Astronomy 1L (Lab Class) Information
Mercury Orbit Project: Part 3a Link for sharing this page on Facebook
Finding the Orbital Period of Mercury
     After plotting Mercury's path, you will find that it is sometimes below (to the South of) and sometimes above (to the North of) the Ecliptic. This occurs because the planet's orbit is inclined (tilted) relative to our orbit, which defines the plane of the Ecliptic. As seen from the Sun, each planet's path appears to be a circle. Ours is the same as the Ecliptic, but Mercury's is tilted relative to ours, by an angle called the inclination of the orbit. Someone of the Sun would see Mercury below the Ecliptic on one half of its path, and above the Ecliptic on the other half, crossing the Ecliptic at one of two places, exactly opposite each other in the sky, called the nodes of the orbit. The one where it is moving northwards is called the Ascending Node, and the one where it is moving southwards is called the Descending Node.
      Because the Earth is not in the same place as the Sun, we do not see the movement in the same way. Mercury's path appears to be much more complicated, having back and forth movements, called retrograde loops, in addition to the North-South movement. But since we are always in the plane of our own orbit, we do share one thing in common with the Sun. When Mercury is above our orbit, we see it above the Ecliptic, and when is below our orbit, we see it below the Ecliptic, and, at the moments when it is at one of its nodes, we see it exactly on the Ecliptic. If Mercury is moving northwards where we see it cross the Ecliptic, it must be at the Ascending Node. If it is moving southwards where we see it cross the Ecliptic, it must be at the Descending Node.
      To find Mercury's orbital period, we need to find a time when Mercury is at a particular place in its orbit, and then find the next time when it is exactly in that same place. For most positions in its motion, the fact that we have a different view from that of the Sun makes being sure exactly what part of the motion we see corresponds to a particular part of the motion as seen from the Sun, but when it is at one of its nodes, we can be absolutely certain of its position, so we can use those places to find its period.
      At the beginning of 1982 Mercury was South of the Ecliptic, moving slowly northwards. On January 17th (Julian Day 2444986.5), it was still below the Ecliptic, but quite close to it, and by January 21st (Julian Day 2444990.5), it had crossed the Ecliptic, and was slightly to the North of it. Therefore, it must have passed through the Ascending Node sometime between those two dates. To find the exact time when it cross, you must divide the arc between the two dots which represent Mercury's position on those two dates into four equal time intervals, corresponding to the four days between the two dates.
      If the motion of Mercury during this four-day period were uniform, then dividing the arc into equal time intervals would be quite easy. You would simply divide it into equal distances, each of which would correspond to the same amount of time. But on the 17th of January Mercury was approaching a stationary point, and its daily motion was rapidly slowing down, so that it moved much further during the four days prior to the 17th than during the four days after the 17th, and still less during the four days after that. Therefore, during the first day or so after the 17th, it would have moved further than during the last day before 21st, meaning that if the arc between the two dots is divided into equal distances, it will not correspond to equal times.
      It is possible to adjust for this, by doing a visual estimate of the changing distance, and trying to guess how far the planet would move each day, but a simpler way would be to use the graph of latitudes which was done in the second part of this project. Since that graph has a uniform time scale, seeing where the latitude curve crosses the Ecliptic (0 degrees latitude) will directly reveal the time when this occurred. Unfortunately, the scale of the latitude graph is much smaller than the original graph, so the distance corresponding to four days may be too small to accurately estimate the time of the crossing, but if you examine the graph, and see approximately where the nodal passage is, then replot just that part of the graph at a larger scale, you can bet a better estimate, and compare it to the one from the original graph, to see whether the two times agree.
      So, what you need to do to find the orbital period is as follows. On both the original graph of the motion of the Sun and Mercury, AND on the graph of latitude, find every place where Mercury crosses the Ecliptic, heading either North or South, and estimate the date of each nodal passage, accurate to the nearest tenth of a day (you will have to estimate the decimal fraction of a day, but if you don't, and just round off to the nearest whole day, you will make the entire exercise completely meaningless). For each graph, make the best estimate of the nodal passage dates that you can, and if the two estimates of the same nodal passage date do not exactly agree, decide whether one of the graphs provides a notably more accurate value, and use only that one, or if they both seem equally good (or bad), take the average. Next, list the dates in chronological order, as shown below, dividing them into northward passages (Ascending Nodes) and southward passages (Descending Nodes), as shown, and subtract each date of a given sort from the next date OF THE SAME SORT. The six nodal passages of each sort will give you five differences, for a total of ten values, as shown below. Please note that the numbers in this table are made up, only for the purposes of illustrating the technique, and are NOT the same as those which you will obtain.

Ascending Node Date
2444987.6
2445075.8
2445161.1
2445253.4
2445341.1
2445429.2
Period

88.2
85.3 ???
92.3 ???
87.7
88.1
Descending Node Date
2445030.6
2445118.2
2445205.6
2445293.5
2445381.5
2445470.3
Period

87.6
87.4
87.9
88.0
88.8
      If your nodal passage dates were perfectly accurate, all of the periods shown above would be exactly the same as the orbital period , but as you can see, this is not likely to occur, because of the difficulty of dividing the intervals into equal time increments, and of estimating times to tenths of days. In addition, there may be some terrible blunders, such as the values marked ??? above. In this case, I have deliberately imitated what would happen if one of the dots for Mercury were marked with the wrong Julian Date, causing a four-day error in the nodal date, which should have been 2445165.1, at least for this example. This sort of error produces an unusually small difference followed or preceded by an unusually large difference. When such a discrepancy occurs, you should check to see if there is some reason for this problem, and correct the value if it proves appropriate to do so. However, you should realize that it is NOT appropriate to change a value just because you don't like it. Data can ONLY be ignored or changed if there is a very good reason to do so. Changing data just because they don't give the values you want is considered intellectually dishonest, and people who have deliberately done this and been caught have had their careers destroyed. In the imaginary case above, however, it is appropriate to correct the bad data, as shown here:

Ascending Node Date
2444987.6
2445075.8
2445165.1
2445253.4
2445341.1
2445429.2
Period

88.2
89.3
88.3
87.7
88.1
Descending Node Date
2445030.6
2445118.2
2445205.6
2445293.5
2445381.5
2445470.3
Period

87.6
87.4
87.9
88.0
88.8

Obtaining an Average and Standard Deviation
      When you have several experimental values which are supposed to represent the same quantity, but are not the same (as is normal), you need a way to find the best possible estimate of the correct value, and some judgment as to how good (or how bad) that estimate is. Under normal circumstances, the best value you can estimate is the average of the data, and the best estimate of the quality of the result is the standard deviation of the values.
      To do this for the orbital periods, take the ten values obtained from the nodal passage tables and put them into a new table (which should be on the same sheet of paper, for easy comparison), as shown below. Take the average by adding the values together and dividing by the number of values (in this case, 10).
      To obtain the standard deviation, subtract the average just obtained from the individual data values. The result is entered as the deviation, as shown below. If we knew that the average was exactly the same as the correct value, we could call this difference the error of the individual values, but since it is not likely that the average is exactly correct, the difference is called the deviation, instead. As a check on your arithmetic, you can find the sum of the deviations, which should be zero unless you rounded off the average, in which case it should be round-off error times the number of values.
      After obtaining the deviations, take the square of each one, and find the sum of the squares, as shown below. Then fine the "mean-square" of the deviations by dividing the sum of the squares by the number of values MINUS ONE. Why you do it this way, instead of dividing by the number of values, has to do with the theory of random errors, as will be explained in class. The standard deviation is the "root-mean-square" of the deviations, or the square root of the mean-square deviation.
      The best value for the period is the average, plus or minus the standard deviation, as shown at the bottom of the table. Note that the standard deviation is rounded off to one or two non-zero digits, and the average is then rounded off to the same number of places as the standard deviation.

Orbital Period

88.2
89.3
88.3
87.7
88.1
87.6
87.4
87.9
88.0
88.8
Deviation   

.07
1.17
.17
-.43
-.03
-.53
-.73
-.23
-.13
.67
Square of Deviation  

.0049
1.3689
.0289
.1849
.0009
.2809
.5329
.0529
.0169
.4489
SUM = 881.30.002.9210
AVG = 88.13 Σ / 9 = .3246
Standard Deviation (σ = √ (Σ / 9) = .5697
ORBITAL PERIOD = 88.1 +/- .6 days
(1 digit for standard deviation)

ORBITAL PERIOD = 88.13 +/- .57 days
(2 digits for standard deviation)

Next: Part 3b