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## PowerPoint Slideshow about ' Measuring the Solar System' - bethany-duffy

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Measuring the Solar System

What we want

to determine the distance to the various planets, asteroids, and comets from the Sun

Basic method:

Take Earth-Sun distance = 1 unit of measure (the AU)

Determine the distances to other planets in AU

Measure the Earth-Sun distance directly:

During transits of inferior planets across Sun’s disk

By radar (see later)

Some terminology

Superior planet - a planet with an orbit greater than Earth’s (Mars Neptune)

Inferior planet - a planet with an orbit smaller

than Earth’s (Mercury and Venus)

Conjunction - planet is directly lined up

with the Sun as seen from the Earth

Opposition - Sun and planet in line with

Earth, but in opposite directions (180o apart)

on the sky (as seen from Earth)

Sun

Elongation

Earth

More Terminology

Elongation:

The angular separation of a planet

from the Sun (as seen from the Earth)

Inferior planet:

0o elongation greatest elongation (GE)

Mercury: GE = 23o

Venus: GE = 46o

Superior planet:

0o elongation 180o

Measured quantities

Conjunction

Opposition

Conjunction

180o

Earth

Opposition

The orbit of an inferior planet

Method of Copernicus

Theory: At greatest elongation the observers line-of-sight is tangential to the planet’s orbit

- Deduction: can determine the size of the orbit
- relative to Earth’s (i.e., in AU) by measuring
- the greatest elongation

I wrote that theory

in 1543…URW

Angle

measurement

RE = radius of Earth’s orbit = 1 AURP = radius of planets orbit

Tangent condition:

Right-angle

LOS

Planet

RP

RE

Earth

Sun

Greatest elongation

(from observations)

Method:

Sin(greatest elongation) = RP / RE

or,

RP = RE x Sin(greatest elongation)

Units: recall, RE = 1 AU

OMG.

It works

“The answer lies in trigonometry”,

sang out Holmes. “Gads… old boy…you’re right” replied Watson

GE

Measured

The first two of Kepler’s laws

1st law:

The planets revolve around the Sun along elliptical orbits with the Sun at one focus

WTF

- 2nd law:
- A line drawn from the planet to the Sun sweeps out equal areas in equal time

Law: the first - the picture

Planets have elliptical orbits, with the Sun at one focus

Planetary orbit - exaggerated

center

Aphelion

perihelion

Sun

“empty” focus

The ellipse - the picture

Two focal points

Definition:

Eccentricity (e)

e = OF1 / a

P

r1

r2

O

F2

F1

Distance

OF1 = OF2

r1 + r2 = 2 a

Semi-major axis (a)

Major axis

Eccentricty (e):

Describes the ‘shape’ of the ellipse

For a ‘closed’ orbit 0 e < 1

A circle = an ellipse of zero eccentricity

Planets:

Earth: e = 0.0167

Moon: e = 0.055

Mars: e = 0.094

Mercury: e = 0.206 – most eccentric planetary orbit

Venus: e = 0.007 – most circular planetary orbit

Halleys comet: e = 0.967 – ‘cigar’ shaped orbit

No need

to know

numbers

a = 17.94 AU

e = 0.966

a = 3.124 AU

e = 0.5

e = 0.05

a = 1700 AU

e = 0.9998

e = 0.009

e = 0.252

Law: the second - the picture

The planet-Sun line sweeps out equal areas in equal time

Time T

D

C

B

E

Time T

Time T

F

G

A

Kepler’s 2nd law says in pseudo-computer code:

if

area AFB = area CFD = area EFG

then

time (A to B) = time (C to D) = time (E to G)

D

C

B

E

Time T

Time T

F

G

A

So, K2 reveals….

Near perihelion - closest point to Sun

large distance to travel in time T

hence high velocity

Near aphelion - greatest distance from Sun

small distance to travel in time T

hence low velocity

The velocity of a planet (comet, asteroid…) varies according to its position in the orbit and distance from Sun

Slow

aphelion

perihelion

The ellipse divided into 14 equal areas (blue and white triangles) and 14 equal travel times (red dot) along the orbital path ---- orbital period = T/14

as measured from

the Sun

Where is a planet after ¼ of its orbital period following perihelion ?

Planet 1/4 of way

around orbital path

if velocity was constant

Planet at 1/4 of

orbital period

P

Aphelion

Perihelion

Sun

Area of entire eclipse = one orbital period

= perihelion to perihelion travel time

Area of half ellipse = half orbital period

= perihelion to aphelion travel time

Area (Sun, P, Perihelion) = Area(Sun, P, Aphelion) = 1/4 area of ellipse

“Eight of thirteen”

Discovered in 1618 and published in Harmonice Mundi (Harmony of the World) as number 8 in a list of 13 points necessary for the contemplation of celestial harmonies

Kepler argued:

“if the Sun has the ‘power’ to govern the motion of the planets, then there must be some relationship (law) between planetary periods and distance”

7 of 9 - of course

Tertiary Adjunct of Unimatrix 01

Do your assignments

resistance is futile

The third law

Say what?

The square of a planets orbital period (P) is proportional to the cube of its orbital semi-major axis (a)

P2 = K a3

OOTETK

- where, P = orbital period of planet
- a = semi-major axis of planets orbit
- K = a constant

P2

Pluto

straight line

Mercury

a3

OOTETK

A clever dodge:

if P(years) and a(AU) then K = 1 and P2(yr) = a3(AU)

Example:

The first asteroid (now dwarf planet), Ceres, was discovered on January 1st, 1801.

Observations reveal that it has an orbital radius (semi-major axis) a = 2.77 AU

So

Hence

Periodic comet Encke has an orbital period of 3.3 years (the shortest of any known active comet)

From K3

Hence

Questions:

How close does HC approach the Sun?

What is the orbital eccentricity?

From Kepler’s 3rd law: P2(yr) = a3(AU)

So, a3(AU) = 76 x 76 = 5776

Hence a(AU) = (5776)1/3 = 17.94

By definition: perihelion distance + aphelion distance = 2a

So, we have:

Perihelion distance = 2a – aphelion distance = 2x17.94 – 35.3

Which gives

Perihelion distance = 0.58 (AU)

Closer than orbit of Venus

to the Sun

Aphelion

(35.3 AU)

perihelion

a

Sun

0.58 AU

Definition:

Eccentricity (e)

e = OF / a

Perihelion distance

a = 17.94 AU

By construction: a = perihelion distance + OF

Hence: OF = a – perihelion distance = 17.94 – 0.58 = 17.26 AU

And, accordingly, eccentricity e = 17.26 / 17.94 = 0.96

Aphelion

perihelion

a

Sun

Last seen in 1986 – back in 2061

Perihelion distance

Summing up – For Halley’s comet

Orbital period = 76 years

Semi-major axis a = 17.94 AU

Perihelion distance = 0.58 AU

Aphelion distance = 35.3 AU

Sun displacement from center OF = 17.26 AU

eccentricity e = 0.96

Newton’s genius

Same “physics’ everywhere

in the lab, in the Solar System and anywhere else in the Universe – in other words we can measure and understand the physics of the cosmos around us

He argued:

The rules describing the acceleration of objects falling on the Earth can also describe the motion of the planets

- Hypothesis: Newton, 1687:
- There is a gravitational attraction between all of the planets and the Sun

Keeping the planets in their place

Newton’s 1st law of motion

A body will remain at rest or in constant motion along a straight line path unless acted upon by an external force

In reality, a planet is continuously accelerated towards the Sun by a gravitational force

It is this continuous gravitational interaction that causes a planet to follow an elliptical orbit rather than a straight line path through space

In each second the Moon ‘falls’ 1.4 mm towards the Earth (away from straight line path) and moves 1 km around its orbit – the Moon is continuously falling towards Earth, but still stays in orbit

Fg

Path of Moon

with gravity

(orbit)

Moon

Path of Moon

without gravity

Earth

See ASTRO page 50

Kepler’s 3rd law…. Newton style

Cutting to the chase - Newton showed:

P2/a3 = K = 4p2 / G(MSun + MPlanet)

In other words, Newton found that the constant

K in K3 is related to the system mass

Units are now kilograms, meters and seconds

(The SI units of measure)

G is the universal gravitational constant

Phobos

Period = 7.656 hours

Orbital radius = 9400 km

Moon diameter is about 20 km

Discovered by Asaph Hall in 1877

Weighing Mars

MMars + Mphobos = (4p2) a3 / G P2

Can safely assume MMars >> MPhobos

so, using SI units (meters, sec., kg)

MMars = (4p2) (9.4x106)3 / G (7.656 x 3600)2

= 6.6 x 1023 kg

= 1/10th MEarth

Not just for planets

Provided a measure of the size of the orbit (a) and the orbital period (P) can be made K3 as formulated by Newton can be used to find the masses of astronomical objects…..

Later on we will weigh the Milky Way Galaxy using K3

We now have a set of tools and laws to describe:

Position on the sky – the celestial sphere (ecliptic)

The distances to the planets and the scale of the Solar System – Copernicus’s method for inferior planets

Orbital shape – semi-major axis and eccentricity

Planetary motion – Kepler’s three laws

The mass of a planet – if it has a Moon – Newton’s refinement to K3

Our next task is to take an inventory of the Solar System – what exactly is it and what kind of objects does it display ?

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