introduction - nlcs maths department · web viewwill explore the properties of ellipses and their...

Caroline Jones 13JLB

Mathematical derivations of Kepler's Laws of Planetary Motion and the equations of

planetary orbitsContents1. Introduction..........................................................................................................1

2. Ellipses...................................................................................................................2

i. Cartesian form...................................................................................................2

ii. The elliptical properties of Jupiter’s orbit..........................................................5

iii. Polar form......................................................................................................6

3. Kepler’s First Law..................................................................................................8

4. Kepler’s Second Law............................................................................................12

5. Kepler’s Third Law...............................................................................................13

6. Conclusion...........................................................................................................15

7. Bibliography........................................................................................................16

1. IntroductionI have always been fascinated by the popular science of Astrophysics, mainly

from watching programs of Stephen Hawking and Brian Cox. From very on in the history of the human race, people have been perplexed by astrological phenomena, therefore it is one of the very oldest branches of Physics, making it a very interesting subject to study historically as well. It was also often associated with Philosophy, another of my favourite subjects, with some of the most ancient Astronomers also studying Philosophy (such as Thales, Anaximander and Aristotle), as both disciplines were considered to be some of the highest branches of academia.

I was introduced to Kepler’s Laws in my Physics lessons, where we learnt a simple derivation for Kepler’s Third Law using equations of motion. I found the concept interesting, but the proof unsatisfying. This is because this particular proof approximates the equation of a planet’s orbit to a circle, whereas Kepler’s First Law tells us that orbits are ellipses. I therefore wanted to find a more general derivation that showed that Kepler’s Third Law is true for all ellipses, not just specific to circles. For this reason, I decided to research how Kepler’s Laws of Planetary Motion1 can be mathematically derived.

These derivations are fascinating for me as they combine two of my favourite topics in maths – geometry and calculus. Researching planetary orbits led me to examine conic sections: the shapes formed from the intersection of a plane and a circular

1 "Kepler's Laws." Hyperphysics. N.p., n.d. Web. 9 Oct. 2014.

Figure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sectionsFigure 1 Diagram of conic sections


cone at different angles, as shown in Figure 12. Kepler’s Laws looks specifically at ellipses, and over the course of this essay I will explore the properties of ellipses and their link to the orbits of planets, using predominately calculus and algebra to prove Kepler’s three laws. In doing so, my aim is to make the mathematics and physics of the derivations clear to understand for an audience of my peers studying Maths and Physics at Higher Level, as this is not covered by the standard curriculum, and not something that I have been able to find in the many sources I consulted. I hope to share my enthusiasm and make the ideas that I describe clear and engaging.

Over the course of this essay, I aim to make sense of the following: How equations of circles and ellipses come about, in different coordinate

systems Kepler’s First Law: The orbit of a planet is an ellipse with the Sun at one of the

two foci. Kepler’s Second Law: A line segment joining a planet and the Sun sweeps out

equal areas during equal intervals of time. Kepler’s Third Law: The square of the orbital period of a planet is

proportional to the cube of the semi-major axis of its orbit.

2. Ellipsesi. Cartesian form

Kepler’s Laws of Planetary motion are rules that describe the orbits of planets, therefore it seems necessary to me to first explore the equations that can model the orbit of planetary motion.

In their most basic form, the orbits of planets are approximated to circles, which are modelled by the equation:

x2+ y2=r2

where r is the radius of the circle, the centre of the circle is the origin, and (x , y ) is any point on the circle. I do not think I will need to consider the more complicated equation for a circle not centred on the origin, with centre (a ,b), because in a planetary system we can define the centre to be anywhere, therefore it is convenient to use the origin. Figure 2 shows an example of a circle with radius of 1.

However, orbits are more accurately defined by ellipses, modelled by the Cartesian equation (named after the mathematician and philosopher René Descartes):

x2

a2 + y2

b2 =1

where a is the horizontal semi-axis (the distance from the centre to the point on the ellipse with the same y-coordinate),

2 "File:Conic Sections.svg." Wikipedia. Wikimedia Foundation, 22 July 2014. Web. 12 Oct. 2014.

Figure 2 Graph of circle


b is the vertical semi-axis (the distance from the centre to the point on the ellipse with same x-coordinate),and the centre of the ellipse is the origin. This can be confusing as “axis” is usually associated with the x-axis, which is a line against which we measure other shapes, whereas here the semi-axes are distances.Although many books derive this by using trigonometry, I think it is possible to derive this equation rather neatly using only transformations of the equation for a circle. In an equation, replacing x

by xa corresponds to a stretch of scale factor a in

the horizontal direction, and replacing y by yb

corresponds to a stretch of scale factor b in the vertical direction.

One of the semi-axes will be the semi-major axis (a), which is the point furthest from the centre, and one will be the semi-minor axis (b), which is the point closest to the centre. It is conventional to draw the horizontal semi-axis as the semi-major axis and the vertical as the semi-minor axis; it is helpful to have conventions likes this in Maths as it aids mathematicians in making descriptions of shapes universally understandable.

Figure 3 shows a graph of an ellipse, with a=3 , b=2.The area of an ellipse is given by

A=πabIf we use the idea of obtaining the shape of an ellipse using coordinate

transformations, this formula is also easy to derive. The area of a circle of radius 1 around the origin is π, and when we stretch the semi-major axis by scale factor a, and the semi-minor axis by scale factor b, this becomes πab.

An ellipse can also be defined by two points called its foci3, which always lie along the semi-major axis. An ellipse with foci F1and F2has the property that for any point P on the ellipse, |P F1|+|P F2|=constant, as depicted in the Figure 4. This constant is equal to 2a. We can understand this by imagining when the point P is at y=0 and x is negative, |P F2|=a+ f , and |P F1|=a−f . Therefore |P F1|+|P F2|=2a. We can deduce from this that the distance from F2 to the positive y-axis is a, as it is half the sum of the distances from the foci to the point on the ellipse.

Furthermore, I wish to prove|P F1|+|P F2|≡2a as follows, using a proof

3"Foci (focus Points) of an Ellipse." Math Open Reference. N.p., n.d. Web. 16 Oct. 2014.

Semi-minor axis, b

Semi-major axis, a

Figure 3 Graph of ellipse

|P F2|

|P F1|

P(x , y )

Figure 4 Ellipse showing formulation for eccentricity

distance, a

distance to F2, f

x


that my Maths teacher helped me to derive. If we define the point P to be (x , y ) then using Pythagoras, we know that |P F1|=√ (−f−x )2+ y2 and |P F2|=√ (f−x )2+ y2

∴|P F1|+|PF2|¿√ ( f +x )2+ y2+√ ( f−x )2+ y2

I shall now try to rewrite the first square root all in terms of a ,b and x, using the Pythagorean relationship that we can see in Figure 4 that f 2=√a2−b2 and

rearranging x2

a2 + y2

b2 =1 to get y=b2− x2b2

a2 :

( f +x )2+ y2= (√a2−b2+ x)2+(b2− x2b2

a2 )This simplifies as follows:

(√a2−b2+x )2+(b2− x2b2

a2 )¿a2−b2+x2+2x √a2−b2+b2− x2b2

a2

¿ x2( a2−b2

a2 )+2x √a2−b2+a2

¿x2 (a2−b2 )+2 x√a2−b2a2+a4

a2

¿(x√a2−b2 +a2 )2

a2

Similarly therefore we know that the second square root will rearrange and simplify to:

(−x √a2−b2 +a2 )2

a2

So |P F1|+|P F2| is equal to:

√ ( x√a2−b2+a2 )2

a2 +√ (−x √a2−b2+a2 )2

a2

¿ x√a2−b2+a2

a+−x √a2−b2 +a2

a

¿ 2a2

a∴|P F1|+|PF2|=2a

These notions of the foci of ellipses will link to Kepler’s First Law of planetary motion, which states that the orbit of a planet is an ellipse with the Sun at one of the two foci.

An important property of an ellipse will be its eccentricity, which can be thought of as how much a conic section deviates from being circular. Eccentricity (let us call it e) is defined by the following equation:

e= fa


as shown in Figure 44. For an ellipse 0 < e < 1, and in the case that e = 0 the equation is a circle, since the focus has a distance of zero from the centre.

Alternatively, the eccentricity can be found using the following property (which will later be used in Kepler’s Third Law):

b2=a2(1−e2)

We can show this beginning with the relation e=fa and (using Pythagoras)

substituting in a2=f 2+b2:√a2−b2=e×aa2−b2=e2a2

b2=a2−e2a2

∴b2=a2(1−e2)

ii. The elliptical properties of Jupiter’s orbit

The links between all of these variables and properties is better seen using an example, so I shall try to model the orbit of Jupiter using these equations and plotting these values on autograph.

Jupiter: 5 Semi-major axis, a = 778.57 x106km = 5.203 AU

(where AU is Astronomical Units, a unit of length defined as the distance from the Earth to the Sun)

Eccentricity, e = 0.048In order to find the distance of the focus to the centre, we can use the

equation e=fa , and substitute in other known equations and values until everything

is in terms of f.

e= fa

∴ f=e×af=0.048×5.203

∴ f=¿0.250 AU to 3dpNow that we have the

distance to the foci, we can also work out the semi-minor axis, since (using Figure 4) we can see that a ,b and f have a Pythagorean relationship:

b=√a2−f 2

¿√5.2032−0.2502

¿5.197 AU ¿3dp

4 "Eccentricity an Ellipse - Math Open Reference." Eccentricity an Ellipse - Math Open Reference. N.p., n.d. Web. 17 Oct. 2014.5 "Jupiter Fact Sheet." Jupiter Fact Sheet. N.p., n.d. Web. 3 Oct. 2014.

Figure 5 Graph of Jupiter’s orbit

Foci (one of which is the Sun)

Semi-major axis, a

Semi-minor axis, b


I have then plotted a graph on Autograph in order to visualise this data, shown in Figure 5. However since the eccentricity is very small, the orbit does not look very noticeably elliptical. Therefore perhaps it would have been better to have chosen a planet with greater eccentricity, for example Pluto (which I am delighted to be able to describe as a planet again having been officially reinstated)!

I have drawn arrows showing the semi-axes, and plotted and marked the points for the foci (one of which would be the Sun, according to Kepler’s First Law).

iii. Polar formHowever, an ellipse can also be defined using the polar equation – which is a

different way of expressing the same shape. Using Figure 66, let us start by defining F1 (or the Sun) to be

the origin, and F2 to be the point (2 f ,0 ) – meaning that it is a distance of 2 f from F1 on the x-axis. Normally mathematicians use the c instead of f , but I thought I would use f for the sake of consistency, as this is how I have previously defined the distance from a focus to the origin, therefore the distance between the foci is 2 f . We also define PF1 to be the vector r, therefore PF2 is r−2 f i (where i is the horizontal unit vector).

We know that |P F1|+|P F2|=2a, therefore:|r|+|r−2 f i|=2a

Since|r|=r , we can work out that |r−2 f i|=r−2a, and so now, to get rid of the modulus signs, we can say that:

(r−2a )2=|r−2 f i|2

The vectorr can also be written by its polar co-ordinates as (r cosθ , r sin θ )(r−2a )2=(r cosθ−2 f )2+(r sin θ )2

Note that the change in the horizontal component of 2 f is subtracted from other horizontal component (ie. r cosθ) and not the vertical component (ie. r sin θ). We can now rearrange to make the equation in terms of r.

r2−4ar+4 a2=r2 (cosθ )2−4 rf cosθ+4 f 2+r2 (sinθ )2

r2−4ar+4 a2=r2 (cosθ )2−4 rf cosθ+4 f 2+r2 {1−(cos θ )2 }−4ar+4a2=−4 rf cosθ+4 f 2

−ar+rf cosθ=−a2+ f 2

r (−a+ f cosθ )=−a2+ f 2

r= −a2+ f 2

−a+f cosθSince |P F1|+|P F2|=2a, we can imagine when the point P is on the positive

y-axis, meaning that the distance a is the hypotenuse of the sides f and b. Therefore b2=a2−f 2. Therefore:

r= b2

a−f cosθ

6 "Kepler's 2nd Law: The Speeds of Planets." - Windows to the Universe. N.p., n.d. Web. 10 Nov. 2014.

Figure 6 Ellipse showing vectors to point P

x2

5.2032 +y2

5.1972 =1


∴r=

b2

a

1− fa

cosθ

We have already defined the eccentricity to be fa , therefore:

r=

b2

a1−ecosθ

The semi-latus rectum (r 0) is the length of the segment perpendicular to the major axis through one of the foci to the ellipse, as shown in Figure 77 The semi-latus rectum is equal to b2

a, as I shall try to prove.

Since the eccentricity is the ratio of the distance to the foci from the origin and the

semi-major axis (ie. e=fa ), the distance to the foci can be said to be f=ae, meaning

that the coordinates for the two foci are at (ae ,0) and (−ae ,0). Since the semi-latus rectum is the segment perpendicular to the the semi-major axis to the ellipse, the equation of this line segment is just x=ae or x=−ae. The intersection of these equations with the ellipse is the end point of the semi-latus rectum, therefore the y-coordinate is the length r0. So we can substitute in x=ae into the Cartesian equation for the ellipse:

(ae )2

a2 + y2

b2 =1

a2 y2=a2b2−a2e2b2

y2=b2−e2b2

y2=b2 (1−e2)As we have already stated that a2=f 2+b2, we can substitute in f=ae to get

a2=(ae)2+b2

∴b2=a2 (1−e2 )

Substituting b2

a2 =(1−e2 ) into the previous equation:

y2=b4

a2

∴ y=b2

a

∴r 0=b2

a

7 "Ellipse." Online Math Help. N.p., n.d. Web. 21 Nov. 2014.

r0

Figure 7 Semi-latus rectum


Therefore substituting this into the previous equation for an ellipse, r=

b2

a1−ecosθ

,

the polar equation for an ellipse can be given by:r=r 0

1−ecosθ

Two other important features of an ellipse are the apoapsis (the longest distance r+¿ ¿from a focus to ellipse) and the periapsis (the shortest distance r−¿¿from a focus to the ellipse). Looking at Figure 88 we can see that both the longest and shortest distances are going to lie on the x-axis. At the apoapsis and periapsis, the distance from the focus to the x-axis is equal to r±=a (1±e ).

This can be proven using some of the properties of an ellipse that I looked at

earlier, rearranging the equation b2=a2(1−e2) to give e=√ a2−b2

a2 and substituting

this in to our equation for r+¿ ¿and r−¿¿:

r±=a (1±√ a2−b2

a2 )r±=a±√a2−b2

r±=a± fThis makes sense as it states that from either focus, the distance to the point on

the ellipse cutting the x-axis is equal to the distance travelled to the centre (∓ f ¿, plus the length of the semi-major axis (a).

After having looked at ellipses and their relation to Kepler, I would now like to look at a mathematical derivation of Kepler’s First Law. I will be basing my derivation on a paper that worked through it with very little detail, whilst adding in more detail in order it make it accessible to other students.

3. Kepler’s First LawThe total energy E for a body in orbit can be given by

E=Kinetic energy+Potential energy

¿ 12mv2−GMm

rwhere m is the mass of the body in orbit, M is the mass of the star being orbited, v is the velocity of the orbit, and G is the universal gravitational constant.

Looking at Figure 99 which depicts this system, we can break the velocity into two components:

8 "Apoapsis." -- from Wolfram MathWorld. N.p., n.d. Web. 10 Nov. 2014.9 Astronomy, A1 Dynamical. Proof of Kepler’s First Law from Newtonian Dynamics (n.d.): n. pag. Web. 1 Nov. 2014.

r−¿¿r+¿ ¿

Figure 8 Apoapsis and periapsis


radial component of velocity=drdt

¿ rperpendicular component of velocity=rω

(where ω is the angular velocity)

¿r dθdt

¿r θThese two components are orthogonal (meaning that they are at right angles

to one another) therefore Pythagoras’ Theorum can be used: the square of the total velocity is equal to the sum of the squares of the componants.

∴E=12m(r2+r2 θ2)−GMm

r

Now if we look at the equation for angular momentum, L, we know thatL=mass × angular velocity

∴L=mr2ω¿mr2 θ

We shall now create the substitution that ρ=1r , so that

L=mθρ2

∴ θ= Lρ2

m

Remembering that θ=dθdt , we can integrate with respect to t to get an expression for

θ:

θ=∫ Lmρ2dt

¿∫ Lmρ2 dtdρ

dρ

Using related rates of change we can simplify this, as drdt

=drdρ× dρdt

and r=1ρ implies that

drdρ

=−1ρ2 , therefore

drdt

=−1ρ2

dρdt

So substituting in this value for drdt (or r):

θ=−∫ Lmr

dρ

This gives us the equation in which we will eventually substitute in an equation for r.

We shall now put this aside for now and return to our equation for the total energy, which rearranges as follows:

E=12m( r2+r2 θ2)−GMm

r

Figure 9 Diagram of planet in orbit of star


2Em

= r2+r2 θ2−2GMr

r2=2 Em

+ 2GMr

−r2 θ2

It is now convenient to substitute in the equation previously established

θ=L ρ2

m in order to get rid of the θ, as this has no place in the final equation for an

ellipse.

r2=2 Em

+ 2GMr

−r2( L ρ2

m)

2

Since r=1ρ , r2 and ρ2 cancel out, we can therefore say:

r2=2 Em

+ 2GMr

− L2ρ2

m2

We now make two new substitutions, as this equation has many variables that are not relevant in the equation of an ellipse in polar co-ordinates, and lack two vital parameters: r0 and e. Therefore we need to use equations connecting some of the current variables to these constants, which should eventually rearrange to produce the equation for an ellipse.

The first equation we can use is

r0=L2

GMm2

and the second equation is

e2=1+2Er0

GMmboth of which can be derived using the Conservation of Energy, although I will not do so, as the proof of these equations are not directly relevant. I had initially hoped to be able to explain them by suggesting how they are intuitively correct, but I could not find any instinctive reason as to why these are true, therefore I will not include any derivations for either equations but have included links to explanations in the footnotes.

Rearranging the equation for r2 into an equation for r with more relevant parameters for an ellipse:

r2=2 Em

+ 2GMr

− L2ρ2

m2

We can take out the L2

m2 and square root the whole equation, as Lm will cancel

out when we eventually plug the equation for r into θ=−∫ Lmr

dρ.

r= Lm {2Em

L2 + 2GMρm2

L2 − ρ2}12

We can use the substitute here that r0=L2

GMm2 ,

r= Lm {2Em

L2 + 2 ρr0

−ρ2}12


r= Lm {2Em

L2 +1r0

2 −1r0

2 +2 ρr0

− ρ2}12

r= Lm {2Em

L2 + 1r0

2 −( ρ− 1r0 )

2}12

Rearranging r0=L2

GMm2 to make L2 the subject, we can substitute again to get

r= Lm { 2 E

r0GMm+ 1r0

2 −(ρ− 1r0 )

2}12

Ideally we now want to make 2 E

r0GMm+ 1r0

2 into one fraction, as this will make

subsitution easy, therefore we can multiple the top and the bottom of the first fraction by r0:

r= Lm { 2Er 0

r02GMm

+ 1r0

2 −( ρ− 1r0 )

2}12

And rewrite this as:

r= Lm { 2 Er0

GMm+1

r 02 −(ρ− 1

r0 )2}

12

And conveniently our other substitution, e2=1+2Er0

GMm, can easily be used here:

r= Lm { e2

r02 −(ρ− 1

r0 )2}

12

r= Lm {( er 0 )

2

−( ρ− 1r0 )

2}12

It is important here that we have written r in the form Lm

{a2−b2 }12 , as now

when we substitute r into θ=−∫ Lmr

dρ, the L and m will cancel out and we can use

the general rule that ∫ 1√a2−x2

dx=cos−1( xa ) as follows:

θ=−∫ L

m× Lm √( er0 )

2

−(ρ− 1r0 )

2dρ

θ=−∫ 1

√( er 0 )2

−( ρ− 1r0 )

2dρ

θ=cos−1( ρ−1r0

er0

)


Finally this can be rearranged to form the polar co-ordinates of an ellipse, with the the origin (defined to be the Sun) at one of its foci, which is Kepler’s First Law.

ρ− 1r0

er0

=cosθ

ρ r0−1e

=cosθ

ρ r0−1=ecos θr0

r−1=e cosθ

r0

r=1+ecos θ

r=r0

1+e cosθ

4. Kepler’s Second LawDue to the nature of an ellipse, at some points in the orbit the planet will be

experience stronger gravitational force than at others, as gravitational force is a function of the distance between the two masses. This causes the planet to travel at a faster velocity at some points (when it is close to the Sun) and slower at others (when it is further away from the Sun). However Kepler’s Second Law is very statisfying in my opinion, as it states that an imaginary line segment joining a planet and the Sun always sweeps out equal areas during equal intervals of time. This can be demonstrated best on a diagram, as shown in Figure 1010.

This seems logical when considered as, when the planet is furthest from the Sun, it will have a smaller velocity, therefore cover a small distance in a given time period. However the distance from the Sun will be greater, so the other dimension of the triangle will be larger. Equally, when close to the Sun, the distance from the Sun is obviously smaller but the velocity will be greater, so the displacement about the Sun will be greater in the same given time period. Therefore it is easy to believe that

10 "Orbital Docking Dynamics." AIAA Journal 1.6 (1963): 1360-364. Web. 17 Nov. 2014.

Equal lengths of time

Equal Areas

Figure 10 Diagram visualising Kepler’s 2nd Law


Kepler’s Second Law should be approximately true once told, but slightly more difficult to prove. I shall try to do so below, using infentissimals and calculus.

In Figure 1111, we see the two successive positions (A and B) of a planet in orbit in a time dt , with a changing distance to the Sun, which we shall call r. In this time it has moved through displacement ds, and therefore travelled through an angle dθ.

A line segment from B can be drawn to a point C such that BC is perpendicular to SB, in order that AC is the change in radius, dr .

Using trigonometry, we can state that BC=rsin dθ. However as ds becomes infinitely small, BC becomes perpendicular to SC as well as to SB, meaning we can approximate the distance BC to r ×dθ, as for any very small angle α , sin α tends toα .

Like in the derivation for Kepler’s First Law, we will resolve the velocity into

two components:radial component of velocity=drdt

= r

perpendicular component of velocity=rωThe area swept out in time dt is the area of a triangle (not sector) ABS

because as curve AB becomes infentessimally small, it will tend to the segment AB for very small dθ. As distance AB (ds) becomes infentissimally small, the area of triangle ABS becomes equivalent to the area of triangle BCS, with base r ×dθ and height r.

∴Area swept out in time dt , dA=Area ofBCS

¿12

base×height

¿ 12r2dθ

¿ 12r2 dθdtdt

¿ 12r2ωdt

∴ dAdt

=12r2ω

As I stated in Kepler’s First Law, the angular momentum L of a system in orbit is equal to mr2ω, so

dAdt

= L2m

∴ dAdt∝L

Newton’s Laws state that “the rate of change of angular momentum is equal to the torque of the forces acting on the system”, where torque is a force causing

11 "Solar System Astronomy, Lecture Number 7." Solar System Astronomy, Lecture Number 7. N.p., n.d. Web. 17 Nov. 2014.

Figure 11 Orbit showing properties of a movement of a planet

r ×dθ drdθ

ds


rotation. For a planet orbitting the Sun, torque is zero, therefore angular momentum

must remain constant.We can therefore define L

2m as a constant, k .

∴ dAdt

=k

As the rate of change of area is constant through time, equal ares are swpt out in equal times, which is Kepler’s Second Law.

5. Kepler’s Third LawFinally I wish to prove Kepler’s Third Law in two ways: first from a simple

rearrangement of equations of forces and motions, which approximates an orbit to circular. The second will be a more complex mathematical derivation from Kepler’s Second Law, which will be more accurate as orbits (as previously discussed) are ellipses not a circles.

Kepler’s Third Law is that the square of the time period T for an object to complete one full orbit is proportional to the cube of its semi-major axis: T 2∝a3.

The first method uses Newton’s law of gravity, which states that the gravitational force F between two bodies can be defined as follows:

F=GMmr2

where G is the universal gravitational constant, M is the mass of the object being orbited, m is the mass of the object in orbit, and r is the distance between the two.

It is also known that the centripetal force of an object in circular motion is:

F=mv2

rwhere m and r are the same paramaters, and v is its velocity. Note that this derivation is using the equation for circular motion – therefore approximating the planetary orbit to be circular, whereas in actual fact it is elliptical. Therefore this is a less reliable proof of Kepler’s Third Law, but does show that the same law would be true for hypothetical planets in circular orbits.

Since the centripetal force is provided by gravity, we know that these two forces stated are equal, therefore:

GMmr2 =mv2

rwhich simplies to:

GMr

=v2

Kepler’s involves time period T, therefore we need to substitute in an

equation with T. We should therefore use v=2πrT , which states that the velocity is

the total distance travelled in one orbit (therefore for a circle, this is 2πr ¿ divided by the time taken to complete this orbit (ie. time period T). Note again that this uses the approximation of an orbit to be a circle, where velocity is constant.

GMr

=( 2 πrT )

2

∴GMr

=4 π2 r2

T2


∴T 2=4 π2 r3

GMBoth G and M are constants for a system of a planet orbitting a star,

therefore T 2∝ r3, where the constant is 4π 2

GM.

This proof clearly has its limitations however, since it only proves that for a circlular orbit Kepler’s Third Law will apply, therefore I now wish to use a different derivation, which starts with Kepler’s Second Law:

dAdt

= L2m

Since the right-hand side of this equation is constant, we know

Total area, A= L2m

T

where T is the time period.As I stated when examining the properties of ellipses, the standard equation

for an ellipse isA=πab

∴ πabT

= L2m

Now we put this equation aside for a while, and look at the equation for an ellipse:

r=r0

1+e cosθ

We can substitute in the equation r0=L2

GMm2 that we used in Kepler’s First

Law, to make:

r= L2

GMm2 (1+ecosθ )At the periapsis, θ=0 and r=a (1−e ), as derived earlier when looking at

ellipses. Therefore:

a (1−e )= L2

GM m2 (1+e )From this, we can get an expression for angular momentum in terms of the

properties of an ellipse and the objects in the system:L2=a (1−e )GMm2 (1+e )

∴ L2

m2 =a(1−e2)GM

Returning to our equation πabT

= L2m , we can square this and combine it with

this new equation as follows:π 2a2b2

T 2 = L2

4m2

π 2a2b2

T 2 =a(1−e2)GM4

Now we can use the relation b2=a2(1−e2),


π 2a2a2(1−e2)T2 =

a(1−e2)GM4

This simplifies as follows:π 2a2a2

T 2 =aGM4

T 2=4 π2a3

GM∴T 2∝a3

This has successfully proven Kepler’s Third Law for ellipses: the square of the time period T is proportional to the cube of the semi-major axis a.

6. ConclusionOverall I have found the derivations for Kepler’s Laws of Planetary motion

incredibly satisfying as they can prove conceptual ideas to a high degree of certainty. Kepler’s Second Law, for example, I think is fairly intuitive that the area should be roughly constant, but I find it fascinating that we can prove that it is precisely constant, which to me really highlights the elegance of maths and physics. These proofs have also emphasised to me how dealing with abstract operators can have physical significance – particularly how the ideas of calculus and geometry that I have studied from a purely mathematical perspective can be linked to concepts in my Physics syllabus. Although the rearrangement of the equations during the process may seem fairly abstract, it is amazing to see how they can link together far more conceptual ideas that tell us how the Universe works. These equations also have practical implications in allowing scientists to predict the movement of planets, and therefore to adjust trajectories of satellites or probes to use the gravitational pull of celestial bodies in order to maximise efficiency. If I had had more time I would have liked to examine this further, perhaps by modelling the slingshot effect, as this would give me an insight into how knowledge of orbits can be useful in conserving fuel.

During my exploration I have also developed a sense for how complicated derivations can be, since it seems that all 3 of Kepler’s equations were reliant on other derivations and ideas that I needed to prove along the way, which was a weakness of this approach. For example many of the parameters of an ellipse required individual examination so that I could be satisified that I understood the proof as a whole. Furthermore there were also a few physics equations that I have not looked at in my Higher Level syllabus that I did not have an opportunity to fully derive, which was a shame. If I were to do my exploration again, I might have spent more time looking at these physics equations. However overall I feel very satisfied that I have understood the maths used and have been fascinated to see how the concepts I study in Maths lessons are applied in the real world, and I have also greatly enjoyed the challenge of teaching myself material about conic sections.

7. Bibliography"5 Minute Lesson: Kepler's 2nd Law." The Very Spring and Root. N.p., n.d. Web. 3 mmmmNov. 2014."Apoapsis." -- from Wolfram MathWorld. N.p., n.d. Web. 10 Nov. 2014.


Astronomy, A1 Dynamical. Proof of Kepler’s First Law from Newtonian Dynamics mmmm (n.d.): n. pag. Web. 1 Nov. 2014."Eccentricity an Ellipse - Math Open Reference." Eccentricity an Ellipse - Math Open mmmmReference. N.p., n.d. Web. 17 Oct. 2014."Ellipse." Online Math Help. N.p., n.d. Web. 21 Nov. 2014."File:Conic Sections.svg." Wikipedia. Wikimedia Foundation, 22 July 2014. Web. 12 mmmmOct. 2014."Foci (focus Points) of an Ellipse." Math Open Reference. N.p., n.d. Web. 16 Oct. mmmm2014."Jupiter Fact Sheet." Jupiter Fact Sheet. N.p., n.d. Web. 3 Oct. 2014.“Kepler's 2nd Law: The Speeds of Planets." - Windows to the Universe. N.p., n.d. mmmmWeb. 10 Nov. 2014."Kepler's Laws." N.p., n.d. Web. 3 Nov. 2014."Kepler's Laws." Hyperphysics. N.p., n.d. Web. 9 Oct. 2014."Solar System Astronomy, Lecture Number 7." Solar System Astronomy, Lecture mmmmNumber 7. N.p., n.d. Web. 17 Nov. 2014."Orbital Docking Dynamics." AIAA Journal 1.6 (1963): 1360-364. Web. 17 Nov. 2014.

introduction - nlcs maths department · web viewwill explore the properties of ellipses and their...

Documents