Chatfield Senior High Physics Department Mars Orbit Lab Chatfield Senior High Physics Department

Elliptical Orbits All planets sweep out elliptical orbits around the Sun For the Earth…. the perihelion (closest distance) is on Jan. 5 the aphelion (furthest distance) is on July 5

Earth rotation The Earth sweeps out 1 o counter-clockwise ellipse per day Mar. 21 Vernal Equinox June 21 Summer Solstice Sept 21 Autumnal Equinox Dec 21 Winter Solstice

Cause of seasons Seasons are not caused by the closeness of the Earth to the Sun but by the 23.5 o tilt of the Earth

Earth Orbit

Mars Orbit Lab Stellar Template

Mars Orbit Stellar Template Overlay template over stellar map

Determining Longitude of Mars from Earth Using a stellar map for 8 pairs of points that are exactly one Martian year apart (687 days) you can create the following data table

Earth position at each point

To plot Earth points 3/21 A directly left 10/11 J 14+11=25o 4/4 O 10+4=14o 11/22 I 20+22=42o 4/12 F 8o 12/9 L 8+9- 17o 4/20 C 8o 1/21 K 22+21 =43o 5/26 E 10+26=36o 2/3 N 10+ 3 = 13o 8/4 H 5+30+31+4=70o 2/5 B 2o 9/16 G 27+16=43o 2/21 P 16o 3/8 D 7+8 = 15o

Plot each pair of Mars longitudes

Finished Mars points

Kepler’s 1st Law Applications The left hand side shows that Mars is at aphelion because the points are farther from the Sun The sun is at one focal point, and the other focal point must be somewhere along that side Two pins and a string can be used to plot the ellipse as shown.

Mars Orbit best-fit curve

Mars Orbit Statistics The radius of Mars orbit is 1.525 AU The eccentricity of the Mars’ orbit is 0.08

Kepler’s 2nd Law A line can be drawn from Mars to the Sun to sweep out the following areas… Area 1 (2/5/33-4/20/33) 74 days Area 2 (8/4/41-11/22/41) 110 days

Kepler’s 2nd Law

Kepler’s 2nd Law You can use the area of a sector of a circle to approximate Area 1 and Area 2 Area = (n / 360) p R 2 If you divide the areas by the Earth day’s swept out, you will find that the Area/day is equal

Kepler’s 3rd Law You can use the fact that Mars has a period of 687 Earth days (1.88 yrs) Mars R av is 1.525 AU K (or Kepler’s constant) can be found by…. k = (1.88)2 = 1 yr 2 / AU 3 (1.525)3

Kepler’s Quote “…a fullness of agreement between my seventeen years of labor on the observations of Brahe and his present study of mine that at first I believed I was dreaming”