Tycho Brahe (1546-1601)people.physics.tamu.edu/papovich/courses/fall11/lecture2_20110906.pdfSep 06,...

Tycho Brahe (1546-1601)

Foremost astronomer after the death of Copernicus. King Frederick II of Denmark set him up at Uraniborg, an observatory on the island of Hveen.

With new instruments (quadrant), Brahe could measure positions of stars and planets to 4 minutes of arc (1/8th the diameter of the full moon).

With new instruments (quadrant, similar to present-day sextant), Brahe could measure positions of stars and planets to better than 4 minutes of arc (1/8th the diameter of the full moon). Made the most accurate observations of planet positions ever recorded.

Tuesday, September 6, 2011

Using the Quadrant (similar to a Sextant) to measure a star’s altitude. An observer

sights a star along the Quadrant while a plumb line

measures the angle.


Tycho Brahe (1546-1601)

Also Credited with observing the supernova of 1572, demonstrating that the stars do change. He coined the term “Nova” for “new star” in a book. Term still used today.

Brahe failed to find evidence for the Earth’s motion and presumed the Copernican model was false. Brahe endorsed the Geocentric cosmology.


Johannes Kepler (1571-1630)

Brahe moved to Prague in 1599. Kepler joined him to develop cosmological model consistent with Brahe’s observations.

Kepler initially obtained excellent agreement with Brahe’s data with the planets moving in spheres and equants, similar to Ptolemy’s model, except for 2 data points which were off by 8 minutes of arc (twice the accuracy of Brahe’s measurements).

Kepler kept struggling and finally rejected Ptolemaic model and eventually realized that that Brahe’s data were only consistent with a model where planets (1) orbit the Sun and (2) their orbits are ellipses.


Observed Planetand uncertainty on the measurement


Observed Planetand uncertainty on the measurement

Position Predicted by Model



Kepler’s Laws: published in 1609 in Astronomica Nova (The New Astronomy).

1st Law: A planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse.

2nd Law: A line connecting a planet to the Sun sweeps out equal areas in equal time intervals.

3rd Law: The square of the orbital period, P, of a planet is equal to the cube of the average distance, a, of the planet from the Sun.


Ellipses

Sun

PerihelionAphelion

defines an ellipse


Sir Isaac Newton (1643-1727)

Developed Universal theory of Gravity, and set his three Laws of Motion, which are framework for classical mechanics, including basis for modern engineering.

-1st Law: An object at rest, remains at rest unless acted on by an outside force. An object with uniform motion, remains in motion unless acted on by an oustide force.

- 2nd Law: An applied force, F, on an object equals the rate of change of its momentum, p, with time.

- 3rd Law: For every action there is an equal and opposite reaction.


Newton’s Theory of GravityStart with Kepler’s 3rd law:

For circular orbit:

Insert this into Kepler’s 3rd law:

Multiply both sides by factors:

Yields: Force !


Newton’s Theory of Gravity

By Newton’s 3rd law, this is the force on mass M from mass m, so the force on mass m should be:

Equating these two formula gives:

where and

Defining we get the “usual” formula:


Define Center of Mass Frame

by Newton’s 3rd Law

for 2 objects

for i=1..N objects

Differentiate with time,

Derive Kepler’s Laws from Newton’s Laws

Differentiate with time, again:


Choose Frame with R=0

Define “Reduced Mass” of the System : μ= m1 m2 / (m1+m2)


=0


r = r’2 - r’1

Now write out total Energy of the System

Insert:

And :



Total Energy Becomes:

And:

Grouping Terms :

Total Energy of the System is the sum of the Kinetic Energy of the reduced mass and the potential energy of the total mass/reduced mass system.


Which Gives:


Introduce Angular Momentum:

Look at rate-of-change of L as a function of t:

= 0 , Conservation of Angular Momentum


Similarly for the Angular Momentum:

Insert:

Leads to:

The Two-Body problem may be treated as a One-body problem with μ moving about a fixed

mass M at a distance r.




Kepler’s Laws: published in 1609 in Astronomica Nova (The New Astronomy).

1st Law: A planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse.

2nd Law: A line connecting a planet to the Sun sweeps out equal areas in equal time intervals.

3rd Law: The square of the orbital period, P, of a planet is equal to the cube of the average distance, a, of the planet from the Sun.

P2 = Constant x a3


Ellipses


Eccentricity


http://faculty.physics.tamu.edu/papovich/courses/animations/eccentrcty_and_semimjr_axis.swf



Eccentricity


Eccentricity=1Parabolic orbit(goes through perihelion only once)

Eccentricity


Kepler’s First Law

Each planet moves on an elliptical orbit with the Center of Mass (the Sun) at one focus:

Perihelion distance = (1-eccentricity) x (semimajor axis)



Each planet moves on an elliptical orbit with the Center of Mass (the Sun) at one focus:

“Perilously close”

Perihelion distance = (1-eccentricity) x (semimajor axis)


Planets Follow Eliptical Orbits

The Inner Solar System The Outer Solar System


http://faculty.physics.tamu.edu/papovich/courses/animations/01_AccurateModelOfSS.swf

http://faculty.physics.tamu.edu/papovich/courses/animations/01_AccurateModelOfSS.swf

Not just planets! Comets have elliptical orbits too.




Halley’s Comet (76 yr orbit)




Halley’s Comet (76 yr orbit)

Halley’s Comet in 2024



Kepler’s Second LawEqual Areas = Equal TimesConservation of Angular Momentum

Near perihelion, in any particular amount of time (such as 30 days) a planet sweeps out an area that

is short by wide.

Near aphelion, in the same amount of time (such as 30 days) a planet sweeps out an area that is long by narrow.

The areas swept out during any 30-day period are all equal

perihelion


Kepler’s Second LawEqual Areas = Equal TimesConservation of Angular Momentum

The areas swept out during any 30-day period are all equal


Velocity Inversely Proportional to Orbital Radius


http://faculty.physics.tamu.edu/papovich/courses/animations/kepler_2_velocty_vs_orbit_r.swf

http://faculty.physics.tamu.edu/papovich/courses/animations/kepler_2_velocty_vs_orbit_r.swf

Kepler’s Third LawSquare of orbital periods (P) is proportional tocube of semi-major axis (a): P

2! a

3

Jupiter

Saturn

Average distance cubed (AU3) from Sun.

Orb

ital P

erio

d sq

uare

d (y

ears

2 )

Semimajor axis

MercuryEarth

Mars

Venus


Kepler’s Third Law


Orbital Period vs Orbital Radius


http://faculty.physics.tamu.edu/papovich/courses/animations/kepler_3_orbit_period_vs_r.swf

http://faculty.physics.tamu.edu/papovich/courses/animations/kepler_3_orbit_period_vs_r.swf

The Two-Body problem may be treated as a One-body problem with μ moving about a fixed

mass M at a distance r.



A x (B x C) = (A·C) ·B - (A·B)·C


Take cross product:


D is a constant

v x L and r lie in the same plane, so must D

D is directed toward Perihelion,

D determines e (eccentricity) of ellipse.

Maximum reached with r and D point in same direction.

^

^



A · (B x C) = (A x B)·C

Kepler’s 1st Law !



^

= Kepler’s 2nd Law !

Integrate from Focus to r: dA = ½ r2 dθ

dAdt

= ½ r2 dθdt

Let : v = vr + vθdrdt

r + r dθdt θ

^=dAdt

= ½ r vθ

r and vθ are perpendicular, therefore : r vθ = |r x v| = | L / μ | = L / μ

dAdt

L2μ = Constant



Integrate 2nd Law with time: A = ½ L/μ ∫dt

A = ½ (L/μ) P, where P = period

A = π ab

P2 = 4π2 a2 b2 μ2

L2

r = L2 / μ2

GM (1 + e cosθ)1st Law: L = μ [ GMa (1-e2) ]½

P2 = 4π2 a3

G (m1 + m2)Kepler’s 3rd Law


a

b


Calculate Orbital Speed at Perihelion and Aphelion

SunPerihelionAphelion

Use Kepler’s 1st Law

Total Energy:

At Perihelion

At Aphelion


Total Energy in a Bound Orbit

Etot =-GMμ

2a =

-Gm1m2

2a = <U> / 2

That is the total energy is 1/2 of the average potential energy.

This is the Virial Theorem.

More in depth derivation by Rudolf Clausius (reproduced in book). It has many applications

in astrophysics, including providing evidence for Dark

Matter in Clusters of Galaxies !

Rudolf Julius Emanuel Clausius (January 2, 1822 – August 24, 1888), was a German physicist. He is one of the central founders of the science of thermodynamics.


High Tide in Bay of Bundy

Low Tide


What Causes Ocean Tides ?


Saturn as seen by Spacecraft Cassini, October 2004


Fc,x = -G Mm / r2 Fc,y = 0

Fp,x = (G Mm/s2 )cosϕ Fp,y = -(G Mm/s2 )sinϕ

ΔF = Fp - Fc = G Mm cosϕ

s2 r21

i - G Mms2 sinϕ j^ ^( ( ))-

s2 = (r - Rcosθ)2 + (Rsinθ)2 = r2 1 - 2Rcosθr

( ) R << r

Tidal Forces

M= mass of Moon; m= test mass

Force of Moon on m at center of

Earth

Force of Moon on m on surface of

Earth


ΔF ≈ G Mm/r2 [ cosϕ 1 +

Tidal Forces

ΔF = Fp - Fc = G Mm cosϕ

s2 r21

i - G Mms2 sinϕ j^ ^( ( ))-

s2 = (r - Rcosθ)2 + (Rsinθ)2 = r2 1 - 2Rcosθr

(

) R << r

2Rcosθr ) -1 ] i

(

^

- G Mm/r2 [ 1 + 2Rcosθr ] sinϕ j

For the Earth-Moon, cosϕ ≈1, r sin ϕ = Rsinθ, sin ϕ = (R/r) sinθ

ΔF ≈ G MmR/r3 [ 2cosθ i - sinθ j ] ^ ^


Tidal Forces


Force of the Moon on the Earth

Differential (relative) force on the Earth,

relative to the center


When will the tidal force equal the force of gravity ? Consider the Earth-Moon system:

Mm, Rm = mass of moon, radius of moon

ME, RE = mass of Earth, radius of Earth

Tidal Forces


G Mmm/Rm2 = ( 2GMEm/r3 ) Rm

Assume constant density: Mm = (4/3)πRm3 ρm

r3 = (2ME/Mm ) Rm3 = (2RE3 ρE) / (Rm3 ρm) Rm3


r3 = (2ME/Mm ) Rm3 = (2RE3 ρE) / (Rm3 ρm) Rm3

Tidal ForcesG Mmm/Rm2 = ( 2GMEm/r3 ) Rm

Assume constant density: Mm = (4/3)πRm3 ρm

r = ( )⅓2ρE

ρmRE

Roche, using a more sophisticated assumption for the density of the Moon calculated fR = 2.456

r = fR ( )⅓ρE

ρmRE, fR = 2 ⅓ =1.2599

or


Comet Shoemaker-Levy 9 after passing through Jupiter’s Roche Limit


Saturn as seen by Spacecraft Cassini, October 2004


Other MoonsJanus

Epimetheus

DioneRhea

Distance from Saturn Center in RS

1 2 3 4 5 6 7 8

RingsRings

Saturn

Roche Limit

Mimas Enceladas Tethys

Rings

Planetary Moons form when a moon crosses the Roche Limit


How do you know the Earth Rotates ?



In 1851, Léon Foucault proved the Earth’s rotation directly.

A pendulum swinging on the Earth feels the rotation due to the Coriolis Force.

FC = -2 m Ω x v

Versions of the Foucault Pendulum now swings in the Panthéon of Paris

and the Houston Museum of Natural Science

http://www.youtube.com/watch?v=CpxDAdtJX2s

http://www.youtube.com/watch?v=49JwbrXcPjc











Coriolis Force affects rotation of weather patterns (cyclones rotate counter clockwise in southern hemisphere; hurricanes rotate clockwise in northern hemisphere).



Hurricane GustavAug 31, 2008



Coriolis Force affects rotation of weather patterns (cyclones rotate clockwise in southern hemisphere; hurricanes rotate counter-clockwise in northern hemisphere).

Hurricane GustavAug 31, 2008


Tycho Brahe (1546-1601)people.physics.tamu.edu/papovich/courses/fall11/lecture2_20110906.pdfSep 06,...

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