Physics 320 

Prof. Dale E. Gary
NJIT 
Orbital Mechanics II
B: Orbits and Energy
1. Conservation of Energy
Recall the concepts of potential energy U, and kinetic energy K, the sum of which gives the total energy2. Total Energy for any OrbitE = K + Uwhere K = 1/2 mv^{2}. But what is the potential energy appropriate to our planetary system?Remember that only differences in potential energy are important. If we raise a mass at the surface of the Earth by a distance h, we do work against gravity
r_{2} W = F ^{.} dr = mgDr = mgh r_{1} and the negative of this work is the change of potential energy
Raising a mass from the floor to the desk raises the potential energy by mgh, and raising it by the same height from the desk to a shelf also raises U by mgh  only the difference DU matters. Thus, we are free to choose our zero of potential energy anywhere we wish.
r_{2} DU = W =  F ^{.} dr = mgh r_{1} We will choose DU = 0 at r = infinity, which means the potential energy anywhere in the system is negative. Now what is the potential energy at position r? We first place a test mass of mass m at infinity, and then move it radially inward to distance r from the center (from mass M) with the force of gravity acting all the way along this path. We have
r GMm U =   r^{ }^{.}dr inf r^{2} which evaluates to
r dr GMm  r U = GMm =   inf r^{2} r  inf or finally,
U = GMm/r (Gravitational potential energy)As advertised, the potential energy is negative for any r, and approaches zero as r approached infinity. The total energy for an elliptical orbit, then, isE = K + U = 1/2 mv^{2} GMm/rwhere we have used the vis viva equation (which is why this is valid only for elliptical orbits), and we have made the approximation that M >> m.
= 1/2 mG(M+m)[2/r 1/a]  GMm/rE =  GMm/2a
Note that the total energy is negative, and is just a constant. Thus, energy is conserved along the orbit, as of course it must be.
Now we will examine the total energy for any orbit, not limited to elliptical orbits. To do this, we need to use conservation of angular momentum. It is possible to show, although we will not do so, that the angular momentum and eccentricity are related by3. Planetary Motion and Effective PotentialL^{2} / GMm^{2}r = 1 + e cos q,so solving for r, we getr = (L^{2} / GMm^{2}) / (1 + e cos q).Note that this has the same form as the general expression for the polar equation for a conic section. Let us now repeat the calculation of total energy in the same way as before:v_{r} = dr/dt = (L^{2} / GMm^{2}) d/dt [(1 + e cos q)]^{1}Recall that L = r x p = mrv_{q}, so that rv_{q} = L/m. But since v_{q} = r dq/dt, we have r^{2}dq/dt = L / m, so. Putting this into the above equation,
= (L^{2} / GMm^{2}) e sin q dq/dt / (1 + e cos q)^{2}v_{r} = (GMm / L) (e sin q).The corresponding perpendicular component of the velocity is even simpler to derivev_{q} = r dq/dt = r^{2}dq/dt / r = L / rm = (GMm / L) (1 + e cos q).So the total velocity isv^{2} = v_{r}^{2} + v_{q}^{2} = (GMm / L)^{2} (1 + 2e cos q + e^{2})and finally the kinetic energy isK = 1/2 mv^{2} = 1/2 m(GMm / L)^{2} (1 + 2e cos q + e^{2})while the potential energy isU = GMm/r = (GMm)^{2}m/L^{2}(1 + e cos q)We finally come to the rather simple expression for total energy, for an orbit of the form of any conic section:
= 1/2 m(GMm / L)^{2} (2 + 2e cos q)E = K + U = 1/2 m(GMm / L)^{2 }(e^{2 } 1) (Total energy for any orbit)It is instructive to solve this equation for the eccentricity, to gete = [1 + 2L^{2}E / (GMm)^{2}m]^{1/2}In particular, note that
 for E > 0, we have e > 1 ==> hyperbola
 for E = 0, we have e = 1 ==> parabola
 for E < 0, we have e < 1 ==> ellipse
We can consider the kinetic energy as having a radial part and an angular part:K = 1/2 mv^{2} = 1/2 mv_{r}^{2} + 1/2 mv_{q}^{2}where v_{q}^{2} = L^{2} / m^{2}r^{2} ==> K_{q} = 1/2 mv_{q}^{2 }=1/2 L^{2} / mr^{2}. Consider a body moving inward through the solar system. As r decreases, the angular kinetic energy increases for smaller r due to conservation of angular momentum. Since the total energy is conserved, this increasing angular kinetic energy comes in part from the radial kinetic energy. The system acts as though there were a radial force opposing the inward motion. Let us write the total energy as
K = K_{r} + K_{q}E = K_{r}+ K_{q} + U(r) = 1/2 mv_{r}^{2} + 1/2 L^{2} / mr^{2}  GMm/r = K_{r}+ U_{eff}so we call this combination an effective potential. Graphically, we have:There is a nice way to interpret orbits using this diagram. Since E = constant on any orbit, if E < 0 we have a bound orbit, with two turning points where K_{r}= 0, r_{min} = r_{peri}Note that any orbit with E < 0 has two turning points. For E = 0, there is only one turning point, and the object reaches infinity with zero energy. Finally, for E > 0, there is again one turning point, but the object reaches infinity with energy left over.
r_{max} = r_{ap}Effective potential in general relativity (strongly curved spacetime).