ASTR 217Name:
Lab 2 Planetary Orbits In Terms of Kepler’s 3 Laws of Planetary Motion
Kepler found that the planets all have elliptical orbits. He was the first to discover this.
- What are the other two laws of Kepler with regard to planetary orbits? Write them down here.
An ellipse is shown below. Draw an ellipse on the next page by putting the page on a piece of cardboard, putting pins or thumbtacks in each focus (big black dot) and tying a loop of thread or string so that the ellipse you draw stays within the left and right boundaries of the grid. Then answer the questions on that page.
Semi- means half of, so the semimajor axis is half of the major axis and the semiminor axis is half of the minor axis. The semimajor axis of a planet’s elliptical orbit is its average distance from the Sun.
Note: To make equal areas in completing the instructions on the page after next, imagine that the long swept area drawn out to the right of the Sun is, to a close approximation, a right triangle, and the area of a triangle equals one half of the base times the height: A = ½(BH).
Draw an ellipse on the grid above. Have the piece of paper on a piece of cardboard backing. Put thumbtacks through the two foci of your ellipse (the two black dots). Tie a loop of string that goes around the thumbtacks and, when stretched tight, does not stretch beyond the right and left-hand-sides of the grid. Draw the ellipse with a pencil, holding the string, with its non-slipping knot, taught around the thumbtacks while you do so.
2. Use a ruler, measure in cm or mm. What is the minor axis of your ellipse?
What is the semiminor axis of your ellipse?
What is the major axis of your ellipse?
What is the semimajor axis of your ellipse?
3. What is the average distance from the ellipse to one of its foci?
Draw a straight line from the Sun (left focus, black dot) towards the letter A. Stop the line from the Sun when at the ellipse. Do the same with a line from the Sun towards the letter B. The area in-between is a “swept area.” Lightly shade in your swept area. (You should be using a pencil, not a pen.)
4. What is the area of your swept area, in terms of grid squares?
Draw another line from the Sun (left-hand dot) to aphelion – the point on the ellipse farthest from the Sun, on the right. Make that the beginning of another swept area equal to the first swept area. Figure out where to draw the line back to the ellipse from the Sun to complete this swept area by assuming this second swept area is, to a close approximation, a right triangle, with area = ½ x base x height.
You already know the area of the (approximate) triangle (it equals the first swept area), and the base (count how many squares there are from the Sun to the aphelion, along the first line of the second swept area).
5. Plug in the numbers (area and base) to the equation for the area of a triangle (see above) and solve the equation for height. Write your complete equation below and, when sure, write the answer in the box here:
Go up from the aphelion that many squares to find the point on the ellipse from which to draw a line back to the Sun. This will complete the second swept area, with area equal to the first swept area. Lightly shade in this second swept area (with a pencil).
Get a protractor and measure the angle swept out in the first area, with the Sun at the apex of the angle of course.
6. What is the angle of the first swept area?
7. What is the angle of the second swept area?
8. How can a planet following such an elliptical orbit, according to Kepler’s second law, sweep out such a small area near its aphelion (farthest from the Sun) in the same amount of time as it swept out such a large area near its perihelion (nearest to the Sun)?
The Earth is closest to the Sun in January, when it is 146 million km (91 million miles) from the Sun at perihelion, and farthest from the Sun in July, when it is 152 million km (94.5 million miles) from the Sun at aphelion.
When it is said that “the Earth is 150 million kilometers from the Sun,” or “the Earth is 93 million miles from the Sun,” that is not really a true statement (except by coincidence on two days each year).
9. What axis of the Earth’s elliptical orbit is 150 million km (93 million miles) in length?
10. By the way, if the distance from the Earth to the Sun varies from 91 to 94.5 million miles due to the elliptical shape of the Earth’s orbit, and the fact that the Sun is at one focus of the ellipse, is that why summer is hotter, because the Earth is closer to the Sun in summertime in the northern hemisphere? Explain
Kepler’s 3rd Law (which he did not list, did not number, and did not call laws) relates the period of a planet to its distance from the Sun.
The period (P) is the length of time it takes to complete one orbit around the Sun. We will use years (the period of the Earth) for the units of periods.
The distance from the Sun (a) is the semimajor axis of the planet’s elliptical orbit. We will use astronomical units (AU) as units for the semimajor axes of the planetary orbits. One AU is, by definition, the semimajor axis of the Earth’s orbit.
Using years for time units and AU for distance units, Kepler’s third law becomes
P2 = a3
The period of a planet (in years), squared, equals the semimajor axis of the planet (in AU), cubed.
- Solve Kepler’s third law for the period of a planet, in the space below:
- Use your equation (previous answers) to calculate the period of each planet (and one dwarf planet, Pluto) from its semimajor axis.
Write your answers in the table below, in the third column.
- In the fourth column, write down the measured period of each planet (and Pluto).
- Are the results based on Kepler’s law very close to the actual periods?
- How close: within greater than 10%, within between 1 and 10%, or within less than 1%?
Planet or Minor Planet / Semimajor Axis (AU) / Calculated Period (yr) / Actual Period (yr)
Mercury / 0.3871
Venus / 0.7233
Earth / 1.0000
Mars / 1.5237
Jupiter / 5.2028
Saturn / 9.5388
Uranus / 19.18
Neptune / 30.0611
Pluto / 39.44