Selected Questions - Answers From All Experts Astronomy Forum (Abolish Kepler's 2nd Law?)

Question:  If we relocated the Sun in one focus of the orbit - say from the diagram in the link below

http://starchild.gsfc.nasa.gov/docs/StarChild/solar_system_level2/activity/song.html

we will discover that the straight line between the Earth and the Sun sweeps the interval of time between March 21 and Jun 21 is much bigger than that swept between September 21 and December 21, and the distance crossed by the Earth from March 21 to Jun 21 is also bigger than that crossed from September 21 to December 21.  This is  contrary to Kepler's second law so I say it must be  abolished.   Try it by yourself.. it is inevitable!  Do you concur?
regards,     Dr. Mohammed Barzaq

Answer:

The foci-adjusted mean daily motion for the Earth, from a celestial mechanics table, is 0.9856874 deg/day. When this is multiplied by the correct time interval for each given “quarter” orbit the same area will be obtained. There is no "violation" of the 2nd law.

With sufficient accuracy each area mapped out in accord with the 2nd law is given by:

½  r ²  (D  Θ)

Where Θ is the angular difference between t2 and t1 in the orbit..

The rate of areal description (mapping)  is then usually divided by (D t) so:

A =   ½  r ²  (D  Θ)    / (D t)

But, since this rate is constant (by the 2nd law) we may write:

h  =   r ²   d ( Θ) / dt

where h, a constant is 2x the rate of mapping of area by the radius vector.  We may also express it in terms of three constants c1, c2 and c3 defining the orientation of the orbital place, e.g.

h =  [c1 ² +  c2 ²   +  c3 ²]  ^½)   

 Thus, n, the mean daily motion is the mean value of  d ( Θ) / dt

 For all points in the orbit.

Re: 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 absolutely  fixed when they are not.

 For example, the date for the winter solstice which you fix as Dec. 21 can actually be any of the dates: 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 (vernal equinox)  – which can also vary, and June 21 (summer solstice) , 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, e.g. for the winter solstice – which would have to be larger (i.e. Dec. 23 >   Dec. 21), to compensate for the lesser r.

Again, the error is in presuming fixed dates to mark the termination and initiation points of your orbital (seasonal)   intervals.

Selected Questions- Answers From All Experts Astronomy Forum (Relative Distances of Superior, Inferior Planets)

Question: Can you explain how one can find the relative distance to an inferior or superior planet, i.e. in the solar system?

Answer:  Let's first define geometrically the inferior and superior planet based on the diagram shown below:
















As seen from the diagrams, an inferior planet defines one which is interior to Earth's orbit, relative to the Sun. A superior planet defines one which is exterior to the Earth's orbit.

While  it's easy to apply Kepler's 3rd law to obtain the basic dimensions of an orbit, namely the semi-major axis (a)  of the orbit - or the mean distance from the Sun, things become somewhat more difficult when we seek to find the distance say, of the Earth to the planet.

So, we consider two cases:

(A) Inferior planet (see the diagram A)

(B) Superior planet (Diagram B)

Case (A):

The maximum elongation occurs when the planet's geocentric radius vector (pE in diagram A) is perpendicular to the planet's heliocentric radius vector, pS. Then,  by a careful measurement of the angle SEp, say over a series of nights around maximum elongation, one can obtain a value for the angle of maximum elongation. At such time the angle SEp is right-angled hence:

Sp/ SE = sin(SEp)

The quantity Sp/SE is therefore the distance of the planet from the Sun in terms of Earth units (or AU, astronomical units, where 1 AU = 1.5 x 10¹¹  m). Let Sp = R and SE = a(E) the semi-major axis for Earth's orbit, then:

Sp = R = sin(SEp) [a(E)]

Example:

If the angle SEp = 60 deg, find the planet's distance from the Sun.

Then: sin(SEp)= sin (60) = Ö3 /2 = 0.866

So: R = 0.866[a(E)]= 0.866 AU  

Case (B):

Here, we let the planet p be in opposition at some given time with the Earth and Sun (e.g. showing the alignment S-E-p in diagram (B). As we know, with opposition, the elongation is a straight angle or 180 degrees. Then after t days have elapsed the Earth's radius vector SE has moved ahead of the planet's as shown in comparing SE1 to Sp1. As can be seen, this reduces the angle of elongation from 180 degrees at opposition to angle SE1p1. This is then measured.

Now, over t days, the angle ESp will have increased from 0 (at opposition) to a value Θ given by:

Θ = [n - n(p)] t

where n, n(p) are the mean daily motions of the Earth and the planet, respectively. Using relations for the periods (from the previous answer) we may write:

Θ = 360 (1/P - 1/P') t

where P and P' are the sidereal periods for the Earth and the planet, respectively.  Then, it follows by the synodic/ sidereal relations seen in previous answer. 

Θ = 360 (1/S)

So, since t and S are both known, Θ can be obtained - that is, angle E1Sp1 is calculated. Hence, angle E1p1S can be found from:

angle E1p1S = 180 - angle SE1p1 - angle E1Sp1

From plane trigonometry we then obtain:

sin(p1E1S)/ Sp1 = sin(E1p1S)/SE1

or:

Sp1/ SE1 = sin(p1E1S)/ sin(E1p1S)

Again, giving the distance from the Sun in terms of Earth's distance unit. 


To fix ideas here is a further  example problem:

Estimate the distance of Venus from the Sun at its most recent maximum elongation, if the angle of maximum  elongation was 46 degrees.

Solution:

The diagram is shown below with the angle of maximum elongation.


We know, from the geometry:

SV/ SE = sin (SEV)

Then Venus' distance is:

SV = (sin(SEV)) SE

where: angle SEV = 46 deg and SE = a(E) = 1 AU

then:

SV = sin (46) 1 AU = (0.719) 1 AU = 0.719 AU