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Extrasolar
planets
and
Kepler’s
Third Law
2
4

T2 
a3
G  M  m
AF. May 2015
Nicolaus Copernicus
1473-1543
Tycho Brahe
1546-1601
Johannes
Kepler
1571-1630
Nose lost in 1566 following a
sword duel with third cousin
Manderup Parsberg over the
legitimacy of a mathematical
formula!
Isaac
Newton
16421727
r
a 1   2 
1   cos 
Polar
equation
of ellipse
2
b Eccentricity of
  1  2 ellipse
a
2
4

P2 
a 3 Orbital
period P
G(M  M )
Johannes Kepler
1571-1630
dA 1
 2 G(M  M ) 1   2  a
dt
Equal areas swept out in
equal times
This is a constant
Radii of planets not to scale!
Kepler’s
Third Law
Pa
Orbital period P /years
Johannes Kepler
1571-1630
P  a 


Yr  AU 
Mercury,
Venus,
Earth,
Mars
Saturn
Jupiter
Neptune
3
2
3
2
Uranus
2
4

P2 
G M  M

a3
Semi-major axis of orbit a / AU
Isaac Newton
(1642-1727) developed
a mathematical model of
Gravity which predicted the
elliptical orbits proposed by
Kepler
Planet and Solar
masses
Force of
gravity
r
GMM
F
r2
G  6.67  1011 m3 kg -1s-2
a 1  
2

1   cos 
Semi-major
axis
2a
M
Polar
equation
of ellipse
b 2 Eccentricity of
  1  2 ellipse
a
2
4

P2 
a3
G(M  M )
Semiminor
axis
2b
Orbital
period P
M
r

F
GMM
r2
Kepler’s Third Law
T
 a 
f ( M , m)  k  

days
AU


M  1.99  1030 kg
G  6.67  1011 Nm 2 kg -2
k
AU  1.49597871  10 m
11
24  3600s  1day
2  AU
3600  24 GM
f ( M , m) 
4
T 
a3
G  M  m
2
2
T
2
 a 
 3600  24 
 AU 

days
 AU 
G  M  m
T
2  AU


days 3600  24 GM
m planet mass
3
2
 a 


M
m  AU 

M
M
1
M
m

M
M
M star mass
3
2
3
2
 365
3
2
3
2
T
2
 a 

 AU 

days 3600  24 G  M  m 
 AU 
3
2
3
2
T orbital period
3
2
a orbital 'radius'
semi-major axis of
elliptical orbit
T
 a 
f ( M , m)  k  

days
 AU 
k
2  AU
3
2
3600  24 GM
f ( M , m) 
3
2
 365
M
m

M
M
Vertical intercept is very
close to ln(365) = 5.90
Gradient is 1.49, not far
off the Kepler #3
prediction of 3/2
 T 

 a 
3
ln  
f
(
M
,
m
)

ln
k

ln


2


days
AU






k
2  AU
3
2
3600  24 GM
f ( M , m) 
 365
M
m

M
M
f ( M , m) 
M
m

M
M
m jupiter  1.898  1027 kg
m jupiter
1

M
1047.9
M star mass
m planet mass
f ( M , m) 
M star mass
m planet mass
M
m

M
M
m jupiter  1.898  1027 kg
m jupiter 
1
M
1047.9
M
m

M
M
M star mass
f ( M , m) 
m planet mass
m jupiter  1.898  1027 kg
m jupiter 
1
m 
m jupiter
317.85
m
 0.0031
m jupiter
1
M
1047.9
M
m

M
M
M star mass
f ( M , m) 
m planet mass
m jupiter  1.898  1027 kg
m jupiter 
1
m 
m jupiter
317.85
m
 0.0031
m jupiter
1
M
1047.9
Earth-like candidates?