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.
A corollary of this result is that the motion of m2 must lie in a single plane determined by
the normal vector L .
Areas swept out are constant (Kepler’s Second Law)
The key challenge here is express the area swept out in a given amount of time. The
most convincing way I know is to express it in polar coordinates based at the origin.
Define r = r (cos  i  sin  j ) . NOTE: r and  are functions of time.  is relative to
some (arbitrary) fixed ray.
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
2
L = v = r × mv
= r (cos  i  sin  j ) × m  r (cos  i  sin  j )  r ( sin  i  cos  j ) 
 mr 2 k
Hence
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
3
r
.
r
 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
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.
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Then
L2
r
GMm2
The orbit is a circle.
Now let
L2
 ed
GMm 2
Then
ed  r (1  e cos  )
Convert to Cartesian coordinates:
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 
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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
e2 d
c 1  e2

e.
ed
a
1  e2
Kepler’s Third Law
Assume The derivative of the area is a constant
L
. Thus over an entire closed orbit of
2m
time T
L
T   ab
2m
Therefore
T
2m
 ab
L
2 m  ed   ed 



L  1  e2   1  e2 
2 m (ed ) 2

L (1  e2 )3/2

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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
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.
m
r2 − r1
r2
M
r1
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Mr1 G
Mm
(r2  r1 )
| r2  r1 |3
mr2  G
Mm
(r2  r1 )
| r2  r1 |3
Or
r1 G
m
(r2  r1 )
| r2  r1 |3
r2  G
M
(r2  r1 )
| r2  r1 |3
Subtract the first equation from the second:
r2  r1 G
M m
(r2  r1 )
| r2  r1 |3
Or
(r2  r1 )  G
M m
(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 m + M.
The corrected analysis is essentially identical as the one above. M is replaced by
M + m. Kepler’s Third Law becomes
4 2
T 
a3
G( M  m)
2
Further analysis (which is essentially the same analysis as above) shows for an outside
observer both M and m orbit the center of mass of the system in ellipses with the same
period.
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See excellent animations at http://en.wikipedia.org/wiki/Center_of_mass
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