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95

Does bottle water rise a little bit on full moon days? No. Tidal forces are about the difference in gravitational pull at different points in the same body. For the oceans, this difference causes water to flow from one region to another, which causes the rise in tides. For example, this is why, even though the sun's gravitational pull is much larger on the ...


14

The color of the sky doesn't just depend on the atmospheric density, but also on aerosols and particulate matter. Mars has about 1% of Earth's atmospheric density, just as you specify in the question. But its sky looks pale orange during the day, whereas sunsets look bluish. So that's almost opposite to what we see on Earth. The reason for these colors is ...


13

Even the most rewarded answer here has missed out on the fact that the entire bottle of water will rise by up to a meter due to the phenomenon of earth tides. The body of earth is deformed by the same lunar and solar tidal forces that cause the sea tides. See https://en.wikipedia.org/wiki/Earth_tide.


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So for the bottle, the difference in gravitational pull from one side of the bottle to the other side of the bottle is extremely small because the distance is extremely small relative to the distance to the moon, and the tidal forces can not be observed. How small? Let's work it out. The acceleration on a bottle of water due to the Moon is... $$\text{...


5

If there is not enough atmosphere or stuff in the atmosphere to scatter incoming light, it will appear blac or very dark.. Think about how you can't see a laser beam middle mid-air unless it bounces off something like fog to Our sky appears blue because blue wavelengths are small relative to the air molecules so bounce off them and get scattered more than ...


3

Short answer: you use the radius of the earth. Long answer: You are correct that it is every particle of the earth pulling on you at a different direction. But, if you use calculus to break up this net force into the combined resultant force of every particle on the Earth pulling on you (and you approximate the shape of the earth as an exact sphere), the ...


3

Strictly speaking, this equation for the force due to gravity only holds between point-like objects. In the case of a point-like object, the notion of "distance" between them is simply the distance between their positions. For extended objects, things get a little more complicated. Really, if you had an extended object (a sphere, cylinder, ...


2

The illuminance of full daylight on the surface of the Earth is about 10,000 lux. Jupiter is about five times as far from the Sun as the Earth, and Pluto is between thirty and fifty times as far from the Sun as the Earth (depending on where it is in its orbit). So daytime illuminance on the surface of Europa will be about 400 lux and on Pluto it will be ...


2

As correctly stated in the deleted answer, the tides have nothing to do with the fact that the moon is full. Tides occur always, regardless of the moon being full or not. Imagine your bottle filled with water is free-falling towards the moon, the bottle parallel to the moon's gravity field. What will happen to the water inside? On one side of the bottle, the ...


2

The sky would look as if you were at a height of near 26 km, because that is where the pressure has dropped to 0.01 atmosphere.


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This question maybe suited for another SE site, and may also be opinion based too, but I find it interesting and would like to answer. I believe you’re thinking is correct. Life started on Earth due to (if you are a theist you may disagree) many different processes occurring simultaneously, such that the first carbon based life forms were introduced. And ...


1

Why not make it simple? Let's say, that here on Earth nothing else changes, just that the atmospheric density decreases to your required level (even if it is impossible for the sake of argument let's say that). What do you see when you look up? Good old blue. Why? The answer is Rayleigh scattering. If nothing else changes, this is the dominant form of ...


1

The object in this law assumed to be points (so the distance is just distance between points). Any two object that are very small compare to the distance between them (no matter from which points on the objects you measure it) can be considered as points. So, for example if you have a two rigid molecule that are small compare to the distance between them, ...


1

$h$ is a change in height, it might help to see the derivation of $mgh$ from a change in $\frac{-GMm}{R}$ The increase in potential energy, when lifting by a small height $h$ is $\frac{-GMm}{R+h} - \frac{-GMm}{R}$ this is approximately $GMm(\frac{1}{R}-\frac{1}{R+h})$ = $GMm(\frac{R+h-R}{R^2})$ so it's $\frac{GMmh}{R^2}$ or $mgh$ and the last expression is ...


1

The formula $E_\text{pot}=mgh$ only holds for objects which are located in the vicinity the surface of the earth where $$F_g=mg\qquad \text{with}\qquad g=\frac{Gm_\text{earth}}{r_\text{earth}^2}$$ The formula $E_\text{pot}=-\frac{Gm_\text{earth}m}{r}$ holds for all distances $r$, not just for distances $r\approx r_\text{earth}$. Also, the reference point ...


1

$mgh$ is an approximate formula that can be used only close to the Earth's surface, i.e., when $h\ll R$. It is obtained from $-GMm/(R+h)$ by Taylor expansion in h.


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