I think that self-reflection alone is a good reason for making the comparison, although I didn’t anticipate such a strong reaction to a statement that everyone now seems to agree is true. I just thought it was an interesting, and little known, fact.

It seems to me that the problem stems from the fact that your comment about the difference between Earth’s actual orbit and your hypothetical circle comes in a thread that is about perihelion, and hence about the way Earth’s distance from the sun changes over the year.

Within that context, people’s minds are already focused on the aphelion versus perihelion comparison, which is a difference of just over 3,000,000 km.

It is therefore natural to assume that the circle to which you alluded would indeed be a circular orbit with a period of 365.25 days, and such a circle would deviate from Earth’s actual elliptical orbit by those distances others have stated. It is not obvious that the circular orbit to which you alluded was simply a circle centred at the centre of Earth’s elliptical orbit with a radius equal to the mean of the semi-major and semi-minor axes. If your hypothetical circle doesn’t have a period of 365.25 days, then what is the point of making the comparison? You might just as well state that Earth’s orbit has an eccentricity of only 0.02 (source: nineplanets.org).

There were no “unstated assumptions” in my post. To find the circle that best fits an ellipse, with the least deviation from the circle, you use the center of the ellipse, and the average of the semi-minor and semi-major axis for the radius. You don’t have to assume anything, just calculate it from the known facts. If you do it any other way, you’ll get a worse fit. That’s a fact.

I think the real problem comes down to this: To what circle are we comparing the ellipse? A circular orbit with the same period as Earth’s (whose orbital radius would be, according to Kepler’s Third Law, equal to the semi-major axis (a) of the actual orbit)? If we position this hypothetical circular orbit’s center at the center (not the focus) of the true (elliptical) orbit, we’d find that the maximum deviations between it and the actual orbit (at the ends of the minor axis) would, indeed, be 20,832 km, which is closer to twice the Earth’s diameter of 12,756 km than it is to the diameter.

A circular orbit (again, concentric with the true orbit) which cuts inside the true orbit at the major axis by the same amount that it swings outside the minor axis would orbit faster than the Earth). Seems like a meaningless comparison, to me.

“nothing you’ve contributed contradicts what I said.”
After you make a clarification that wasn’t in your original comment. It was a poorly phrased statement if you intended it to be unequivocal. You also have a radius that no physicist who teaches this stuff would ever use.

“The center of earth’s orbit is not the sun”
Didn’t say that it was. I said the Earth orbits the Sun, which in Newtonian terms and Keplerian terms is what any reasonable physicist would assume. It’s an orbit around the center of mass of the system, and that is the more important center vs the geometrical center of an ellipse. And yes, I realize that the center of mass moves as the Earth orbits, but you work the problem using a reduced mass and fixed CM.
” anyone who objects to that has a problem with Newtonian physics.”
Vice-versa. The geometrical center may not be the sun, but the gravitational and angular momentum center IS the center of mass of the system which is in the Sun, not too far from the center. Angular momentum about the center of the ellipse is NOT conserved, and that presents a problem for Newtonian physics, unless you believe that central forces don’t conserved angular momentum.
” The center of an ellipse is different than the focus of an ellipse”
OK, that’s geometrical and obvious.
“for the earth, the focus lies within the sun–which is not necessarily the case, Jupiter’s is outside the sun.”
And your point is ?

(((much study of axis of ellipses & such to be sure I’m not making a fool of myself)))

OK, I suppose the orbit itself is quite circular as you point out, the Sun is just not in the center of the circle/ellipse. Indeed, if you look at the shape of the orbit Earth makes, it would fit into a very fine line on a piece of 8.5″ paper, as in, .002 inches, So I learned sumthin here. Also, the difference in radius on the ellipse is 21,000 KM, the radius of the Earth is 6,375 KM, so as others have mentioned the Earth does go outside itself, but certainly not by 3.1 million miles.

So lets agree on this…Put a 1/16th inch dot on a piece of paper (Sun) then draw an incredibly fine line .002″ (couldn’t see it without a microscope) in a perfect 8.5″ circle centered 1/8th inch from the 1/16th inch dot.

You thought you would have your way
But orbital mechanics has the final say.
So take your cold and snow and ice
And hit the road: we want the weather to be warm and nice.

It will be near 22.5 degrees South, about 11 hours ahead of Greenwich, so about 165 degrees E.? Which would put it near New Caledonia in the Coral Sea (E. of Australia, NW of New Zealand).

The center of earth’s orbit is not the sun, anyone who objects to that has a problem with Newtonian physics. The center of an ellipse is different than the focus of an ellipse; for the earth, the focus lies within the sun–which is not necessarily the case, Jupiter’s is outside the sun.

That’s a poor definition of your circular orbit. Its radius is neither perihelion, aphelion, semi-major, not semi-minor, so it wouldn’t have the same revolutionary period as Earth. Sorry, what you have is a fictoid.

This is a prime example of a poorly stated idea. There are too many unstated assumptions, making it ill-defined. These types of statements lead to BAD ASTRONOMY.

Misleading at best, here are two reasons I consider it wrong.
1) You shouldn’t “split the difference” between the semi-major (a) and semi-minor(b) axes because they are already “split.” If you match a of the ellipse with r of the circle then when you get to b the difference is (like Tim Gaede says) about 21000 km.
2) While the path defined by Earth’s orbit, if centered on a circle with same radius as the semi-major axis would have the 21000 km difference, the ORBIT of the Earth is NOT around the mid-point of the major axis. The orbit is around the Sun at the focus. Matching the Sun of Earth’s orbit with the Sun of a circular orbit, there are millions of km of difference.

Kepler determined that planetary orbits were elliptical based on Tycho Brahe’s very precise observations of Mars, whose orbit is much more elliptical (eccentricity = 0.093) than the Earth’s orbit.

The Earth is 3.4% (2*0.017) further from the Sun in July than in January and, by the inverse square law, the heat received from the Sun 6.8% greater in January than in July.
By the fourth-power relationship between radiation energy and temperature, the Earth, if it were airless like the Moon, would have a mean temperature 1.7% greater, on the Kelvin scale, in January than in July. With the mean temperature being around 280K, this would
be 4.8 degrees Kelvin warmer in January than in July, which is equal to 4.8 degrees Celsius or 8.6 degrees Fahrenheit.

As has been pointed out, the oceans moderate this difference, especially in the southern hemisphere.

But if you look at climate of towns and cities in Africa, near the equator, you can usually tell that July is a few degrees cooler than January. It would be a good science project to plot the difference versus latitude for a number of such towns and cities, and fit the trend line to see what the mean difference is right at the equator.

The “width” (double the semi-major axis) of the ellipse should be only 0.014% more than the “height” of the ellipse. The Earth is 3.4% farther from the Sun at aphelion than at perihelion.

The difference between the semi-major and semi-minor axis is actually 21,000 km, which is more than the diameter of the Earth but still small compared to the distance to the Sun.

I wonder if Kepler actually considered off-center circles before looking into ellipses.

The semi-major axis of the earth’s orbit is 149598000 km, and it’s semi-minor axis is 149577000 km, so splitting the difference between them means a little over 10,000 km, much less than the diameter of the earth.

You could draw a circle with a .0005 inch line and it would fit inside.

If you draw a circle on a piece of paper with a 1/4 inch (7mm) line the Earth does stay inside that circle.