Proof of the Kepler’s Three Laws of Planetary Motion.

LAW II.

[1]The areas, which revolving bodies describe by radii drawn to an immovable centre of force do lie in thesame immovable planes, and are proportional to the times in which they are described.

For suppose the time to be divided into equal parts, and in the first part of that time let the body byits innate force describe the right line AB. In the second part of that time, the same would (by Law I.),if not hindered, proceed directly to c, along the line Bc equal to AB; so that by the radii AS, BS, cS,drawn to the centre, the equal areas ASB, BSc, would be described.

S A

B

C

DE

F

f e d

V

Z c

But when the body is arrived at B, suppose that a centripetal force acts at once with a great impulse;and, turning aside the body from the right line Bc, compels it afterwards to continue its motion along theright line BC. Draw cC parallel to BS meeting BC in C; and at the end of the second part of the time,the body (by Cor. I. of the Laws) will be found in C, in the same plane with the triangle ASB. Join SC,and, because SB and Cc are parallel, the triangle SBC will be equal to the triangle SBc, and thereforealso to the triangle SAB. By the like argument, if the centripetal force acts successively in C, D, E. &c.;and makes the body, in each single particle of time, to describe the right lines CD, DE, EF, &c., theywill all lie in the same plane; and the triangle SCD will be equal to the triangle SBC, and SDE to SCD,and SEF to SDE. And therefore, in equal times, equal areas are described in one immovable plane: and,by composition, any sums SADS, SAFS, of those areas, are one to the other as the times in which theyare described.

Now let the number of those triangles be augmented, and their breadth diminished in infinitum; and (byCor. 4, Lem. III.) their ultimate perimeter ADF will be a curve line: and therefore the centripetal force,by which the body is perpetually drawn back from the tangent of this curve, will act continually; andany described areas SADS, SAFS, which are always proportional to the times of description, will, in thiscase also, be proportional to those times. �

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LAW I. [1][2]

If there is no external force, a planet P travels along a straight line from P to R in a certain time intervaltPQ, But, when it is enacted at P by a central force exerted in the direction of PS, it is deflected by RQparallel to the direction of the force, i.e., PS.

From Galileo’s law of falling bodies, the distance RQ by which it is deflected from the straight line is

RQ =1

2× accel.× t2PQ =

1

2× accel.× t2PQ

By Kepler’s second law, tPQ = (sector SPQ)(areal speed) , and sector SPQ≈ 1

2 × (SP ×QT ),

∴ accel. ≈ 8× (areal speed)2 ×QRSP 2 ×QT 2

=(a constant)(distance)2

× QR

QT 2.

We will show that, if the curve PQ is an ellipse with a focus at S, QRQT 2 is a constant, In fact it is the

reciprocal of the latus rectum of the ellipse, i.e., QT 2

QR = l.r. = 2b2

a , which is constant throughout thewhole path, where a and b are the semimajor and semiminor axes, respectively, of the ellipse.

(a) on the diameter parallel to a tangent that touches the ellipse at P , we can easilly show that

PE = a (1)

From the similarity of triangles SCE and SHI, and from SC = CH.

Since SE = EI and SP +PH = 2a, and the triangle PIH being an isosceles (proof is left as an exercise),PI = PH,

2

PE = EI + IP =1

2(SE + EI + IP + PH) = a �

(b) Let there be a circle DGKP with two diameters DK and GP , perpendicular to each other. From apoint Q on the circle, draw a line parallel to DK to a point v on GP , then,

CP = CD

Pv : Qv = Qv : vG

Pv × vG = Qv2

∴Pv × vGCP 2

=Qv2

CD2

When stretched uniformely in horizontal direction, the circle DGKP is transformed into an ellipseD′G′K ′P ′, and the two diameters DK and GP are transformed into a pair of conjugate diametersD′K ′ and G′P ′:

And in the ellipse the similar relation holds for the transformed line segments, i. e.,

∴P ′v′ × v′G′

C ′P ′2=

Q′v′2

C ′D′2(2)

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because the segments on parallel lines are stretched in the same propertion,

Cv : C ′v′ = vG : v′G′ = CP : C ′P ′ = 1 : α

andQv : Q′v′ = CD : C ′D′ = 1 : β,

therefore,P ′v′ × v′G′

C ′P ′2=α(Pv)× α(vG)

α2(CP 2)=Pv × vGCP 2

andQ′v′2

C ′D′2=

β2(Qv2)

β2(CD2)=

Qv2

CD2.

thus the relation (2) also holds true for the ellipse. �

(c) The area of any parallelogram inscribed by the ellipse:

S = πab (3)

the proof is left as an exercise.

(d) The latus rectum:

MH +MS = 2a (4)

MH2 −MS2 = 4SC2 = 4(a2 − b2)

(MH +MS)× (MH −MS) = 4(a2 − b2) (5)

dividing (5) by (4),

MH −MS = 2a− 2b2

a, (6)

and (4)−(6) gives

2MS =MN (latus rectum) =2b2

a. � (7)

Now, expanding the quantity in question,

4

P

E

S C

F

x

vT

Q R

G

K

D

B

A

QT 2

QR=Px

QR× Pv

Px× Qv2

Pv×(Qx

Qv

)2

× QT 2

Qx2

(e) Px = QR from the parallelogram, therefore

Px

QR= 1

(f) Pv : Px = CP : PE from the similarity of the triangles, and from (1),

Px

Pv=CP

a

(g) from (2),Pv × vGCP 2

=Qv2

CD2,

∴Qv2

Pv=CD2 × vGCP 2

=CD2

CP× vG

CP,

and since vGCP approaches 2 as Q approaches R,

∴Qv2

Pv≈ 2CD2

CP.

(h) also QxQv approaches 1 as Q approaches R,

∴Qx

Qv≈ 1.

(i) QT : Qx = PF : PE from the similarity of triangles QxT and PEF ,

∴QT

Qx=PF

PE=PF

a.

QT 2

QR≈ 1× CP

a× 2CD2

CP× 1× PF 2

a2

=2(CD × PF )2

a3,

5

where

CD × PF =1

4(area of the parallelogram inscribed by the ellipse)

= ab,

finally, we get

∴QT 2

QR=

2b2

a= l.r.

Therefore, if the planet moves along an elliptical orbit by the central force towards S, the force acting onthe planet is inversely proportional to the square of the distance from S. �

LAW III.

If a planet moves along an ellipse ABP with eccentricity ε, the semiminor axis b and the semimajor-axisa are related by

b2 = a2(1− ε2). (8)

The velocities vP and vA of the planet at the perihelion rP = mµ a(1−ε) and at the aphelion rA = m

µ a(1+ε),respectively, are perpendicular to the major axis PA, thus they are related by the equation,

1

2× a(1− ε)× vP =

1

2× a(1 + ε)× vA = (areal speed) =

πab

T, (9)

where µ = MmM+m is the reduced mass, and T is the orbital period of the planet.

i) in case ε 6= 0,

From the conservation of mechanical energy,

−GMµ

r+

1

2mv2 = const.,

− GMµ2

ma(1− ε)+

1

2mv2P = − GMµ2

ma(1 + ε)+

1

2mv2A.

Substituting for vP and vA,

− GMµ2

a(1− ε)+

2π2b2m2

T 2(1− ε)2= − GMµ2

a(1 + ε)+

1

2

2π2b2m2

T 2(1 + ε)2.

Rearranging, (1

1− ε− 1

1 + ε

)GMµ2

a=

{1

(1− ε)2− 1

(1 + ε)2

}2π2b2m2

T 2,

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and multiplying both sides by T 2a and dividing by(

11−ε −

11+ε

),

GMT 2 =

(1

1− ε+

1

1 + ε

)2π2b2a

m2

µ2=

4π2b2a

1− ε2m2

µ2

Using (8),

T 2 =4π2

GM

(1 +

m

M

)2a3 ∝ a3.

ii) in case ε = 0,a = b =

m

µr, and vP = vA = v.

From the centripetal acceleration for a circular motion

ac =v2

r

and the speed being constant for a circular motion

v =2πa

T,

GMm

r2= m

4π2a

T 2,

∴ T 2 =4π2

GMar2 =

4π2

GM

(1 +

m

M

)2

a3 ∝ a3.

i.e., The square of the orbital period of a planet is proportional to the cube of the simi-major axis, or ofthe orbital radius in case of a circular orbit. Note that the circular orbit is a special case where ε = 0;where a = b = r.�

Reference:

[1] Newton, Sir Isaac, “ The Mathematical Principles of Natural Philosophy ”, American edition, 1846

[2] 大上雅史 & 和田純夫, ‘‘ が解き明かした物理の法則(The Elegance of Mathematics Has RevealedLaws of Physics)’’, ベレ出版, 2003

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