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  = Gm 1 m 2, 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(  )  / l 2, 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

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 r min =perihelion (perigee) r max = 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 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  O a b c d planet star

November 5, 2009  Statement of the problem: Halley’s comet follows a very eccentric orbit, with  = 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 r max /r min = (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

November 5, 2009  Squaring this and using the definitions of b and c given earlier:  Recall that, for the Sun,  = Gm 1 m 2  G  M sun, 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

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), r min, the total energy E = U eff (r min ) (of course, we could alternatively use r max ):  Recalling that r min = c/(1 +  ), and substituting c = l 2 / , we have:  Putting this into the above equation for energy, after some algebra: Relation between Energy and  valid 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  >1 =1 <1 0

November 5, 2009  Important relations of Kepler orbits are: Summary of Kepler Orbits  = 0 (circle)  < 1 (ellipse)  = 1 (parabola)  > 1 (hyperbola) path equation energy equation eccentricityenergyorbit  = 0 E < 0circle 0 <  < 1 E < 0ellipse  = 1 E = 0parabola  > 1 E > 0hyperbola 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 c 2 and  2. The new orbit and the old orbit have to agree for some particular r o 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 v 1 to v 2 = v Changes of Orbit

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 new orbit has a higher eccentricity, and a higher angular momentum and higher energy.  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 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 transfer orbit that takes the least energy. Let’s calculate the velocity changes needed to do this transfer.  There are three orbits involved, two with zero eccentricity (  1 = 0,  3 = 0 ), with orbital radii R 1 and we’ll assume R 3 = 2R 1.  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 P P’ Need 15% of Earth orbital speed