I was confused by your diagram but I believe it represents where the Earth is day and where it's night in it's orbit around the Sun (please correct me if I'm wrong.
My question is more related to the number of solar middays (noon) and the number of solar midnights.
Are they equal in numbers in the arc I and arc II or after the completion of 4 years?
IMPOV - NO.
Yes. There are an equal number of noons and midnights.
QuoteYes. There are an equal number of noons and midnights.
it's only possible when the length of arc I = length of arc II but arc I > arc II as shown in a diagram
Can you explain why the length of arc 1 > the length of arc II?
resulting in a larger area.- Area is irrelevant
Number of solar middays (noon) = Number of solar midnights;- Always
The following diagram (not to the scale) depicts the path traced by earth in its orbit (either circular or elliptical) around the sun.
Any point on the outer circle represents solar midnight while on inner circle solar noon. The length of an outer circle is greater than the length of the inner circle (or Arc I > Arc II). And hence
Anti noon (midnight points) are more than solar noon (middays points) when the earth revolves around the sun in its orbit – Any reason
(https://s26.postimg.cc/m7t80onnt/IMG_3066.jpg)
Number of solar middays (noon) = Number of solar midnights;- AlwaysCertainly "Arc I > Arc II" simply because the orbital radius at the outside (midnight side) is very slightly more than that on the inside (midday side).
The following diagram (not to the scale) depicts the path traced by earth in its orbit (either circular or elliptical) around the sun.
Any point on the outer circle represents solar midnight while on inner circle solar noon. The length of an outer circle is greater than the length of the inner circle (or Arc I > Arc II). And hence
Anti noon (midnight points) are more than solar noon (middays points) when the earth revolves around the sun in its orbit – Any reason
(https://s26.postimg.cc/m7t80onnt/IMG_3066.jpg)
Doesn't any point on the outer circle represents solar midnight while on inner circle solar noon?All you are saying is that the:When there is a noon, there is a midnight
Simply, open up and straighten out aforementioned 3 circles (outer, middle and inner) to line segments. You will notice thatMid Nights (Anti Noons) > Middays (Noons)
Piesigma,
You just confused yourself with numbers.
The Face of Arc I > Face of Arc II while time is same.
Face and Time of both Arcs must equal (Arc I = Arc II) if we have equal appearances of solar noon and solar midnight.
Hope I have explained it clearly enough to understand.
The average orbital radius on the midnight side = (149,597,871 + 6,378 km) and The average orbital radius on the midday side = (149,597,871 - 6,378 km)
Just as an exercise, let's put the figures in (it's really a waste of time because . . . ):QuoteThe average orbital radius on the midnight side = (149,597,871 + 6,378 km) and The average orbital radius on the midday side = (149,597,871 - 6,378 km)
Here is the caption of Question
"Does time dilate during nighttime when the earth orbits the sun?" - never indicated by our mechanical clocks
Average orbital velocity of earth, v_{orb} = | 29,786 m/s | |
Average rotational velocity of earth at the equator, v_{rot} = | 465 m/s | |
Hence: | ||
Midnight velocity of earth, v_{night} = | 30,251 m/s | |
Midday velocity of earth, v_{day} = | 29,321 m/s | |
Now the time dilation factor, γ = 1/sqrt(1 - (v^{2}/c^{2}) | ||
So: | ||
Midnight time dilation factor, γ_{night} = | 1.0000000051 | |
And the | ||
Midday time dilation factor, γ_{day} = | 1.0000000048 |
OK, here is the confusion - Since the orbital velocity of the earth is greater than its axial rotational speed, therefore, Noon/ Midnight is the DURATION (time span). Noon or Midnight doesn't occur instantly at a particular point on the surface of the earth due to the combined effect of orbital and axial motion of the earth. Sorry, a couple of more questionI don't know even what you mean by the "DURATION on the inner and outer circles".
Can you calculate aforesaid DURATION on the inner and outer circles?
Would the said DURATION affect sidereal and solar time?I don't know what you mean by "during noon" = there is no "duration of noon" or "duration of midnight".
Since shadow of any vertical pole would not grow during noon (DURATION) therefore would we be able to notice or measure such DURATION on earth?
OK, here is the confusion - Since the orbital velocity of the earth is greater than its axial rotational speed, therefore, Noon/ Midnight is the DURATION (time span). Noon or Midnight doesn't occur instantly at a particular point on the surface of the earth due to the combined effect of orbital and axial motion of the earth. Sorry, a couple of more question
Can you calculate aforesaid DURATION on the inner and outer circles?
Would the said DURATION affect sidereal and solar time?
Since shadow of any vertical pole would not grow during noon (DURATION) therefore would we be able to notice or measure such DURATION on earth?
IMPOV
Instant Noon occurs when orbital velocity = or < axial speed. Shadows growth doesn’t stop in instant noon.
Duration Noon occurs when orbital speed > axial speed.
Noon point is dragged forward in the orbit but on the constant orbital radius (on the inner circle) for some DURATION before it spun again to afternoon due to the greater orbital velocity. Shadow growth is stopped when the sun is overhead during this "constant noon".
Would the position of earth change in its orbit for seasons due to aforementioned "Duration Noon" (when the earth (or each noon) is dragged and forward in orbit)?
Imagine the said diagram (3 circles) is a circular disc. The center of this disc is the center of the sun. The disc is rotating about its center such that the rotating speed of middle circle on a disc is equivalent to an orbital speed of earth.
The earth as shown in the diagram is another small disc. This small disc rotating about its own axis – spinning of earth
Now imagine both the bigger disc (for orbital speed) and smaller disc (for axial speed) at the same time – would you be able to find “duration noon” now?
The most simplified form of "Instant Noon" is when the earth spins around its axis and is stationary (it doesn't revolve around the sun (i.e orbital speed is zero) - instant Noon occurs when the orbital speed < speed of axial rotation of the earth or equal to zero.
Draw a radius r of earth at any Noon. Let R1 is the radius of the inner circle in the diagram. Both r and R1 are in a straight line at any noon. Since the orbital speed of earth > its axial rotation speed, therefore, the line r of earth transfers to another R2 but =R1 on the inner circle without axial spin (Earth doesn't have a chance to spin at that point due to the greater orbital speed). Now r is in a straight line with R2 not R1. So the earth simply moves forward in its orbit on the inner circle b/w R1 and R2 but w/o spin. This means Noon is a DURATION not Instant if it drags forward in its orbit w/o any spin. it happens when the orbital velocity of earth is greater than the speed of its axial rotation.
I pose a challenge for you to demonstrate what it is you are claiming with the noon point having duration. Again, don’t worry about scale. Just show a point having a noon position
I pose a challenge for you to demonstrate what it is you are claiming with the noon point having duration. Again, don’t worry about scale. Just show a point having a noon position
I'm novice in astronomy. It's not about the challenge but it's all about positive discussion and IMPOV. I may be wrong unless someone agrees.
Figure on the Top Right:) - NOON is an INSTANT when the stationary earth (zero orbital speed) rotates around its axis. "When there is an instant noon, there is an instant midnight as well"
Disc #1: It rotates about its own center which is the center of the sun. The rotation speed of this disc at R=R1=R2... is equivalent to the orbital speed of the outer surface of the earth at the equator (noons on inner circle). Disc#2 rests on Disc#1 as shown.
Disc #2 represents earth with radius r. It also rotates about its axis. The surface of the earth at the equator moves at a speed of 460 meters per second - approx
Figure on bottom Right:)
There are two types of motion of the earth
1. Earth's rotation w.r.t its axis (nearly 460 m/s) -
2. Earth's rotation w.r.t sun (nearly 30,000 m/s) -
Both r and R1 are in a straight line at any noon. Since 30,000 m/s is way greater 460 m/s, therefore, an earth orbits around the sun faster than its spin around its axis. Thus line r of earth transfers to another R2 but =R1 on the inner circle but without axial spin (Earth doesn't have a chance to spin at this particular moment due to the greater orbital speed) before it spun to an afternoon when the the earth moves forward in its orbit. Now r is in a straight line with R2 not R1. Hence,
Noon is a "DURATION" between B and A when an earth doesn't have a chance to spin around its axis while simply moves/ drags fast forwards in its orbit.
---:)) Even if a noon is an instant, midnight should also be an instant not multiples midnights for single/instant noon - Isn't midnight a "DURATION" due to its longer path for instant noon according to you in the previous discussion.
(https://s26.postimg.cc/mvdro267d/IMG_3089.jpg)
Do you agree midnight (Arc I in the diagram) is a "DURATION" due to its longer path in space and greater velocity in the earth's orbit for any instant noon in/on Arc II if not then why there is a drag of midnight OR Arc I > Arc II?I'm afraid no-one has a clue what you are talking about with your " 'DURATION' due to its longer path in space".
Midnight and midday don't have durations- if they don't have DURATIONS, then my question is why Arc I is greater than Arc II
Because Arc I is at a greater radius than Arc II and circumference 2 x pi x radius.QuoteMidnight and midday don't have durations- if they don't have DURATIONS, then my question is why Arc I is greater than Arc II
Because Arc I is at a greater radius than Arc II and circumference 2 x pi x radius.
QuoteBecause Arc I is at a greater radius than Arc II and circumference 2 x pi x radius.
My question is in reference to Midnight and Noon -
Each point on the outer circle represents the MIDNIGHT on an outer surface of the earth at the equator - Right
Each point on the inner circle represents the MIDDAY on an outer surface of the earth at the equator - Right
There is a Midnight for every Noon if both Noon and Midnight are instant - Right.
Therefore, shouldn't Arc I = Arc II if both Midnight and Midday are instant and equal in numbers but Arc I > Arc II - Any special reasons???
By equating Arc I and Arc II you are referring to their measure or extent (Length). Each point has no measure so no quantity of points will give you measure. Each point is an instant in both position and time and has no duration. Each point is undergoing a continuous change in time and position.
To me it’s like asking how many apples equal an orange?
No, each 24 hour period has one occurrence of midnight and one occurrence of midday - end of story.QuoteBy equating Arc I and Arc II you are referring to their measure or extent (Length). Each point has no measure so no quantity of points will give you measure. Each point is an instant in both position and time and has no duration. Each point is undergoing a continuous change in time and position.
To me it’s like asking how many apples equal an orange?
"The occurrence of Midnights (Anti-Noons) on outer circles are more than the Middays (Noons) on inner circle"
The speed the surface of the earth is travelling and the orbital circumferences have nothing to do with the number of occurrences of anything
I still cannot see any problem. What has "touching the outer circle instantly" got to do with the number of midnights?QuoteThe speed the surface of the earth is travelling and the orbital circumferences have nothing to do with the number of occurrences of anything
I think you didn't follow the post properly - As mentioned earlier the speed of the earth's spin is 30 m/sec and earth's traveling speed is 30,000 m/sec in its orbit. Midnights are touching the outer circle. Middays are touching the inner circle - Right?
So you agree Midnight is traveling in the orbital circumference for a while (as you quoted) not just touching the outer circle instantly but not Midday - Right?
Number of solar middays (noon) = Number of solar midnights;- Always
The following diagram (not to the scale) depicts the path traced by earth in its orbit (either circular or elliptical) around the sun.
Any point on the outer circle represents solar midnight while on inner circle solar noon. The length of an outer circle is greater than the length of the inner circle (or Arc I > Arc II). And hence
Anti noon (midnight points) are more than solar noon (middays points) when the earth revolves around the sun in its orbit – Any reason
(https://s26.postimg.cc/m7t80onnt/IMG_3066.jpg)
Number of solar middays (noon) = Number of solar midnights;- Always
The following diagram (not to the scale) depicts the path traced by earth in its orbit (either circular or elliptical) around the sun.
Any point on the outer circle represents solar midnight while on inner circle solar noon. The length of an outer circle is greater than the length of the inner circle (or Arc I > Arc II). And hence
Anti noon (midnight points) are more than solar noon (middays points) when the earth revolves around the sun in its orbit – Any reason
(https://s26.postimg.cc/m7t80onnt/IMG_3066.jpg)
Wow, that's really deep. So, in RE model every sunset and sunrise are not just passing the terminator line, but a drastical change in the speed of every object on Earth, about 1 kilometer per second.
;D
And yet - nobody feels that.
;D ;D
Isn't that amazing?
;D ;D ;D
A tangent line to a circle is a line that touches the circle at exactly one point – Wikipedia
Fig #1: when the earth rotates on its axis but has zero orbital motion.
Tangent A touches C exactly at x or Anti-Noon x
Tangent B touches C exactly at y or Noon y
A common point of C and tangent A is x
A common point of C and tangent B is y
Noon and Midnight are happening in this case INSTANTLY.
Fig #2: When the earth rotates on its axis and has also orbital motion
As said
Tangent A touches C exactly at x or Anti-Noon x
Tangent B touches C exactly at y or Noon y - Similarly,
Tangent A1 touches C exactly at x1 or Anti-Noon x1
Tangent B1 touches C exactly at y1 or Noon y1
Tangent A2 touches C exactly at x2 or Anti-Noon x2
Tangent B2 touches C exactly at y2 or Noon y2
Tangent A3 touches C exactly at x3 or Anti-Noon x3
Tangent B3 touches C exactly at y3 or Noon y3 .... and son on
When there is a midnight x , there is a noon y and hence
When there is midnight x1,, there is a is noon y1
When there is midnight x2,, there is a is noon y2
When there is midnight x3,, there is a is noon y3 and so on
Original Diagram of the question
The outer circle is traced by all the Midnights (x, x1, x2, x3……) on earth in space due to the spin of earth on axis and its orbital motion
The inner circle is traced by all the Middays (y, y1, y2, y3……) on earth in space due to the spin of earth on axis and its orbital motion
If both Noons and Midnights were instant and equal in numbers then, the
Circumference of outer circle must be equal to the Circumference of inner circle but
Circumference of outer circle > Circumference of inner circle and hence
# of Midnights > # of Middays or
Occurrence of midnights > Occurrence of middays
Since the instantaneous
Orbital velocity of Point x > Orbital velocity of Point y
Orbital velocity of Point x1 > Orbital velocity of Point y1
Orbital velocity of Point x2 > Orbital velocity of Point y2
Orbital velocity of Point x3 > Orbital velocity of Point y3 and so on
Therefore, the
Drag of Point x > Drag of Point y in space
Drag of Point x1 > Drag of Point y1 in space
Drag of Point x2 > Drag of Point y2 in space
Drag of Point x3 > Drag of Point y3 in space and so on
The earth spins on its axis at the speed of 30 m/sec.
The travelling speed of earth is 30,000 m/sec in its orbit.
As 30,000 m/sec >>>> 30 m/sec , therefore either # of Midnights > # of Middays (which is impossible)
or there must be the happening of drag of midday and midnight in the orbit of earth if the
Circumference of outer circle > Circumference of an inner circle
I called this Drag or gentle ride of midnight (or midday) in space DURATION between the happening of x and x1 (or y and y1)
Although the orbital speed of midnight > middays but the DURATION between the two successive midnights (say x1 and x2) is same and = DURATION between the two successive middays (say y1 and y2)
Shadow growth stops (or very very slow) due to the gentle ride or transferring of noon or midnight to next point on earth during this DURATION
(https://s22.postimg.cc/52azocye9/IMG_3529.jpg)
No, each 24 hour period has one occurrence of midnight and one occurrence of midday - end of story.
The speed the surface of the earth is travelling and the orbital circumferences have nothing to do with the number of occurrences of anything.
I still cannot see any problem. What has "touching the outer circle instantly" got to do with the number of midnights?
Number of solar middays (noon) = Number of solar midnights;- Always
The following diagram (not to the scale) depicts the path traced by earth in its orbit (either circular or elliptical) around the sun.
Any point on the outer circle represents solar midnight while on inner circle solar noon. The length of an outer circle is greater than the length of the inner circle (or Arc I > Arc II). And hence
Anti noon (midnight points) are more than solar noon (middays points) when the earth revolves around the sun in its orbit – Any reason
(https://s26.postimg.cc/m7t80onnt/IMG_3066.jpg)
Wow, that's really deep. So, in RE model every sunset and sunrise are not just passing the terminator line, but a drastical change in the speed of every object on Earth, about 1 kilometer per second.
;D
And yet - nobody feels that.
;D ;D
Isn't that amazing?
;D ;D ;D
How did you arrive at the conclusion there is a "drastic change in speed of every object on earth, about 1 kilometer per second"?
What you are claiming is inconsistent. A Change in Position would have units of Length/time (ex. a constant Speed of 1 kilometer/sec). You claimed a "Change in Speed" which is an Acceleration and should have units of L/t^2 (a change in L/t per every unit time).
Until you resolve the problem with your first statement I'm not sure how your proceeding claims or expectations of feeling something or being amazing follow?
So a couple questions:The diagram is the top view
1. Is the outer midnight circle as represented or traced out by (x0,x1,x2,...) larger in circumference than the inner circle as represented or traced out by (y0, y1, y2,...)?
2. If yes to question 1. Would there be more midnights than noons because the Circumference of outer circle > Circumference of inner circle and hence the # of Midnights > # of Middays?
Mid day and midnight, are determined by rotation of the earth, not by orbit, and occur simultaneously, with points opposite each other. Therefore there Count Will always be equal.QuoteSo a couple questions:The diagram is the top view
1. Is the outer midnight circle as represented or traced out by (x0,x1,x2,...) larger in circumference than the inner circle as represented or traced out by (y0, y1, y2,...)?
2. If yes to question 1. Would there be more midnights than noons because the Circumference of outer circle > Circumference of inner circle and hence the # of Midnights > # of Middays?
Only Midnights (x,x1,x2,x3………) are touching the outer circle
Only Middays (y,y1,y2,y3………..) are touching the inner circle
Any point on OUTER circle represents Midnight
Any point on INNER circle represents Midday
The circumference of outer circle MUST EQUAL TO the circumference of inner circle if the occurrence of Midnights is EQUAL TO the occurrence of Middays BUT -
Circumference of outer circle is greater than the Circumference of inner circle – RIGHT. So
So there are two possibilities as said earlier
1- The # of Midnights are more than the # of Middays OR the occurrence of Midnights is more than the occurrence of Middays
2- There is a orbital DRAG of Midnight in the space
1 of 2 is not possible – we all know
So if there is an orbital DRAG of Midnight then there is an orbital DRAG of Midday as well
Since orbital velocity of Midnight (x,x1,x2...) > orbital velocity of Midday (y,y1,y2....).Therefore orbital GRAG of Midnight is more than the orbital DRAG of Midday in order to make the occurrence of Midnight = occurrence of Middays.
Shadow growth on earth is either stopped or very very slow during this drag.
Question:
Would this DRAG affect SEASONS due to the change in the position of the earth in its orbit as asked earlier?