Kaley_Sechler_Lab05

Mercury’s Orbit Name:_____ Kaley Sechler _______________________ Date: 11/2/23 _____________ Kepler’s First Law 1. Circle any three points on the orbit plot that are widely spaced out and label them P1, P2 and P3. Use a ruler to draw a line from F1 to P1 and then draw another line from P1 to F2. Measure the exact length of each line segment in centimeters. Add the two lengths to find the total distance. Repeat this procedure for points P2 and P3. Record your measurements and sums in the table below. Poin t Distance (cm) F1 to Point Distance (cm) Point to F2 Total Distance P1 8 8.5 16.5 P2 9.5 7 16.5 P3 7 9.5 16.5 2. Now measure the length of the major axis in centimeters. ___ 16.5 ______________ 3. What can you conclude from these measurements? The points are all the same distance from the longest line in the shape, then the total distance they cover is the same length as that longest line. The Eccentricity of an ellipse is a measure of its "flatness". Eccentricity is defined as the ratio of the distance between focal points to the length of the major axis. Page 1

eccentricity = distance between focus points length of major axis The numerical range of eccentricity can have values between 0 and 1. An orbit with an eccentricity of 0 is a perfect circle. A value of say, 0.9, is a very flat ellipse which is typical of a comet’s orbit. 4. Determine the eccentricity of Mercury’s orbit below by measuring the distance between f1 and f2, and dividing it by the length of the major axis. Eccentricity = 3.5/16.5= .212121 5. How would you describe Mercury’s orbit (circle best description)? Low Eccentricity (close to a circle) / Moderate Eccentricity / High Eccentricity Kepler’s Second Law Time from A to B = Time from C to D Area 1 = Area 2 The point at which a planet is closest to the Sun is called perihelion and the greatest distance is called aphelion. Both points are located on the major axis. Circle and label the perihelion and aphelion points for Mercury’s orbit. 6. At what point in its orbit is the planet Mercury moving fastest? Page 2

Area = ½ (base x height). To form a triangle draw a line from the Sun’s center to point “A” and another line from the Sun’s center to point “B”. Connect points “A” and “B” to complete the first triangle and label it “Area 1”. Repeat the procedure for the other two areas and label them “Area 2” and “Area 3”. The base distance is the longest side of the triangle. The height of the triangle is a perpendicular line from the base to the vertex opposite the base. Measure the length of the base and the height for each triangle and record those values in the table below. The final step is to calculate the area using the formula above. Area 1 Base = _____ 7 ______ Height = _____ 3 ______ Area = ____ 10.5 _______ Area 2 Base = _____10______ Height = _____ 2 ______ Area = ____ 10 _______ Area 3 Base = ____ 9 _______ Height = _____ 2 ______ Area = _____ 9 ______ 8. Do these measurements support Kepler’s 2 nd Law – meaning, are the areas equal, or very close to equal? The areas of the three triangles are very close to the same values, so it supports kepplers 2 nd law. 9. Explain your results, and discuss why there might be slight differences in the measured areas. There might be slight differences in area values because of false measurements of the height, I was confused on how to draw it on the triangle. Kepler’s Third Law Complete the following exercise. Show your work and use the proper units and significant figures. 10.Measure the semi-major axis of Mercury’s orbit in both AUs and in kilometers using the scales provided. Record both values below. 61 x 10^6 km 0.41 AU 11.Convert the semi-major axis from kilometers into meters. (1000 m = 1 km) Page 4

61000000 x 1000= 61 x 10^9 m 12.Using Kepler’s Third Law, P = √ a 3 , calculate the orbital period of Mercury in years - ‘a’ is the value above (in AU’s), Sqrt(.41^3)=.2626 years 13.Convert the orbital period from years into days. (1 yr = 365 days) .2626 x 365 = 95.8125 days 14.Convert the orbital period from years into seconds. (1 yr = 3.16 X10 7 s) .2625 x (3.16 x 10^7)= 8.295 x 10^6 seconds Newton’s Version of Kepler’s 3 rd Law: Now let’s “weigh” the sun! For any two bodies orbiting each other Newton generalized Kepler’s third law to read: P 2 = 4 π 2 G ( m + M ) ⋅ a 3 , where G is the gravitational constant, M the Sun’s mass and m the planet’s mass. This form of Kepler’s Third Law is frequently used to find the total mass of any two objects orbiting around each other. For example, Mercury’s orbit can be used to find the Sun’s mass. Since the mass of a planet is insignificant compared to the Sun’s mass, we can set m + M equal to M. Solving the above equation for M (mass of Sun) gives: M sun = 4 π 2 G ⋅ a 3 P 2 , 15.Using the values for Mercury’s semi-major axis (a) in meters and the orbital period (P) in seconds , calculate the Sun’s mass using the expression for Kepler’s 3 rd Law shown below. Page 5

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