881 time bigger than earth in surface area.  I am currently neglecting this tiny hiding and blocking effects in calculation.

In other word, every one particle received on an earth like planet at its night time, there would be another 11881 particles received by the star of the planet from half of the universe. Another 11881 particles are also reaching its sun at the opposite side of the sun.

Total particles reaching the sun is 23762 (11881x2) for every particle reaches the earth at night from outside the solar system. One other particle will reach the earth surface facing the local sun at day time.

Assuming majority of particles are absorbed in the receivers instead of reflected out.

N = equivalent number of stars in the universe that our sun can reach and fill up 100% of the night or day sky (facing one half of universe).

Y = number of particles receive from each star on earth at night in any given time T.

NY = number of total particles receive at night on earth at a given time T.

Number of particles receive by the sun from the whole universe = 23762NY at a given time T.

If our sky is totally covered 100% with stars as far as we can see, every particle transmit from the sun will be intercepted by something in other solar system, then no particle will leak outside universe from our sun. Every particle come from the stars surrounding our sun also ends up inside the universe in our surrounding systems.

This is the part we can get away from spherical attenuation.

I call this method: Equal transmission from stars within a fully meshed lattice.

Figure 1

By neglecting earth like planets around the each solar systems which only account for a very tiny amount equal to K of total transmission that going to earth like planets.

We can simplify the calculation starting with a 2D figure 1 above. Assumingly every red sun can only see 8 other red suns and totally covered by 8 other suns.  Therefore every red sun can only exchanging particles with 8 others through the black transmission lines. The particles transmit out to each 8 red sun will also have the same number of particles come back in each Black line. Every sun in the figure 1 will have the same number of transmit particles and receive particles at any time. The transmission from all suns are equal and the reception are also equal in every transmission lines.

In the 3D universe of much larger scale, assuming all stars are evenly spaced,  every sun only see N other suns and can only exchange particles with N others because the sky is already 100% fully covered by N number of stars. Therefore particles send out from our sun to one of the N stars out there will receive the same number particles back to our sun from the same star and no particles will miss solar systems out there.

In this setting that our sky is 100% all covered with shining stars, then:

The total transmit particles from our solar system is also equal to the total number of received particles in our solar system.

It is because total particles comes out from all the stars that we can see at night are assumingly evenly divided by the same number of star systems surrounding it to receive it.

The total number of particles transmit from our sun will also be 23762NY at a given time T, this number do not change even if sky is barely covered with a few shining stars. It is a constant basically.

But if on average only one half of the particles transmit from the suns are intercepted by other solar systems with shining stars, then the sun will lost one half of total particles transmitted into the void or dead stars or absorbed by the dust and same for all the stars surround us, and the particles transmit from our sun will remain constant. We can tell easily if only half of the night sky covered by stars should be by darkened by one half compared to night sky fully 100% covered by stars.

A night and day sky fully 100% covered with stars is N stars.

A half covered night and day sky with  stars creates a half brightness with N Stars

Each star is still transmitting 23762NY particles at a given time T.

Let z =

Also Z =

Z N Y = night light brightness with void or invisible dust in the sky

NY = night light brightness with night sky fully 100% covered with stars

K = 4.54407877 x

The total energy from the sun is the number of particles transmitting out everywhere spherically at any given time and the earth only takes a small portion of it based on spherical attenuation. Previously we have calculated a K factor based on this spherical attenuation which can determine the brightness of light we get from the sun at daytime by knowing the number of transmitting particles.

Therefore:

23762NYK = the power (energy at a given time) we get on daytime from local sun on earth. Our sun always transmit the same number of particle no matter what the value of z will be.

If z = 1 then, the night sky is fully 100% covered with shining stars.

On earth:

= R =  = 23762K

23762k = 0.00001079763997

R = 0.00001079763997

The night light from stars in universe will be = 92,612.83 times more brightness than the light from the local sun alone.

Moreover even at day time, the background star light has the same intensity of night light, and therefore make no difference between night light and daylight and burn out everything on earth with so many times of our local sun light energy, the earth will have nothing left for burning.

Put z factors into it:

R =  =

Make R = 1, then:

Z = 0.00001079763997

R = 1 means local sun energy to earth is equal to total night light energy to earth, it also means light brightness at day time is twice the brightness to night light because the stars at the background of universe at day time is also shinning the same rate at night light.

Z is the ratio of number of particles reaching stars, planets instead of lost in void or absorb by invisible dust in our universe transmitting from a star to the total number of particles transmitting from the star.

It means our universe would have to leak out 99.999% of energy to outside of our universe to make light from our sun give out the same brightness at day time as night light from stars all over universe on earth.

It also means there is no star light shining in 99.999% of area in the night sky,

Or

99.999% of the stars are dead and are not shinning so that sunlight has the same brightness of night light from stars.

Or

99.999% of the sky is not covered with stars which could be empty space or something tends to absorb any light in any directions.

Or

A combination of all above or any similar setting can cause transmitted light not reaching any stars and planets.

Moreover, if the night sky is only  of our day light brightness……

If you want to make the night sky as dark as what we are actually getting at  of our sun’s brightness, you will need to increase the void, the number of dead stars, or some invisible things that absorb light and decrease the light particles reach any solar systems by a factor of.

The new value of Z = 0.00001079763997 ×  = 1.079763997 ×

Z can become ratio of area when z= 1 is the area of the whole sky.

If you draw a stars map of night sky within one  area of paper, the total area of shining stars with the new Z value will only able to add up to about   or  .

The new Z value for the current brightness at night could probably make many of the current night sky stars disappear if spherical attenuation is the only factor that governing particles propagation.

In other word, if our world is not burning, the Spherical attenuation rule does not be look very good enough to explain a lot of things in our vast universe setting.

Alternate method to calculate Z:

The Area of the sky at any location is depending on the distance of the sky and the optical magnifications of the receivers being used to measure.

To compare the area of the sun with the whole sky, I shall use the distance of the sun to Earth as a radius of sphere and use this sphere to surround earth with earth as the center of sphere.

The effective receiving area is also the effective transmitting area too, because the sun is transmitting with an area in the sky that we see also determined by the optical magnification factor of our eyes or cameras (warning : Must always use proper protection when looking at the sun. The sun energy could cause huge irreversible damages to our eyes if looking at it without proper protection, just do not look at it directly for the health of our eyes). In this case K factor is also applicable. It is because K factor is the comparison of the surface area of an extended sphere from the center of a sphere with the extended surface at a distance object to the effective receiving area of the distance object, and this time the center of the sphere is at the center of our earth and also the distance object is the sun.

The effective receiving surface area of the sun is 11881 times of the earth.

If the sun has the same size of Earth, then the k factor will remain as 4.54407877 × .

But the area of the sun is 11881 times the earth, the ratio of effective transmitting area of the sun to the sphere will become 11881 x 4.54407877 x  = 0.000005398819986637…

The sun is covering 0.000005398819986637 area of the sphere surrounding earth.

But the day sky can only cover half the sphere, the result is base a complete sphere around earth, the sun will be 2 times bigger if only compare to half of the sphere at day time. Therefore 2 times K:

0.00000539881986637 x 2 = 0.00001079763997…

It means the sun is only covering 0.00001079763997 of the whole day.

Every double the distance from earth, you need four times the number of suns create the same brightness on earth. Now you we can add the area of the stars together at the night sky to compare the brightness from the day sun alone. The previous calculation for Z is actually based on number of stars, z (0-1) is the rate of number of N stars where ZNY is the brightness of the night sky. In here Z is the direct rate of area of our sun that covering our day sky. It turns out both are the same result.

Z = 0.00001079763997, this is the rate of area at night sky covered by stars measure with the same optical magnifications will make the night sky as bright as the local sun alone.

This will exactly match the previous result with a different way of calculation.

Kevin Lung

Nov 18, 2014

Note:

Every stars can have different size and different brightness.

Orbits of planets are elliptical.

Stars are not evenly distributed.

Not every stars are shining stars.

Solar systems can have different combinations of planets.

Stars are in clusters and in galaxies formations, and not many standalone stars in space.

Every galaxies have some dust factors.

All these varying factors are not able to make significance difference in the calculation of how bright the sky should be. By measuring the total area of all shining stars should always give us a very good idea of how much star light is missing at night.