Brahe’s Data, Kepler’s Analysis & The Orbit of MarsData Analytics Course 2015

Earth’s Rotation

Earth

Star or Planet

Counterclockwise spin when viewed from the north

How Do We Know The Earth Rotates?

W E

N

Pole Star barely moves

The sun, the stars and planets trace a daily circle of radius that depends on their latitude and that of the observer

The Yearly Movement of the Sun Relative to Fixed

StarsIt appears to move counter-clockwise against these stars from north, completing a cycle in a year, 1 degree a day roughly

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.

W E

NThe Sun

W E

N

The Sun

Initially

After a month

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

α

β

x,y

This turns out to be just less than 1, i.e., the turns happens on either side of β = α (Opposition)

β - α

Assuming both orbits are circular with radii M and E, respectively, and co-planar

Retrograde Twists on Either Side of an

Opposition

Are the Orbits Truly Co-Planar?

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

αβ

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?

How else can we determine motion relative to the Sun?

Eliminate the Earth using Oppositions

αβ

δ

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!

αβ

δ

Two Suns: Actual and Average!

• The Sun appears to move faster in Dec-Jan, slower in Jun-July, so non-uniform in angular speed

• This causes the Analemma below

• Average Sun: A hypothetical point around which there is uniform angular speed

W E

NThe Sun

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N

The Sun

Initially

After a month

The Average Sun/The Equant for Mars

SunOff-Center

Average Sun

• 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 non-uniform 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.

12 Oppositions in Brahe’s Data β S corresponds

to units of 30°

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

From http://www.keplersdiscovery.com/CorrectedTable.html

Modeling the Orbit

SunOff-Center

Average Sun

Center C

• Assumption is still circular motion, possibly non-uniform motion

• Though motion simple enough that average sun exists• Prove average sun need not

always exist

• 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

• 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°

0°

Fitting the Circle

C

• 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.

• Or we could try to get to the right x,y by a procedure called regression.

• 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

1

x

0°

0°

y°

Circle of radius r

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.

Ecliptic

The Orbit of Mars

Node

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

• 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

S E

M

X

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?

Sun 1

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 eclipticFrom http://www.pafko.com/tycho/observe.html