Physics 430: Lecture 19 Kepler Orbits

Dale E. Gary

NJIT Physics Department

November 5, 2009

Last time we derived a general equation for the path of a body in the 2-body central force problem:

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 = Gm1m2, we have

Inserting this into the path equation, we have the simpler, linear equation

The solution can be found by one last substitution, w() = u() /l2, which transforms the equation into our old friend

again with solution w() = A cos(). We will choose coordinates for which = 0, so the final solution, then, is

8.6 The Kepler Orbits

2 2( ) ( ) ( ).

( )u u F r

u

l

22

( ) .F r ur

2( ) ( ) / .u u l

( ) ( ),w w

2

1( ) cos 1 cos .u A

c

l

2A

l2

c

l

November 5, 2009

Finally, substituting for u = 1/r, we have

Bounded Orbits The dimensionless constant is going to play a big role in the

shape of

the orbit, depending on whether it is greater or less than 1. If < 1, then the denominator is always positive for any value of . If > 1, there is a range of values of 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 oscillates between 1 and 1, the

orbital distance r varies between

The Final Kepler Path 1

( ) 1 cos ( ) .1 cos

cu r

c

2A

l

min max and .1 1

c cr r

rmin =perihelion (perigee)rmax = aphelion (apogee)

November 5, 2009

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

can be written in the form:

where

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

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).

Bounded Orbits, cont’d

( )1 cos

cr

2 2

2 21.

x d y

a b

2 2; ; .

1 1

c ca b d a

21 .b

a

Oa

b

c

d

planet

star

November 5, 2009

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

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

Since the total area of an ellipse is A = ab, the period is

Example 8.4: Halley’s Comet

max min min min

1 1.96760 35 AU.

1 0.033r r r r

.2

dA

dt

l

2.

/

A ab

dA dt

l

November 5, 2009

Squaring this and using the definitions of b and c given earlier:

Recall that, for the Sun, = Gm1m2 GMsun, which then gives Kepler’s third law:

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).

Orbital Period-2

3 2 32 2 2

24 4 .

a c a

l

22 3

sun

4.a

GM

2 62 3 Earth

Earth 2Earth

4 6.38 10 m2 2 5070 s 85 min.

9.8 m/s

RR

GM g

November 5, 2009

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):

Recalling that rmin = c/(1 + ), and substituting c = l2/, we have:

Putting this into the above equation for energy, after some algebra:

Relation between Energy and

2

min 2min min

2

min min

( )2

12 .

2

effE U rr r

r r

l

l

2

min .(1 )

r

l

2

22

1 .2

E

lvalid for any eccentricity

November 5, 2009

Going back to our original equation for the path ,

let’s now consider the case 1 (which corresponds to E 0). In this case, the denominator blows up at some values of , hence the orbits are unbound.

For the special case = 1 (which corresponds to E = 0), we can convert the above to the cartesian form 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

For > 1, the denominator blows up for some other value of , such that

In this case, it can be shown that the cartesian form is a hyperbola:

where the legs go out at angles ±max. (the angles of the asymptotes). The geometrical relationships are shown in the summary plot.

8.7 The Unbound Kepler Orbits

2 2 2 ,y c cx

2 2

2 2

( )1,

x y

( )1 cos

cr

.

maxcos 1.

>1

=1

<10

November 5, 2009

Important relations of Kepler orbits are:

Summary of Kepler Orbits

( )1 cos

cr

= 0(circle)

< 1(ellipse)

= 1(parabola)

> 1(hyperbola)path equation

2

22

1 .2

E

lenergy equation

eccentricity

energy orbit

= 0 E < 0 circle

0 < < 1 E < 0 ellipse

= 1 E = 0 parabola

> 1 E > 0 hyperbola

2

1 2

cGm m

l

scale factor for orbit

November 5, 2009

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 path equation is

where is some inclination angle of the elliptical orbit, and the constants c, and 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 o, where the spacecraft was when its velocity changed, so we can calculate the remaining quantity 2 by

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 = v1.

8.8 Changes of Orbit

1

1 1

( )1 cos

cr

1 2

1 1 2 2

.1 cos 1 coso o

c c

November 5, 2009

At perigee, cos( – ) = 1, so the “continuity” between orbits gives

Because l is proportional to velocity, and the constant c is proportional to l, we have Therefore

If > 1 (increase in velocity), the neworbit has a higher eccentricity, and ahigher angular momentum and higherenergy.

Likewise, if < 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?

Tangential Thrust at Perigee

1 2

1 2

.1 1

c c

22 1.c c

2 22 1 1 .

P P

November 5, 2009

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.

This is called a Hohman transfer ellipse, and is the transferorbit that takes the least energy. Let’s calculate the velocitychanges needed to do this transfer.

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

The second match between orbits 2 and 3 requires

This is easily solved for to give

Example 8.6: Changing between Circular Orbits

221 2 1

1 21 2 2

1.1 1 1

c c cc

PP’1

2

3

2 2

32 1 13 1 22

2 3

2 .1 1 21 1

cc c Rc R

4 / 3 1.15. Need 15% of Earth orbital speed