STEM 698 Derivation of Kepler’s Laws from Newton’s Laws
Kepler’s Laws
1. A planet orbits the sun in an elliptical path with the sun at one of the foci.
2. The area swept out by a planet radially from the sun over a fixed period of time is
constant. (Hence the planets vary their speed as orbit the sun; planets travel faster when
they are closer to the sun.)
3. The square of the orbital period is proportional to the cube of the semimajor axis of the
orbit.
Newton’s Laws
1. F = ma
2. Suppose a point mass M is located at the origin (0, 0) and a point mass m is located
at ( x, y ) . If we let r denote the vector ( x, y ) and r denote the length of this vector (i.e.,
r  x 2  y 2 ), then the gravitation force that M exerts on m is given by
Mm
r
r3
where G is a universal constant, independent of M , m , and r.
F  G
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Angular Momentum of planet about the origin is conserved
Suppose a point mass M is located at the origin (0, 0) and a point mass m is located at
( x, y ) . Suppose that at all times, there is a central force acting on m (i.e., a force
always in the direction of r = ( x, y ) . Note r is the position of mass m relative to M. By
Newton’s law of universal gravitation, gravity is such a force. Let v = r  = ( x, y) be the
velocity vector of m .
Define the angular momentum of m2 to be L = r × mv . Then
L  0
Proof. L' = (r × mv)' = (r' × mv)  (r × mv') = (v × mv)  (r × ma)
where a is the acceleration. The first term is 0, because
v × v = 0 . As for the second term, F = ma is in the direction of r so because r × r = 0 , it
is also 0.
A corollary of this result is that the motion of m2 must lie in a single plane determined by
the normal vector L , {c 
3
| L  c  0} .
Areas swept out are constant (Kepler’s Second Law)
The key challenge here is to express the area swept out in a given amount of time.
Define r = r (cos  i  sin  j ) . NOTE: r and  are functions of time.  is relative to
some (arbitrary) fixed ray.
Here is an elementary way to get the formula for the area swept out in a given amount of
time.
y
Q
R
r(θ + Δθ)
P
Δθ
r(θ)
θ
x
O
2
A 
1
r (   )r ( ) sin(  )
2
So
A 1
sin( ) 
 r (   )r ( )
t 2

t
So
A
1
sin( ) 
 lim r (   ) r ( )
t  0 t
t  0 2

t
1 2 d
 r 1 
2
dt
lim
So
dA 1 2 d
 r
dt 2 dt
Another way I know is to express it in polar coordinates based at the origin.
Now the area swept out by m2 from time 0 to time t is given by
A
 (t )
r d
1 2
 (0) 2
Taking the derivative with respect to time we get
A  12 r 2 
Now taking the derivative of r = r (cos  i  sin  j ) , we find
v = r (cos  i  sin  j )  r ( sin  i  cos  j )
So
L = v = r × mv
= r (cos  i  sin  j ) × m  r (cos  i  sin  j )  r ( sin  i  cos  j ) 
 mr 2 k
Hence
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A  12 r 2  
L
 constant
2m
where L denotes the magnitude of the constant angular momentum vector.
Another way of approaching this is to think about the incremental area swept out as one
half of the area of the parallelogram formed by r and Δr , i.e.,
ΔA  12 | r  r |
Then
A 1
r
 2 r
t
t
In the limit,
dA 1
L
 2 r v 
dt
2m
Differentiate the unit radius vector
One way to derive the first law is to differentiate the vector
 r  v r 
    2r
r r r
1
= 3 (r 2 v  rr r )
r
1
 3 ((r  r )v  (r  v )r )
r
r
 (r  v )  3
r
L
a
 
m
GM
1

(a  L)
GMm
1

(v  L)
GMm
4
r
.
r
Hence,
1
r
(v  L)     e = a constant vector
GMm
r
Note that e is in the plane of motion; the plane of motion is the plane whose normal is L
and e  L = 0 .
Taking the dot product with r,
1
r r
r  (v  L) 
  r e
GMm
r
L2
 r  re cos 
GMm2
or
L2
 r  re cos 
GMm2
where  is the angle between e and r now.
Kepler’s First Law
Case 1. e = 0.
Then
r
L2
GMm2
The orbit is a circle.
Now let
L2
 ed
GMm 2
Then
ed  r (1  e cos  )
Convert to Cartesian coordinates:
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ed  er cos   r
ed  ex  x 2  y 2
e2 (d  x) 2  x 2  y 2
e 2 d 2  2de 2 x  e 2 x 2  x 2  y 2
e 2 d 2  x 2 (1  e 2 )  2de 2 x  y 2
We now obtain Case 2: If e = 1, the orbit is a parabola.
Also if e > 1, then we get Case 3: If e > 1, the orbit is a hyperbola.
Now assume e  1 . Complete the square to get
2

e2 d 
y2
e2 d 2


x

1  e 2  1  e 2 (1  e 2 ) 2

2

e2 d 
x



1  e2 
y2


1
e2 d 2
e2 d 2
(1  e 2 ) 2
(1  e 2 )
e2 d 2
e2 d 2
2
,
. Then distance from the center to the ellipse to the focus
b

(1  e2 )2
(1  e2 )
is c where c 2  a 2  b 2 .
Let a 2 
c2  a 2  b2

e2 d 2
e2 d 2

(1  e 2 ) 2 (1  e 2 )

e4 d 2
(1  e 2 ) 2
 e2 d 

2 
 1 e 
So c 
2
e2 d
. So (0,0) is a focus of the ellipse. Note that the eccentricity of the ellipse is
1  e2
6
e2 d
c 1  e2

e.
ed
a
1  e2
Kepler’s Third Law
The derivative of the area is a constant
L
. Thus over an entire closed orbit of time T
2m
L
T   ab
2m
Therefore
2m
 ab
L
T
2 m  ed   ed 



L  1  e2   1  e2 
2 m (ed ) 2

L (1  e2 )3/2

But
L2
 ed or L  edGM m
GMm 2
So
T
2 m
(ed ) 2
2 3/2
edGM m (1  e )
2
(ed )3/2
2 3/2
GM (1  e )
2

a 3/2
GM

Or
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4 2 3
T 
a
GM
2
Newton’s Correction
Newton realized that mass m will pull on M as well. A more complete analysis goes as
follows.
m2
r2 − r1
r2
m1
r1
m1r1 G
m1m2
(r2  r1 )
| r2  r1 |3
m2 r2  G
m1m2
(r2  r1 )
| r2  r1 |3
Or
r1 G
m2
(r2  r1 )
| r2  r1 |3
r2  G
m1
(r2  r1 )
| r2  r1 |3
Subtract the first equation from the second:
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r2  r1 G
m1  m2
(r2  r1 )
| r2  r1 |3
Or
(r2  r1 )  G
m1  m2
(r2  r1 )
| r2  r1 |3
Therefore, taking into account that the larger mass is also affected by the smaller mass,
we see that the distance between bodies actually satisfies the same equation as above with
M replaced by m1  m2 .
The corrected analysis for the relative motion r2  r1 is essentially identical as the one
above with M replaced by m1  m2 , so Kepler’s Third Law becomes
4 2
T 
a3
G (m1  m2 )
2
Further analysis (which follows) shows for an outside observer both m1 and m1 orbit the
center of mass of the system in ellipses with equal periods.
The center of mass is defined to be rc 
m 1r1  m 2 r2
.
m1  m2
By adding the two equations
m1r1 G
m1m2
(r2  r1 )
| r2  r1 |3
m2 r2  G
m1m2
(r2  r1 )
| r2  r1 |3
we see that rc ″= 0. This means that the center of mass moves in a straight line at a
constant speed. It is an inertial reference system. Now let us change coordinates so that
the origin is at rc . So let r1c  r1  rc and r2c  r2  rc . The goal is to find the equations
that r1c and r2c satisfy.
First notice that r2  r1  r2c  r1c , so r2c  r1c satisfies exactly the same equation as r2  r1 ,
m  m2
(r2  r1 )
namely (r2  r1 )  G 1
| r2  r1 |3
.
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Next notice that
m1r1c  m2 r2 c  m1 (r1  rc )  m2 (r2  rc )
 m1r1  m1rc  m2 r2  m2 rc
 m1r  m2 r2  (m1  m2 )rc
 m1r  m2 r2  (m1  m2 )
m 1r1  m 2 r2
m1  m2
0
Also notice
r1c  r1  rc  r1 
r1  r2 
m 1r1  m 2 r2 m 2 r1  m 2 r2
m2


(r1  r2 ) and so
m1  m2
m1  m2
m1  m2
m1  m2
r1c .
m2
Also
r2 c  r2  rc  r2 
r2  r1 
m 1r1  m 2 r2 m 1r2  m 1r1
m1


(r2  r1 ) and so
m1  m2
m1  m2
m1  m2
m1  m2
r2 c
m1
So
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r1c''  (r1  rc )''
 r''
1  r''
c
= r''
1
=G
m2
(r2  r1 )
| r2  r1 |3
G
m2
m1  m2
r1c
m2
 G
 m1  m2 

r1c 
3 
m2


m 32
r
3 1c
(m1  m2 ) 2 r1c
and
r2 c''  (r2  rc )''
 r2''  r''
c
= r2''
= G
 G
 G
m1
(r2  r1 )
| r2  r1 |3
m1
m1  m2
r2 c
m1
3
 m1  m2 
r1c 

 m1

m 13
(m1  m2 ) 2 r2 c
3
r2 c
Notice that the original set of coupled differential equations are decoupled in this
reference system.
The conclusion is that the center of mass moves in a straight line at a constant speed and
the first object moves with respect to the center of mass as if a fictitious object of mass
m 32
were located at the center of mass and the second object moves with respect to
(m1  m2 )2
m 13
were located at the center
(m1  m2 )2
of mass. Notice that in practice you really only need to solve one equation because
m1r1c  m2 r2c  0 .
the center of mass as if a fictitious object of mass
See excellent animations at http://commons.wikimedia.org/wiki/File:Orbit1.gif
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