Type of diagram the questioner used to try to prove Kepler's 2nd law of planetary motion was violated for Earth.
One thing that makes volunteering at 'All Experts' interesting is a particular class of questions received: usually when a person believes he or s he has disproven a fundamental physical or astronomical law or principle. Or, alternatively, a phenomenon reserved for stars - like the Sun - also occurs on Earth. See e.g.
Recently this came up again when a 59 year old ENT specialist claimed he'd shown Kepler's 2nd law of planetary motion doesn't hold for Earth. To achieve that, he used a conventional diagram such as shown above - sketched by Janice.
Anyway, he asked me to look at two time intervals: Mar. 21 to Jun. 21 and Sept. 21 to Dec. 21. He then made reference to the areas not being equal, i.e. the first area was "much larger" than the second. His other complaint was that for the given diagram the Sun was shown at the center instead of at a focus. As we know all planets have elliptical orbits, e.g.
With the Sun (S) at one focus. For such an orbit we have a varying radius vector, r, instead of a constant radius as for a circle. For example, the value of r at point P = 0.98 AU and at point A = 1.02 AU.
Thus the planet is farthest from the Sun and point P when closest. At P there is the perihelion velocity:
V_P = h/ a(1 - e)
where a is the semi-major axis (1 AU = 149.6 million miles), and e is the eccentricity.
Similarly, at A we have aphelion velocity:
V_A = h/ a (1 + e)
Where the value of e for Earth is 0.0167, or scarcely different from a circle (0).
As I tried to explain to the person, such conventional 'season' diagrams - as depicted at the top - aren't drawn to scale because the emphasis is on showing the axial tilt remaining constant through the orbit. So one is not going to try to put the Sun at the focus of an ellipse or Earth's elliptical orbit given that's of secondary import. Indeed, a much larger divergence is in the relative sizes of Earth and Sun! (The Sun is 864,000 miles in diameter vs. the Earth's 7,900 miles) and also the distance of Earth to Sun.
Now, If Kepler's 2nd law holds at every point (equal areas swept out in equal intervals of time) we also have:
r2 (2π/T) = h
where 'h' is a constant ('specific relative angular momentum') which is twice the rate of area description (i.e. by the radius vector). Thus, if the radius vector is r1, then h = 2A1, when the radius vector is r2, then h = 2A2.
A neat animation showing the law can be found at this link:
Note how the area is narrower (as a sector of the ellipse) at greater distance and wider (greater angle q = q2 - q1) at lesser distance.
So what was the error made by the person? As I pointed out to him:
From your specific examples of time intervals, i.e. March 21 to Sept. 21, and Sept. 21 to Dec. 21, it is evident to me that your construal of “unequal” areas is based on the misconception that the dates of solstices and equinoxes are fixed when they are not.
For example, the date for the winter solstice which you fix as Dec. 21 can actually be any of: Dec. 20, 21, 22 or 23. That is as much as a 3 day spread which will make a significant difference in your computations. The same applies to your dates from March 21 – which can also vary, and June 21, ditto.
My point is that when the proper specific dates are used (for the given year – they change year to year) you will find the same areal ‘map out’ for each quarterly interval of the orbit – even with the differing distances factored in.
Thus, it follows that when the greater distance (radius vector) is entered for one interval, the comparison interval (in days, and hence degrees per day) will be counter -balanced by a different date spread, which would have to be larger to compensate for the lesser r.
Again, the error is in presuming fixed dates to mark the termination and initiation points of your orbital intervals.