1. The eccentricity of the Mars Elliptic Orbit is e = 0.092 and the distance from the sunto the aphelion is Ra = 1.665 AU. A figure (drawn to scale!) of this orbit is shownbelow. It is easy to see (from the figure) that Mars does not orbit in a circle aroundthe sun. It is much more difficult to see that Mars does not move in a circle about thepoint “C” (as is clear from the figure). Indeed, Kepler played with this as a possibilitybut eventually ruled it out.

To show that this figure is not a circle we need to show thatRCM = the distance to mars 6=RCA.

(a) What is an AU? What is the distance to the Mars aphelion in km?

(b) What is R̄ = (Rp +Ra)/2 in AU.

(c) Explain why for an ellipse L = R̄. (Hint, what is an ellipse?)

(d) Show that sin θ ' 0.092 (Hint, find the distance between the center and the Sunfirst, and then find the angle)

(e) Determine RCM in AU (use geometry and cos θ).

(f) Show that RCA/RCM = 1.0042 as claimed in lecture.

An approximate formula (which goes beyond the math of this course) shows that theratio RCA/RCM ' 1 + e2/2. The fact that the orbit is almost circular (though notaround the sun) is what made the Ptolemaic system quite successful.

Solution

(a) 1 AU = 1.5 × 108 km. The aphelion is Ra = 1.665 AU = 1.665 × 1.5 × 108 km =2.5 × 108 km

(b) UsingRa = (1 + e)R̄

We have with e = 0.092

1.665 AU = (1.+ 0.092)R̄ ⇔ R̄ = 1.52 AU

(c) see note

(d) see note

(e) see note

(f) see note

2. Describe qualitatively the funny way that the planets move in the sky. Give a qualita-tive explanation as to why they move this way.

• Solution. Relative to the stars the planets seem to move throughout the year.Most of the time they seem to go forward with respect to the stars. But sometimes

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Copernican Explanation for the Motion of Mars (Source hyperphysics)

• Video due to Dr. Steven J. Daunt Figure 2:

3

In general see different constellations at different times

The twelve zodiac signs (”houses”) are the constellations that you see overhead at

“midnight” at different times of the year

This image was copied from Nick Strobel’s Astronomy Notes. Go to his site at www.astronomynotes.com for the updated and corrected version.

Figure 3: The signs of the zodiac

they seem to undergo a backward motion relative to the stars known as retrogrademotion. A picture of the path of Mars relative to the stars is shown in Fig. ?? Todescribe retrograde motion we use the figure Fig. ?? from the lecture: As earthgoes around the sun, since it moves faster than mars, the angle it makes in thechanges with respect to the stars.

3. Why do all the heavenly bodies (the earth, the sun, the moon, and the planets, as wellas the zodiac signs) move on the ecliptic.

• Solution. This is because all of the sun, the earth, and the moon all lie on thesolar plane. The zodiac signs are those constellations which also lie in the solarplane. As the year evolves and the earth goes around the sun the signs that yousee in the sky change too. The relevant picture which explains the motion of thezodiac signs is in Fig. 1 .

4. Draw a set of pictures approximately to scale showing the sun, the earth, the moon,α-centauri, and the milky way.

• Solution. See notes

5. We plot objects of very different size on a log scale. In a log scale, one plots the log(base 10) of the distance on the x axis. (However, the labels indicate the number itself.Making it easy to plot.) If we have 5 objects of size 0.1 AU, 1 AU, 10 AU, 100 AU,1000 AU, these size are equally placed on a log scale, since they differ by a common

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multiplicative factor. Formally, the log (base 10) of these numbers is -1, 0, 1, 2, 3,which are equally spaced. This is shown in fig (a) However 4 objects of size 1, 2, 3, 4AU, do not appear equally spaced on a loga scale, see fig (b). This is because the logof these numbers is, 0, 0.301, 0.477, 0.60, which are not equally spaced. The (uneven)small-tick marks show the locations of 1,2,3,4 on the log scale. Plot the size of thefollowing objects on the log scale below, fig (c): (i) the radius of the earth, (ii) theradius of the sun, (iii) the earth-moon distance, (iv) the earth-sun distance, (v) thesun-Saturn distance, the distance to the nearest star in our galaxy, α-centauri.

• Solution. These things have size

(a) RE = 6000 km = 4× 10−5 AU

(b) R� ' 700, 000 km ' 4.6× 10−3 AU

(c) The earth moon distances REM ' 400, 000 km ' 2.6× 10−3 AU

(d) The earth sun distance 1 AU = 1.5× 108 km.

(e) Saturn 10 AU.

(f) α-centauri is 4.2 ly. Using 1 year' π × 107s and c = 3 × 108 m/s we get4× 1016 m = 2.6 × 105 AU

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