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64

Improbable. It is well known that the Coriolis force is needed to form a hurricane, and the figure of 5oN/S as the minimum for formation is widely publicized. You can also find record of tropical storm formation near India as far south as 1.4oN. The problem of crossing the Equator isn't one of hurricane formation though, it is one of hurricane motion. ...


13

As noted in the comments, this answer applies to things like sun-bathing and solar panels, but it does not apply so much to a specific point-receptor like an eyeball. If all objects in question are pointing directly at the sun, then the angle of incidence is equal for all of them and this answer does not apply. For an optic facing its target, the amount of ...


12

Quoting from the Wikipedia article on the Solar Storm of 1859; On September 1–2, 1859, the largest recorded geomagnetic storm occurred. Aurorae were seen around the world, those in the northern hemisphere even as far south as the Caribbean; those over the Rocky Mountains were so bright that their glow awoke gold miners, who began preparing breakfast ...


7

To start with a definition, the tropopause is the boundary between troposphere and stratosphere; within the troposphere temperature decreases with increase in altitude (the temperature profile is dominated by radiative heat from the Earth's surface), whereas in the stratosphere temperature increases with altitude (the temperature profile is dominated by ...


6

The Coriolis effect is necessary for FORMATION but NOT for MAINTENANCE of a tropical cyclone. Once formed, in a full-fledged tropical cyclone of hurricane intensity the wind balance is cyclostrophic, between the pressure-gradient and centrifugal force, with the Coriolis effect negligible by comparison. This is especially true if the tropical cyclone is ...


6

The ecliptic path, is a well defined trajectory when displayed on the background of the fix stars like in the following figure (taken from physics.csbsju.edu) However, there is not such thing as an ecliptic path on the surface of the Earth. If you thought of it as the "Ground track" of a satellite but applied for the Sun, you have to consider that the ...


5

To further clarify, assume we are on the equator, I want to know how long a time, as a percentage, you could consider to be nighttime on Earth, with the points in time separating night and day being within sunrise and sunset. I personally wouldn't consider that the dividing line between night and day. I consider night to be the period between dusk and dawn ...


5

The way "I see the diagram" it looks weird, but it is probably correct. My perception is that it is about frames of reference. We are used to seeing the Earth with Earth's north at the top, not at 23.5 deg from the vertical. This picture has the ecliptic pole at the top. As you state, Earth's equator matches the celestial equator. The Earth's poles would ...


1

As I see it, the Earth is well oriented, but due to (over)simplification the drawing of the geographical features may be misleading. The aspects that show that the North Pole is not upwards but tilted to the right are: The Mediterranean sea extends basically in an East-West direction. The (simplified) Mediterranean here is clearly tilted and not horizontal. ...


1

The figure depicts the Northern hemisphere summer solstice, given that, there is no information missing: At which width circle does the sun shines perpendicularly? The one labeled "13 1/2 uur" that corresponds to the Tropic of Cancer How long is a day south of the southpole circle? If we understand "day" as the length of daytime, the answer is zero: The ...


1

No. There is a difference in daylight according to latitude, but it is not because of proximity to the sun, and nor is there a significant difference in the total number of daylight hours in a year. What is different is that at the equator, the length of the day is nearly equal all year round. The higher your latitude (the further from the equator and the ...


1

The Earth's equator is about 0.006% closer to the sun than the Earth's mean spin axis, so the effect of proximity to the sun on daylight hours is so small that my calculator - which has a register of 10 significant figures - can't even discern any difference. A rough back of an envelope hand calculation indicates less than a millionth of a second difference ...


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