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PH12 LAB 12


Mercury, the innermost planet, is never very far from the sun in the sky.  It can be seen only close to the horizon, just before sunrise or just after sunset, and viewing is made difficult by the glare of the sun.  Except for Pluto, which differs in several respects from the other planets, Mercury has the most eccentric planetary orbit in the solar system (e = 0.206).  The large eccentricity of Mercury’s orbit has been of particular importance, since it has led to one of the tests for Einstein’s general theory of relativity.  For a planet with an orbit inside the earth’s there is a simple method to plot its orbit which you will use here.

Assume a heliocentric model of the solar system.  Mercury’s orbit can be found from Mercury’s maximum angle of elongation east and west from the sun as seen from the earth on various known dates.  The angle (see figure 1) between the sun and Mercury, as seen from the earth, is called the elongation.  Note that when the elongation reaches its maximum value, the sight lines from the earth are tangent to Mercury’s orbit. 

Since the orbits of Mercury and the earth are both elliptical, the greatest value of the elongation varies from revolution to revolution.  Table 1 gives the angles of a number of these greatest elongations.                                                                                      


Plotting the orbit:                                                                                  

Start by drawing the earth’s orbit.  Since the earth orbit eccentricity is very small (e = 0.017) we can use a circle with the sun at the center without introducing much error.  Draw the earth’s orbit as a circle of radius 10 cm.  Draw a reference line horizontally from the center of the circle to the right.  Label the line 0 degrees.  This line points toward the vernal equinox and is the reference from which the earth’s position in its orbit on different dates can be established.  The point where the 0 degree line from the sun crosses the earth’s orbit is the earth’s position on its orbit about September 22.

The earth takes about 365 days to move once around its orbit (360 degrees).  Use the rate of approximately 1 degree per day, to establish the position of the earth on each of the dates given in the table.  You should make a table in your labtop (Excel, of course) where you can record these angles.  Note that the earth moves around this orbit in a counterclockwise direction, as viewed from the north celestial pole.  Label each position with the date and draw a radial line from the sun to the earth position for each date you have located.

From these positions for the earth, draw sight lines for the elongation angles.  Be sure to note, from figure 1, that for an eastern elongation, Mercury is to the left of the sun as seen from the earth, while for a western elongation, Mercury is to the right of the sun.

You know that on a date of greatest elongation Mercury is somewhere along the sight line, but you do not know exactly where on the line to place the planet.  You also know that the sight line is tangent to the orbit.  A reasonable assumption is to put Mercury at the point along the sight line closest to the sun.  The shortest distance between a point and a line is along a line perpendicular to the original line.  Use your protractor to draw the perpendicular line for each sight line.

You can now find the orbit of Mercury by drawing a smooth curve through, or close to these points.  Remember that the orbit must touch each sight line without crossing any of them.

Finding ravg.  The average distance from the sun of a planet in an elliptical orbit is equal to one half the long diameter (the major axis) of the ellipse.   To locate the major axis of the orbit you must first find the aphelion and perihelion points of the orbit.  You can use a drawing compass with its point on the sun to find these points on the orbit that are farthest from and closest to the sun.  Mark and label them!  Measure the greatest diameter of the orbit along the perihelion-sun-aphelion line (this is the major axis).  Since 10 cm corresponds to 1 AU, you can now obtain the semimajor axis of Mercury’s orbit in AU.  Show your calculations and answer.  Calculate your % difference from the accepted value. (Which you can find in your text).

Recall that eccentricity is defined as e = c/a. (See figure 2.)  Since 2c, the distance between the foci is small, you lose accuracy if you try to measure it directly.  Instead, note from figure 2 that c is equal to the major axis a minus the perihelion distance rp ( c = a - rp).  So e = c/a = (a - rp)/a = 1 - rp/a

You can measure rp and a with reasonable accuracy from your plotted orbit.  Remember that a is one half the long diameter.  Compute e and compare your value with the accepted value, e = 0.206. (Find the % difference.)

Table 1:  Some Dates and Angles of Greatest Elongation for Mercury




Jan 4, 1963

19o E

Feb 14

26o W

Apr 26

20o E

Jun 13

23o W

Aug 24

27o E

Oct 6

18o W

Dec 18

20o E

Jan 27, 1964

25o W

Apr 8

19o E

May 25

25o W