KEPLERS LAWS AND PLANETARY ORBITS - · PDF filethen show that this translates into...

KEPLER’S LAWS AND PLANETARY ORBITS

1. Selected properties of polar coordinates and ellipses Polar coordinates:

I take a some what extended view of polar coordinates in that I allow for a z direction (cylindrical coordinates) even though the motion is entirely in the xy-plane. From the drawing we observe that:

dθ rr

rdr rr+

rdr θθ ˆˆ rdrdrrd +=

r [1.1] Note also that:

kdrrdr ˆ2 θ=×rr [1.2]

and et. cyc. kr ˆˆˆ =×θWe can write: jyixr ˆˆ +=

r from which we have immediately jir ˆ)sin(ˆ)cos(ˆ θθ += . [1.3]

jid

rd ˆ)cos(ˆ)sin(ˆ

θθθ

+−= .

but jirk ˆ)cos(ˆ)sin(ˆˆ θθ +−=×

so: θθ

ˆˆˆˆ=×= rk

drd [1.4]

Ellipses: The ellipse shown below has the form:

)cos(1 θeRr

+= [1.5]

where R and e are constants. The quantity e has the property that the focus, f, is displaced from the origin by the amount ea. From this we see that:

2a

f f

⎟⎠⎞

⎜⎝⎛

−+

+=

eeRa

11

11

2

21 eR

a−

= [1.6]

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2. Newton’s laws of motion and gravity In this section I will set forth the fundamental laws and definitions that will be used to derive the results of the following sections. I begin with Newton’s law of gravitation:

rrD

rr

mMGF ˆˆ 22 −=−=

r where: [2.1]

D = G m M [2.2] D is a constant with no physical meaning. It has dimensions of Kg m3/s2

Newton’s 2nd Law states:

dtpd

Frr

= [2.3]

The definition of angular momentum gives:

kdtd

mrdtrd

rmprL ˆ2 θ=×=×=

rrrrr

[2.4]

Where we have used equation [1.2]

3. Kepler’s 2nd Law In this section I show that the angular momentum is conserved for a central force field and then show that this translates into Kepler’s 2nd Law. Since the force in a central-force situation is directed directly from one body to the other, the quantity Fr

rrr×=τ is identically zero and hence the angular momentum is constant.

Kepler’s 2nd Law In polar coordinates the incremental area swept out by an incremental change in direction is:

θrdrdA21

= since r is the height of the

incremental triangle and r dθ is the base. dθ

r rdθ

Hence:

dtd

rdtdA θ2

21

= [3.1]

Therefore by [2.4] the quantity on the right of [3.1] is constant which is Kepler’s 2nd Law.


4. Kepler’s 1st Law In this section I derive the relationship between r and θ and show that the trajectory of an object in an inverse-square-law central-force field is a conic section. Consider the cross product: LF

rr×

kdtd

mrrrD

Ldtpd ˆˆ 2

2θ

×−=×rr

where on the left side we have substituted from [2.3] for Fr

and on the right side [2.2] and [2.4] for and F

rLr

respectively. Simplifying the two sides of this equation independently gives:

krdtd

mDdt

Lpd ˆˆ)(

×−=× θrr

where on the left we have used the fact that Lr

is a constant. We can simplify the right hand

side further by noting that θdrd

krˆˆˆ −=× [1.4] and

dtrd

drd

dtd ˆˆ

=θ

θ to give

dtrd

mDdt

Lpd ˆ)(=

×rr

[4.1]

Now we can integrate both sides of [4.1] to get: )ˆ( ermDLp rrr

+=× [4.2] where e is the constant of integration. It will turn out to be a vector parallel to the semi-major axis of the ellipse with magnitude equal to the eccentricity, e.

r

If we dot rr into both sides of [4.2] we get: )ˆ( errmDLpr rrrrr

+•=×• ( )θcosrermDLpr +=•×

rrr ( θcos12 emDrL += ) [4.3]

θcos112

emDL

r+

= [4.4]

This is the desired result. This is the equation for a conic section, If e < 1 we have an ellipse, if e = 1 a parabola and if e > 1 a hyperbola. We can define the quantity

mDL

R2

= [4.5]

and write [4.4] as

θcos1 eR

r+

= [4.6]

Here we see that R has the dimensions of length and the physical meaning of the radius of the orbit if e = 0.


5. Potential, Kinetic and Total Energy In this section I show the relationship between energy and angular momentum for planetary motion. Potential Energy: For an inverse square law we can write the potential energy:

( θcos12

2

eL

mDrD

U +−=−= ) [5.1]

Kinetic Energy: Consider the quantity Lv

rr× . From the definition of angular momentum these two

vectors are perpendicular to each other so from [4.2]: ( )222

cos21 eeDLv ++=× θrr [5.2]

( )22

22 cos21 ee

LD

v ++= θ

( 22

2

cos2121

eeL

mDK ++= θ ) [5.3]

Total Energy: Thus the total energy is:

( 22

2

121 e

LmDUKE −−=+= ) [5.4]


6. Kepler’s 3rd Law In this section I relate the semi-major axis of the planetary ellipse with the Period of the planet. From [2.4] we have:

2mrL

dtd

=θ

[6.1]

Substituting for r from [4.4]

( 23

2

cos1 θθ

eL

mDdtd

+= ) [6.2]

We can define the constant : 3

2

0 LmD

=ω [6.3]

which has units of time-1 and the physical interpretation of the angular velocity of the circular orbit if e = 0. Then we can write:

( 20 cos1 θω )θ

edtd

+= [6.4]

( ) 20 cos1 θω

θed

dt+

= [6.5]

( )∫+

=π

θθ

ω2

0 20 cos1

1ed

T [6.6]

This integral can be found in Meyer zur Capellan 3.4.3 page 162 and 3.1.3 page 161. For our parameters it becomes:

( ) 2/320 1

21e

T−

=π

ω [6.7]

( ) 42

6

32

22

1)2(

DmL

eT

−=

π [6.8]

Now from [1.6] and [4.5]:

2

2

11emD

La

−= [6.9]

( )3233

63

11eDm

La

−= [6.10]

eliminating L between [6.8] and [6.10] gives:

mDT

a 2

23

)2( π= [6.11]

22

3

)2(T

GMa

π= [6.12]


7. Energy and Angular momentum in terms of the Semi-major axis

In this section I relate the total energy and the angular momentum of a planet to the semi-major axis of the elliptical orbit. Combining [6.9] and [5.4] to eliminate e, we find:

aGmM

aD

E21

2−=−= [7.1]

Rearranging terms in [6.9] gives:

( ) ( )2222 11 eMaGmemDaL −=−= [7.2]

MRGmL 22 =

MGmLR 2

2

=


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