KNOWLEDGE TO “GET”

FROM TODAY’S CLASS MEETING

Class Meeting #6, Monday, February 1st, 2016

1) GRAVITY: finish up from Fri, Jan 29th (pages 111-112, 123)

2) Isaac Newton’s LAWS of MOTION (briefly) (pages 115-117)

3) Distances of interest (pages 2, 4, 15, 70, A-15)

4) Planet Orbit characteristics (shapes) (pgs 65-67)

5) Kepler’s 3 Laws of Planetary Motion (these are important !!) (pgs 66-70)

Your Lab Sections Wednesday & Thursday in Room 232 in Walden Hall will conduct the

“DENSITY” Lab Read through the lab BEFORE your section meets!!

ANY QUESTIONS?

You should have a printed copy of the DENSITY Lab with you when you arrive at your Lab Section meeting

The PDF of the Lab is available within the LAB-MANUAL folder at the class website:

http://astronomy.nmsu.edu/murphy/ASTR105G-M040506-Spring2016/LAB-MANUAL/LAB02-Density-Feb03-04.pdf

QUIZ #2 WILL OCCUR:

this coming Friday, February 5th (the final ~20 minutes of class) Topics:

i) Kepler’s 3 Laws of Planet Motion ii) initial telescopic observations which supported the Sun-centered theory of Solar System structure iii) Eratosthenes’ determination of Earth’s size iv) concepts of Mass, Volume, Density, & Gravity v) the 8 planets in our solar system and their ‘order’ of increasing Orbital Semi-Major Axis value: Table E1, pg A-15

listed on the next slide are text sections and questions for quiz study

Kepler’s Laws: text pgs 66-68, 70, 73,79 Text questions: Chapt 3 #s 9, 20, 25, 26, 27

Initial Observations that supported Copernicus’ idea: text pg 67-70

Text questions: Chapt 3 #s 723, 24, 29

Eratosthenes: text pg 63 Text questions: Chap 3 #s 45, 46 (if you can do the math, great, but the idea is

what’s important)

Mass, Volume, Density: text pgs 113-114; 143-146 (in Chap 5) Text questions: Chap 4 #s 4, 11, 15, 26 ; Chap 5 #s 6, 7, 30

Gravity (including comparing mass to weight: text pages 123-125

Text questions: Chap 4, #s 9, 18, 19, 32, 33, 42

You should by the end of this week have read through

CHAPTER 4 in the text….. and pages 143-147 in Chapt 5

(with special emphasis upon the topics we have discussed here in class PLUS

Eratosthenes’ method for estimating Earth’s circumference, and Galileo’s Copernicus-

supporting telescope observations, both in Chapter 2)

RESULTS OF QUIZ #1 Average = 16.9/20 (84.5%) Median = 17

LET’S RETURN TO OUR TOPIC OF GRAVITY

from our previous class…

but first

ANY QUESTIONS ????

1) The Gravitational attraction (force) between YOU (70 kg) and the Earth is:

FGrav = G MassEarth MassYou / (Earth’s radius)2

= 6.67 x 10-11 N m2/kg2 x 6 x 1024 kg x 70 kg (6,378,000 meters)2 = 688.66 Newtons (equivalent to 154 pounds)

When we calculate the gravitational force between the Earth and another object, we

assume that all of Earth’s mass is concentrated at its center:

It is sometimes easier to think in terms of:

Gravitational Acceleration = FGrav / MassYou if one mass is much, much larger than the other…

Earth’s downward acceleration at its surface: 688.66 Newtons / 70 kg = 9.8 (m/s)/s

Earth’s Surface Gravitational Acceleration = 9.8 (m/s)/s Which indicates that if you, or anything else, is dropped here at the Earth’s surface, its FALLING SPEED will increase by 9.8 meters per second (~20 mph) during each second that you fall

Time = 0 seconds, Speed = 0 m/s (time to fall 1 meter = 0.45 seconds) Time = 1 second, Speed=9.8 m/s m/s Distance fallen = 4.9 meters ( (= ~16.5 feet) Time = 2 seconds; Speed=19.6 m/s Distance fallen=19.6 meters (= ~66 feet) Time = 3 seconds; Speed=29.4 m/s Distance fallen=44.1 meters

Acceleration (9.8 m/s)/s

44.1 meters

(= ~150 feet)

What is the Gravitational Acceleration rate between two people seated side-by-side in this classroom?

FGrav = G MassNeighbor MassYou / (1 meter)2

= 6.67 x 10-11 N m2/kg2 x 70 kg x 70 kg (1)2

= 0.000000327 Newtons (=3.27 x 10-7 Newtons)

Gravitational Acceleration between you two:

3.27x10-7 Newtons / 70 kg = 5 x 10-9 (m/s)/s (equal to ~0.00000001 miles per hour per second)

SO, due to gravity: i) an object dropped from 1 meter above the ground will require 0.45 seconds to complete its fall

BUT ii) Two people 1-meter apart from each other and far out in space away from any large mass (source of gravity) would require approximately 20,000 seconds (approximately 5 hrs 33 minutes) before their centers collide

For Gravity Acceleration conditions: DISTANCE = ½ x Acceleration x TIME2

So, how can these concepts be used to understand the Planets (or moons or stars)?

1) you want to know what material(s) some planet in the solar system is composed of; you can see it, and know its size (radius) (from which you can determine its VOLUME), but don’t know much more….

2) if a moon orbits the object, or a spacecraft passes by, you can determine the gravitational pull of the planet upon the moon or upon the spacecraft; a measure- ment of the GRAVITY can then permit you to calculate the MASS of the planet

Consider a planet in the Solar System; a spacecraft is going to ‘pass by’ the planet The spacecraft will ‘feel’ the planet’s gravitational attraction (‘acceleration’ induced upon the craft) The magnitude of the gravitational attraction felt, and the known distance between the planet and the spacecraft, allow us to determine the planet’s MASS

A moon orbiting a planet can also provide information from which the planet’s mass, and density can be determined…

3) If you know the VOLUME of the planet and its MASS, you can determine the DENSITY of the planet: DENSITY = MASS / VOLUME 4) Planets are not pure substances, rather they are a mixture of many materials, but a planet’s DENSITY can tell you if it is rocky:metallic (like Earth) or gaseous hydrogen & helium (like Jupiter), WITHOUT EVER HAVING TO GO TO THAT PLANET !!

This technique allowed astronomers to learn Jupiter’s density, ~1.3 grams per cubic centimeter, well before we ever sent a spacecraft to Jupiter..

The telescopically observed “apparent” size of Jupiter and our knowledge of its distance from the Sun, PLUS the orbit characteristics of Jupiter’s four large moons, permitted determination of Jupiter’s volume, its gravitation attraction upon its moons, and thus Jupiter’s mass, and.. DENSITY = MASS / VOLUME

ANY QUESTIONS ABOUT GRAVITY?

ORBIT MOTIONS (planets around the Sun, the Moon around Earth, etc.) are controlled by the Gravity Force between the central object and the orbiting object…

Let’s start to take a look at this control…. with Isaac Newton’s ‘3 Laws of Motion’

Isaac NEWTON’S 3 Laws of Motion (pg 115) (these help us understand orbit motions….)

1st Law: an object continues at the same speed (which can be zero) and in the same straight direction unless acted upon by an outside force 2nd Law: the acceleration (change in speed OR direction) an object experiences is directly related to the magnitude of the force applied upon the object & the mass of the object 3rd Law: each action (force) occurs in the presence of an equal magnitude but opposite-direction action (force) [ “action:reaction” ] .. we will in a few minutes discuss how gravity and Newton’s laws of motion control orbit motions

OK, let’s recall DISTANCE UNITS for a moment… METRIC ENGLISH

Units of length 1 centimeter (10 millimeters; 0.3937inch) 1 inch (2.54 centimeters) 1 foot(12 inches; 30.48 cm) 1 yard (3 feet; 91.44 cm) 1 meter (100 cm; 39.37 inches) 1 kilometer (1000 meters; 0.621 miles) 1 mile (1.61 kilometers) 5 kilometers (3.1 miles)

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ASTRONOMICAL DISTANCES

Within our Solar System, the Standard Distance is the average “Sun-to-Earth” distance:

~93,000,000 miles = ~9.3 x 107 miles

[ How many kilometers is this equal to? ]

9.3 x 107 miles x 1.61 km per mile ~1.5 x 108 km (which is 391 times the Earth-to-Moon distance) This distance (150,000,000 km or 93,000,000 miles) is called: ONE ASTRONOMICAL UNIT (1 AU)!!! 1 AU

Earth

In Our SOLAR SYSTEM, Planets’ “Average Distance from the Sun”

range from: Closest to the Sun

MERCURY: 0.387 AU (= 6 x 107 km = 37 million miles)

to

Farthest from the Sun PLUTO: 39.48 AU

(= 6 x 109 km = 3.7 billion miles)

(and remember, Earth is 1 AU from the Sun)

While AUs are a very nice length scale to employ when talking about distances within our Solar System…(better than meters, or feet, or leagues)

AU’s are not very useful when discussing distances BETWEEN stars within our Milky Way Galaxy

FOR EXAMPLE: The Sun’s nearest stellar neighbor, a star named ALPHA CENTAURI, is located:

~250,000 AU (= 2.5 x 105AU) from the Sun

What other Astronomical unit(s) are you familiar with?

PARSEC= 3.1 x 1013 kilometers (=210,000 AU)

LIGHT YEAR= 9.5 x 1012 km (= 63,000 AU) (so…. 1 PARSEC = 3.3 LIGHT YEARS)

What does a LIGHT YEAR really mean? ( is it a time duration, or a distance, or…. ? )

A “TIME” = DISTANCE EXAMPLE: How far is it from Las Cruces to Deming?

97 kilometers (60 miles) -------------------------------------------------------- But, what if I said, ‘It is one hour to Deming’.

How would you interpret this?

“Travelling at freeway speed, on I-10, a time interval of one hour will get me to Deming”

--------------------------------------------------------- So, a ‘time interval’ at a known speed = a distance

This is the idea behind the distance unit of a LIGHT YEAR

ONE LIGHT YEAR = the distance light travels in one year

= SPEED OF LIGHT times a time interval of 1 YEAR

= 3 x 108 meters per second x 365.25 days per year x 86,400 seconds per day

= 9.5 x 1015 meters (= 9.5 x 1012 km = 5.9 x 1012 miles) .. and one AU = 1.5 x 108 km or 93,000,000 miles, so ..

ONE LIGHT YEAR = ~ 63,000 AU

Speed of light = 186,000 miles per second

= 3 x 108 meters per second

How far apart are the Sun and Earth?

1 AU = 1.5 x 108 kilometers = 93,000,000 miles

How about their distance in LIGHT TIME?

~8 minutes !

So, sunlight striking the Earth’s surface here in Las Cruces right now left the Sun 8 minutes ago

.. and, the nearest star beyond the Sun is 4 Light Years away !!

Since the Speed of Light is the fastest speed physics allows, if the Sun was to ‘shut off’ right now, we would

not know about it for ~8 minutes……

Mercury (0.4 AU) is 3.2 light minutes from the Sun , Pluto (40 AU) 320 minutes or 5.33 hours from the Sun

ANY QUESTIONS ABOUT GRAVITY, or DISTANCES?

NOW, on to Johannes Kepler’s determination of three (3) characterizations of planetary

motions around the Sun

THESE ARE VERY IMPORTANT CONCEPTS!

Kepler developed these concepts BEFORE the

mathematical understanding of gravity was appreciated…

Kepler developed these ‘Laws’ in the late 1500’s and early 1600’s

JOHANNES KEPLER’s Early 1600’s

3 LAWS OF PLANETARY MOTION

Kepler developed his ‘Laws’ by analyzing many years of eyeball observations of planets’ changing positions relative to the background pattern of stars (Mars’ motions were especially useful !!)

Kepler was a ‘believer’ in Copernicus’ idea that the Earth traveled (orbited) around the Sun (and so did the other known 5 planets: Mercury, Venus, Mars, Jupiter, Saturn)

Copernicus’ Sun-centered theory arose in the 1540’s

The Sun-centered structure of the Solar System, and measured angles of planet positions, enabled Kepler to calculate the distance between a planet and the Sun, relative to the Earth-Sun distance, equal to 1 AU Mars Mars’ Orbit 1.52 AU Earth 1 AU Earth’s Orbit The angle at the Sun is a 90-degree angle

Measure this angle… calculate that length

When Mars and Earth are at different orbit locations, another measure of Mars’ distance from the Sun can be determined (and another, and then another,…): Mars’ orbit Earth’s orbit Mars

1.48 AU Earth

1 AU

Measure this angle… calculate that length

This is a 90-degree angle

1) Kepler’s 1st Law of Planetary Motion:

The shape of a planet’s orbit around the Sun is an ELLIPSE (a circle is one type of an ellipse, but no planet orbit in our Solar System is a perfect circle) A B C Orbit A: circular (eccentricity = 0) Orbit B: ellipse (eccentricity = ~0.5) Orbit C: ellipse (eccentricity = ~0.9)

All 3 of these planets have the

same Orbital Semi-Major Axis,

which is the measure of the

AVERAGE distance between

the planet and the Sun

An ELLIPSE is defined by its ECCENTRICITY (‘non-circularity’)

and its SEMI-MAJOR AXIS (SMA) length

Eccentricity values range from ZERO (for a circle) to 0.99999 (for a really stretched out ellipse) [Figure 3.18 in text] Major Axis The Semi-Major axis (SMA) length is equal to ½ of the Major Axis Length …and for a planet orbiting the Sun, it’s SMA is that planet’s AVERAGE distance from the Sun

F1 F2

In the process of determining his 1st Law, Kepler also determined that for a SINGLE PLANET travelling along an elliptical (non-circular) orbit, that single planet traveled fastest when nearest to the Sun, and slowest when farthest from the Sun. This understanding is now:

Figure 3.20 in your text KEPLER’S 2ND LAW OF PLANETARY MOTION: while travelling on their elliptical orbits, an individual planet travels fastest when nearest to the Sun, and slowest when farthest from the Sun (this is also known as the ‘Equal Area in Equal Time’ Law) [demonstrated on-line with your text, Figure 3.18]

Kepler’s 2nd Law A B C Cp Bp Location of slowest speed in orbit Orbit A: circular (orbital speed is constant) Orbit B: ellipse (planet travels fastest at point Bp) Orbit C: ellipse (planet travels fastest at point Cp)

We can animate Kepler’s 2nd law at:

http://astro.unl.edu/naap/pos/animations/kepler.swf

See Figs 3.16 and 3.18 in your text

But.. KEPLER’S 2ND LAW

DOES NOT

tell us anything directly about how long a given planet takes to complete one orbit…. For ORBIT PERIOD information we must use KEPLER’S 3RD LAW OF PLANET MOTION

Using the available observations, Kepler was also able to determine that: planets farther from the Sun on average (planets with larger Orbital Semi-Major Axes) require more time to complete one trip (one orbit) around the Sun than do planets nearer to the Sun (planets with smaller OSA’s). Using his knowledge of the orbital semi-major axes of the planets (relative to Earth’s = 1 AU) and also their orbital periods, Kepler worked to find a mathematical relationship between Orbital Semi-Major Axis and Orbital Period

… let’s consider an incorrect idea first…..

IF ORBITAL PERIOD depended directly upon ORBITAL SEMI-MAJOR AXIS, (‘twice as far = twice as long’) then: ORBITAL PERIOD OSA Estimated Earth 1 AU 1 Earth Year

Mercury 0.4 AU 0.4 EY(146d) Venus 0.7 AU 0.7 EY (255d) Mars 1.52 AU 1.52 EY Jupiter 5.2 AU 5.2 EY Saturn 9.5 AU 9.5 EY These results would arise IF all of the planets traveled at the same speed… BUT actually the larger a planet’s OSA, the smaller its average orbit speed…

The OBSERVED orbit periods for planets indicate that planets farther from the Sun travel more slowly (on average) than does Earth, and planets closer to the Sun (on average) travel faster than Earth ORBITAL PERIOD OSA Estimated Observed Earth 1 AU 1 Earth Year 1 Earth Year

Mercury 0.4 AU 0.4 EY(146d) 0.25 EY (88d) Venus 0.7 AU 0.7 EY (255d) 0.58 EY (224d) Mars 1.52 AU 1.52 EY 1.88 EY Jupiter 5.2 AU 5.2 EY ~12 EY Saturn 9.5 AU 9.5 EY ~30 EY These results indicate that planets do not all travel at the same speed, rather planets farther from the Sun than Earth travel more slowly than does Earth…

This Relationship between distance from the Sun and orbital speed, and thus Orbit Period, is quantified in Kepler’s 3rd Law, which relates: i) how far a planet is from the Sun on average

(its Orbital Semi-major Axis distance, OSA)

VS.

ii) the length of time for that planet to complete one complete orbit around the Sun,

the planet’s ORBIT PERIOD:

KEPLER’s 3rd LAW of Planetary Motion:

(Orbit Period)2 = OSA3

Period x Period = OSA x OSA x OSA

with PERIOD in units of Earth years and

OSA in units of Astronomical Units -----------------------------------------------------------

GRAVITY is responsible for Kepler’s Laws…

For Example: Orbital Period vs. Orbital Semimajor Axis

P2 = OSA3 which leads to P = \ OSA3 ) OSA | Period (Earth years) Mercury: 0.39 AU ___________

Venus: 0.72 AU ___________

Earth: 1.0 AU ___________

Mars: 1.52 AU ___________ Jupiter: 5.20 AU ____________ Saturn: 9.54 AU ____________ Uranus: 19.2 AU ____________ Neptune: 30.1 AU ____________ Pluto: 40 AU ____________

For Example: Orbital Period vs. Orbital Semimajor Axis

P2 = OSA3 which leads to P = \ OSA3 ) OSA | Period (Earth years) Mercury: 0.39 AU ___0.24___

Venus: 0.72 AU ___0.62__

Earth: 1.0 AU ___1.0_____

Mars: 1.52 AU ___1.88____ Jupiter: 5.20 AU ___11.86____ Saturn: 9.54 AU ___29.46____ Uranus: 19.2 AU ___84.1_____ Neptune: 30.1 AU ___165______ Pluto: 40 AU ___253______

Notice: we DO NOT use Eccentricity for Period !!!!

NOTE: Kepler’s 3rd Law DOES NOT CARE about the eccentricity of the orbit in question, it CARES ONLY about the Orbital Semi-Major Axis (average distance to the Sun during an orbit) A B C Orbit A: (eccentricity = 0; OSA=1AU; Period=1EY) Orbit B: (eccentricity=0.5; OSA=1AU; Period=1EY) Orbit C: (eccentricity=0.9; OSA=1AU; Period=1EY)

All 3 of these planets have the

same Orbital Semi-Major Axis length,

which is the measure of the

average distance between the planet

and the Sun

TABLE of PLANET INFORMATION

KEPLER’S 3 LAWS OF PLANETARY MOTION 1) Planet orbits are elliptical in shape (includes circular) 2) When traveling along its elliptical orbit, a planet

travels at its fastest speed when it is closest to the Sun (and slowest when farthest from the Sun)

3) P2 = OSA3 with P in units of Earth Years and

OSA in units of Astronomical Units

1 Astronomical Unit = 1 AU 1 AU = Earth’s orbital semi-major axis

Let’s work through an exercise to illustrate these concepts

Now, back to… Isaac NEWTON’S 3 Laws of Motion

(these help us understand orbit motions….)

1st Law: an object continues at the same speed (which can be zero) and in the same direction unless acted upon by an outside force 2nd Law: the acceleration (change in speed OR direction) an object experiences is directly related to the magnitude of the force applied upon the object and the mass of the object 3rd Law: each action (force) occurs in the presence of an equal magnitude but opposite direction action (force) [ “action:reaction” ]

Isaac Newton’s “3 Laws of Motion” help us understand Kepler’s Laws…

NEWTON’S FIRST LAW “An object moves at a constant velocity (speed and direction) if there is no net force acting upon it.” The constant velocity can be zero speed, so an object held in your hand has a velocity of zero (relative to the ground) because the upward force provided by your hand is negating the dominating downward force of Earth’s gravity (no NET force is acting upon the object)

NEWTON’S SECOND LAW “Force = Mass times Acceleration”

A potentially easier way to think of this is: Acceleration = Force / Mass

If you drop the object in your hand (remove the upward force you are providing), the object will accelerate downward, due to the force of Earth’s dominating gravitational force attraction, and the object’s acceleration will be 9.8 (m/s)/s, which we discussed during our previous class meeting

NEWTON’S THIRD LAW

“For any force, there is always an equal and opposite reaction force”

This law is not as easily ‘demonstrated’ as are Newton’s first two laws… the Earth is exerting a gravitational force upon you AND you are exerting a gravitational force upon the Earth.. If this is true (and it is), then WHY when you step off of a tall building do you appear to fall to the Earth’s surface, rather than having the Earth’s surface ‘rise’ to you?

this discussion is continued on next slide…

As Newton’s 2nd Law tells us:

Acceleration = Force / Mass

As we calculated previously, the FORCE between 70 kg you and the Earth is: 688 Newtons

and the gravitational acceleration you feel due to the Earth is: 688 N / 70 kg = 9.8 (m/s)/s (~20 mph/second)

BUT the acceleration the Earth feels’ from you is only:

688 N / (6 x 1024 kg) = 1.1 x 10-22 (m/s)/s

(which is equal to two tenth billionths of one trillionth of a mile per hour per second !!!)

.. this is Newton’s 3rd Law !!

The launching of a rocket is an example of Newton’s 3rd Law of Motion in action:

Hot gas molecules accelerated downward by the ‘burning’ of the fuel result in a ‘reaction’ that is the upward acceleration of the rocket

HOT GAS ATOMS and MOLECULES moving downward

THE ROCKET’S MASS MOVING UPWARD

So, how can NETWON’S 3 Laws and Gravity help us understand Kepler’s Laws?

Newton’s 1st and 2nd Laws tell us that a planet

would travel along a straight line path (constant speed and direction) IF no external

(unbalanced) force was acting upon the planet

BUT We know planets travel along curved ellipse-shaped paths (orbits), indicating that some force must be acting upon the planet…. the

force is GRAVITY between planet and Sun

GRAVITY

We can think of orbital motion as a ‘balance’ (Newton’s 3rd Law of Motion) between the:

GRAVITY FORCE vs ‘CENTRIPETAL FORCE’ FG = FC

FG FC

x Speed2 = MP Orbit Radius

NEWTON’S MODIFIED VERSION OF KEPLER’S 3rd LAW

Kepler’s 3rd Law is: P2 = OSA3

It “WORKS” only for an object orbiting our Sun, OR for an object orbiting around another star which possesses the exact same mass as (same number of protons and neutrons as) the Sun Isaac Newton determined, after he conceived of the physical understanding of GRAVITY (~100 years after Kepler determined his 3rd Law), that Kepler had made a slight error……

Newton determined that the true form of Kepler’s 3rd Law should be: (MassSun + MassPlanet) x P2 = OSA3

MassSun --------------------------------------------------------------------- Why did KEPLER miss this mass issue?

MassSun + MassJupiter = 1.001

MassSun ….. so the mass of the planet(s) is inconsequential, and so

is its “pull” upon the Sun !!

..because the Sun and the planet exert a gravitational attraction upon EACH OTHER !.. it is NOT just one way !!

Not much different from a

value of 1.0 !!

One item of information this MODIFIED version of KEPLER’S 3RD LAW provides is the following:

If the Sun was more massive than it actually is, than the

Earth’s orbital period would be < 1 Earth year,

and

if the Sun was less massive than it actually is, than the Earth’s orbital period would be > 1 Earth Year

We can also use this MODIFIED version of KEPLER’S 3RD LAW to calculate some orbital information,

and more….

[(MStar + Mplanet) / MSun] x P2 = OSA3

Knowing that the Earth’s mass is << the Sun’s Mass: (so we can ignore Mplanet above)

If the Sun’s mass was twice its actual value: 2 MSun x PEarth

2 = OSAEarth

3 MSun

1 2 x PEarth

2 = OSAEarth3

P2 = 1/2 P= 1/2 PEarth = 1/1.414 = 0.707 EY

If MSun was only 1/2 of the Sun’s mass, P = 1.414 EY

As it turns out, the modified form of Kepler’s 3rd Law is a very powerful tool that can be

applied to any two objects orbiting each other (not just planets around stars)….

GENERAL FORM OF MODIFIED 3rd LAW

(Mcentral + Morbiter) x P2 =OSAorbiter

3 MSun 1 AU

We can use this equation to calculate the Moon’s orbital period around the Earth (assuming we know the Moon’s OSA)

(applicable when Mcentral >> Morbiter)

What is the Moon’s orbital period around Earth? (MEarth + MMoon) x P2 = OSAMoon

3 MSun 1 AU

MEarth = 6 x 1024 kg; MMoon = 7.35 x 1022 kg

MSun = 2 x 1030 kg; OSAMoon= 384,00 km= 0.00257 AU

[(6 x 1024 + 7.35 x 1022) /(2 x 1030)] x P2 = (0.00257)3

P2 = .00559 Earth years PMoon = .00747 Earth years = 27.3 days

Compare this to information on page A16 (Appendix E) in your text..

KEPLER’S 3 LAWS OF PLANETARY MOTION 1) Planet orbits are elliptical in shape 2) When traveling along its elliptical orbit, a planet travels at its fastest speed when it is closest to the Sun (and slowest when farthest from the Sun) 3) P2 = OSA3 with P in units of Earth Years and

OSA in units of Astronomical Units

1 Astronomical Unit = 1 AU 1 AU = Earth’s orbital semi-major axis

GRAVITY is the force that dictates these Laws !

Kepler’s 3rd Law and Gravity The following steps illustrate how Kepler’s 3rd Law of Planetary Motion is related to Newton’s formulation of Gravity; both GRAVITATIONAL force and CENTRIPETAL force (related to curved, rather than straight, motion): 1) Sun’s Gravitational force:

Fgrav = G MSun Mplanet / [ OSA(in meters)2 ] 2) Planet’s Centripetal (think ‘centrifugal’) force:

Fcent = Mplanet * Speedplanet2 / OSA (in meters)

We can think of orbital motion as a ‘balance’ between the:

GRAVITY FORCE vs ‘CENTRIPETAL FORCE’ FG = FC

FG FC

Speed = Distance around orbit circle

Orbit Period

Speed2 = MP Orbit Radius

3) For balance: Fgrav = Fcent (G MSun Mplanet / OSA2 ) = Mplanet * Speedplanet

2 / OSA

4) Speed = distance around orbit divided by time Speed = 2 * π * OSA / (Orbit Period)

5) So: Gravitational Force = Centripetal Force

MPlanet MPlanet

= G MSun / OSA2 = (4 π2 OSA2/ Orbit Period2) / OSA 6) Rearranging:

(G MSun / 4 π2) x (Orbit Period)2 = OSA3

Constant x P2 = OSA3

7) (G MSun / 4 π2) x (Orbit Period)2 = OSA3

Up to this point, we have been using units of kilograms, meters, and seconds; if we stick in the correct values for G and MSun and π , and convert Orbit Period to Earth years and OSA to AU, we get: P2 = OSA3 with Period in Earth years, OSA in AU

SO, KEPLER’S LAWS “WORK” IN

EXPLAINING PLANET ORBIT MOTIONS BECAUSE OF

THE NATURE OF THE GRAVITATIONAL FORCE