Anonymous at Wed, 28 Aug 2024 20:19:14 UTC No. 16349833

>Wouldn't the moon appear brighter as you approached it?
No, because it gets bigger in angle. So the rays are more spread out. The solid angle the Moon takes up rises with r^2 (in the limit of small angles). But you have the inverse square law too, which goes as r^-2. These two effects cancel each other out, extended bodies like the Moon have constant surface brightness (flux/angular area).

Anonymous at Wed, 28 Aug 2024 20:20:59 UTC No. 16349836

>>16349813
Here is an experiment to try right now. Stand on one side of your bed room, now walk towards the wall, is the wall getting brighter? How about you place your eye right up against the wall (obviously not in shadow), the wall should be blindingly bright according to the inverse square law, and yet the brightness is unchanged. This is because your eye creates a focused image, focused images get smaller with distance rather than loosing brightness. To further clarify, making the image smaller counteracts the loss in available power since its being spread over a smaller area of your retina.

Anonymous at Thu, 29 Aug 2024 04:22:27 UTC No. 16350460

>>16349813
Math fail!
Moon distance: 385 Mm
Moon radius: 1.7Mm
Ratio of the two: 226
Number of times brighter at surface: 226^2=513000.
Number of times brighter our eyes can endure from minimum intensity to maximum endurable: about 1x10^14.

Anonymous at Thu, 29 Aug 2024 05:18:30 UTC No. 16350512

>>16349813
Camera exposure

Anonymous at Thu, 29 Aug 2024 06:07:19 UTC No. 16350537

>>16349833
This is a retarded answer.

>>16349813
So, the moon measurable intensity of the moon on a surface on earth during a clear night and full moon is 0.1 lux.
You can read a book by it.

Now the problem is, according to the models the moon is not a light source, but is reflecting. The moon is not a mirror hence some of the light gets absorbed.
White paint diffusly reflects ~85% of the light.
Lets assume the moon is a uniform white painted ball.

In order to reflect light in such a manner that 0.1lx reach earth, the incoming suns luminosity onto the moon has to be in the least ~400,00,000 lux in order to reflect enough light to earth with 0.1lux.

It's not even the surface that would be the problem, but the illumination + reflection, that must occur to have this effect.

Each m^2 would emit due to reflection 384,400,000 lumen, and actually even more because the moon is a uneven dusty dirty ball and a lot of its light would get reflected in a diffuse manner.

To compare that with how bright the sun is on earth: 120,000 Lux

Meaning the sun must be ~120 times brighter on the moon than on earth, in order to reflect that much light back to us.

Which indee raises the question:
How did they take the picture in picrel, and how did they even manage to see anything?

Anonymous at Thu, 29 Aug 2024 07:41:43 UTC No. 16350603

>>16350537
>In order to reflect light in such a manner that 0.1lx reach earth, the incoming suns luminosity onto the moon has to be in the least ~400,00,000 lux in order to reflect enough light to earth with 0.1lux.
Numbers pulled out of your ass are meaningless.

>Meaning the sun must be ~120 times brighter on the moon than on earth, in order to reflect that much light back to us.
Nonsense conclusions because you based on garbage assumptions. The Moon is the same average distance from the Sun, so the solar flux is the same on the Moon as at the top of the atmosphere. That is only 40% higher than what makes it to the Earth's surface. 1.36 kW/m^2.

Anonymous at Thu, 29 Aug 2024 09:55:16 UTC No. 16350714

>>16349813
Incorrect! It's the exact opposite, but for the reason you describe
When you're looking from afar, you see a lot of light from a lot of ground in the "same" space. Say you pick a 1mm by 1mm square of the sky, that has kilometers of moon surface projecting it, and all of that reflected light hits a tiny are of your eye
As you get closer, the mapping from actual moon area to visual are gets more and more sparse, so each cell of your eyes receives less total light from the moon
It's like your screen. You look at it and see a complete and bright image, but if you take a magnifying glass and look closely it's all made from tiny and not very bright RGB lights. The light just gets all summed together in your eyes to give the illusion of brightness

Anonymous at Thu, 29 Aug 2024 13:01:39 UTC No. 16350964

>>16350537
>White paint diffusly reflects ~85% of the light.
>Lets assume the moon is a uniform white painted ball.
lets not, since the measured reflectivity of the lunar surface is between 8 and 14% depending on location.

Anonymous at Thu, 29 Aug 2024 13:53:54 UTC No. 16351035

>>16350537
Indee you like talking schizo

Anonymous at Thu, 29 Aug 2024 14:08:20 UTC No. 16351048

>>16351035
its a flat earther/moonhoaxer who will never accept anything you explain to him.

Anonymous at Thu, 29 Aug 2024 18:08:41 UTC No. 16351368

>>16350603
>Numbers pulled out of your ass are meaningless.
https://academic.oup.com/astrogeo/article/58/1/1.31/2938119

The intensity of moonlight varies greatly depending on the lunar phase, but even the full moon typically provides about 0.05–0.1 lux illumination.When a full Moon around perigee (a "supermoon") is viewed around upper culmination from the tropics, the illuminance can reach up to 0.32 lux.

ot


How bright is the full moon during a clear night on earth measured in lux?


The brightness of a full moon on a clear night on Earth is typically around 0.1-0.25 lux. To put this into perspective, a normally lit office or living room might be around 300-500 lux, while a cloudy day outdoors might be around 10,000-20,000 lux.

So, the full moon on a clear night is actually quite dim, but our eyes can adapt to the low light level, making it seem brighter than it actually is!


How far is the moon away?


The average distance from the Earth to the Moon is about 384,400 kilometers (238,900 miles). This is called the "lunar distance" or "lunar mean distance." However, the Moon's orbit is not a perfect circle and its distance from Earth varies slightly over the course of a month.


Let's use the inverse square law to solve for the required surface brightness of the Moon, so that we can detect 0.1lux!

We know:

The luminosity of the Moon on Earth during a full moon is ~0.1 lux (I_Earth)
The distance from the Moon to Earth is approximately 350,000 km (r)
We want to find the surface brightness of the Moon (I_Moon):

Using the inverse square law:

I_Earth = I_Moon / r2

Rearranging to solve for I_Moon:

I_Moon = I_Earth × r2

Plugging in the values:

I_Moon = 0.1 lux × (350,000 km)2
I_Moon = 0.1 lux × 1.225 × 1017 km2
I_Moon ≈ 12,250,000 lux

So, the surface brightness of the Moon must be approximately 12,250,000 lux for the luminosity of 0.1 lux to reach Earth during a full moon!

Anonymous at Thu, 29 Aug 2024 19:06:32 UTC No. 16351457

>>16351368
>So, the surface brightness of the Moon must be approximately 12,250,000 lux for the luminosity of 0.1 lux to reach Earth during a full moon!
Can you see the whole Moon while standing on it? No. So the total luminosity is irrelevant.

I_Moon = 0.1 lux × 1.225 × 1017 km2
I_Moon ≈ 12,250,000 lux
Unit error. When you integrate over the sphere it's not flux any more, it's a power. lux × m^2 =/= lux. So what you calculated cannot even be compared to the Solar flux.

Anonymous at Thu, 29 Aug 2024 19:20:10 UTC No. 16351480

>>16349813
the brightness per unit of area on the celestial sphere is constant, ignoring obstructions such as smoke or haze.
if you got closer to it, yes, the moon would take up more and more area of your celestial sphere, but the brightness per unit of area remains the same.
if you put a mirror behind yourself and engulfed yourself in a 360 degree envelopment of lunar light, you still would not even be bothered.

Anonymous at Thu, 29 Aug 2024 19:31:42 UTC No. 16351497

>>16351480
if he thought of it as, say, a wall covered in flashlights that you gradually step closer to it might help.

Anonymous at Thu, 29 Aug 2024 19:33:38 UTC No. 16351502

>>16351457
Units are irrelevant. The inverse square law does not deal with absolute distance.
But with the factor of the distance. Here I assumed 1m from the surface vs the distance of the moon. ~350,000 km.
Hence it's 350,000,000 times.

Anonymous at Thu, 29 Aug 2024 21:09:04 UTC No. 16351655

>>16351368
>1.225 × 1017 km2
where do these values come from? what are they please?

Anonymous at Thu, 29 Aug 2024 21:27:45 UTC No. 16351687

>>16351655
>>16351368
>I_Moon = 0.1 lux × (350,000 km)2
>I_Moon = 0.1 lux × 1.225 × 1017 m2

I messed up my exponents on my phone.

(350,000km)2 = (350,000,000m)2 =
1,225*10^17


I missed some numbers and typos. Because phone fag.
Let's redo the calculation:

I_Earth = I_Moon / r^2

We want to find the surface brightness of the Moon (I_Moon) when it's 1 meter away from the surface, given the luminance at 350,000 km:

I_Earth = 0.1 lux (at 350,000 km)
r_1 = 350,000 km (distance from Earth to Moon)
r_2 = 1 m (distance from surface to Moon's surface)

First, let's find the ratio of distances:

r_12 / r_22 = (350,000 km)2 / (1 m)2 = (350,000,000 m)2 / (1 m)2 = 122,500,000,000,000,000

Now, rearrange the formula to solve for I_Moon:

I_Moon = I_Earth × (r_12 / r_22)
I_Moon = 0.1 lux × 122,500,000,000,000,000
I_Moon ≈ 12,250,000,000,000,000lux

Anonymous at Thu, 29 Aug 2024 21:36:48 UTC No. 16351694

>>16351687
>I messed up my exponents on my phone.
ooh ok got it.

i mean, even assuming that its the exact same calculation for a large non-point source like the moon and that your end number is correct, thats just the sum of the brightness for the entire area of the moon you're seeing.

For a full moon, thats about 9.5 millions sq km

Anonymous at Thu, 29 Aug 2024 22:40:29 UTC No. 16351783

>>16351502
>r_2 = 1 m (distance from surface to Moon's surface)
You are here assuming that you are 1 meter from the whole surface. Which is of course, nonsense.
Your calculations are wrong. You are assuming it is a point source, it is not. The smallest radius you can consider is the Moon's radius. And you're assuming you can see the whole thing.

Anonymous at Thu, 29 Aug 2024 22:48:36 UTC No. 16351789

>>16349813
All you would need to do is use a large lens

Anonymous at Fri, 30 Aug 2024 05:39:11 UTC No. 16352222

Good thread

🗑️ Anonymous at Fri, 30 Aug 2024 06:28:32 UTC No. 16352240

>>16351694
>>16351783
Distance between the Moon and Earth: approximately 384,400 km (238,900 miles)
Illuminance at Earth's surface: 0.1 lux
Reflectance of the Moon's surface: average reflectance is around 0.12 (albedo)
Let's do some calculations:

Convert lux to watts per square meter (W/m2):

0.1 lux × (1 lumen / 1 lux) × (1 watt / 683 lumens) ≈ 0.00147 W/m2

Calculate the distance factor:
Since the Moon is approximately 384,400 km away, we need to calculate the solid angle (ω) subtended by the Earth at the Moon's surface:

ω ≈ (Earth's area) / (4 × π × distance2) ≈ 6.17 × 10^(-5) sr

Now, we can calculate the required irradiance at the Moon's surface:

E_Moon = E_Earth × distance2 × ω
E_Moon ≈ 0.00147 W/m2 × (384,400 km)2 × 6.17 × 10^(-5) sr ≈ 13.37 W/m2

Calculate the required brightness of the Moon's surface:
To achieve this irradiance, we need to calculate the required radiance (L) of the Moon's surface, considering its albedo (0.12). This value represents the amount of power per unit area per unit solid angle:

L = E_Moon / (π × albedo)
L ≈ 13.37 W/m2 / (π × 0.12) ≈ 35.46 W/m2·sr

Now, to express this value in candelas per square meter (cd/m2), we can use the conversion factor:

1 W/m2·sr ≈ 683 cd/m2

L ≈ 35.46 W/m2·sr × (683 cd/m2 / 1 W/m2·sr) ≈ 24,249 cd/m2

Finally, to convert this value to lux, we can use the fact that 1 cd/m2 is equivalent to π lux:

Brightness of the Moon's surface ≈ 24,249 cd/m2 × (π lux / 1 cd/m2) ≈ 76,321 lux

The Moon's albedo (average reflectivity) for visible light is around 0.12, which means it reflects about 12% of the incident sunlight.
For infrared radiation (thermal infrared), the albedo is slightly higher, around 0.15-0.18.
This means that, of the total solar irradiance (TSI) incident on the Moon's surface, approximately:

12% is reflected as visible light (about 42 W/m2, considering the average TSI of 342 W/m2)

(...)

🗑️ Anonymous at Fri, 30 Aug 2024 06:31:08 UTC No. 16352241

>>16352240
(...)
Reflected intensity on the Moon's surface: 76,321 lux
Albedo of the Moon's surface: 0.12 ( average reflectance)
We can set up a proportionality relationship between the reflected light and the incoming sunlight:

Reflected lux = Albedo × Incoming lux

Incoming lux = Reflected lux / Albedo
= 76,321 lux / 0.12
≈ 635,843 lux

To help put the value of 635,843 lux into perspective, here are a few comparisons:

Comparison 1: Sunlight on Earth

Direct sunlight on Earth's surface during peak hours can reach up to 100,000 lux.
635,843 lux is equivalent to approximately 6.36 times the intensity of direct sunlight on Earth during peak hours.

Comparison 2: Stadium Lighting

Professional sports stadiums are often illuminated with high-intensity lighting, which can reach levels of around 5,000-10,000 lux.
635,843 lux is equivalent to approximately 63-127 times the lighting intensity of a well-lit sports stadium.

Comparison 3: Work Environments

A typical office environment has an illuminance of around 300-500 lux.
635,843 lux is equivalent to approximately 1,272-2,119 times the lighting intensity of a typical office environment.

Anonymous at Fri, 30 Aug 2024 07:05:37 UTC No. 16352270

>>16351694
>>16351783
To be more accurate, the calculation should take into account the Moon's surface area, will will use the total power required to immuniate and then clculate the required incoming sunlight while accounting for the inverse square law.

Let's revisit the calculation, considering the Moon's surface area:

• Convert lux to watts per square meter (W/m2):

0.1 lux × (1 lumen / 1 lux) × (1 watt / 683 lumens) ≈ 0.00147 W/m2

• Calculate the total power required to that represent 0.1 lux on Earth's surface (remember here atmospheric absorbtion is excluded which would):

P_total = E_Earth × Earth's_area
P_total ≈ 0.00147 W/m2 × 4 × π × (6,371 km)2 ≈ 1.43 × 10^8 W

• Calculate the required emission power from the Moon's surface, considering its surface area:

P_Moon = P_total / Moon's_surface_area
P_Moon ≈ 1.43 × 10^8 W / (4 × π × (1,738 km)2) ≈ 33,400 W/m2

Now, let's convert this value to lumens per square meter (lm/m2), and then to lux:

33,400 W/m2 × (683 lumens / 1 watt) ≈ 22,841,000 lm/m2
22,841,000 lm/m2 ÷ π ≈ 7,281,510 lux

So, the Moon's surface would need to emit approximately 7,281,510 lux to produce an illuminance of 0.1 lux on Earth's surface, considering its surface area.
For sake of simplicity, I used the full surface, of earth and moon. Because halving both has no effect here.

To find the required incoming sunlight intensity, we'll need to consider the Moon's albedo and the absorption factor.

Let's assume an average albedo of 0.12 for the Moon's surface.

Reflection calculation:

Reflected lux = Incident lux × Albedo
7,281,510 lux = Incident lux × 0.12
Reflection calculation:

Reflected lux = Incident lux × Albedo
7,281,510 lux = Incident lux × 0.12

Now, let's solve for the incident lux:

Incident lux = Reflected lux / Albedo
= 7,281,510 lux / 0.12
≈ 60,685,850 lux of the incoming sun is required in order to reflect light in such a manner that, 01lux reach earth. (...)

Anonymous at Fri, 30 Aug 2024 07:09:31 UTC No. 16352274

>>16352270
(...)
To help put the value of 60,685,850 lux into perspective, here are a few comparisons:

Comparison 1: Sunlight on Earth

• Direct sunlight on Earth's surface during peak hours can reach up to 100,000 lux.

• 60,685,850 lux is equivalent to approximately 606-607 times the intensity of direct sunlight on Earth during peak hours.

Comparison 2: Industrial Lighting

• High-intensity industrial lighting, like those used in manufacturing or welding, can produce up to 10,000-50,000 lux.

• 60,685,850 lux is equivalent to approximately 1,213-1,213 times the intensity of high-intensity industrial lighting.

Comparison 3: Stadium Lighting

• Professional sports stadiums are often illuminated with high-intensity lighting, which can reach levels of around 5,000-10,000 lux.

• 60,685,850 lux is equivalent to approximately 6,068-12,137 times the lighting intensity of a well-lit sports stadium.

>If you want to say: but moon is a sqhere therefore surface needs to be calculated differently

Yes. But it would make the required incoming light and reflection even brighter. Because of the angle and diffusion.

And the requirement to include the atmospheric absobtion and reflection. Would also make it brighter.

No matter how you twist and turn. The moons surface and the required incoming light has to be incredibly bright in order that 0.1lux can illuminate earth over the distance of 350,000km.

Anonymous at Fri, 30 Aug 2024 07:39:09 UTC No. 16352294

>>16352270
>P_Moon = P_total / Moon's_surface_area
>P_Moon ≈ 1.43 × 10^8 W / (4 × π × (1,738 km)2) ≈ 33,400 W/m2

Nope. The surface of the Moon is 3.7958532e+13 m^2. 10^8/10^13 =/= 33000. You fucked up, I don't know how because this isn't even a simply a unit error.


>P_total ≈ 0.00147 W/m2 × 4 × π × (6,371 km)2 ≈ 1.43 × 10^8 W
This one is also wrong. So apparently you can't use a calculator.

Anonymous at Fri, 30 Aug 2024 07:45:04 UTC No. 16352298

we babl'i'n

Anonymous at Fri, 30 Aug 2024 14:13:09 UTC No. 16352651

>>16352274
>60,685,850 lux
but, even if correct, this would how much the entire full moon surface area (half of 38 million sq km) would be reflecting. Its not a point source!

Anonymous at Fri, 30 Aug 2024 14:16:08 UTC No. 16352656

>>16352274
>Direct sunlight on Earth's surface during peak hours can reach up to 100,000 lux.
sure, but this is a measurement taken at one point, which means the sun is outputting enough energy to create this light level across a tiny part of its sky, 93 million miles away.

by comparing the total reflected light needed for the moon to create 0.1lux on the surface of the earth to the level the sun can create on the earth, you're entirely confusing the issue!

i feel that you're merely trolling now anyway.

Anonymous at Sat, 31 Aug 2024 00:20:33 UTC No. 16353354

>>16349813
it's funny that you people never bother leaning that basics of cameras

Anonymous at Sat, 31 Aug 2024 16:01:50 UTC No. 16354274

>>16353354
its a flat earther/moonhoaxer thing. not a one of them ever knows even basic exposure concepts.

Anonymous at Mon, 2 Sep 2024 08:14:18 UTC No. 16356363

No clue send help i want the answer

Anonymous at Mon, 2 Sep 2024 13:31:57 UTC No. 16356604

>>16354274
You're conflating two things here. People who believe we went to the moon are just as stupid as flat earthers.

Anonymous at Mon, 2 Sep 2024 13:39:49 UTC No. 16356621

>>16356604
>People who don't believe we went to the moon are just as stupid as flat earthers.
fixed it for you. its all the same level of 'evidence' and argumentation.

Anonymous at Mon, 2 Sep 2024 13:47:51 UTC No. 16356632

>>16356621
Your naivety is so laughable.

Anonymous at Mon, 2 Sep 2024 14:07:41 UTC No. 16356654

>>16356632
as is the ignorance of moonhoaxies and flat earthers. Anyway, im sure you must be getting badk to watching American Moon so cheerio.

Anonymous at Mon, 2 Sep 2024 14:13:14 UTC No. 16356666

>>16356654
https://www.youtube.com/watch?v=9HQfauGJaTs

Anonymous at Mon, 2 Sep 2024 14:14:29 UTC No. 16356667

Since most stuff is wrong....
We need to do some assumption to estimate the surface brightness of the moon:
- pretend that the moon is a spherical lambertian surface (reflecting light everywhere eqaually and assume the value as the average)
- then take the average and apply known hotspots, highs and lows and absorption of the moon surface to get the brightness of the incoming sun.
- we must exclude the atmopheric absoption and distortion, meaning we have to deduct, infer or retrieve the brightness of the moon measured from earths most outer layer of the atmosphere
> https://www.physicsforums.com/threads/luminance-of-a-lambertian-sphere-formula.449703/ Luminance: The luminance of the full Moon is approximately 0.25 lux (lx) at the Earth's surface Source: https://stjarnhimlen.se/comp/radfaq.html
>Full Moon overhead
>Stellar magnitude: -12.5 Illuminance Lux: 0.267
Extraterrestrial luminance: We need to calculate for the atmospheric absoption of moonlight which peaks 400nm waveleght. According to Atmospheric Administration (NOAA), the transmission coefficient for 400 nm light at sea level is approximately 0.56 and with moon light it averages around 0.58
>s: https://repository.library.noaa.gov/view/noaa/41797/noaa_41797_DS1.pdf

meaning: Extraterrestrial luminance is ~0.45 lux
First, we need to calculate the solid angle (Ω) using the formula: Ω = 2 * π * (r / d)2, where r is the radius of the sphere (1740000 m) and d is the distance (384400000 m). Plugging in the values, we get: Ω = 2 * π * (1740000 / 384400000)2 = 0.00000763 sr.
Next, we use the formula: luminance (cd/m2) = (L * Ω) / π to calculate the luminance. We are given the target detector measured luminance (0.45 lux), so we can set up an equation: L * (Ω / π) = 0.45 lux / π.

L = 0.45 lux / (Ω / π) = 0.45 lux / (0.00000763 / π) = 5891.19 cd/m2 is the average brightness.
The radiance of the hotspot is related to the radiance of the sphere by the Lambert's cosine law, now do the rest.

Anonymous at Mon, 2 Sep 2024 14:21:16 UTC No. 16356669

>>16356654
>>16356666
And since you're too stupid to infer things with your own brain, I'll make it crystal clear for you. Never mind the 70s tier special effects in the video. Listen to the audio.
>We're on our way houston!
There's not a single sound of an explosion or rocket noise coming from inside the craft. You're so fucking retarded that you don't even notice something that obvious. Think about that for a second. NASA relies on retards like you refusing to admit the obvious.

Anonymous at Mon, 2 Sep 2024 15:58:33 UTC No. 16356812

>>16356669
i love how quickly you people are moved to insults. especially telling is the constant mantra of 'you're stupid' and 'you're a dumb consumer/follower/npc". its very funny, and says a great deal about what all this really means for you.

you've read nothing about the subject and have fed yourself on lying memes. you rely on your own understanding which is woefully uninformed.

Anonymous at Mon, 2 Sep 2024 18:40:34 UTC No. 16357006

>>16356812
Notice how you don't refute my central point, but resort to an ad hominem attack on how I'm insulting you. You deserve to be insulted for being so stupid.

Anonymous at Mon, 2 Sep 2024 19:02:19 UTC No. 16357032

>>16356667
> now do the rest.
How is anyone else supposed to figure you what you mean this time, when you can't even be bothered to finish it?
How many times have you done this same "calculation" and got a completely different answer?

>Notice how you don't refute my central point
Literally your first comment in this thread was baselessly rejecting the correct answer, without any argument whatsoever.
>This is a retarded answer.
No thought at all, no attempt to even understand it. Cause you say so. What a hypocrite.

Anonymous at Mon, 2 Sep 2024 19:06:54 UTC No. 16357041

https://www.youtube.com/watch?v=O9y_AVYMEUs

Anonymous at Mon, 2 Sep 2024 19:40:05 UTC No. 16357104

>>16357032
I am a different anon.
I simply demonstrated and explained that the moons luminosity can only be estimatimated as a spherical lambertian surface.

This retard:
>>16352270
>>16351687

Did absolute nonsense without any sources. And dumb units.

I gave the formular and the sources for the values.

Since the moon is not a lambertian sphere, and I am too lazy to calculate the brighness gradient from edge to center and all the craters, I said do it yourself.
Also 5891.19 cd/m2 is not even that bright that it may be worth to discuss.
Its a camera suitable brighness. No special optics are required. To answer OP

Anonymous at Mon, 2 Sep 2024 19:47:26 UTC No. 16357124

>>16357006
>sound of an explosion or rocket noise coming from inside the craft.
was that your 'central point'?

why do you assume there should be a massive explosion? there is a constant rumble on other recordings, and ive distinctly heard it during powered descent because the engine was much more powerful....

But the ascent engine wasn't a very powerful rocket anon, and there was no air to help with transmitting the sound more fully to the cabin. just the steady gentle rumble/roar conducted through the floor.

if they squelched it a bit more, or adjusted what the mic picks up in terms of noise cancellation, on this liftoff then it could well not be picked up as much as on others.

take a listen to the liftoffs from the other missions using original recordings on ALSJ if you like.

there you see; i replied to you without any of the silly insults you like so much.

Anonymous at Mon, 2 Sep 2024 19:52:29 UTC No. 16357133

>>16357104
>Also 5891.19 cd/m2 is not even that bright that it may be worth to discuss.
and i think its right about what you'd expect for the moon which receives slightly more insolation due to the lack of atmosphere.

Anonymous at Tue, 3 Sep 2024 12:32:20 UTC No. 16358234

>>16349813
That's only for points
Extended objects don't get brighter point for point. They just get bigger

Anonymous at Tue, 3 Sep 2024 12:47:26 UTC No. 16358256

>>16356604
Humans can launch into space in a rocket ship then land on earth. That's like 90% of the technology needed to get to the moon. We have robots that land on the moon so the technology for getting objects to the moon is there. What technology do we lack that precludes a manned moon landing? We can keep humans alive at the bottom of the mariana trench.

Anonymous at Tue, 3 Sep 2024 13:31:15 UTC No. 16358316

>>16358256
Possible doesn't mean it happened already. I suspect Musk will get some people to the Moon but they'll suffer severe radiation poisoning. They'll make some excuses and then they'll start sending robots there instead.

Anonymous at Tue, 3 Sep 2024 19:40:25 UTC No. 16358830

>>16358316
>they'll suffer severe radiation poisoning.
what do you base that claim on?

Anonymous at Tue, 3 Sep 2024 20:34:30 UTC No. 16358912

>>16358830
The fact that our atmosphere protects us from severe radiation poisoning 24/7.

Anonymous at Tue, 3 Sep 2024 21:16:00 UTC No. 16358947

>>16358912
thats another claim. do you have support for either of your claims, or just more unsubstantiated emoting?

Anonymous at Tue, 3 Sep 2024 21:23:21 UTC No. 16358963

>>16358947
This is gaslighting.

Anonymous at Tue, 3 Sep 2024 22:01:09 UTC No. 16359001

>>16358947
Go spend a year in geostationary orbit and tell me how good your kidney function is.

>then try again with N-acetylcysteine, see if it helps.

Anonymous at Tue, 3 Sep 2024 22:16:26 UTC No. 16359024

>>16359001
So if I'm spending a year on the moon I'll take precautions. Being on a planet or moon immediately cuts your radiation in half compared to free space.

Anonymous at Wed, 4 Sep 2024 10:26:25 UTC No. 16359927

>>16358963
it's not. you're making claims without anything to support them. thats very easy to understand.

>>16359001
so which is it - in orbit, or on the surface of the moon?
you dont seem to have any data for either situation.

Anonymous at Wed, 4 Sep 2024 11:19:45 UTC No. 16359981

>>16358963
how can you say it's gaslighting when first of all the atmosphere is transparent, and second of all if it was than how would it protect us from severe radiation poisoning, because the light is, by definition, radiation???

Anonymous at Wed, 4 Sep 2024 12:10:45 UTC No. 16360067

>>16359981
You mean to say that the radiation is by definition light, but that statement is false.

Anonymous at Wed, 4 Sep 2024 12:19:54 UTC No. 16360085

>>16360067
correct. the main radiation danger in space is from fast heavy particles rather than EM radiation

Anonymous at Wed, 4 Sep 2024 12:30:34 UTC No. 16360100

>>16349813
>Wouldn't the moon appear brighter as you approached it?
no you can see from your picture that the total number of photons is constant
>Wouldn't the moons surface be too bright to take a photograph on?
No because you can't see the entire moons surface when you on on the moon, it should be darker