Brahe’s Data, Kepler’s
Analysis &
The Orbit of Mars
Data Analytics Course 2015
Earth’s Rotation
Star or Planet
Counterclockwise
spin when viewed
from the north
Earth
How Do We Know The Earth
Rotates?
The sun, the stars
and planets trace a
daily circle of radius
that depends on their
latitude and that of
the observer
W
Pole Star
barely
moves
N
E
The Yearly Movement of the
Sun Relative to Fixed Stars
It appears to move counterclockwise against these
stars from north, completing
a cycle in a year, 1 degree a
day roughly
Initially
The
Sun
N
E
W
Q: Isn’t the sun too bright for
spotting the star
background?
• Use the setting/rising
trick
So the sun goes around the
earth or the earth goes
around the sun.
The
Sun
After a month
N
W
E
Resolving Sun Around Earth
or Earth Around Sun?
Stellar parallax: Nearby stars move
relative to far away stars when observed
months apart, indicating that the earth
moves —very subtle, difficult to measure
in Kepler’s time
Use Venus Instead [Galileo]
• Very bright in the western sky at dusk,
phases visible with a telescope
• Full Venus appears smaller than
crescent Venus (unlike the Moon)
• At least Venus orbits the Sun, not the
Earth
• How about Mars?
The Movement of Mars
Against Fixed Stars
The orbit of
Mars. But why
the zig-zag??
Shouldn’t be
there if Mars
goes around
the Earth.
Could it
happen if Mars
goes around
the Sun?
Epicycles?
Or could epicycles explain it? [Ptolemy]
Both Mars and Earth Around
Sun Explains It
Assuming both orbits are circular with radii M and E, respectively, and co-planar
x,y
β
α
This turns out to be just less
than 1, i.e., the turns
happens on either side of β
= α (Opposition)
β-α
Retrograde Twists on Either
Side of an Opposition
Are the Orbits Truly CoPlanar?
Over the year, as seen from the
earth, the sun seems to move
against the backdrop of the 12
Zodiac constellations
• Observe just after sunset or
before sunrise
• Or check which constellation
is highest in the sky at
midnight for the diametrically
opposite constellation
Broadly, these define the
Ecliptic, the plane of revolution
of the earth, angled at 23.5
degrees to the Earth’s equator
Mars’ Orbit Relative To The
Ecliptic
Mars stays close to the Zodiac, so its plane of revolution is only slightly tilted
relative to the ecliptic plane
Slight Non-Co-Planarity
Causes the Curve We See
The Orbit of Mars and Earth
 So Earth and Mars possibly go around the Sun like
Venus does
 What are their exact orbits?
 We have assumed a circle so far centered on the Sun,
the simplest possible model. Is it true? What are the
dimensions?
 Can these be inferred from positional observations of
Mars?
Positional Observations of
Mars
 Tycho Brahe made
positional observations over
a period of 20 years
 Radii cannot be directly
observed
 Brahe and his team
observed anglesαandβ (up
to 2 min) against the
backdrop of the zodiac
constellations

Mars’ orbit being at a slight angle to the
ecliptic poses a small problem that
Kepler had to resolve carefully
α
β
Angular Motion of Mars
Around Sun
 For motion around the sun,
we need the angle of Mars
relative to the Sun?
 Can δ be computed
givenαandβ?
α
 How else can we determine
motion relative to the Sun?
 Eliminate the Earth using
Oppositions
δ
β
Oppositions
 At oppositions, δ=β!
 How do we determine
moments of oppositions?
 When Mars rises as Sun
sets
 Observation at the horizon
suffer from refractive
effects, which Kepler
corrected after an year of
work!
α
δ
β
Two Suns: Actual and
Average!
•
•
Initially
The Sun appears to move faster
in Dec-Jan, slower in Jun-July,
so non-uniform in angular speed
This causes the Analemma
below
The
Sun
N
E
W
The
Sun
• Average Sun: A hypothetical
point around which there is
uniform angular speed
After a month
N
W
E
The Average Sun/The Equant
for Mars
•
Still circular orbit for the Mars, too early to give
that up
•
The Sun is not in the center; that would explain
non-uniform motion of the sun
•
The Average Sun is at the center, and the motion
of Mars around this center is uniform speed
•
It is possible that the motion is actually nonuniform in which case the Average Sun could be
off-center (or even nonexistent Prove this but
lets us assume it exists)
•
Given the period of revolution of Mars and
assuming uniform angular motion around the
average sun, time alone determines longitude
from the average sun, so given the date/time,
longitude from average sun (if it exists) can be
determined.
Average Sun
Sun
Off-Center
12 Oppositions in Brahe’s
S corresponds
Data β to units of 30°
From
http://www.keplersdiscovery.com/CorrectedT
able.html
Longitude relative
to the actual sun
Longitude relative
to the average sun
for Mars
Longitudes are along the ecliptic, latitudes are relative to ecliptic,
Brahe’s observations are very accurate, usually within a few min
Modeling the Orbit
•
Assumption is still circular motion, possibly
non-uniform motion
•
Though motion simple enough that
average sun exists
• Prove average sun need not always
exist
•
Average Sun
Center C
With center lying on the line between the
sun and the average sun for Mars
• Assumes angular speed is
asymmetric primarily due to the sun
being off-center
Sun
Off-Center
0°
•
For each opposition, we can draw the
longitude angle, relative to 0, from the sun
and the average sun, to locate Mars at that
opposition
0°
Fitting the Circle
•
Parameters x, y and r
•
Does y matter? If we change y alone, does
the circle change?
•
For a fixed value of x and y, any 3
oppositions can determine r and hence the
circle. A good fit if the other 9 oppositions
too lie on or close this circle.
•
We could try x,y exhaustively and see
which combination of values provides the
best fit.
•
Circle of radius r
x
C
1
Or we could try to get to the right x,y by a
procedure called regression.
y°
•
Write the positions of all opposition
locations as functions of x,y
•
Define a notion of goodness of fit to
the best circle
•
Maximize this fit
0°
0°
Finding the Nodes
•
Now that we have the orbit, what is its
slant relative to ecliptic (the angle
between the two planes)
•
Nodes: points where Mars’ circle
intersects with the ecliptic, points of 0
latitude
•
Kepler had several observations, at each
node 687 days apart, which is a Mars
year.
•
The nodes were at longitude 15° 44’
Scorpio and 16° 48’ Taurus, with respect
to the Sun, and 27° 14’ Scorpio and 5° 13’
Taurus from the average Sun.
The Orbit
of Mars
Node
Ecliptic
Node
A Method for the Slant
•
Kepler had a few observations where the Sun
and Earth were in line to the nodes and the
angle SEM was near orthogonal
M
•
Then the latitude of Mars w.r.t the ecliptic as
observed from Earth is the required slant
•
Which comes out to be 1° 51’
•
Kepler cross-checked this with other clever
arrangements as well
X
S
E
Identifying Distances
•
So we have the shape of the orbit, now
we need the orbit dimensions, i.e., the
radius of the circle. How do we get this, if
all our measurements are just angles?
•
Take two observations 687 days apart. So
we know Mars is in the same place, but
the Earth is not. Kepler had 5 such pairs
of observation
From http://faculty.uca.edu/saustin/3110/mars.pdf
Identifying Orbit Dimensions
•
Assume Earth’s orbit is circular with radius 1
•
Of course, the Sun is not in the center, so one would
have to assume a position for the Sun
•
Given the longitude of the Earth as seen from the Sun,
•
And the longitude of Mars as seen from Earth
•
One can triangulate to get a position for Mars
•
So one gets 5 such positions. Each is a position on the
ecliptic, the z-coordinate coming out of the page is not
obtained here.
•
Now find the best position of the Sun for which the 5
points are on a circle
•
Does a circle fit? If not, then what fits?
1
Sun
0°
The Final Step
• What shape for the orbit fits the
above data, assuming Tycho
Brahe’s measurements after
Kepler’s corrections were accurate
to a few min?
Interpolating Declinations
Declination is latitude from the equator rather than from the the ecliptic
From http://www.pafko.com/tycho/observe.html