# Deriving Keplers Laws of Planetary Motion

**Topics:**Kepler's laws of planetary motion, Orbit, Semi-major axis

**Pages:**10 (757 words)

**Published:**March 22, 2013

Tanner Morrison

November 16, 2012

Abstract

Johannes Kepler, a world renowned mathematician and astronomer, formulated three of today’s most inﬂuential laws of physics. These laws describe planetary motion around the sun. Deriving these laws (excluding Kepler’s First Law) will stress the concept of planetary motion, as well as provide a clear understanding of how these laws became relevant.

1

Kepler’s First Law

Kepler’s First Law states: The orbit of every planet is an ellipse with the Sun at one of the two foci.

2

Kepler’s Second Law

Kepler’s Second Law states: A line joining a planet and the Sun sweeps out equal areas during equal time intervals. In more simpler terms, the rate at which the area is swept by the planet is constant ( dA = constant). dt

2.1

Derivation Of Kepler’s Second Law

To start this derivation, we will need to know how to ﬁnd the area that is swept out by the planet. This area is equal to

θ

r

A=

rdrdθ =

0

r2

θ

2

(1)

0

The position can be deﬁned by the planetary motion.

r = r cos θˆ + r sin θˆ

i

j

(2)

The velocity can then be found by taking the derivative of the position. r = (−r sin θ

dθ dr

dθ dr

+

cos θ)ˆ + (r cos θ

i

+

sin θ)ˆ

j

dt

dθ

dt

dθ

(3)

As noted during the derivation of Kepler’s First Law, h is a constant, due to the fact that r × r is a constant.

h = r × r = constant

To ﬁnd the constant vector h evaluate the determinate that is given by the cross product of r × r .

ˆ

ˆ

ˆ

i

j

k

h=

r cos θ

r sin θ

0

dr

dθ

dr

dθ

−r sin θ dt + dθ cos θ r cos θ dt + dθ sin θ 0

Once the determinate is evaluated it can be simpliﬁed to

h = r2

1

dθ ˆ

k

dt

(4)

The magnitude of this vector being (the same).

|h| = r2

dθ

dt

(5)

by the deﬁnition of h this value is a constant. Recall that the area swept out by the planet can be described as.

θ

r

A=

rdrdθ =

0

r2

θ

2

0

The area swept through a little change in time (dt) is then equal to r2 dθ

dA

=

dt

2 dt

Notice

dA

dt

(6)

looks alot like h = r2 dθ

dt

h

dA

=

dt

2

Showing that

a constant.

3

dA

dt

is constant. Showing that the area swept out by the planet is

Kepler’s Third Law

Kepler’s Third Law states: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. This derivation will show that

4 π 2 a 2 b2

T2 =

h2

3.1

Deriving Kepler’s Third Law

From the derivation of Kepler’s Second Law we know that

h

dA

=

dt

2

By using integration we can ﬁnd the area swept out during a certain time interval (T), the period. The fundamental theorem of calculus states that the integral of the derivative is equal to the integrand,

T

T

dA =

0

h

2

dt

0

2

by simplifying we get the area of the planetary motion

h

T

2

A=

(7)

recall that A = πab, inputting this into our area equation we get πab =

h

T

2

Solving for the period (T), we get

2πab

h

T=

By squaring this period we get,

4 π 2 a 2 b2

h2

T2 =

(8)

2

Recall the directrix of an ellipse is (d = h ) and the eccentricity of an ellipse is c

c

(e = GM ). Multiplying these together and simplifying we get ed =

h2 e

h2

=

eGM

GM

(9)

Also recall that the square of half of the major axis of an ellipse is a2 = and the square of half of the minor axis is b2 =

√

Consider

√

a2 =

e2 d2

(1 −

e2 ) 2

e2 d 2

(1−e2 ) .

=a=

e2 d2

(1−e2 )2

Solving for a

ed

1 − e2

2

b

a

b2

e2 d2 (1 − e2 )

=

= ed

a

(1 − e2 ) ed

(10)

Equating equations (9) and (10) yields

h2

b2

=

GM

a

Simplifying this we get

h2 =

recalling T 2 =

4π 2 a2 b2

,

h2

b2 GM

a

(11)

inserting the new found h we get

T2 =

4π 2 a2 b2 a

4π 2 a3

=

h2 GM

GM

(12)

Showing that the square of the period (T 2 ) is proportional to the cube of the semi-major axis...

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