WHY PLANETS HAVE ELLIPTICAL ORBITS

REYNALDO B. VEAPHILIPPINE SCIENCE HIGH

SCHOOLNOVEMBER 10, 2008

QUIZ

THE EARTH IS CLOSEST TO THE SUN IN THE MONTH OF

JANUARY

THE EARTH IS FARTHEST FROM THE SUN IN THE MONTH OF

JULY

ADDITION OF VECTORS

A A

B

BC C

A + B = C

TIP-TO-TAIL ADDITION PARALLELOGRAM LAW

VELOCITY IS A VECTOR QUANTITY

NOT HERE

BUT HERE

CURRENT

RELATIVE TO WATER

RELATIVE TO GROUND

DUE TO CURRENT

FORCE IS A VECTOR QUANTITY

2 MULES PULLING A BOAT ON EITHER BANK

FORCE IS A VECTOR QUANTITY

SAME AS ONE RIVER TUG PULLING THE BOAT STRAIGHT UP THE MIDDLE

MURAL QUADRANT SEXTANT

URANIBORG

TYCHO BRAHE

Kepler’s Laws:

I. Each planet moves around the sun in an ellipse with the sun at one focus.

II. The radius vector from the sun to the planet sweeps out equal areas in equal intervals of time.

III. The squares of the period of any two planets are proportional to the cubes of the semimajor axes of their respective orbits: T a a3/2.

IN ONE OF HIS EXPERIMENTS GALILEO FOUND THAT IF A BALL WAS ALLOWED TO ROLL DOWN ONE PLANE AND BACK UP ANOTHER IT WOULD TEND TO KEEP ROLLING UP THE SECOND PLANE UNTIL IT REACHED THE SAME HEIGHT IT STARTED AT. WITH A LEAP OF THE IMAGINATION, AGAIN, HE STATED THE IF THE SECOND PLANE WERE HORIZONTAL THE BALL WOULD NEVER STOP ROLLING (BECAUSE IT WOULD NEVER REACH THE ORIGINAL HEIGHT). THUS HE CONCLUDED THAT THE NATURAL STATE OF AN OBJECT IN HORIZONTAL MOTION WAS TO KEEP ON MOVING HORIZONTALLY, AT CONSTANT SPEED, FOREVER. THIS LATER WOULD LEAD TO THE LAW OF INERTIA. THIS IS A RADICAL DEPARTURE FROM ARISTOTELIAN PHILOSOPHY IN WHICH ANY HORIZONTAL MOTION REQUIRED A PROXIMATE CAUSE.

RENE DESCARTES GENERALIZES GALILEO’S LAW OF INERTIA TO NOT ONLY APPLY TO HORIZONTAL MOTION THUS: IN THE ABSENCE OF ANY EXTERNAL FORCE ACTING ON IT A BODY AT REST WILL REMAIN AT REST AND A BODY IN MOTION WILL REMAIN IN MOTION, AT CONSTANT SPEED, IN A STRAIGHT LINE.

NEWTON’LAWS OF MOTION

LAW I. EVERY BODY CONTINUES IN ITS STATE OF REST, OR OF UNIFORM MOTION IN A STRAIGHT LINE, UNLESS IT IS COMPELLED TO CHANGE THAT STATE BY FORCES IMPRESSED UPON IT.

LAW II. THE CHANGE IN MOTION IS PROPORTIONAL TO THE MOTIVE FORCE IMPRESSED; AND IS MADE IN THE DIRECTION OF THE STRAIGHT LINE IN WHICH THAT FORCE IS IMPRSSED.

LAW III. TO EVERY ACTION THERE IS ALWAYS OPPOSED AN EQUAL REACTION; OR, THE MUTUAL ACTION OFBTWO BODIES UPON EACH OTHER ARE ALWAYS EQUAL, AND DIRECTED TO CONTRARY PARTS.

Kepler’s 2nd Law: The Law of Areas All forces are

directed towards the sun

Kepler’s 3rd Law: T a 3/2

Forces are inversely proportional to the square of the distances

Force of gravity

Kepler’s 1st Law: The Law of Ellipses

Galileo’s law of inertia

Descartes law of Inertia

PREDICTION

Newton’s 1st Law: The Law of Inertia

Newton’s 2nd Law: F = d(mv)/dt

Newton’s 3rd Law: Action and reaction

F’ F

P

F’ F

P

q

Q

F’ F

ALL LIGHT RAYS STARTING AT ONE FOCUS WILL BE FOCUSED TO A POINT AT THE OTHER FOCUS.

P

ELLIPSE OR ANY CURVE

TANGENT LINE

P

LIGHT RAY MIRROR-REFLECTED FROM THE CURVE

ELLIPSE OR ANY CURVE

TANGENT LINE

light ray

reflected light ray

P

LIGHT RAY MIRROR-REFLECTED FROM THE TANGENT LINE

TANGENT LINE

ELLIPSE OR ANY CURVE

light ray

reflected light ray

A LIGHT RAY, MIRROR-REFLECTED FROM FROM THE CURVE AT ANY POINT, FOLLOWS THE SAME PATH IF IT WERE MIRROR-REFLECTED AT THAT POINT FROM THE TANGENT LINE.

P

TANGENT LINE

ELLIPSE OR ANY CURVE

light ray

reflected light ray

THE REASON THAT LIGHT REFLECTS FROM THE CURVE JUST AS IT WOULD FROM THE TANGENT LINE AT THE SAME POINT IS THAT THE TANGENT LINE INDICATES THE DIRECTION OF THE CURVE AT EXACTLY THAT POINT.

MIRROR

LIGHT RAY

REFLECTED LIGHT RAY

THIS ANGLE

EQUALS THIS ANGLE

LAW OF REFLECTION FROM A FLAT MIRROR

F’ F

P

THIS ANGLE

EQUALS THIS ANGLE

THE INCIDENT RAY FROM F TO P MAKES THE SAME ANGLE WITH THE TANGENT LINE AT P AS DOES THE REFLECTED RAY, WHICH GOES TO F’. PROVE THAT THIS STATEMENT IS EQUIVALENT TO SAYING THAT THE DISTANCE F’P PLUS THE DISTANCE FP IS THE SAME FOR ANY POINT P ON THE ELLIPSE.

INCIDENT RAY

REFLECTED RAY

FF’

G’

P

START OF PROOF

T

F’

G’

P

G’T = F’T

TP = TP

ANGLE GTP=ANGLE F’TP = RT ANGLE

SAS CONGRUENCE

T

F’

G’

P

CONGRUENCE G’P = F’P

ANGLE F’PT = ANGLE G’PT

T

FF’

G’

P

ANGLE F’PT = ANGLE G’PTANGLE G’PT = ANGLE FPS

ANGLE F’PT = ANGLE FPS

S

therefore

perpendicular bisector line ST will reflect light form F to F’ at point P

CONGRUENCE F’P = G’P

therefore F’P + PF = FG’

T

perpendicular bisector

FF’

G’

straight string

FF’

same string bent

FF’

G’

P

perpendicular bisector

FF’

G1’G2’ G3’

G4’

RECAPITULATION

WE ALSO HAVE AN ELLIPSE THAT OBEYS THE STRING-AND-TACKS CONSTRUCTION, i.e., F’P + PF IS THE SAME ALL THE WAY AROUND THE STRING.

WE HAVE A LINE THAT REFLECTS LIGHT THAT ARRIVES FROM POINT F AT POINT P WITH EQUAL ANGLES OF INCIDENCE AND REFLECTION BACK TO POINT F’. THIS LINE HAPPENS TO BE THE PERPENDICULAR BISECTOR OF F’G’.

FF’

G’

PT

(reflecting line)

perpendicular bisector

FF’

G’

PT

WE KNOW THAT EACH POINT ON THE ELLIPSE HAS THE SAME MIRROR-REFLECTING PROPERTIES AS A TANGENT LINE AT THAT POINT.

THUS IF THE RELECTING LINE AT P IS ALSO TANGENT TO THE ELLIPSE AT P THEN WE HAVE PROVED THAT THE TWO PROPERTIES (STRING-AND-TACKS CONSTRUCTION AND REFLECTING LIGHT FROM ONE FOCUS TO THE OTHER) ARE EQUIVALENT.

ALL THAT’S LEFT TO PROVE, THEREFORE, IS THAT THE REFLECTING LINE AT POINT P IS ALSO TANGENT TO THE ELLIPSE AT POINT P.

Also tangent?

perpendicular bisector

(reflecting line)

FF’

P

TANGENT: EVERY POINT BUT P IS OUTSIDE THE ELLIPSE

NONTANGENT: THIS SEGMENT IS INSIDE THE ELLIPSE

FF’

G’

PT

Q

F’Q = G’Q

F’Q +QF = G’Q + QF

FF’

G’

P

Q

G’Q + QF > G’P + PF

since G’PF is a straight line

FF’

P

Q

G’Q + QF > G’P + PF

F’Q +QF = G’Q + QF

F’P + PF = G’P +FP

earlier it was shown that:

therefore:

F’QF > F’PF

IN OTHER WORDS IF WE WANTED TO REACH POINT Q WITH A STRING STRETCHED FROM TACKS AT F AND F’, THE STRING WOULD HAVE TO BE LONGER THAN THE ONE NEEDED TO REACH THE UNIQUE POINT P. IT WAS

SHOWN EARLIER THAT THIS MEANS ALL SUCH POINTS ARE OUTSIDE THE

ELLIPSE. THUS THE LINE IS TANGENT TO THE ELLIPSE AT POINT P. QED !

for any such point q on that line except for the one coincident with point P

F’QF = G’QF > G’PF = F’PF

or

SA

DURING A CERTAIN INTERVAL OF TIME

B

PART II

SA

DURING NEXT INTERVAL OF TIMEwith no force acting

c

Bc = ABB

SA

DURING NEXT INTERVALS OF TIMEwith Sun’s force acting

c

V

Sun’s force represented by an impulse applied at Bresults in a component of motion directed toward the Sun, BV

C

compounding of Bc and BV using parallelogram law yields “actual” motion BC

note: Cc is not directed toward the Sun but is parallel to BV

B

SA

DURING NEXT INTERVALS OF TIMEwith Sun’s force acting

c

V

Sun’s force represented by an impulse applied at Bresults in a component of motion directed toward the Sun, BV

C

compounding of Bc and BV using parallelogram law yields “actual” motion BC

same procedure is repeated at each point

dD

B

note: Cc is not directed toward the Sun but is parallel to BV

SA

SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

cC

SHOW THAT: area of SAB = area of SBC

B

SA

SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

cC

SHOW THAT: area of SAB = area of SBC

First step is to show that area of SAB = area of SBc

B

SA

B

SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

ccommon base

altitudes

SHOW THAT: area of SAB = area of SBC

First step is to show that area of SAB = area of SBc

x y

AB = Bc ; rt angles equal ; angle ABx = angle cBy congruence of 2 small ’s, by ASA

Ax = cy or equal altitudes area of SAB = area of SBc QED !

SA

SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

cC

SHOW THAT: area of SAB = area of SBC

First step is to show that area of SAB = area of SBc

B

Next show that area of SBc = area of SBC

Cc parallel to SB altitudes are equal

common base SB & equal altitudes area of SBc = area of SBC

by transmissivity, area of SAB = area of SBC

QED !

SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

SA

cC

D

B

The path Bc would have been taken if there were no force at all. Instead, there is a force, directed toward S. That force changes the direction from path Bc to path BC, but it cannot change the area swept out during the fixed interval of time.

The same idea could be applied to successive triangles swept out in equal intervals of time. We have thus succeeded in proving Kepler’s Second Law of planetary motion!

We have thus far used Newton’s first Law (the law of inertia), Newton’s second law (any change in motion is in the direction of the impressed force), and the idea that the force of gravity on the planet is directed toward the Sun. Nothing else. For example, we have not used the idea that the force of gravity is inversely proportional to the square of the distance. So the inverse-square-of-the-distance character of gravity has nothing to do with Kepler’s second law. Any other kind of force would have produced the same result provided only that the force is directed toward the sun. What we have learned is that if Newton’s first and second laws are correct, then Kepler’s observation that planets sweep out equal areas in equal intervals of time means that the gravitational force of the planet is directed towards the sun. Incidentally, we have also used Newton’s first corollary to his laws – that the net motion produced by both tendencies in the time interval is given by the parallelogram of the separate motions that would have occurred.

The above demonstration is an exact copy of the one in Principia Mathematica by Newton.

NEWTON’S DRAWING

FEYNMAN’S DIAGRAM

sun

planet 1

a1

a2

planet 2T1/T2 = (a1/a2)3/2

Let T = time for planet to complete one orbit

FF’

a

b

R

R

A circle is a special case of an ellipse that can be constructed by moving the two foci to the center. Kepler’s laws allow planetary orbits to be circles but don’t require it. In reality the orbits of the planets in our solar system are very nearly circles.

T R3/2

SA

C

D

B

SUCCESSIVE POSITIONS OF A PLANET IN SPACE

POSITION DIAGRAM

****THE FOLLOWING PROOF OF THE INVERSE SQUARE LAW ARE ATTRIBUTED TO DAVID AND JUDITH GOODSTEIN OF CALTECH.

SA

C

D

B

VELOCITY OF THE PLANET DURING EACH SEGMENT

vAB

vBC

vCD

SA

C

D

B

VELOCITY OF THE PLANET DURING EACH SEGMENT

vAB

vCD

vBC

VELOCITY DIAGRAM

vBC

SA

C

D

B

VELOCITY OF THE PLANET DURING EACH SEGMENT

vAB

vCD

CHANGE IN VELOCITY

vB

vC

SIMPLEST SPECIFIC EXAMPLE

CIRCULAR ORBIT OF RADIUS R

RS

A

B

C

D

SA=SB=SC=SD

IMPULSE AT A,B,C,D AND SO ON SAME NO MATTER HOW FORCE DEPENDS ON DISTANCE

vab=vbc=vCD

AB=BC=CD

POSITION DIAGRAM IS A REGULAR POLYGON

1

2

3

4

5

< 2 = < 5

REGULAR POLYGON

< 1 + < 2 + < 3 = 180

< 1 + < 5 + < 3 = 180

< 4 + < 5 + < 3 = 180

THEREFORE

< 1 = < 4

ALL THESE ANGLES ARE EQUAL

S

A

B

C

D

POSITION DIAGRAM

vAB

vBC

vCD

VELOCITY DIAGRAM

v

ALL VELOCITIES ARE OF EQUAL MAGNITUDE AND EQUAL ANGLE APART.

ALL THE CHANGES v ARE OF EQUAL MAGNITUDE.

THUS THE VELOCITY DIAGRAM IS ALSO A REGULAR POLYGON.

AS FINER AND FINER INCREMENTS ARE TAKEN THE POSITION DIAGRAM AND THE VELOCITY DIAGRAM TRANSFROM FROM REGULAR POLYGONS INTO CIRCLES.

POSITION OR ORBIT DIAGRAM VELOCITY DIAGRAM

R

v

v

THIS IS A SNAPSHOT AT A POINT IN TIME.

R

v

v

A SNAPSHOT AT ANOTHER POINT IN TIME

v

v

POSITION OR ORBIT DIAGRAM

R

v

DISTANCE AROUND IS 2R

TIME TO TAKE TRIP IS T

SPEED v = 2R/T

R

v

R

v

POSITION OR ORBIT DIAGRAM VELOCITY DIAGRAM

F v/T

v/T = 2v/T

but v = 2R/T

therefore F (2)2R/T2

but from Kepler’s 3rd Law:

T R3/2

therefore F 1/R2

THE INVERSE SQUARE LAW!

Sun

IN TIME t, MOVING RAPIDLY WHEN IT IS CLOSE TO THE SUN, THE PLANET SWEEPS OUT THIS AREA

IN EQUAL TIME t, MOVING SLOWLY WHEN IT IS FAR FROM THE SUN, THE PLANET SWEEPS OUT THIS AREA.

THE TWO AREAS ARE EQUAL.

PART IV

FEYNMAN’S APPROACH: DIVIDE THE ORBIT INTO EQUAL ANGLES INSTEAD OF EQUAL AREAS.

These two angles are equal.

BECAUSE THE PLANET MOVES RAPIDLY WHEN IT IS CLOSE TO THE SUN, THE TIME TAKEN TO TRAVERSE THIS SEGMENT IS SMALL.

HERE THE PLANET MOVES MORE SLOWLY, SO THE TIME TO TRAVERSE THIS PART IS LONGER.

A

B

C

D

E

F

S

vv

THE PLANET MOVES FASTER ALONG BC THAN ALONG EF. TO SEE HOWMUCH FASTER, THE AREAS OF ’s SBC AND SEF MUST BE COMPARED. THE TIME t IS PROPORTIONAL TO THE AREA SWEPT BECAUSE EQUAL AREAS ARE SWEPT IN EQUAL TIMES.

DIGRESSION: FOR EQUAL ANGLES SWEPT THE AREA IS PROPORTIONAL TO THE SQUARE OF THE DISTANCE FROM THE SUN. (ATTRIBUTED TO GOODSTEIN)

S

H

G

W

X

These two angles are equal.

S

H

G

W

X

LAY SWX ON SGH

h

g

Z

z

DRAW hg PARALLEL TO HG SUCH THAT Whz HAS THE SAME AREA AS Xgz Sgh HAS THE SAME AREA AS SWX AND IT IS SIMILAR TO SGHCALL SZ AND Sz THE DISTANCES FROM THE SUN TO THE ORBITFOR SIMILAR ’s THE BASE AND ALTITUDE INCREASE AS THE SIZE AND THEREFORE THE AREA IS PROPORTIONAL TO THE SQUARE OF THE SIZE.THE SIMILAR ’s SGH AND Sgh HAVE AREAS IN PROPORTION TO THE SQUARES OF THE LENGTHS SZ AND Sz. BUT SWX HAS THE SAME AREA AS Sgh, SO THE AREA OF SWX IS ALSO IN PROPORTION TO THE SQUARE OF Sz.NOW SHRINK THE CENTRAL ANGLE DOWN SMALLER AND SMALLER. THE LENGTH Sz ULTIMATELY BECOMES EQUAL TO SW OR SX, THE DISTANCE TO THE SUN. QED!

A

B

C

D

E

F

S

vv

S B

C

E

FREORIENT & OVERLAY SEF ON TOP OF SBC.

THE TIME IT TAKES TO GO THROUGH ANY PORTION OF THE ORBIT IS PROPORTIONAL TO THE AREA SWEPT OUT, WHICH IS PROPORTIONAL TO THE SQUARE OF THE DISTANCE FROM THE SUN.

t R2, WHERE R IS THE DISTANCE OF THE PLANET FROM THE SUN.

A

B

C

D

E

F

S

vv

v

v

v

v

THE BIGGER THE F, THE BIGGER THE vTHE LONGER THE t, THE GREATER THE v, OR

v F t

BUT SINCE F 1/R2 AND t R2,

v (1/R2) x (R2) = 1

v DOES NOT DEPEND ON R AT ALL! EVERYWHERE IN THE ORBIT, NO MATTER HOW CLOSE OR FAR AWAY FROM THE SUN, THE v PRODUCED, FOR EQUAL ANGLES, IS THE SAME.

ORBIT DIAGRAM WITH EQUAL-ANGLE SEGMENTS

ORBIT DIAGRAM VELOCITY DIAGRAM

S

J

vj

vj

v K

L

vK

v

START AT J

MOVE FROM J TO K WITH VELOCITY vj

AT K THERE IS Av, PARALLEL TO SK

THIS DETERMINES vk

IN ORBIT DIAGRAM, FROM K, DRAW A LINE PARALLEL TO vk

DRAW A LINE SUCH THAT < 1 = < 2

THIS DETERMINES POINT L

AND SO ON …v

vK

1

2

M

ALL v’S EQUAL IN MAGNITUDE

jk

l

m3

vl

vl

v

ORBIT DIAGRAM VELOCITY DIAGRAM

S

J

K

L

1

2

M

jk

l

m3

o

oj // JK

KS // jk

4

5

678

<7 = <4

Ok // KL<8 = <5

9

<9 = <6

kl // LS

10

<8 + <9 + <10 = 180

<5 + <6 + <2 = 180

<2 = <10

THEREFORE

SINCE ALL LINES FROM THE SUN ARE CONSTRUCTED WITH EQUAL ANGLES, THE SIDES OF THE VELOCITY DIAGRAM ALSO HAVE EQUAL EXTERNAL ANGLES , SUCH AS < 10

EQUAL EXTERNAL ANGLES

EQUAL SIDES

WHEN THE VELOCITY DIAGRAM IS COMPLETE IT WILL BE A FIGURE WITH EQUAL SIDES AND EQUAL EXTERNAL ANGLES OR A REGULAR POLYGON!

AS THE ANGLES ARE MADE SMALLER AND SMALLER THE POLYGON APPROACHES A CIRCLE.

S

J

Pvj

vp

ORBIT DIAGRAM VELOCITY DIAGRAM

O

C

j

p

’s are equal because orbit diagram is divided into equal angles and the velocity diagram as a regular polygon must have the same number of equal angles. But both are dividing 360 deg. Therefore the individual angles, or the ’s, must all be equal.

OP // vp

CORRESPONDENCE OF THE ORBIT AND VELOCITY DIAGRAMS

ANY POINT

C

CONSTRUCTION OF THE ORBIT DIAGRAM FROM THE VELOCITY DIAGRAM

VELOCITY DIAGRAM

ANY ORBIT PERMITTED BY NEWTON’S LAWS AND THE FORCE OF GRAVITY, OR THE R-2 LAW IN GENERAL, WILL HAVE THIS SAME CIRCULAR VELOCITY DIAGRAM.

THE EXACT SHAPE OF THE ORBIT WILL DEPEND ON WHERE WE CHOOSE TO PLACE THE ORIGIN OF THE VELOCITIES.

PICK A POINT, ANY POINT

FOR FAMILIARITY TURN THE WHOLE DIAGRAM UNTIL THE CHOSEN POINT LIES DIRECTLY BELOW C.

C

CONSTRUCTION OF THE ORBIT DIAGRAM FROM THE VELOCITY DIAGRAM

THE CHOSEN POINT SERVES AS THE ORIGIN OF VELOCITIES

THE LINE FROM THE ORIGIN THRU THE CENTER OF THE CIRCLE’S PERIMETER IS THE LONGEST LINE AND REPRESENTS THE POINT OF THE ORBIT WHERE THE PLANET IS MOVING FASTEST.

ORIGIN

ANY ORBIT PERMITTED BY NEWTON’S LAWS AND THE FORCE OF GRAVITY, OR THE R-2 LAW IN GENERAL, WILL HAVE THIS SAME CIRCULAR VELOCITY DIAGRAM.

THE EXACT SHAPE OF THE ORBIT WILL DEPEND ON WHERE WE CHOOSE TO PLACE THE ORIGIN OF THE VELOCITIES.

PICK A POINT, ANY POINT

FOR FAMILIARITY TURN THE WHOLE DIAGRAM UNTIL THE CHOSEN POINT LIES DIRECTLY BELOW C.

S

J

Pvj

vp

C

j

p

CONSTRUCTION OF THE ORBIT DIAGRAM FROM THE VELOCITY DIAGRAM

vj

ORIGIN

THEREFORE AT EACH ANGLE WE KNOW THE DIRECTION OF THE TANGENT TO THE CURVE OR ORBIT. HOW THEN DO WE CONSTRUCT THE ORBIT?

DRAW THE ORBIT SO THAT THE LINE TO THE POINT CLOSEST TO THE SUN, SJ, IS HORIZONTAL.vj IS VERTICAL AS SHOULD BE.DRAW A LINE FROM THE ORIGIN TO ANY OTHER POINT ON THE CIRCLE, p.

ORBIT DIAGRAM VELOCITY DIAGRAM

THIS POINT CORRESPONDS TO A POINT P ON THE ORBIT WITH THE FOLLOWING PROPERTIES; THE LINE FROM THE ORIGIN TO p IS PARALLEL TO THE TANGENT AT POINT P, AND THE ANGLES jCp AND JSP ARE EQUAL.

S

J

Pvj

vp

C

j

p

CONSTRUCTION OF THE ORBIT DIAGRAM FROM THE VELOCITY DIAGRAM

vj

ORIGIN

ORBIT DIAGRAM VELOCITY DIAGRAM

THE TRICK IS TO ROTATE THE VELOCITY DIAGRAM CLOCKWISE ( A LA FEYNMAN) SO THAT THE DIRECTIONS ON IT ARE THE SAME AS THOSE ON THE ORBIT DIAGRAM:

S

J

Pvj

vp

C

j

p

CONSTRUCTION OF THE ORBIT DIAGRAM FROM THE VELOCITY DIAGRAM

ORIGIN

THE TRICK IS TO ROTATE THE VELOCITY DIAGRAM CLOCKWISE ( A LA FEYNMAN) SO THAT THE DIRECTIONS ON IT ARE THE SAME AS THOSE ON THE ORBIT DIAGRAM:

“vp”

WE NOW KNOW, FROM THE ROTATED VELOCITY DIAGRAM, THE DIRECTION TO THE POINT P ON THE ORBIT, REPRESENTED BY .

AND WE KNOW THE DIRECTION OF THE TANGENT TO THE ORBIT AT THAT POINT. IT IS PERPENDICULAR TO THE LINE MARKED “vp.” BUT WE DON’T KNOW YET WHERE POINT P IS. vp COULD TRANSLATE IN THE PLANE. WHAT IS SJOWN IS JUST ONE POSSIBLE POSITION FOR P.

C

j

p

CONSTRUCTION OF THE ORBIT DIAGRAM FROM THE VELOCITY DIAGRAM

ORIGIN

EASIEST WAY TO DRAW THE ORBIT IS TO DRAW IT RIGHT ON TOP OF THE VELOCITY DIAGRAM.

PERPENDICULAR BISECTOR

THE SIZE OF THE ORBIT WILL BE ARBITRARY, BUT ALL THE DIRECTIONS AND THEREFORE THE SHAPE OF THE ORBIT WILL BE CORRECT.

THE PERPENDICULAR BISECTOR WILL BE PARALLEL TO vp.

INTERSECTION POINT P

AS SHOWN BEFORE THE LOCUS OF SUCH POINTS WILL BE AN ELLIPSE.

QED

ELLIPSE

THE INTERSECTION OF THE PERPENDICULAR BISECTOR AND LINE Cp SATISFIES THAT OF THE CORRESPONDING POINT P ON THE ORBIT. IT IS IN THE SAME DIRECTION , AND THE TANGENT, WHICH IS THE PERPENDICULAR BISECTOR ITSELF, IS PARALLEL TO vp.

THEREFORE, AS p MOVES AROUND THE CIRCLE, THE LOCUS OF SUCH INTERSECTION POINTS CONSTITUTES THE ORBIT.

Kepler’s 2nd Law: The Law of Areas All forces are

directed towards the sun

Kepler’s 3rd Law: T a 3/2

Forces are inversely proportional to the square of the distances

Force of gravity

Kepler’s 1st Law: The Law of Ellipses

Galileo’s law of inertia

Descartes law of Inertia

PREDICTION

Newton’s 1st Law: The Law of Inertia

Newton’s 2nd Law: F = d(mv)/dt

Newton’s 3rd Law: Action and reaction

DISTANCE BETWEEN FOCI AS PERCENT OF DIAMETER

EARTHEARTH 1 %1 %

MARSMARS 9 %9 %

MERCURYMERCURY 20 %20 %

PLUTOPLUTO 20 %20 %

HALLEY’S HALLEY’S COMETCOMET

97 %97 %

ORBITS

James Clerk Maxwell provided a similar proof in his book Matter and Motion.

THANK YOU !