Proof of Kepler's Third Law arising out of the Equation

Proof of Kepler’s Third Law arising out of the Equation of Path of a Planetary Body about a Local Star We seek to solve the following integral: 2𝜋

∫0

𝑇 4

𝑑𝜙 1 3𝜋 2 [1+ cos(𝜃+ )] 4 √2

= ∫0

𝑟02

𝐺𝑀𝑟0

√

2

𝑑𝑡

( 1)

which comes from the equation of path: 𝑢=

2 1 3𝜋 [1 + cos (𝜃 + )] 𝑟0 4 √2

for a planet (or a comet) orbiting its local star. This is also the particular-solution to the ordinary differential equation (ODE): 𝑑2𝑢 𝐺𝑀 + 𝑢 = 𝑑𝜃 2 ℎ2 governing the body’s motion for an orbit of eccentricity 𝑒 = 1/√2 within the given integral limits (see ‘A First Course in Mechanics’, by Dr Mary Lunn, Oxford University Press, 1992; also see ‘Theory and Problems of Theoretical Mechanics’, by Murray R. Spiegel, Schaum’s Outline Series, International Edition, 1982). Note, this theory does not work for Mercury, due to the relativistic gravitational precession of its orbit of 43 seconds of arc per century. We write the integral on the left-hand-side of equation (1) as: 2𝜋

∫0 where we have set 𝜙 = 𝜃 +

2𝜋

𝑑𝜙 𝑞 𝑝

(1+ cos 𝜙)

2

= ∫0

𝑝2 𝑑𝜙 (𝑝+𝑞 cos 𝜙)2

( 2)

3𝜋 4

, √2 = 𝑝 and 1 = 𝑞 for the general case.

Put: 𝜙

𝑡 = tan ( ) 2

( 3)

so that: 1

𝜙

𝑑𝑡 = 2 sec2 ( 2 ) 𝑑𝜙 where: 1 𝜙 𝑑𝑡 = (1 + tan2 ( )) 𝑑𝜙 2 2 or:

Page | 1

𝑑𝑡 =

1 (1 + 𝑡 2 )𝑑𝜙 2

so that: 2𝑑𝑡

𝑑𝜙 = (1+𝑡 2 )

( 4)

From the trigonometric relationship implied by equation (3): 𝜙

𝑡

𝜙

1

sin ( 2 ) = √1+𝑡 2

( 5)

and: cos ( 2 ) = √1+𝑡 2

( 6)

Now, since: 𝜙 𝜙 sin(𝜙) = 2 sin ( ) cos ( ) 2 2 then: 2𝑡

sin(𝜙) = (1+𝑡2 )

( 7)

Also, since: 𝜙 𝜙 cos(𝜙) = cos 2 ( ) − sin2 ( ) 2 2 then 𝑡2

1

1−𝑡 2

cos(𝜙) = 1+𝑡 2 − 1+𝑡 2 = 1+𝑡 2

( 8)

Substituting: 2𝑑𝑡

𝑑𝜙 = (1+𝑡 2 )

( 9)

and: cos(𝜙) =

1−𝑡 2

( 10 )

1+𝑡 2

into equation (2), we have: ∫

𝑝2 𝑑𝜙 =∫ (𝑝 + 𝑞 cos 𝜙)2

2𝑝2 𝑑𝑡 2

1 − 𝑡2 (1 + 𝑡 2 ) (𝑝 + 𝑞 ( )) 1 + 𝑡2

or: 𝑝2 𝑑𝜙

∫ (𝑝+𝑞 cos 𝜙)2 = ∫

2𝑝2 𝑑𝑡 2 𝑝(1+𝑡2 )+𝑞(1−𝑡2 ) (1+𝑡 2 )[ ] 1+𝑡2

=∫

2𝑝2 𝑑𝑡 2 (𝑝+𝑞)+(𝑝−𝑞)𝑡2 ] 1+𝑡2

(1+𝑡 2 )[

Page | 2

Factoring out the (𝑝 − 𝑞) term, we obtain: 2𝑝2 𝑑𝜙

2𝑝2

∫ (𝑝+𝑞 cos 𝜙)2 = (𝑝−𝑞)2 ∫

(1+𝑡 2 )𝑑𝑡

( 11 )

(𝑎+𝑡 2 )2

𝑝+𝑞

where 𝑎 = 𝑝−𝑞. The integral can now be expanded, via partial fractions, to yield: (1 − 𝑎)𝑑𝑡 2𝑝2 𝑑𝜙 2𝑝2 𝑑𝑡 ∫ {∫ ∫ } = + (𝑝 + 𝑞 cos 𝜙 )2 (𝑝 − 𝑞)2 ( 𝑎 + 𝑡 2 )2 𝑎 + 𝑡2 or: ∫

𝑝2 𝑑𝜙 2𝑝2 𝑑𝑡 𝑑𝑡 ) {(1 ∫ ∫ } = − 𝑎 + (𝑝 + 𝑞 cos 𝜙 )2 (𝑝 − 𝑞)2 (𝑎 + 𝑡 2 ) 2 𝑎 + 𝑡2

The integral on the far right-hand-side is straightforward, where: 𝑑𝑡

1

𝑡

∫ 𝑎+𝑡 2 = √𝑎 tan−1 √𝑎

( 12 )

For the first integral on the right-hand-side, we use another trigonometric substitution: 𝑡 = √𝑎 tan 𝑠

( 13 )

𝑑𝑡 = √𝑎 sec2 𝑠 𝑑𝑠 = √𝑎(1 + tan2 𝑠)𝑑𝑠

( 14 )

so that:

Thus, we may now write: 3

(1 −

𝑑𝑡 𝑎) ∫ (𝑎+𝑡 2 )2

= (1 −

√𝑎 sec2 𝑠 𝑎) ∫ 𝑎2 (1+tan2 𝑠)2 𝑑𝑠

= (1 −

√𝑎 sec2 𝑠 𝑎) ∫ 𝑎2 (sec2 𝑠)2 𝑑𝑠

(√𝑎−𝑎2 )

=

𝑎2

sec2 𝑠

∫ sec4 𝑠 𝑑𝑠

Henceforth: (1 − 𝑎) ∫

𝑑𝑡 1 1 1 1 1 = ( 3/2 − ) ∫ 𝑑𝑠 = ( 3/2 − ) ∫ cos 2 𝑠 𝑑𝑠 2 2 2 (𝑎 + 𝑡 ) 𝑎 sec 𝑠 𝑎 √𝑎 √𝑎

Using the identity: 1

1

cos2(𝑠) = 2 + 2 cos(2𝑠)

( 15 )

we may write: (1 − 𝑎) ∫

𝑑𝑡 (𝑎+𝑡 2 )2

=(

1 𝑎3/2

−

1 1 1 ) ∫ { + cos(2𝑠)} 𝑑𝑠 𝑎 2 2 √

=(

1 2𝑎3/2

−

1 )𝑠 2√𝑎

+(

1 4𝑎3/2

−

1 ) sin(2𝑠) 4√𝑎

So that: ∫

2𝑝 2 𝑑𝜙 2𝑝 2 1 1 1 1 1 𝑡 = {( 3/2 − ) 𝑠 + ( 3/2 − ) sin(2𝑠) + tan−1 } 2 2 (𝑝 + 𝑞 cos 𝜙) (𝑝 − 𝑞) 2𝑎 4𝑎 2√𝑎 4√𝑎 √𝑎 √𝑎 Page | 3

Since, from equation (13), we have: 𝑠 = tan−1

𝑡 √𝑎

then, by trigonometry and use of the relationship sin(2𝑠) = 2 sin(𝑠) cos(𝑠), we have: sin 2𝑠 = 2 (

𝑡

2√𝑎𝑡 √𝑎 )( )= 𝑎 + 𝑡2 √𝑎 + 𝑡 2 √𝑎 + 𝑡 2

such that: ∫

2𝑝2 𝑑𝜙 2𝑝2 1 1 𝑡 1 1 2√𝑎𝑡 1 𝑡 = {( − + ( 3/2 − + tan−1 } ) tan−1 )( 2) (𝑝 + 𝑞 cos 𝜙)2 (𝑝 − 𝑞)2 2𝑎3/2 2√𝑎 𝑎 + 𝑡 4𝑎 4√𝑎 √𝑎 √𝑎 √𝑎

or: ∫

2𝑝2 𝑑𝜙 2𝑝2 1−𝑎 𝑡 1 − 𝑎 2√𝑎𝑡 1 𝑡 = {( ) tan−1 + ( 3/2 ) ( )+ tan−1 } 2 (𝑝 + 𝑞 cos 𝜙)2 (𝑝 − 𝑞)2 2𝑎3/2 𝑎+𝑡 √𝑎 √𝑎 √𝑎 4𝑎

and, simplifying further, the above becomes: ∫

2𝑝2 𝑑𝜙

=

(𝑝 + 𝑞 cos 𝜙)2

(1 − 𝑎 ) 𝑡 2𝑝2 𝑎+1 𝑡 {( 3/2 ) tan−1 } + 2 (𝑝 − 𝑞 ) 2𝑎 √𝑎 2𝑎(𝑎 + 𝑡 2 )

From equation (3): 𝑡 = tan

𝜙 2

so: 2𝑝2 𝑑𝜙 ∫ (𝑝+𝑞 cos 𝜙)2

2𝑝2

= (𝑝−𝑞)2 {(

2𝑎

Substituting in the value 𝑎 = ∫

𝑑𝜙 2 1 (1+ cos 𝜙) 2 √

𝑎+1

𝑝+𝑞 𝑝−𝑞

−1 ( 3/2 ) tan

1

𝜙

√𝑎

tan ) + 2

𝜙 2 𝜙 2𝑎(𝑎+tan2 ) 2

(1−𝑎) tan

}

( 16 )

+𝐶

( 17 )

, where 𝑝 = √2 and 𝑞 = 1, we get: 𝜙

2√2sin(𝜙)

= 4√2tan−1 [(√2 − 1)tan ( 2 )] − 2+

√2cos(𝜙)

However, at 𝜙 = 0, we note that the integral, ∫

dϕ 2 1 (1+ cos ϕ) 2 √

is equal to a multiple of 𝜋,

or: ∫

dϕ 2 1 (1 + cos 𝜙) √2

=𝑛𝜋+𝐶

But, if 𝑛 = 0, indicating the starting point on the ellipse of 0 radians, then 𝐶 = 0. This means that the integral must be solely equal to 𝑛 𝜋, which leaves us to conclude that:

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𝑑𝜙

∫ (1 +

1 cos 𝜙) √2

2

𝜙 2√2sin(𝜙) = 4√2tan−1 [(√2 − 1)tan ( )] − + 𝑛𝜋 2 2 + √2cos(𝜙)

If, instead, we simply work out the integral in equation (17), between the limits of 𝜙 = 0 and 𝜙 = 2𝜋, we adopt the following procedure: 2𝜋

𝜋

𝑑𝜙

∫

2 1 (1 + cos 𝜙) √2

0

=4

[4√2tan−1 [(√2

𝜙 𝜋/2 2√2sin(𝜙) ] − 1)tan ( )]] − 2[ 2 −𝜋/2 2 + √2cos(𝜙)

−𝜋

Note that we have changed the limits in each of the above terms, since we need to ensure – according to the law of areas, the rules of contour integration, and other 𝜋 analytic reasons associated with discontinuities at 𝜙 = – that the symmetry of the 2

ellipse is preserved. Thus, we have: 2𝜋

𝑑𝜙

∫ 0

(1 +

2 1 cos 𝜙) √2

𝜋 𝜋 = 4 [4√2tan-1 [(√2 − 1)tan ( )] − 4 [(√2 − 1)tan (− )]] − 2[0 − 0] 4 4

2𝜋

𝑑𝜙

∫ 0

(1 +

2 1 cos 𝜙) √2

𝜋 𝜋 = 4 [4√2tan-1 [(√2 − 1)tan ( )] + 4√2tan-1 [(√2 − 1)tan ( )]] 4 4 − 2[0 − 0] 2𝜋

𝑑𝜙

∫ 0

(1 +

1 cos 𝜙) √2

2

𝜋 = 32√2 ( ) 8

or: 2𝜋

𝑑𝜙

∫

2 1 (1 + cos 𝜙) √2

0

= 4√2𝜋

Since we know from equation (1) that: 2𝜋

∫0

2𝜋

𝑑𝜙

= ∫0

2 1 (1+ cos 𝜙) √2

𝑇 4

2𝑑𝜙 2

(√2+1 cos 𝜙)

= ∫0

𝑟02

𝐺𝑀𝑟0

√

2

𝑑𝑡

( 18 )

then: 4√2𝜋 =

4 𝐺𝑀𝑟0 √ 𝑇 2 𝑟02

So, Kepler’s Third Law becomes: Page | 5

𝑟03 =

𝐺𝑀 2 𝑇 4𝜋 2

Note that we directly substituted angular values for 𝜙 into equation (18), and not the relationship, 𝜙 = 𝜃 +

3𝜋 4

. This does not matter, as the result is still exactly the same,

whether one does so or not. Had we chosen to substitute 𝜙 = 𝜃 +

3𝜋 4

at that point,

there would simply have been more tedious and protracted algebra to negotiate in accomplishing the same result of Kepler’s Third Law.

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