Copy of Lab 6_ast101

AST 101 Lab 6 Kepler’s Laws: Part II Lab 6 Kepler’s Three Laws of Planetary Motion: Part II PURPOSE This laboratory exercise will allow the student to study Kepler’s laws of planetary motion by examining the motion of Mercury. REFERENCES ● none BACKGROUND Kepler’s Second Law of Planetary Motion states that planets sweep out equal areas in equal times. In general terms, it means that if a body is moving in an elliptical path under the influence of a central force, a line drawn from the source of the central force to the body will sweep out equal areas in equal intervals of time (Figure 6-1). The gravitational attraction between the Sun and the Mercury is an example of a central force in operation. The force on Mercury is directed toward the Sun, which is at one focus of the orbit of the Earth. Figure 6-1: Equal areas swept out in equal times. Q P E Q P

AST 101 Lab 6 Kepler’s Laws: Part II Kepler’s Third Law of Planetary Motion, sometimes referred to as his “Harmonic Law", relates the periods of the orbits of planets to their average distances from the sun. In general, it means that the square of the amount of time an object requires to complete an orbit in an elliptical path under the influence of a central force is directly proportional to the cube of the semi-major axis of the ellipse. Therefore the greater the distance a planet is away from the Sun, the longer the period. The Harmonic Law may be stated as a 3 = kP 2 (6-1) where a is the semi-major axis of the orbit, P is the sidereal period and k is the constant of proportionality called the Kepler constant. The Kepler constant depends only on the body being orbited and not the orbiting body. In practice, since the period of an orbit and its semi-major axis are both observable quantities, the Harmonic Law is used to find the mass of the object supplying the central force. We will use the orbit of Mercury to do three tasks. First, we will plot the orbit itself and describe the properties of the elliptical path Mercury follows. Second, we will verify Kepler’s Second Law by calculating the areas of two separate wedges swept out by the planet. EQUIPMENT ● Graph paper ● Ruler ● protractor PROCEDURE Exercise #1: Plotting the Orbit of A Planet Using the data in Table 6-1, plot the orbit of Mercury. The orbital data is given in terms of an angle relative to the Sun and its distance from the Sun in millions of miles Draw a line segment from the star straight up the page and label it as 0 degrees. Angles are measured counter-clockwise. 1) From the center of the page, mark a point that is 3.75 cm above the center along the 0 degree line. This point will represent the focus of the ellipse were the Sun is located. All orbital data for Mercury will be measured from this point. 2) Plot the data in Table 6-1 by using a protractor to establish the angle and a ruler to measure the corresponding distance along this angle. Note that each point represents 4 days of motion for Mercury.

AST 101 Lab 6 Kepler’s Laws: Part II Exercise #2: Verify Kepler’s Second Law We will verify Kepler’s Second Law by measuring the area of wedges swept out during equal amounts of time by Mercury. To simplify calculations, we will approximate the wedges with right triangles. The area, A , of a right triangle is simply (6-2) where b is the base of the triangle and h is the height of the triangle. Pick a point along the ellipse. Draw a line from this point to a neighboring point. The amount of time taken by Mercury to travel from one point to the next is always four Earth days. Now draw a line from each point to the Sun. The three lines represent a triangle, but not a right triangle (Figure 6-2). To get a right triangle, choose a point exactly in- between the two points previously chosen, and draw a line from this point to the Sun. This represents the height of the triangle. Next draw a line through this middle point at a right angle to the height line. This represents the base of the triangle. Measure both the height and the base and record the results in the data section. Figure 6-2: Approximating the area of a wedge with a right triangle Repeat this process for another set of points. For the best results, choose a pair of points far away from the previous pair. Record the height and base of the second triangle in the data section. Using the area formula, calculate the area of each of the triangles. P star h b P 2

AST 101 Lab 6 Kepler’s Laws: Part II Exercise #3: Using Kepler’s Third Law Assume that the Kepler constant for Mercury is 6.03 x 10 18 miles 3 /days 2 . Use equation 6- 1 to find the period of Mercury. Compare this value to the accepted period of 87.969 Earth days. LAB 6

AST 101 Lab 6 Kepler’s Laws: Part II Exercise #2: Verify Kepler’s Second Law For triangle #1: Height: _______10 cm ________ Area of triangle 1: _______20 cm^2 _____ Base: _______4 cm ________ For triangle #2: Height: _______6.1 cm ________ Area of triangle 2: ______17.4 cm^2 _____ Base: _______5.7 cm ________ Answer the following questions: 1) Are the areas of the triangles exactly equal? Why or why not? Not quite, because this is a scale drawing. It is not 100% accurate. Also, we forgot to extend the base line outside of the circle on triangle #2, so it would’ve been even closer. 2) Does this verify Kepler’s Second Law? Briefly explain your reasoning. Yes, Kepler’s 2nd law states these values should be equal, and they are close enough that I would say it verifies it.

AST 101 Lab 6 Kepler’s Laws: Part II Exercise #3: Using Kepler’s Third Law Period of Mercury: ____ 86.0449 days ___________ (show your calculation below) (35,475,000)^3 = 4.46444228 E 22 (4.46 E 22) / (6.03 E 18) = 7403.72 Sqroot(7403.72) = 86.04487 days Recall that the Earth is 1 AU from the Sun and requires 1 year to complete its orbit. Is the period of Mercury consistent with Kepler’s third law? That is, does the period seem reasonable given its distance from the Sun? Yes.