Physics 320 

Prof. Dale E. Gary
NJIT 
Orbital Mechanics I
A: Physical Interpretation of Kepler's Laws
Kepler's first law states that the planets move in elliptical orbits around the Sun, with the Sun at one focus. Elliptical orbits are indeed a property of inverse square law central forces, as we will show shortly.
Let us examine Kepler's second and third laws in view of Newton's Law of Universal Gravitation.
1. Law of Areas and Angular Momentum
Kepler's second law states that the radius vector between the Sun and an orbiting planet sweeps out equal areas in equal times. Consider a planet moving along its elliptical orbit at a distance r, with velocity v, as in the figure below.2. Kepler's Third LawAfter a time Dt, it moves an angular distance from point P to point Q of Dq = v_{q }Dt / r .During this time, the radius vector sweeps out the triangle FPQ, whose area is DA = 1/2 rv_{q }Dt, soAccording to Kepler's law, dA/dt = constant, and in particular after one complete period P, the area swept out is the total area of the ellipse,
DA dA r v_{q} lim = = = 1/2 r^{2}q' t > 0 Dt dt 2 dA/dt = A/P = pab/P = constant = rv_{q}/2.There are two places in its orbit where the radial velocity, v_{r}, of a planet goes to zero, and it has only v = v_{q }these are at aphelion and perihelion. At these locations, the speeds obey the relationv = 2A/Pr = 2pab/Prbut at perihelion, r = a(1  e) and at aphelion r = a(1 + e), sov_{peri} = 2pab/Pa(1  e)but remember our relation b = a (1  e^{2})^{1/2}, so these become
v_{ap} = 2pab/Pa(1 +e)v_{peri} = (2pa/P)[(1 +e)/(1  e)]^{1/2}v_{ap} = (2pa/P)[(1  e)/(1 + e)]^{1/2}
Example: What are v_{peri}and v_{ap} for Earth orbit?P = 365.26 days = 3.156 x 10^{7} sAgain, this result shows that planets move faster near the Sun, but the Earth's orbit is so nearly circular that the speed does not change much. App for planetary orbits.
e = 0.0167, a = 1 AU = 1.496 x 10^{8} km
v_{peri }= (2pa/P)[(1 +e)/(1  e)]^{1/2}= 30.28 km/s
v_{ap }= (2pa/P)[(1  e)/(1 + e)]^{1/2}= 29.28 km/sAngular Momentum:
What does all of this have to do with angular momentum? Recall that angular momentum is a measure of rotational motion about a center of rotationusually the center of mass (but if an object is "pinned," the center of rotation is about that pinning point).
This system has zero angular momentum This system has nonzero angular momentumThe angular momentum is given by
L = r x pwhere p = mv is the linear momentum. In magnitude, this is L = L = rp sin q = rp_{perp}. But in polar coordinates, p_{perp}= mv_{q}, soso this is the appropriate expression for the angular momentum of a planet about the Sun. The key is to examine how the angular momentum changes around the orbit, i.e.,
but v x p = mv x v = 0
dL/dt = r x F (this is the torque on the planet)For any central force, in particular for Newton's Law of Universal Gravitation, where F =  (GMm/r^{2}) r, we are going to have r x F = 0 also! Thus,dL/dt = 0; so L = constant.In fact, from the above expression, L = mrv_{q}. Finally, we see that the statement of Kepler's second law is that same as the statement of conservation of angular momentum:dA/dt = L/2m = constant
For the general case of two masses interacting according to Newton's Law of Universal Gravitation, the two masses actually orbit about the center of mass of the system, not necessarily the center of the more massive body.Recall the equation for center of mass r_{cm} = Sm_{i}r_{i }/ Sm_{i}
For a two mass system, we will refer to the separation of the two masses as a = r_{1} + r_{2}, where r_{1} is the distance of mass m_{1} from the center, and r_{2} is the distance of mass m_{2} from the center. Consider the case when the two masses are in circular orbits. During their motion, the two planets must be acted on by a centripetal force given by3. Orbital VelocityF_{1} = m_{1}v_{1}^{2}/r = 4p^{2}m_{1}r_{1 }/ P^{2}andF_{2} = m_{2}v_{2}^{2}/r = 4p^{2}m_{2}r_{2 }/ P^{2}where we have used v = 2pr / P. Now, by Newton's third law, these two forces must be equal in magnitude (and opposite in direction), which means m_{1}r_{1 }= m_{2}r_{2 }. This actually proves that the center of the circular motion is the center of mass. From this and a = r_{1} + r_{2}, we haver_{1} = [m_{2}/(m_{1}+ m_{2})]a.Also, by Newton's Law of Universal Gravitation we have the expression for the force:F_{1} = F_{2} = F = Gm_{1}m_{2}/a^{2},so using the expression for F_{1}, we haveGm_{1}m_{2}/a^{2} = 4p^{2}m_{1}r_{1 }/ P^{2} = (4p^{2}m_{1 }/ P^{2}) [m_{2}/(m_{1}+ m_{2})] a,orP^{2} = [4p^{2}/G(m_{1}+ m_{2})]a^{3}which, as promised, is the expression corresponding to Kepler's third law.Note that the center of mass is also called the barycenter. The two masses orbit the barycenter with the same periodyou use the separation between the masses, a, not the distances of the masses r_{1} and r_{2} from the center of mass, to determine the period.
We will now use these results to derive a particularly simple equation for the orbital velocity for any point on an elliptical orbit. Since most orbits are elliptical, this will be a very useful equation.We decompose the velocity into its two components:
v_{r} = dr/dt = r' and v_{q} = r (dq/dt) = rq' Going back to our equation for an ellipse:
r = a(1  e^{2}) / (1 + e cos q)we can explicitly take the derivative and get the radial component of the velocity asv_{r} = dr/dt = a(1  e^{2}) d/dt [(1 + e cos q)]^{1}But note that earlier we had rv_{q} = r^{2}dq/dt = 2pab/P = 2pa^{2}(1  e^{2})^{1/2}/P, so
= ae(1  e^{2}) sin q dq/dt / (1 + e cos q)^{2}dq/dt = 2pa^{2}(1  e^{2})^{1/2}/Pr^{2}
Substitution of this into the equation for v_{r}, givesWhat have we learned?v_{r} = ae(1  e^{2}) sin q [2pa^{2}(1  e^{2})^{1/2}/Pr^{2}]/ (1 + e cos q)^{2}The corresponding perpendicular component of the velocity is
= [2pa / P(1  e^{2})^{1/2}] (e sin q).v_{q} = r dq/dt = 2pa^{2}(1  e^{2})^{1/2}/PrWe simply sum the squares of these components to get the total magnitude of the velocity
= [2pa / P(1  e^{2})^{1/2}] (1 + e cos q).v^{2} = v_{r}^{2} + v_{q}^{2} = (2pa / P)^{2} (1 + 2e cos q + e^{2}) / (1  e^{2}).It is useful to substitute from the equation of an ellipse for the quantity e cos q:e cos q = a(1  e^{2})/r  1which gives:v^{2} = (2pa / P)^{2 }[(2a/r)(1  e^{2}) + e^{2}  1] / (1  e^{2}) = (2pa / P)^{2 }(2a/r  1).Finally, from Kepler's third law, P^{2} = [4p^{2}/G(m_{1}+ m_{2})]a^{3}, we havev^{2} = [(4p^{2}a^{2}) G(m_{1}+ m_{2}) / 4p^{2}a^{3}](2a/r 1) = G(m_{1}+ m_{2}) (2/r  1/a)This final equation for the velocity of an elliptical orbitis called the vis viva equation.
v^{2}=G(m_{1}+ m_{2}) (2/r  1/a)
We found that Kepler's second law (Law of Equal Areas), is equivalent to conservation of angular momentum L = mrv_{q}, so that dL/dt = 0 for any orbit. This is a consequence of the central force nature of the gravitational forceonly a perpendicular force could change a bodies' angular momentum, and since there is none, the angular momentum cannot change. We obtained simple expressions for the speed of a planet or other orbiting body at perihelion and aphelion:v_{peri} = (2pa/P)[(1 +e)/(1  e)]^{1/2}We also noted, using Newton's third law (Law of equal action and reaction), that two bodies orbit their combined center of mass (the barycenter) rather than the center of either body. From this and Newton's Law of Universal Gravitation (F = Gm_{1}m_{2}/a^{2}), we proved Kepler's third law in its quantitative form:
v_{ap} = (2pa/P)[(1  e)/(1 + e)]^{1/2}.P^{2} = [4p^{2}/G(m_{1}+ m_{2})]a^{3}.Applying Kepler's third law, we were able to obtain a more general equation for orbital speed, valid at any point in the orbit, the vis viva equation:v^{2}=G(m_{1}+ m_{2}) (2/r  1/a).