Physics 430: Lecture 19
Kepler Orbits
Dale E. Gary
NJIT Physics Department
8.6 The Kepler Orbits

Last time we derived a general equation for the path of a body in the 2-body
central force problem:
u(f )  -u (f ) -
m
F (r ).
2
2
l u (f )
where, through a change of variables we substituted u = 1/r.
 This is obeyed for any central force F(r), but let’s look specifically at the
gravitational case (the Kepler problem), where, using g = Gm1m2, we have
F (r )  -

g
 -g u 2 .
2
r
Inserting this into the path equation, we have the simpler, linear equation
u(f )  -u(f )  gm / l 2 .

The solution can be found by one last substitution, w(f) = u(f) - gm/l2, which
transforms the equation into our old friend
Al 2
l2

c
w(f )  - w(f ),
gm
gm
again with solution w(f) = A cos(f - d). We will choose coordinates for which
gm
1
d = 0, so the final solution, then, is
u (f )  2  A cos f  1   cos f  .
l
c
November 5, 2009
The Final Kepler Path

Finally, substituting for u = 1/r, we have
u (f ) 
1
1   cos f 
c

r (f ) 
c
.
1   cos f
Bounded Orbits





Al 2
The dimensionless constant  
is going to play a big role in the shape of
gm
the orbit, depending on whether it is greater or less than 1.
If  < 1, then the denominator is always positive for any value of f.
If  > 1, there is a range of values of f for which the denominator vanishes,
and r blows up (the object is unbound).
So  = 1 is the demarcation between bound and unbound orbits. Because we
want to talk about bound orbits, we will first take  < 1.
In the above equation, as cos f oscillates between -1 and 1, the orbital
distance r varies between
c
c
rmin =perihelion (perigee)
rmin 
and rmax 
.
rmax = aphelion (apogee)
1 
1- 
November 5, 2009
Bounded Orbits, cont’d
The shape of the orbit, then, looks like the figure at right.
 We now want to prove that this shape is an ellipse, and
to do that you will show in HW Prob. 8.16 that
planet

r (f ) 
can be written in the form:
c
1   cos f
x  d
where
star
a
c
f
d O
b
2
y2
 2  1.
a2
b
c
c
a
;
b

; d  a .
2
2
1- 
1- 
The graphical meanings of a, b, c and d are shown in the figure. Here a is
called the semi-major axis (half the longer axis) and b is the semi-minor axis.
 The constant  is the eccentricity of the ellipse, and can be determined
from
b
2


a
 1-  .
Notice that as   0, d goes to zero, a and b become equal, and the ellipse
becomes a circle. As   1 , d  a , a   and b/a  0, and the ellipse
grows long and skinny (i.e. very eccentric).
November 5, 2009
Example 8.4: Halley’s Comet

Statement of the problem:


Halley’s comet follows a very eccentric orbit, with  = 0.967. Given that the closest
approach to the Sun (perihelion) is 0.59 AU (astronomical units), what is its
greatest distance from the Sun?
Solution:

Notice that rmax/rmin = (1 + )/(1 – ). Therefore
rmax 
1 
1.967
rmin 
rmin  60 rmin  35 AU.
1- 
0.033
Orbital Period (Kepler’s third law)


Recall that Kepler’s second law (Chapter 3) states that the line between the
Sun and a planet sweeps out equal areas in equal times, and is related to
angular momentum l by
dA
l

.
dt 2m
Since the total area of an ellipse is A = pab, the period is
A
2p abm


.
dA / dt
l
November 5, 2009
Orbital Period-2

Squaring this and using the definitions of b and c given earlier:

3
a3 c m 2
2 a m
  4p
 4p
.
2
g
l
Recall that, for the Sun, g = Gm1m2  GmMsun, which then gives Kepler’s third
2
law:
4
p
2 
a3 .
GM sun
2

2
What is interesting is that this does not depend on the mass of the satellite,
so the law is obeyed for all bodies (planets, comets, asteroids) so long as
they do not get too massive relative to the Sun.
Example 8.5: Period of Low-Orbit Earth Satellite

Use Kepler’s third law to estimate the period of a satellite in a circular orbit
close to Earth (a few 10’s of miles up).
2
6
R
4
p
6.38

10
m
3
2 
REarth
   2p Earth  2p
 5070 s  85 min.
2
GM Earth
g
9.8 m/s
November 5, 2009
Relation between Energy and 

There is an important relation between the eccentricity and the energy of the
orbit. To find this, think about the effective potential energy curve that we
saw last time. At the closest approach (inner turning point), rmin, the total
energy E = Ueff(rmin) (of course, we could alternatively use rmax):
g
l2
E  U eff (rmin )  
2
rmin 2m rmin
1

2rmin

 l2

2
g

.
 m rmin

Recalling that rmin = c/(1 + ), and substituting c = l2/gm, we have:
rmin

l2

.
gm (1   )
Putting this into the above equation for energy, after some algebra:
g 2m 2
E  2  - 1 .
2l
valid for any eccentricity
November 5, 2009
8.7 The Unbound Kepler Orbits

Going back to our original equation for the path r (f ) 
c
,
1   cos f
let’s now consider the case   1 (which corresponds to E  0). In this case,
the denominator blows up at some values of f, hence the orbits are
unbound.
 For the special case  = 1 (which corresponds to E = 0), we can convert the
above to the cartesian form y 2  c 2 - 2cx, which is an equation for a
parabola. For a parabola, the legs of the parabola eventually go parallel and
at infinite distance they approach but never quite reach f  p .
 For  > 1, the denominator blows up for some other value of f, such that
 cos fmax  -1.

In this case, it can be shown that the cartesian form is a hyperbola:
( x - d )2 y 2
- 2  1,
2


where the legs go out at angles ±fmax. (the angles of the asymptotes).

>1
=1
The geometrical relationships are shown in the summary plot.
November 5, 2009
0
<1

Summary of Kepler Orbits

Important relations of Kepler orbits are:
c
r (f ) 
1   cos f
E 
g m 2
 - 1 .
2 
2l
>1
path equation
(hyperbola)
=1
2
eccentricity
energy
orbit
=0
E<0
circle
0<<1
E<0
ellipse
=1
E=0
parabola
>1
E>0
hyperbola
l2
c
Gm1 m2 m
(parabola)
energy equation
=0
(circle)
<1
(ellipse)
scale factor for orbit
November 5, 2009
8.8 Changes of Orbit

Let’s first give the general approach to finding changes in the elliptical orbit
of, say, a spacecraft orbiting Earth. The most general way of writing the
c1
path equation is
r (f ) 
1  1 cos f - d1 
where d is some inclination angle of the elliptical orbit, and the constants c, 
and d are written with subscript 1 to indicate their initial values.
 To change its orbit, a spacecraft can fire its engine in some particular
direction for a brief time, thus causing an instantaneous change in velocity.
From the change in velocity, we can calculate its new total energy and
angular momentum, and thus calculate a new c2 and 2. The new orbit and
the old orbit have to agree for some particular ro and fo, where the
spacecraft was when its velocity changed, so we can calculate the remaining
c1
c2
quantity d2 by
1  1 cos fo - d1 

1   2 cos fo - d 2 
.
Hopefully you can see that this is straightforward, although tedious.
 We can do a simple but interesting problem—a tangential thrust at perigee.
Say the velocity changes from v1 to v2 = lv1.

November 5, 2009
Tangential Thrust at Perigee

At perigee, cos(f – d) = 1, so the “continuity” between orbits gives
c1
c
 2 .
1  1 1   2

Because l is proportional to velocity, and the constant c is proportional to l,
we have c2  l 2 c1 . Therefore
 2  l 21   l 2 - 1 .
If l > 1 (increase in velocity), the new
P
orbit has a higher eccentricity, and a
higher angular momentum and higher
energy.
 Likewise, if l < 1 (decrease in velocity), the new orbit is more circular, has a
smaller angular momentum, and lower energy.
 But what happens when the initial orbit is already circular (1 = 0), and we
decrease the velocity?

November 5, 2009
P
Example 8.6: Changing between
Circular Orbits
If we want to go from, say, Earth to Mars, we have to boost out of an
essentially circular Earth orbit onto an ellipse that takes us to Mars, and then
go into another circular orbit to match that of Mars. The situation is shown
in the figure at right.
2
 This is called a Hohman transfer ellipse, and is the transfer
P
orbit that takes the least energy. Let’s calculate the velocity P’
1
changes needed to do this transfer.
3
 There are three orbits involved, two with zero eccentricity
(1 = 0, 3 = 0), with orbital radii R1 and we’ll assume R3 = 2R1.
 The first match between orbits 1 and 2 requires

c1
c2
l 2 c1

 c1 
  2  l 2 - 1.
1  1 1   2
1  2

The second match between orbits 2 and
3 requires
2
c3
c2
l c1
l 2 R1

 c3 
 2 R1 
.
2
2
1 - 2 1  3
2-l
1- l -1



This is easily solved for l to give l  4 / 3  1.15. Need 15% of Earth orbital speed
November 5, 2009