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September 24[edit]

Questions about the Earth's orientation and seasons[edit]

Yesterday, I got to thinking. The reason we have seasons and a day-night cycle is because the Earth's axis is tilted and the planet rotates around its own axis.

It's a well-known fact that on opposite hemispheres, the seasons are opposite. When it's summer on the northern hemisphere it's winter on the southern hemisphere and vice versa. Does this mean that the equator experiences the same season throughout the year? I mean, if the seasons are opposite on opposite hemispheres, they've got to cancel out somewhere. They can't just magically flip around.

And suppose the Earth's axis was straight. Consider three hypothetical scenarios:

The Earth's axis is straight, but otherwise it's perpendicular to the orbit as it is now. Now there are no seasons, but instead the warmness of the weather depends solely on how far away from the poles you are. The equator is in permanent summer and the poles are in permanent winter. There is a normal day-night cycle.
The Earth's axis is straight, but parallel to the orbit. Now the seasons match the day-night cycle around the world. No matter where you are, there is a regular day-night cycle where the day is warm and the night is cold.
The Earth's axis is straight, and parallel to a radius from the Sun, i.e. one end pointing to the Sun and the other end pointing away from it. Now there are no seasons and no day-light cycle. The weather remains constant as long as you don't move to another place. One hemisphere experiences permanent summer day and the other experiences permanent winter night. The poles are the hottest and coldest places, as well as the lightest and darkest, respectively. The equator experiences medium weather and medium sunlight.

This is of course and over-simplification, not accounting for things like air current and rain. But have I otherwise understood the basic principle correctly? JIP | Talk 13:16, 24 September 2017 (UTC)[reply]

Your scenarios are physically impossible because the centrifugal forces of the earth rotation will, like a Gyroscope, not allow the axis to turn along its orbit nomatter parallel, perpendicular or parallel to a radius from the sun.

There are no seasonal difference worth mentioning in the equatorial band because the sunlight at 90 degree or 80 degree does not make a climatic difference. The difference gets bigger the further you go to the poles and there you even have a permanent night for 6 month and a similar day the other half year. On top we have the Coriolis force that makes hurricans, vortexes etc. turn clockwise on the northern half and counterclockwise on the southern half. --Kharon (talk) 15:30, 24 September 2017 (UTC)[reply]

@Kharon Citation requested for your claimed length of permanent Polar night. Blooteuth (talk) 15:58, 24 September 2017 (UTC)[reply]

@Blooteuth: Cite from article Midnight sun"The sun remains continuously visible for one day during the summer solstice at the polar circle, for several weeks only 100 km closer to the pole, and for six months at the pole." --Kharon (talk) 04:20, 26 September 2017 (UTC) [reply]

Articles: Equatorial climate, Solar irradiance, Effect of Sun angle on climate. As the angle between the Sun and the Earth's surface departs from directly overhead, the insolation is reduced in proportion to the angle's cosine. The OP's 3rd scenario that eliminates the day-light cycle would have gross geophysical effects such as boiling away the life-giving atmosphere to leave a bipolar wasteland resembling planet Mercury (long thought to be tidally locked to the Sun but now known to have a 3:2 spin–orbit resonance). Blooteuth (talk) 15:48, 24 September 2017 (UTC)[reply]

I agree that scenarios 2 and 3 are impossible, because it's like trying to change the direction of rotation of a spinning top. Scenario 1 is certainly possible, and the reduced temperature variation at each point on Earth may be even better for life, as no energy would be wasted in animal migrations, hibernation, etc. Life would also be possible closer to the poles, as they would get some light each day.

A tidally locked Earth is quite unlikely, as at this distance from the Sun, it would need to have almost no spin to be captured in this way. But, if it happened, then water would evaporate from the light side (I'm not sure if it would actually reach boiling temp before the water was gone), then fall as snow on the dark side. Carbon dioxide would also fall as dry ice. Then we'd have absolutely huge glaciers, miles thick, redistributing that ice back to the twilight area until it gets hot enough to melt and evaporate again (or sublimate, in the case of the dry ice). There would be permanent, extreme storms in this zone, due to the temperature extremes on both sides. Would life be possible on such a planet ? I should think so, yes, for some extremophiles.

BTW, there are still seasons near the equator on our present Earth, they are just different. For example, they may have a dry season and a rainy season, as global weather patterns change with the seasons. See Quito#Climate for an example. For the least variation, you want a small island at the equator, where the ocean moderates both the temps and humidity. StuRat (talk) 17:18, 24 September 2017 (UTC)[reply]

Note that although a tidally locked Earth is ~~impossible~~unlikely, many exoplanets have been found around red dwarf stars, the most common type, and are expected to be tidally locked. For example, TRAPPIST-1 is thought to have Earth-sized planets with water in the habitable zone, but they are likely tidally locked. Wnt (talk) 18:32, 24 September 2017 (UTC)[reply]

It's extremely unlikely, but is it completely impossible ? Let's say Earth was hit by a giant planetoid, such as the one which may have formed the Moon, in such a way as to almost cancel the Earth's rotation, in the Earth's early history. StuRat (talk) 18:35, 24 September 2017 (UTC)[reply]

Interestingly, Venus rotates hundreds of times slower than Earth (4 miles per hour) Sagittarian Milky Way (talk) 19:29, 24 September 2017 (UTC)[reply]

We'd have to get rid of our moon somehow before we could become tidally locked to the sun. ApLundell (talk) 16:21, 25 September 2017 (UTC)[reply]

How about if the Moon was moved to the L1 Lagrange point ? Nevermind, I see it would drift towards the Earth or Sun from there, staying on the line between them. StuRat (talk) 22:51, 25 September 2017 (UTC)[reply]

I was stunned to see I'd written "impossible" there. Brain fart. Incipient mental decay? I dunno. Definitely not impossibility though. Wnt (talk) 22:03, 26 September 2017 (UTC)[reply]

Relating to the original question, there are in fact seasonal differences at the equator. At the equinoxes, the Sun passes through the zenith (the top of the sky), but at the solstices, it misses the zenith by 23 degrees. This results in substantially less solar heating at the solstices than at the equinoxes. In many equatorial locations, however, a more important factor is seasonal changes in wind patterns, often resulting in a rainy season and a dry season. Looie496 (talk) 21:19, 24 September 2017 (UTC)[reply]

Feynman Lectures. Exercises. Exercise 14-21 JPG[edit]

. .

...
With what minimum speed must an interstellar probe be launched from near the Earth's surface in order to escape from the solar system with a residual speed of 10 mi sec^-1 relative to the sun? The speed of the Earth in its orbit is 18.5 mi sec^-1 .

— R. B. Leighton , Feynman Lectures on Physics. Exercises

In Solutions they write:
$v$ - start speed (relative to the sun);
$v_{\text{min}}$ - start speed (relative to the Earth);
$v'$ - residual speed (relative to the sun);
$v_{1}$ - escape velocity from the Earth;
$v_{0}$ - Earth speed.

Conservation of energy for the probe:
$0.5v^{2}-G{\tfrac {M_{\text{earth}}}{R_{\text{earth}}}}-G{\tfrac {M_{\text{sun}}}{R_{\text{earth orbit}}}}=0.5v'^{2}$
Note that
$G{\tfrac {M_{\text{earth}}}{R_{\text{earth}}}}=0.5{v_{1}}^{2}$
$G{\tfrac {M_{\text{sun}}}{R_{\text{earth orbit}}}}={v_{0}}^{2}$

Therefore:
$v^{2}={v_{1}}^{2}+2{v_{0}}^{2}+v'^{2}$

$v=29.2{\text{ mi/sec}}$
$v_{\text{min}}=v-v_{0}=10.6{\text{ mi/sec}}$

But the answer from Exercises is 11.8 mi/sec. Why?
Second, in article the author thinks that energy needed is (escape the sun energy in Earth system) + (escape the Earth energy) . How this formula is derived?? And why it differs The Solutions formula??Username160611000000 (talk) 16:42, 24 September 2017 (UTC)[reply]

I try to solve in the sun system.
In the sun system the energy needed to escape from sun at $R_{\text{earth orbit}}$ distance = $G{\tfrac {M_{\text{sun}}\cdot m}{R_{\text{earth orbit}}}}$ . This is true when there is no Earth and the probe is at rest. If the Earth is present, then the probe must have additional energy to overcome earth gravity. Additional energy in the sun system is $0.5m(v_{0}+v_{1})^{2}-0.5m(v_{0})^{2}$ .We therefore have:
${\text{Total energy needed}}=G{\tfrac {M_{\text{sun}}\cdot m}{R_{\text{earth orbit}}}}+0.5m(v_{0}+v_{1})^{2}-0.5m(v_{0})^{2}=8.888\cdot 10^{8}m+8.487\cdot 10^{8}m-4.5\cdot 10^{8}m=1.288\cdot 10^{9}m{\text{ }}[{\text{joule}}]$ .

Converting this to kinetic energy: $v={\sqrt {2\cdot 1.288\cdot 10^{9}}}=5.075\cdot 10^{4}{\text{m/sec}}=50.7{\text{km/sec}}\equiv 31.5{\text{mi/sec}}$ . So in Earth system the start speed is $31.5{\text{mi/sec}}-18.5{\text{mi/sec}}=13{\text{mi/sec}}\equiv 20.9{\text{km/sec}}$ .
So we have 3 different answers with apparent correct reasoning:
1) the solution from Mephi Solutions with answer (putting v' = 0) 13.88 km/sec ~ 8.6 mi/sec
2) the solution in the sun system (from above) 13 mi/sec;
3) the solution in the Earth system (from the article [1]) 16.6 km/sec ~ 10.3 mi/sec.
3*) unknown solution from Feynman that gives 11.8 mi/sec with v'=10mi/sec.
How is it possible?
Username160611000000 (talk) 13:43, 25 September 2017 (UTC)[reply]

I can show how "Exercises" came up with 11.8 mi/sec.

Using:

v_{e_Earth} = 11.2 km/sec = 6.96 mi/sec

v_{e_Sun@1AU} = 42.1 km/sec = 26.2 mi/sec

v_Orbital@1AU = 29.8 km/sec = 18.5 mi/sec

and v_∞² = v_initial² - v_e², the formula from Escape velocity relating initial speed to the hyperbolic excess speed.

Working backwards, first find the initial speed from 1AU necessary to achieve v_∞ = 10.0 mi/sec as sqrt(v_∞²+v_{e_Sun@1AU}²) = 28.0 mi/sec.

Assume that this is tangential (and prograde) to earth's orbit, so subtract v_Orbital@1AU to get 9.5 mi/sec and put us in Earth's reference frame.

Finally, find the initial speed from Earth's surface to achieve that much excess speed via sqrt(v_∞'²+v_{e_Earth}²) = 11.8 mi/sec.

Note that this solution does not take into account any initial speed from Earth's rotation.

I haven't yet verified the energy balance method from "Solutions", but if it disagrees, then there is a wrong assumption somewhere. If no one addresses this, I will try to give it a look tomorrow. -- ToE 13:39, 26 September 2017 (UTC)[reply]

Regarding the system escape V_te = 16.6 km/sec from Escape velocity#List of escape velocities, that will be for a final excess velocity of 0, not the 10 mi/sec specified in your problem. If you compute it using the method I just showed above, you get v_∞' = v_{e_Sun@1AU} - v_Orbital@1AU = 12.3 km/sec, and an initial velocity of sqrt(v_∞'²+v_{e_Earth}²) = 16.6 km/sec, matching the value in our table. -- ToE 13:56, 26 September 2017 (UTC)[reply]

Before looking into issues with the energy balance approach, I'll pick one small nit with the "Solutions" solution of v_min=10.6 mi/sec. Even using their numbers of v_∞=10 mi/sec, v_{e_Earth}=7.0 mi/sec (per exercise 14-20), and v_Orbital@1AU=18.5 mi/sec, with v_Orbital@1AU = v_{e_Sun@1AU}/sqrt(2), and their formula of v_initial² = v_{e_Earth}² + 2v_Orbital@1AU² + v_∞², I calculate v_initial=28.9 mi/sec (not their 29.2 mi/sec) and v_min=10.4 mi/sec (not their 10.6 mi/sec). Those are small differences, and some of the values only had two significant figures to start with, but I just don't see how they came up with the numbers they showed. (And putting aside the 1% error in v_initial, how did they subtract 18.5 from 29.2 and get 10.6?) -- ToE 22:57, 26 September 2017 (UTC)[reply]

@Thinking of England:
Assume that this is tangential (and prograde) to earth's orbit, so subtract v_Orbital@1AU to get 9.5 mi/sec and put us in Earth's reference frame.
Finally, find the initial speed from Earth's surface to achieve that much excess speed via sqrt(v_∞'²+v_{e_Earth}²) = 11.8 mi/sec.
-- Why should we go into the Earth reference frame at this moment? And why do you think that excess energy is 0.5mv_∞'² ? Excess energy is a difference E₂-E₁. So why not calculate the excess energy next way 0.5m(28mi/sec)² - 0.5mv_Orbital@1AU² ??
Energy needed to accelerate an object from 0 to 1 m/sec and from 100 to 101 m/sec is different, so this moment is very important.
Username160611000000 (talk) 15:29, 27 September 2017 (UTC)[reply]

Keep in mind that an object does not have a kinetic energy independent of it's reference frame. More insidiously, not only is kinetic energy not invariant under changes of reference frame (obviously), but change in kinetic energy (ΔKE) isn't either. (A 2kg mass accelerating from 1000 m/s to 1001 m/s gains 2001 J, but in the reference frame where the mass was initially at rest, that same acceleration only increased its kinetic energy by 1 J. At first that would seem to make conservation of energy difficult to work out, but objects don't just accelerate without forces acting on other objects, and when all affected objects are taken into account, then energy is conserved in every non-accelerating reference frame.)

The escape velocity from Earth is defined in its reference frame, and that from the Sun in its. If you had been given independent exercises of what speed from 1AU (but outside Earth's gravity well) the probe needed in order to escape the Sun with a residual speed of 10 mi/sec and then what speed from the surface of the Earth the probe needed to escape the Earth with that residual speed, the individual steps would have been obvious. When asked the two problems combined in one, that stepwise solution seemed the obvious way to go for me, but when presented with the overall energy balance solution from "Solutions", my first reaction was not that they were wrong, but instead of how elegant it was. Only after seeing that the answers didn't match did I think, "Ooh, you need to be careful with mixing kinetic energies in different reference frames, and kinetic energy in the reference frame of the object being escaped from is inherent in the definition of escape velocity." That is my gut feeling about what is wrong with the energy balance method from "Solutions", but it is not a very satisfying feeling. Give me a week to mull this over and I will either post a followup to this in the archives (and ping you), or I will post a new question seeking a better answer here on RDS. -- ToE 18:03, 28 September 2017 (UTC)[reply]

testosterone question[edit]

Why does low testosterone / hypogonadism in adult men cause depression and fatigue but does not in male children? 116.58.227.253 (talk · contribs)

The feeling that one is "abnormal" may cause depression, and this is abnormal for adults but normal for pre-adolescents. Also, post-adolescents are more likely to face dissatisfaction from significant others. And fatigue can be a consequence of depression. On the other hand, low testosterone might actually be handy in some circumstances, such as monks who have taken a vow of celibacy. I should expect less depression in such a case. (It could also be useful for adolescents males who are not sexually active, as it would be advantageous to be able to talk to a girl without getting an erection.) StuRat (talk) 18:26, 24 September 2017 (UTC)[reply]

Stu, just what is the antecedent of "this" in "this is abnormal for..."? Is "this" the feeling of abnormality? Are you saying that the feeling of abnormality itself is abnormal for adults or what? Could you write more clearly, please?144.35.45.70 (talk) 03:11, 29 September 2017 (UTC)[reply]

Who says it "causes depression"? ←Baseball Bugs ^{What's up, Doc?} carrots→ 19:19, 24 September 2017 (UTC)[reply]

There does seem to be a correlation, but the exact relationship is unclear: [2]. StuRat (talk) 21:03, 24 September 2017 (UTC)[reply]

[Edit Conflict] Our article Androgen deficiency says it, for one. (Testosterone is, of course, an androgen.)

For pre-pubertal male children, a low level of testosterone is the normal state. Testosterone levels usually increase before, during and after puberty to their normal higher adult levels, which the rest of the adult body's physiology is adapted to work with: if they fail to do so due to hypogonadism or for several other possible reasons there are knock-on effects, including the symptoms and signs the OP mentions. Like any other (very) complicated machine, the body doesn't work as well if some of its components are missing or insufficiently present. {The poster formerly known as 87.81.230.195} 90.201.115.180 (talk) 21:20, 24 September 2017 (UTC)[reply]

Is it absolute? ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:52, 24 September 2017 (UTC)[reply]

Well, some of the components can be missing, like tonsils, adenoids, the appendix, and wisdom teeth. The body may even work better without some, like wisdom teeth. StuRat (talk) 22:02, 24 September 2017 (UTC)[reply]

Which is why I said "some", not "any": on reflection "certain" would have been a better choice. {The poster formerly known as 87.81.230.195} 2.217.210.199 (talk) 13:08, 26 September 2017 (UTC)[reply]

Puberty and it's subsequent hormones profoundly affect the brain and other organs. Castrati singers, when they were common for example, are physically distinct from men and women. --DHeyward (talk) 02:00, 25 September 2017 (UTC)[reply]