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 counterclockwise 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
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Overlay template over stellar map
Determining Longitude of Mars from Earth
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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
10/11 J 14+11=25o
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3/21 A directly left
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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
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Plot each pair of Mars longitudes
Finished Mars points
Kepler’s 1st Law Applications
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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
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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

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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”