STEM 699 2014 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. (Newton’s Second Law) F = ma
2. (Newton’s Law of Universal Gravitation) Suppose a point of mass M is located at the
origin (0, 0) and a point of mass m is located at ( x, y ) . We assume (for time being) that
the point of mass M is fixed and doesn’t move. Let r denote the vector ( x, y ) , r denote
the length of this vector (i.e., r  x 2  y 2 ), r̂ be the unit vector in the direction of r (so
r
rˆ = ). Then the gravitation force that M exerts on m is given by
r
Mm
F  G 2 rˆ
r
where G is a universal constant, independent of M , m , and r.
Putting these two equations together we get
ma = G
Mm
GM
rˆ or equivalently a =  2 rˆ
2
r
r
1
or a  
GM r
GM
 3 r
2
r r
r
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, a 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
3
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
Conservation of Energy
Prop. For an object moving in with a central force whose magnitude varies according the
inverse square law, the quantity 12 mv 2  GMm is constant.
r
Proof. Note the dynamical variables v and r are scalars in the expression. v  || v ||  || r  ||
and r  || r || . We are going to differentiate, and one needs to be a little careful in doing
so.
d 1 mv 2  GMm  d  1 m || v ||2  GMm 
dt 2
r
dt  2
|| r || 




 d  12 m(v  v )  GMm1/2 
dt 
|| r  r || 
 1 m  2(a  v )  1 GMm3/2 2  r  v 
2
2 || r  r ||
 m(r  v )  GMm3/2  r  v 
|| r  r ||

  GM
 GM
3
r
r3
0
r  v
4
So the quantity 12 mv 2  GMm is constant. We will denote this quantity E.
r
Another Proof using x, y, and z coordinates.
d
dt

1
2



GMm

mv 2  GMm  d  12 m( x2  y2  z 2 ) 
2
2
2 
r
dt 
x

y

z


 12 m  2  ( xx  yy  zz)  12  2  2 GMm
( xx  yy  zz)
( x  y 2  z 2 )3/2

 ( x, y, z )   mx  2 GMm
x, my  2 GMm
y, mz   2 GMm
2
2 3/2
2
2 3/2
(
x

y

z
)
(
x

y

z
)
( x  y 2  z 2 )3/2

But
GM
x , y   2 GM
y , and z    2 GM
z.
2
2 3/2
2
2 3/2
(x  y  z )
(x  y  z )
( x  y 2  z 2 )3/2
So the second vector is 0 and the derivative is zero.
x  
2
A Derivation of Kepler’s First Law
We will use the previously derived result L  mr 2 k is constant.
denot the length of L , i.e., L  mr 2  .
As usual let use L to
We also need an expression for v 2 in polar coordinates:
r  r cos  i  r sin  j
So
v  dtd (r cos  )i  dtd (r sin  ) j
So
v2 = v  v
 ( dtd (r cos  )) 2  ( dtd (r sin  )) 2
 (r  cos   r sin   ) 2  (r  sin   r cos   ) 2
 r 2 cos 2   2rr   cos  sin   r 2 2 sin 2   r 2 sin 2   2rr   cos  sin   r 2 2 cos 2 
 r 2  r 2 2
Substituting L  mr 2  and v 2  r 2  r 2 2 into the expression E  12 mv 2  GMm .
r
5

z

E  12 mv 2  GMm
r
 12 m(r 2  r 2 2 )  GMm
r
2
 12 mr 2  12 mr 2 L2 4  GMm
r
mr
2
 12 mr 2  L 2  GMm
r
2mr
Now solve for r  :
2
r 2  2 E  2GM  L2 2
m
r
mr
Notice that if we take the square root of the previous expression we get a differential
equation with time as independent variable that one could solve. However we are
interested in the equation of the orbit, not the equation as a function of time. We would
like d or dr .
dr
d
L
d
d  dt 
mr 2
dr
dr
2 E  2GM  L2
dt
m
r
m2r 2
L
mr 2
d 
dr
2 E  2GM  L2
m
r
m2 r 2
This is an integral one can evaluate actually using simple, if awkward, Calculus. It
doesn’t look that way at first, but it really is straightforward. Notice that if you let u  1
r
the integral becomes
d 
adu
b  cu  du 2
for some constants a¸b, c, and d. This is an arccos integral. I have left the details of
evaluating in the integral in the appendix.
You get
6
1   cos(  0 )  
r
where

L2
GMm 2
and
2EL2  1  
G 2 M 2 m3
We usually take θ0 = 0 because θ0 is an arbitrary angle from which we start.
Kepler’s First Law
Case 1. e = 0.
Then
L2
r  
GMm 2
The orbit is a circle.
Now we convert to Cartesian coordinates:
r  er cos   
x 2  y 2  ex  
x 2  y 2  (  ex ) 2
x 2  y 2   2  2 ex  e 2 x 2
x 2 (1  e 2 )  2 ex  y 2   2
We now obtain Case 2: If e = 1, the orbit is a parabola.
Now assume e  1 . Complete the square to get
7
x 2 (1  e 2 )  2 ex  y 2   2
2 ex 

(1  e 2 )  x 2 
 y2   2
2 
1 e 


2 ex
 2e2  2
 2e2
2
(1  e 2 )  x 2 


y




1  e 2 (1  e 2 ) 2 
1  e2

e 
(1  e 2 ) 2   2 e 2

2
(1  e 2 )  x 

y


1  e2 
1  e2

2
e 
2

2
(1  e )  x 
 y 
1  e2 
1  e2

2
2
e 

x

y2
1  e2 


2
2
2
1


(1  e 2 ) 2
1  e2
If e > 1, then this last equation is the equation for a hyperbola. If e < 1, the equation is
the equation of an ellipse. Let us assume e < 1, the ellipse case, for the rest.
Let a 2 
2
, b2 
2
1  e2
(1  e2 )2
is c where c 2  a 2  b 2 .
. Then distance from the center to the ellipse to the focus
c2  a 2  b2


2
(1  e 2 ) 2

2
(1  e 2 )
 2  (1  e) 2  2
(1  e 2 ) 2
e2 2

(1  e 2 ) 2
e

1  e2
So c 
e
1  e2
. So (0,0) is a focus of the ellipse. Note that the eccentricity of the ellipse is
e
c 1  e2

e.

a
1  e2
8
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
T
2m
 ab
L
2 m      



L  1  e2   1  e2 
2 m
2

L (1  e2 )3/2

But
L2
  or L  GM m
GMm2
So
T
2 m
2
 GM m (1  e 2 )3/2
2
 3/2
2 3/2
GM (1  e )
2

a 3/2
GM

Or
4 2 3
T 
a
GM
2
9
Newton’s Correction
Newton realized that mass m will pull on M as well. A more realistic 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:
r2  r1 G
m1  m2
(r2  r1 )
| r2  r1 |3
10
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 . This describes the motion of one body relative to the other.
The corrected analysis for the relative motion r2  r1 is therefore essentially identical as
the one above with M replaced by m1  m2 , so Kepler’s Third Law becomes
T2 
4 2
a3
G (m1  m2 )
Aside: The equation
(r2  r1 )  G
m1  m2
(r2  r1 )
| r2  r1 |3
can be written as
m1m2
m1m2
(r2  r1 )  G
(r2  r1 )
m1  m2
| r2  r1 |3
m1m2
is known as the “reduced mass” and is typically denoted by μ. So
m1  m2
mm
the relative motion can be described a force of magnitude G 1 2 3 (r2  r1 ) on orbiting
| r2  r1 |
object of mass μ, the reduced mass, i.e.,
The quantity
 (r2  r1 )  G
m1m2
(r2  r1 )
| r2  r1 |3
11
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
.
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
12
Also
r2 c  r2  rc  r2 
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
r2  r1 
So
r1c''  (r1  rc )''
 r''
1  r''
c
= r''
1
=G
m2
(r2  r1 )
| r2  r1 |3
G
m2
m1  m2
r1c
m2
 G
3
 m1  m2 
r1c 

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 

m

1

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.
13
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
14
Appendix: Details of evaluating the integral
1
2
r
d 
dr
m 2 E  2GM  L2
L m
r
m2 r 2
1
r2

dr
2mE  2GMm2  1
L2
L2 r
r2
Complete square in the denominator
1
2
r
d 
2mE  G 2 M 2 m 4   1  GMm 2 
r

L2
L4
L2 

2
dr
Let

L2
GMm 2
d 
1
r2
2mE  1  1  1
L2
2 r 


2
dr
Finally let
2mE  1   2
L2
2 2
so the integral becomes
d 
1
r2
2  1  1
2 r 


2
dr
Note for future reference, that
15
2mE  1   2
L2
2mE
L4
1   2
2
2
2 4
L G M m
2 EL2  1   2
G 2 M 2 m3
2 EL2  1  
G 2 M 2 m3
Going back to the integral:
1
2
r
d 
dr
2
2
  11
2 r 
1
r2
d 
dr
2
 1  1  1

 r 




Let
u 11
 r 
so
du    12 dr
 r
The integral becomes:
d  du
1 u2
So
  0  arccos(u)
Or
cos(  0 )  u
cos(  0 )   1  1
 r 
 cos(  0 )    1
r
16
1   cos(  0 )  
r
which is the polar equation for a conic with eccentricity ε
It turns out that as mentioned above,
2
1
a  2  L 2
  GMm
2E
1 
GMm  2 EL2
2
2 3
G M m
so
E   GMm
2a
i.e., the total energy of an object in orbit is a function only of its semi-major axis.
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