1

# 8 light minutes to the sun, so where is it now?

Wed 25 Jun, 2003 11:47 am
Since it takes about 8 minutes for the light from the Sun to reach us, then when we point at where we see the Sun, we are actually pointing to where it was 8 minutes ago.

If we could point to where it is, and not where it appears to be, how much further along its path would we be pointing? One sun width? Two sun widths? A half a sun width?

In other words, how far does the Sun travel across our sky in 8 minutes (expressed in Sun-width units)?

I figure that if you're on the equator at the equinox, then the sun will travel across the sky in 12 hours or 720 minutes. 720 divided by 8 minutes is 90. So if I break the entire path of the sun across the sky into 90 units, then each unit would represent the difference between appearance and actuality. But how much of the path of the sun in the sky is 1/90'th of the total? I don't know how big the 12 hour arc is.

Thanks,
• Topic Stats
• Top Replies
Type: Discussion • Score: 1 • Views: 13,835 • Replies: 38
No top replies

littlek

1
Wed 25 Jun, 2003 11:55 am
uh. I have no idea! Good question.
0 Replies

sozobe

1
Wed 25 Jun, 2003 12:10 pm
What do you mean when you say "sun-widths"? Actual diameter of the sun, or how we see it? Because if you mean the latter, it's not a steady measurement -- pollution et al can make the sun look bigger or smaller.
0 Replies

fishin

1
Wed 25 Jun, 2003 12:16 pm
Hmmm.. Using some quick calculations and your 12 hour figure I calculated that in 8 minutes the sun would move approx. 2 degrees across it's arc in the sky. I have no idea right off hand how you'd convert that into "apparent width of the sun" though.
0 Replies

patiodog

1
Wed 25 Jun, 2003 12:38 pm
Hmmm. I could be off here, but if we ignore the effects of the atmosphere (and thus render any calculation useless in practical terms): since the radius of the sun is about 435,000 miles and the distance to the sun is 93,000,000 miles, then the width of the sun in degrees for an observer on earth is roughly twice the arctangent of 435000/93000000, or about half a degree.

I've probably got something fouled up, though, because that doesn't sound at all right.
0 Replies

ebrown p

1
Wed 25 Jun, 2003 12:57 pm
As fishin' implied the logical way to measure the "apparent width" of the Sun is in degrees (or radians).

The width of the sun is 1.4 X 10^6 Km.
The sun is 1.5 X 10^8 Km from the Earth.

So the angular width of the sun is the width divided by the distance or .0093 radians. This can be converted to degrees by multiplying by 180/pi. This gives an apparent width of .54 degrees.

The stuff about equators and equinoxes etc. are not relevent to this problem. If you are not on the equator during the equinox the sun will not be in the sky for 12 hours, nor will it cover a distance with an arc length of 180 degrees.....

What is important is that no matter where you are on the Earth, the apparent daily motion of the sun is a circle (it is just the angle you are viewing the circle that changes.) Becuase of this you can just say the sun appears to move 360 degrees in 24 hours.

As fishin said, the sun moves 2 degrees in 8 minutes. (Since there are 1440 minutes in a day, 8 minutes is .005556 of a day. .005556 X 360 is 2 degrees)

According to these calculations, if you want to point to where the sun is now*** you must point ahead about 4 sun widths.

Notes:

1) The calculations for the apparent width of the sun are based on the actual width of the sun. The sun will probably "appear" to be bigger than this based on glare etc. The .0053 radians is probably correct for what you would see looking through a very dark filter.

2) As a physics nerd, I must point out that this question touches on relativity. Talking about "where the sun is located 8 minutes ago" is problamatic as in the real universe, time and space are both measured relative to an observer. This question is not very well defined.
0 Replies

fresco

1
Wed 25 Jun, 2003 01:41 pm
Aside from the mathematics an interesting philosophical question is the status of the word "is". It might be argued that "current position" is relative to "particular purpose" and that "actual position" is merely "for the purposes of a mathematical exercise".
The "eight minutes ago" concept is philosophically significant for its "looking at history" aspect rather than its positional implications.
0 Replies

rosborne979

1
Wed 25 Jun, 2003 01:41 pm
Thanks Ebrown. For a question which is not very well defined, you gave me exactly the answer I was looking for.

I'm aware of the implications of relativity (as you probably know from our past interactions). I was just curious to know what the 8 minute travel time "looked" like in day to day approximate movement of the sun. My logic skills are pretty good, but my math skills stink so I couldn't do the calculations.

I've heard many times that when we look deeply into space we are actually not looking across space, but back in time. And I know this is true, but it's fun trying to fit that concept in to a smaller scale (like the Earth/Sun system).

Would it be fair to say that we cannot "see" distance at all, but merely deeper and deeper slices of time (relates to Fresco's interesting comment)?

Best Regards,
0 Replies

Frank Apisa

1
Wed 25 Jun, 2003 02:37 pm
I guess one way to figure this out would be to set things up in a way that will dramatically show how far the sun travels in 8 minutes.

However far it travels in 8 minutes -- is how far it travels in 8 minutes.

A sundial setup might work -- a pole might work.
0 Replies

patiodog

1
Wed 25 Jun, 2003 02:44 pm
Well, let me go climb into my camera obscura and I'll do that for ya!
0 Replies

ebrown p

1
Thu 26 Jun, 2003 06:35 am
Ah shucks!

It turns out my analysis was a bit wrong. These types of problems tend to keep me up at night (thanks Rosborne).

I am convinced that *where* you are on the Earth is irrelevent, but the time of year does matter. The key is the elevation of the sun above the equator.

My first thought experiment (that made me realize I was wrong) was to imagine that the Sun were directly above the North Pole. In this case the Earth would spin underneath it, but the sun would appear to be motionless.

Of course the sun never makes it above the pole. Twice a year the sun is directly over the equator. It moves between this position, I call 0 degrees, and where it was last Saturday, at 23 degrees above or below the equator.

The previous analysis is correct for when the Sun is above the equator (i.e. an equinox). Any other time the apparent distance the sun travels in the sky is reduced by a factor of the cosine of the angle of elevation. Note that in my thought experiment above with a sun over the pole would have an elvation of 90 degrees and cos(90) = 0.

On a solstice (and we just had one) the real apparent motion of the sun would be reduced by a factor of .92 (the cosine of 23 degrees).

So, this time of year you should point ahead about 3.4 sun lengths.
0 Replies

Thomas

1
Thu 26 Jun, 2003 12:16 pm
Maybe I'm missing something about the previous answers. But as best I understand it, Galileo taught us 400 years ago that the sun doesn't travel around the Earth at all. The sun is really stationary, and if we think it moves, that's because of an illusion generated by the rotation of the Earth. As a consequence, the angle under which we observe the sun is determined by the current orientation of the Earth, which is information that reaches us almost immediately.

With this in mind, my answer would be that the sun is exactly where we see it, even if you account for the 8 minutes its light takes to reach us.

-- Thomas
0 Replies

patiodog

1
Thu 26 Jun, 2003 12:42 pm
The apparent position still holds, though. If, somehow, you were able to launch a rocket that would travel in a straight line from your position to the sun, and you initially aimed it where you saw the sun in the sky, you would miss. You would need to know what adjustment to make in your launch angle or whatever to make sure you were on target.
0 Replies

ebrown p

1
Thu 26 Jun, 2003 12:57 pm
Thomas,

Of course you are right about Galileo.

We are thinking about the "apparent" motion of the sun around the Earth becuase it is a bit easier to picture the Sun moving around the Earth than the Earth moving around the Sun.

But the answer would come out the same if we did this problem the correct way as you suggest.

We said the sun would travel eight minutes while the light we see if travelling to us. Looking at things your way, the Earth would revolve eight minutes worth of revolution while the light we see is travelling to us. Thus you will still be pointing in the wrong direction.

Either way the math is the same, and the answer is "correct" given the caveats I gave above.

The ancients viewed Space as a "celestial sphere". The stars were painted on this sphere with the Earth at the center. The Sphere rotated once every day (with all the stars etc.) The sun and the planets moved very slowly about the celestial sphere.

Of course the ancients used this model becuase it matched with what they observed. This model was simple, yet it explained and predicted what they observed in the sky.

This celestial sphere is still used by astronomers today, even though we now understand the new model. Astronomers still talk about the celestial sphere, and measure motion in the sky relative to the "celestial equator" and the ecliptic (the apparent path of the sun through the celestial sphere over the course of a year).

Modern astronomers do this becuase the motion of the celestial sphere much easier to think and talk about than the simultaneous multiple motions of the rotation of the Earth, its orbit aound the sun, and the Earths precession etc. As long as we understand and keep in mind that this is a simplification, it is a helpful shortcut.

For questions of celestial geometry it comes up with the right answers and makes life a little easier.

When I was teaching astronomy in high school I used a globe that was a model of the celestial sphere. It was transparent and had the stars painted on the inside. There was a small model of the Earth in the center and a little sun that moved along the ecliptic. This model is invaluable for explaining the motions in the sky. It is also instructive to compare these "apparent" motions to the "real" galilean model.

http://inkido.indiana.edu/a100/celestialsphere.html
0 Replies

Thomas

1
Thu 26 Jun, 2003 01:12 pm
Patiodog and ebrown:

I would agree with you if your complaint was that I ignored the Earth's 8 minutes worth of rotation around the sun. (Not that it's a bad approximation -- 8 minutes isn't much on the calendar.) But if you believe that the mismatch would hold if you consider just the Earth's rotation around its own axis, I'm afraid I'll have to disagree with you. If you ignore the Earth's rotation around the sun, you will find that a rocket fired in the sun's current direction will indeed reach the sun.

-- Thomas
0 Replies

ebrown p

1
Thu 26 Jun, 2003 02:07 pm
Ah Thomas, I believe you are correct. And, I have sheet of paper filled with doodles of circles men and lines to prove it. (I love these sorts of problems!)

My current way of thinking is that the 8 minute old light will come from the same direction that the new light will come from. The new light will show up with a different "apparent" angle from the horizon. But I think, as Thomas argues, that this will not matter when you are pointing (or sending a rocket) at the sun.

I will post a more complete version of my thoughts later.

Thanks Thomas for your points. Patio et al. I would like to hear your response.
0 Replies

patiodog

1
Thu 26 Jun, 2003 02:12 pm
That's absolutely correct; I wasn't thinking of the problem spatially. The light from the sun is coming from the same point even if it takes 1000 years to reach me.
0 Replies

patiodog

1
Thu 26 Jun, 2003 02:14 pm
(which is why i try to remind people regularly not to put any faith in anything i say or post....)
0 Replies

rosborne979

1
Thu 26 Jun, 2003 02:36 pm
In a "clean" thought experiment which ignores gravitational distortion and other intangibles, the light travels in a straight line, but the rotation of the Earth (and an observer) on its axis causes a deflection between apparent Sun position and actual Sun position. So if one were to draw the apparent flow of light (not the actual flow of light), it would be a curve not a straight line.

I believe there is a difference between actual sun position and apparent position in the sky. Are we in agreement on this? Or are we talking about something else now?

I think I'm getting confused.
0 Replies

Thomas

1
Thu 26 Jun, 2003 02:38 pm
To clarify, here's what I think will happen if you fire a rocket toward the sun. After 8 minutes, the sun will have changed its apparent position somewhat, just as you were saying. At the same time though, the rocket will have changed its apparent direction, and it will have changed in just the right way to make it still fly towards the sun. It's easy to see why if you look at it from the sun's perspective. From the sun's perspective, only the Earth's orientation has changed during these 8 minutes. The sun's position and the rocket's direction haven't.
0 Replies

### Related Topics

Evolution 101 - Discussion by gungasnake
Typing Equations on a PC - Discussion by Brandon9000
The Future of Artificial Intelligence - Discussion by Brandon9000
The well known Mind vs Brain. - Discussion by crayon851
Scientists Offer Proof of 'Dark Matter' - Discussion by oralloy
Blue Saturn - Discussion by oralloy
Bald Eagle-DDT Myth Still Flying High - Discussion by gungasnake
DDT: A Weapon of Mass Survival - Discussion by gungasnake

1. Forums
2. » 8 light minutes to the sun, so where is it now?