1. I worked on this most of the day today. I plan to use some of this in my class presentation. Jim Honeycutt

2. 2a is major axis semi-major axis = a 2 times a is the length of major axis Properties of Ellipses a 2a semi-major axis

3. aphelion perihelion a = aphelion + perihelion _______________________ 2 Sun Most planetary orbits are fairly close to being a circle. Comet orbits are the most elliptical. Perihelion is when the object is closest to the Sun, and Aphelion is when the object is farthest from the sun. Add these two numbers together, and divide by 2 gives you a, which is the average distance. If a comet came as close to the Sun as.5 au but travels as far away as 7.8 au, then a= (.5 + 7.8)/2 a= 4.15 au

4. Using Kepler’s Third Law P 2  a 3 The square of the period of a revolving objects equals the distance of one of the objects from the other cubed. This is a simple problem that only works if you do the following: P must be in earth years, and must be in au. An au is the distance from the earth to the Sun (93 million miles). Mars is 1.5 au, and Jupiter is 5 au. So lets use the formula on the comet that was 4.15 au from the Sun. a= 4.15, find the period. a The period is 8.45 years

5. Lets work in the opposite direction : The period of Halley’s Comet is 75 years. What is the comet’s average distance from the Sun? P2 a3P2 a3 P= 75, find a Since in my text I find that Uranus is 19.19 au from the Sun, Halley’s Comet goes almost all the way out to Uranus’ orbit. This formula works not only in our solar system, but for binary systems of stars, or star, and planet.

6. Orbits and Masses of Binaries The primary importance of binaries is that they allow us to measure stellar parameters (especially mass). Both stars in a binary system revolve around their center of mass. The center of mass is the place where a fulcrum would be placed to balance the two stars. This location depends upon how much more massive one object is over the other. For our solar system the center of mass is inside the Sun.

7. This formula is Newton’s modification of Kepler’s Law. A must be in au, P in years, and Mass (M) must be in solar masses. So, when we calculate mass we will be comparing it to the mass of the Sun. This simplifies things. Epsilon Eridani has a planet orbiting around it.The planet is 3.4 au from the star, and has a period of 7.1 years. Find the mass of the star.First, since the mass of the planet, compared to the star is extremely small we can ignore the planet.

8. Sirius is orbited by a White Dwarf star.The star is 20 au from Sirius, and has a period of 50 years. Find the mass. Since the mass of the WD can not be neglected, we will find the combined mass. We could now look up information about bright stars.A spectrum of Sirius reveals it is a A0, and plotted on an HR diagram its mass corresponds to 2.6. So, the White Dwarf has a mass of 3.2-2.6 = 0.6 solar masses.

9. The Masses of Stars Spectral Type Mass in Solar Masses O540 B57.1 A52.2 F51.4 G50.9 K50.7 M50.2

10. 12 years... Jupiter (Meters/sec) Need Telescope... Clevis in the program, Marcy said that the short curve after Jupiter was due to Saturn, this should be it.

11. 16 Cygni B Mass = 1.7 M JUP (Min) Orbit Period 2.2 yr Not Sinusoidal VelocityWobble Here is one of the graphs from last night 16 Cygni B *Note p= 2.2 years

12. Let’s do our calculations, and see what happens. P =2.2 years P 2  a 3 On the internet I found the mass of 16 Cygni to be 1 solar mass. I need a better accurate number than 2.2 years.( some more decimals) Other information stated that the closet the planet comes to the star is.6 au and the farthest is 2.7 au. If you add.6 + 2.7 = 3.3 this is the peri, and aphe. Divide this number by 2 you get 1.65. Our answer for a was 1.69, not bad for a hand calculator. They probably used more accurate data..