moon - keplers...nevil maskelyne 1732-1811 charles hutton 1 1737-1823 schiehallion, perthshire,...

Newton, the apple and the moon

By looking at the sizes of shadows, the Ancient Greeks had reasoned the Moon must be about sixty Earth radii distant. This is impressively close to the modern measurements,

which take into account that the orbital distance varies from perigee to apogee over time due to a slight eccentricity (i.e. not perfect circularity) of the orbit. Note also that the Moon

is currently receding from the Earth at a rate of about 2.91 cm/year due to the tidal gravitational interaction between the moving water on Earth and the Moon.

Isaac Newton

1642-1726

66.371 10 mR

60M

R R

Not to scale!

Orbit of the Moon

(about 28 days)

Earth radius

An apocryphal apple

d

363,104km 406,696km

57.0 63.8

M

M

R

R R R

Galileo had shown experimentally that all falling bodies on Earth should accelerate

a constant rate. In modern terms, we would say that if air resistance, lift, upthrust etc

can be ignored, all objects fall at about g = 9.81 m/s2

From Galilean kinematics, one can predict how far an apple would fall in 1 second.

Galileo Galilei

1564-1642

2 -212 9.81ms 1s

4.9m

x gt g t

x

Newton then asked the question: “How far does the Moon fall in 1 second?”

If one assumes a circular orbit, the diagram on the left (scale highly exaggerated)

represents one second of movement. From Pythagoras’ Theorem, we can calculate

the fall distance in terms of the Earth’s Radii and the distance travelled by the Moon

in one second. We shall assume over such short timescales the Moon travels in a straight line.

2 2 2

2 2 2 2

2 2

2

2

60 (60 )

(60 ) 120 (60 )

120

120

120

R d R

R R d R

R d

R d

d

R

Expect the square of the fall distance to

be negligible compared to the other terms

In one second, the Moon will travel

2 60

28 24 3600

993m

Rd

d

Hence in one

second the Moon

will fall

2

2

6

120

992.81

120 6.371 10

1.3mm

d

R

Newton reasoned that to keep the Moon in orbit, a central force

must act. He postulated that an inverse-square law would be

appropriate, perhaps based upon how light rays might diverge

from a circular source.

From a modern perspective, we might refer to how the radioactive

power per unit area F received from the Sun follows an inverse

square law, since the power per unit area times the area of a

sphere surrounding the Sun must equal the total power output (or

Luminosity L)

2

24

4

LL R

R

F F

212

-2

12 2

9.81ms 1s

9.811.3mm

60

Mg t

g t

Based on the inverse

square law

2

1

60

Mg

g

Hence:

which matches the expected

distance the Moon falls

to maintain its orbit.

i.e. constant acceleration motion

means a quadratic dependence

of displacement upon time

Earth-Moon separation

Newton therefore

proposed a Universal

Law of Gravitation for

the force acting between

two masses separated

by distance r

2

GMmF

r

i.e. gravitational force or weight is

mass x gravitational acceleration*

*It is assumed here that ‘gravitational mass’ in mg is the same as inertia

in force = mass x acceleration. This deep connection shall be discussed later!

Note for the Sun

26

-2

3.846 10 W

1,368Wm

L

F

Mathematics topic handout - Mechanics – Kepler’s Laws and Universal Gravitation. Dr Andrew French. www.eclecticon.info PAGE 1

http://www.eclecticon.info/

Measuring G via the Cavendish experiment The Schiehallion experiment

Analysis adapted from https://en.wikipedia.org/wiki/Schiehallion_experiment

Nevil

Maskelyne

1732-1811

Charles

Hutton

1737-1823

Schiehallion, Perthshire, Scotland

1,083m

d

F

W

T

x

y

By Newton II, assuming

equilibrium of pendulum bob:

2

2

// : 0 sin

// : 0 cos

MGmM

x Td

GmMy T

R

2

GmMW

R

2

MGmM

Fd

2

2

2

2

2

2

sin

cos

tan

tan

M

M

M

GmMT

d

GmMT

R

M R

d M

RM M

d

In the published report it was the

density of the earth that was

reported

243

M

R

Henry Cavendish

1731-1810

2

2

W mg

GMg

R

gRG

M

11 3 -1 -26.67384 10 m kg sG

2 21 1

2 2

2 212

2

2 2

2

2

2

4

2

IP

I m L mL

P mL

mL

P

To make use of Newton’s Law of Universal Gravitation requires us to calculate the

constant G, unless we are content with ratios as in the ‘moon-fall’ example.

2

GMmF

r

The Schiehallion experiment, conducted by

the Royal Society in 1774, measured the

defelection of a long pendulum by the

gravitational attraction of a mountain. If the mass

of the mountain could be calculated then this

experiment could be used to calculate the mass of

the Earth. From this one can readily find G.

Centre of mass

of mountain

Pendulum

Stars

The deflection was measured

by comparing the line of the

pendulum to the positions of

several stars. These observation

were performed by Nevil

Maskelyne.

Applying Newton’s Law of Gravitation

Earth mass

and radius

Mountain

mass

Calculating the mass of Schiehallion

required a detail survey of the

mountain. Its density was taken as

2500 kgm-3, so an accurate calculation

of volume was required. Charles Hutton

performed this arduous mapping task,

and invented contour lines in the

process!

The Schiehallion experiment

reported that the Earth average density

was about 4500kgm-3. A modern (2007)

repeat of the experiment using a digital elevation model

yielded 5480kgm-3. The actual value is 5515kgm-3.

This means the Earth is not hollow, and must

contain denser material at depth, possibly metallic.

Note g can readily

be measured, and a reasonable

estimate of the radius of the Earth

has been known since the time of

Eratosthenes (276-195BC)

A more accurate value of G can be

found by performing a sensitive

experiment in the laboratory, using a

torsion pendulum.

2

2

GmML

r

rG

LmM

Balance the torsion force (the twist) on

the wire with the torque resulting from the

gravitational attraction of masses M and m

2 2 2

2

2 2

2

2

2

mL rG

P LmM

LrG

MP

The torsion constant can be

found by measuring the period P

of small oscillations of the

pendulum

Moment of inertia I of the

pendulum about the wire

axis

We can now combine the expressions to find G Contour lines

In the original

experiment

(see diagram

below)

12

0.73kg, 158kg

1.8m, 230 4.1 mm

875.3s, 4.1mm

m M

L r

P L

http

s://e

n.w

ikip

edia

.org

/wik

i/Cavendis

h_experim

ent

2m

Pk

For a mass

on a spring of

stiffness k

Angle in radians

Cavendish measured

11 3 -1 -26.74 10 m kg sG

245.972 10 kgM


https://en.wikipedia.org/wiki/Schiehallion_experiment

https://en.wikipedia.org/wiki/Cavendish_experiment


Kepler’s Three Laws of Orbital Motion

Inspired by the heliocentric model of Copernicus, and using the astronomical data obtained by Tycho Brahe, Kepler discovered three laws of planetary motion.

Johannes Kepler

1571-1630

Tycho Brahe

1546-1601

Nicolaus

Copernicus

1473-1543

r

2a

2b

M

M

Semi-

minor

axis 2

GMMF

r

2

2

2

2

2 3

1

1 cos

1

4

( )

ar

b

a

P aG M M

Eccentricity of

ellipse

Polar equation of

ellipse

Kepler III

Semi-major axis

Distance /AU

Distance /AU

212

G( ) 1dA

M M adt

Kepler II: Equal areas swept out in equal times

Applying Newton’s Law of Gravitation we can show the rate

of change of area A swept is a constant.

1. The orbit of every planet in the solar system is an ellipse with the Sun at one of the two foci.

2. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time

3. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

The wording of Kepler’s Laws implies a specific application to the solar system. However, the laws are more generally

applicable to any system of two masses whose mutual attraction is an inverse-square law.

Kepler I: Orbits of the planets are ellipses with the

Sun at one focus

i.e. a circle has zero

eccentricity. As eccentricity

tends to unity, the ellipse

becomes more elongated.

Assume a bwithout loss of generality – since

we can rotate the ellipse!



Using Newton’s Law of Universal Gravitation to characterize circular orbits

2

2

2 2

2

3 2

4

4

GMmma

r

r GMmm

P r

r PGM

If a planet orbits a massive object such as a star, to a good approximation

the orbit is a perfect circle centred on the centre of the star. (In general

in a two-mass closed system where relativistic effects can be ignored,

both objects will orbit in an elliptical fashion about their common centre of mass

or barycenter*).

The only force binding the planet to the star is gravity, which is a central

force i.e. acts entirely radially. If we ignore any mass asymmetries for the

planet and the star, we can conclude that there will no tangential forces

which might speed up the orbital rotation rate.

The orbital velocity is therefore a constant. If the period is P and

the orbital radius r, the orbital velocity is

2 rv

P

If the planet is executing circular motion its acceleration is

radially towards the center of the star and has magnitude

2

2 2

2

2

2

4 1

4

va

r

ra

P r

ra

P

Applying Newton’s Second Law, and using

the Universal Law of Gravitation

Which is Kepler’s Third Law. Indeed the form

is quite general. For a two body system the M is

the total mass and the r is the maximum

separation (the ‘semi-major axis’).

rv

m

M

Orb

ital perio

d P

/years

Semi-major axis of orbit a / AU

32

Yr AU

P a

22 34

P aG m M

Radii of planets not to scale!

Mercury, Venus, Earth, Mars

Jupiter

Saturn

Uranus

Neptune Kepler’s Third Law in action for the planets in the Solar System

Kepler’s Third Law

30

11 2 -2

11

24

1.99 10 kg

6.67 10 Nm kg

AU 1.49597871 10 m

24 3600s 1day

332,837

5.972 10 kg

M

G

M m

m

An Astronomical Unit (AU)

is the average Earth-Sun

separation.

Planet T / years r / AU m / Earth

masses

Rotation

period

/days

Orbital

eccentricity

Mercury 0.241 0.387 0.055 58.646 0.21

Venus 0.615 0.723 0.815 243.018 0.01

Earth 1.000 1.000 1.000 1.000 0.02

Mars 1.881 1.523 0.107 1.026 0.09

Jupiter 11.861 5.202 317.85 0.413 0.05

Saturn 29.628 9.576 95.159 0.444 0.06

Uranus 84.747 19.293 14.5 0.718 0.05

Neptune 166.344 30.246 17.204 0.671 0.01

Pluto 248.348 39.509 0.003 6.387 0.25

Sun mass

[Note Pluto orbits in a different plane to the other planets, and is officially a ‘dwarf planet’, not a planet]

*Note the barycenter of the Pluto-Charon system is actually outside Pluto

Earth

mass

These orbit in the “plane of the ecliptic”

2

3 2

2

3 2

23

4

4R

R

r PGM

R PGM

r P

R P

Kepler III as a ratio

between different

planets

since m M



Gravitational field strength and gravitational potential

2

GMm

r

r

m r

Newton’s Law of Universal Gravitation tells us the

force F on a mass m at distance r from the centre of a mass M acts radially inwards along the line joining the centres of the masses.

The gravitational field strength g is defined to be

mF g

2ˆ

m

GM

r

F g

g r

The Newtonian model of gravity

is that of a force which

permeates all space, and whose

magnitude and direction is

computable from the spatial

distribution of mass.

Gravity is therefore a field of vectors – at any point in

space we can draw an arrow pointing the direction of

gravitational force, and with a length proportional to the

strength.

For many calculations it is useful t p compute the

Gravitational Potential Energy, that is a measure of

the work done against gravity to move a particle to

a particular point. This is useful when combined

with the Law of Conservation of Energy, as we can

work out the speed of a gravitationally bound object

based upon scalar parameters, rather than needing to

worry about directions of the vector quantities involved

like force, velocity and displacement.

The work done in moving

an object of mass m from

distance a to b against

gravity is

2

1 1

b

a

b

a

GMmW dr

r

GMmW

r

W GMma b

The maximum work

done is when b is infinite:

max

GMmW

a

If mass m is launched

radially from distance a

with kinetic energy

E, we would expect gravity

to slow it down. If at an

infinite distance away the

mass has zero speed, then

by conservation of energy:

max

GMmE W

a

But the total energy ‘at infinity’ must be zero

since the mass has no speed and will not be

affected by gravity.

Therefore in order to conserve energy

everywhere, the total energy

at any radius must be zero everywhere.

We can therefore define a gravitational

potential energy (GPE) mf such that

0E m

GMmE

a

GM

a

f

f

In general the mass may not have enough

energy to escape to infinity, or indeed have

more than enough. Let the total energy be U

i.e. it makes

sense for GPE to

be negative

212

GMmU mv

r

Escape velocity

212

20

GMm GMmv v

R R

For Earth, the escape velocity is:

11 24

6

1

2

2 6.67 10 5.97 10

6.38 10

11.2kms

escape

escape

GMv

R

v

It is interesting to work out the radius of a star

of mass M such that the escape velocity

exceeds that of the speed of light. Since this is

not possible, the star becomes a Black Hole.

This inequality defines the maximum radius of a

Black Hole, which is called the Schwarzschild

radius. Alternatively, this is the event horizon,

or ‘point of no return’ from the centre of a Black

Hole.

For the Sun to become a Black Hole ( M = 2 x 1030 kg,

R = 6.96 x 108 m ) its radius would have to shrink to

less than 2.97 km. This is a mind-blowing density of

1.8 x 1019 kgm-3

As enormous as this sounds, it is not entirely

outrageous given the density of

the nucleus of a typical atom is approximately

The definition of GPE we have adopted allows us to make a very general connection between field

strength and potential

2,

ˆ

GM GM

r r

d

dr

f

f

g

g r

unit

vector

This provides up with a powerful tool if we wish to generalize the problem

to many masses. We can sum the gravitational potentials and then

take the negative gradient to find the field strength.

In 2D or 3D we need the vector operator “grad” as potential f might vary

with all x,y,z coordinates

ˆ ˆ ˆV V V

x y z

f

g x y z

g

In order to escape, the total energy of

the system must be positive at an

infinite distance from the body. In other

words, it will have some kinetic energy

and will never be gravitationally

attracted back towards the body.

For a mass m blasting off with velocity

v, it will escape the gravitational

influence of M if:

2

2

2

2

2

GMc

R

GMc

R

GMR

c

3

43 2

6

3 2

2

3

32

Black hole

Black hole

M

GM

c

c

G M

27

17 -3

3154

3

2 105 10 kgm

10



Gravitational field strength inside and outside a uniform sphere


R

r

Spherical planet

with uniform

density of total

mass M

Now m(r) is the mass enclosed

within radius r, hence

To construct a generic plot define:

2

2

0 1

1

R

R

R

GMg

R

rx

R

g x x

g gx

x

We can generalize our definition of

gravitational potential to be

Gauss’ Law of Gravity

4S

d Gm g S

where m is the mass enclosed

within closed surface S, whose

surface normal area vector is dS

343 34

3

3

3

3

2

3

3

0

4

ˆ

ˆ

4 4

S

r R

Mm r

R

Mrm

R

d Gm

g

d dS

Mrg r G

R

GMrg

R

g S

g r

S r

2

r R

GMg

r

R

g

g

rx

R

ˆ ˆS S S

d g dS g dS g S r r

This means that if a tunnel could be drilled

through a planet of uniform density, Newton’s

Second Law means for a mass dropped into the

tunnel at rest at the surface

3

GMr r

R

This is the equation of Simple Harmonic Motion (SHM)

22

r rP

where P is the period of

the resulting oscillatory

motion

2

3

3

2

2

GM

P R

RP

GM

3

2cos

cos

tr R

P

GMr R t

R

For the Earth this would be a period of

3

6

11 24

6.371 102

6.67384 10 5.972 10

5061s 84mins

P

P

6

11 2 -2

24

6.371 10 m

6.67384 10 Nm kg

5.972 10 kg

R

G

m

So a 42 minute trip

to Australasia from Europe

without any jet fuel required...


Two-body Kepler problem with circular orbits

The essential features of the more general Kepler problem can

be obtained by considering circular orbits of gravitationally bound

masses about a common centre of mass.


Centre

of mass

m

M

r

R

2

P

Since mass M gravitationally attracts mass m with a force along the lines connecting

the centres (and vice versa), there is no net torque on the system. Hence the

angular acceleration is zero and therefore the angular speed of (both masses) is a

constant. Let P be the orbital period.

By Newton’s second law 2 2

2 2,

GMm GMmMR mr

a a

a R r

2 2

2 2

2 2

2

2

2

,Gm GM

R ra a

G m MR r

a

G m MR r

a

2

3

2

2 3

2

2 3

4

4

G m M

a

G m M

P a

P aG m M

Kepler’s

Third Law

We can now work out the energy of the combined system:

2 21 12 2

2 2 212

GMmE M R m r

a

GMmE MR mr

a

By Newton’s second law

2 2

2 2

2 2 2 2

2 2 2 2

2 2

4 4 4 4

,

,

,

GMm GMmMR mr

a a

Gm GMR r

a a

G m G MR r

a a

Hence

2 2 212

2 2 2 2

212 4 4 4 4

2

12 4 2

GMmE MR mr

a

G m G M GMmE M m

a a a

G Mm GMmE m M

a a

Now from above

2

3

G m M

a

2 3

12 4

12

( )

2

G Mm a GMmE m M

a G m M a

GMm GMmE

a a

GMmE

a

Therefore:

Note the total negative energy is indicative

of a bound orbit. Parabolic or hyperbolic

trajectories will have a positive total energy

The rate of area swept by the masses is

212

212

M

m

dAR

dt

dAr

dt

d

RRd

dt Rd

e.g. 21

2

2 21 12 2

M

M

dA R d

dA dR R

dt dt

From above:

2 2 2 2

2 2

4 4 4 4,

G m G MR r

a a

Hence:

2

3

G m M

a

2 2 2 2

21 1 12 2 24 4 4 3

2 2 2 2

21 1 12 2 24 4 4 3

M

m

dA G m G mR

dt a a

dA G M G Mr

dt a a

The rate of area swept is therefore

a constant for each mass, i.e.

Kepler’s Second Law

Note:

32

32

1 1 12 2 2

2 2

12 4 3

2 2 3

12 4

2 2 3

12 4

1 12 2

( )

m

m

m

m

dA G M

dt a

dA G M a

dt a G m M

m M

dA G M a

dt a GM

dAG M a GMa

dt

This is consistent with

the more general result

for elliptical orbits 21

2( ) 1-

dAG m M a

dt

Include expression for

radial acceleration

for circular motion


Two body Kepler problem

2 1

12 2

2 12 1

12 2 13

2 1

12 21

GMm

GMm

r rF

r rr r

F r rr r

F F

x

y2

r1

r

m M12F 21

F

Newton II

1 12

1 2 13

2 1

2 21

2 2 13

2 1

m

GMmm

M

GMmM

r F

r r rr r

r F

r r rr r

Define centre of mass vector

m M

m M

1 2r r

R

and separation vector

2 1

r r r

Hence:

1

2

M

m M

m

m M

r R r

r R r

3 2

( ) ( )ˆ

G M m G M m

r r

2 1r r r

r r r

m M

m M

m M m M

1 2

1 2

r rR

R r r

From above

1 12

2 21 12

1 2

m

M

m M

r F

r F F

r r 0

Therefore constantR

which means the centre of

mass of the system moves at

a constant velocity. Without loss

of generality we can define a reference

frame co-moving with the centre of mass.

So from now on we will set

Using the previous

Newton II expressions.

This means the two body problem

is basically a one body problem,

with the separation vector r being

the displacement from a total

mass m + M

Angular Momentum

1 2 2m M

M M m mm M

m M m M m M m M

1J r r r r

J R r r R r r

R 0

R 0Note

3

mM mM

m M m M

G M mmM

m M r

J r r J r r r r

J r r r r 0 J J

is a constant for

the system both zero

rr

rθ

2

ˆ

ˆˆ

ˆˆ 2

r

r r

r r r r

r r

r r θ

r r θ

We will now solve the Kepler problem using plane polar coordinates

2

2 222 4 2

2

ˆˆ ˆ

ˆ

mM

m M

mMr r r

m M

mMr

m M

m MJ r

m M

J r r

J r r θ

J r θ

J

Since angular momentum is a

constant

2 2

2

2 2 4

m M J

m M r

2

2

2

2

2 2

2 2 4 2

( )ˆ

ˆˆ 2

2

( )

( )

G M m

r

r r r r

r

r

G M mr r

r

m M J G M mr r

m M r r

r r

r r θ

To simplify let us define 1

ur

2

2 2 2

2

2 2 2

1 1

du du dt u

d dt d

du d du dt d dtu u

d u d d dt d dt d

d

u u u ud u

d

2

2 2 3

1 2,

r ru r u

r r r

2 2

2 2 3 2 2

3 2

2 22

3 2

3 2 2 2

2 22

5 4

2 2

2 2, 2

2( )

2( )

r u ur r u r ur r u ur

r u u

m M Ju uu u G M m

u u m M

m M Juu u u G M m

u m M

2 2, 2

r r uru ru

r r u

From above:

2

2

2 2 2

2uu u u

d u u

d

2 2

2 4

2 2

m M Ju

m M

2 22 2

4

2 2 2

2 m M Ju d uu u

u m M d

by the Chain Rule

Hence:

2 22 22

4 5 4

2 2 2 2 2

2 2 2

2 2

( )

( )

m M J m M Jd uu u u G M m

m M d m M

d u Gm Mu

d M m J

x

y

ˆ ˆ ˆcos sin

ˆ ˆ ˆsin cos

r x y

θ x y



2 2 2

2 2( )

d u Gm Mu

d M m J

If the orbits are ellipses, the equation of

an ellipse in polar coordinates is

21

1 cos

ar

r

2a

2b

Semi-

minor

axis

Semi-major axis

(1 )a

21 1 1(0) 1

1 1

a ar a

(assume use left focus)

2

2

2

2 2

1 cos

1

sin

1

cos

1

ua

du

d a

d u

d a

2 2 2

2 2

2 2

22 2

2 2

22

2 2 2

2

( )

cos 1 cos

( )1 1

1

( )1

1

( )

d u Gm Mu

d M m J

Gm M

M m Ja a

Gm M

M m Ja

Gm M aJ

M m

Which is certainly a constant i.e.

independent of polar angle.

Since ellipses are solutions of

2 2 2

2 2( )

d u Gm Mu

d M m J

we have therefore proved Kepler’s First Law

21-b a

Area dA swept

out by an orbit

radial in time dt is

212

212

dA r d

dAr

dt

2 2

2

2 2 4

2 2 2

2

2 2 2 2

2

2 2 4

2

2

4

2 2

212

1

( )

1

( )

1

1

1

m M J

m M r

Gm M aJ

M m

Gm M am M

m M r M m

G m M a

r

r G m M a

dAG m M a

dt

So equal areas are swept out in equal times

Kepler’s Second Law

2 2

212

2

2 3

1

1

4

ab aP P

dA G m M adt

P aG m M

Since equal areas are swept out in equal times, the orbital period is the area of the

ellipse divided by the rate of area sweep

Kepler’s Third Law: The

square of the orbital

period of a planet is

directly proportional to

the cube of the semi-

major axis of its orbit.

Summary of orbital dynamics

1

2

2

2

2

2

2 2

2

2

2

1

1 cos

ˆ

ˆˆ

1 1 ˆˆsin1 cos

1 sin 1 ˆˆ1 cos

sinˆ ˆ1 cos

1 cos sinˆ ˆ11 cos1

M

m M

m

m M

ar

r

r r

ar

r

r rr r

r

r

G m M aa

r R r

r R r

r r

r r θ

r r θ

r r θ

r θ r

r θ r

r

2

2

sinˆ ˆ1 cos1 cos1

ˆ( 0) 11

G m M

a

G m M

a

θ r

r θ

2 21r G m M a

2

( )ˆ

G M m

r

r r

0

0

0

0

2 2

2 2

22 2

22

22 2

22

33 2

2

1

1

1

1 cos1

1

1 cos1

1

1 cos

dr G m M a

dt

r d t G m M a

a dt

G m M a

a dt

G m M a

a dt

G m M

Evaluate this numerically

Displacement

Velocity

Acceleration



Example two-body simulation: Pluto and Charon

Charon 21

22

1.586 10 kg

606km

19,596km

1.303 10 kg

1187km

6.387days

0.00

C

P

m

R

a

M

R

P

https://en.wikipedia.org/wiki/Charon_(moon) https://en.wikipedia.org/wiki/Pluto

http://nssdc.gsfc.nasa.gov/planetary/factsheet/plutofact.html

Pluto

Alternative simulation

22

22

1.0 10 kg

20,000km

3.0 10 kg

3.981

0.6

m

a

M

P

Planet Y

Planet X

It looks like the orbits might collide...

But the ‘intersection points’ will occur at

different times for each planet. Plotting

the separation magnitude r vs time shows

how far apart the planets get over each

orbit.

For orbits with zero eccentricity

the separation will be constant

i.e. the separation vector r races a

circular orbit.

The Spacecraft New Horizons

made a 12,500km approach of Pluto

on July 14 2015.


https://en.wikipedia.org/wiki/Charon_(moon)

https://en.wikipedia.org/wiki/Pluto

http://nssdc.gsfc.nasa.gov/planetary/factsheet/plutofact.html


Energy in the Kepler problem

x

y2

r1

r

m M

1

2

2

2

2 222

22

1

1 cos

sinˆ ˆ1 cos1 cos1

sin1 cos 1

1 1 cos

M

m M

m

m M

ar

G m M

a

G m M

a

r R r

r R r

r θ r

r

2 21 12 21 2

1

2

212

2 22

12 22 2

2 22 2

22 2 2

sin 1 cos1 cos 1

1 11 cos

1 cos 1 cos 2 1 cossin

2 1 1 11 cos

GMmE m M

r

M

m M

m

m M

mM GMmE

m M r

G m MmME GMm

m M a a

GmME

a

r r

r r

r r

r

2 2 2

2

2 2 2 2

2

2 2 2 2

2

2 2 2

2

1 cos sin 2 1 cos

2 1

1 2 cos cos sin 2 2 cos

2 1

1 2 cos cos sin 2 2 cos

2 1

1 cos sin

2 1

2

GmME

a

GmME

a

GmME

a

GmME

a

GmME

a

Recap of previous results

Total energy is the sum of the kinetic

energy of the masses and the gravitational

potential energy

r

2 1r r r

r

r


Hence

212

212

2

2

1 1

2

22

2

2

GmM mM GMm

a m M r

mMGMm

m M r a

a rG m M

ar

G m M a r

ar

r

r

r

r


moon - keplers...nevil maskelyne 1732-1811 charles hutton 1 1737-1823 schiehallion, perthshire,...

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