Draft:Gravitational Orbits

Gravitational Orbits For a single planet orbiting a fixed Sun (Kepler orbit) the acceleration of the planet is given by Newton’s Law of Gravity as(1) a = –GMr/r3 (1) where G is the constant of gravity, M is the mass of the Sun, and r is the position of the planet relative to the Sun. Equation (1) is a second order differential equation for the position. To solve this equation look for constants of motion. The first is the specific angular momentum vector, h, (angular momentum per unit mass) of the planet relative to the Sun: h = r×v (2) where v is the velocity of the planet relative to the Sun. Conservation of the specific angular momentum follows from equation (1): dh/dt = v×v + r×a = 0 + r×(-GMr/r3) = 0 (3) Both r and v are perpendicular to h so the motion is in a plane, the orbit plane: r·h = 0 v·h = 0 (4) The second constant of motion is the eccentricity vector, e, which is the dimensionless form of the Laplace-Runge-Lenz vector(2): e = v×h/GM – r/r (5) Conservation of the eccentricity vector follows from equations (1) and (3): de/dt = a×h/GM + v×(dh/dt)/GM – v/r – rd(1/r)/dt 

   	= –r×(r×v)/r3 + 0 – v/r – r[d(1/r)/dr](dr/dt)

= –(r·v)r/r3 + (r·r)v/r3 – v/r + (r/r2)(v·r/r) = 0 (6) Note that e is perpendicular to h h·e = h·v×h/GM – (r×v)·r/r = 0 (7) so the eccentricity vector is in the orbit plane. To obtain the path equation, define the orbit angle, , as the angle between the position vector and the eccentricity vector. Then r·e = recos and r·e = r·v×h/GM – r·r/r = h2/GM – r (8) which requires r + recos = h2/GM (9) Equation (9) yields the path equation r = p/(1+ecos) p  h2/GM (10) The eccentricity, e, is the magnitude of the eccentricity vector and p is the semi-latus rectum (perpendicular radius); the radius for  = 90o. For  = 0 the radius is a minimum (periapsis) and then the position vector is parallel to the eccentricity vector. So, the eccentricity vector points toward periapsis. Using h×(e+r/r) = h×(v×h)/GM = h2v/GM – (h·v)h/GM = pv – 0 v = h×(e+r/r)/p (11) relates the velocity vector to the position vector.

To find the relation between orbit angle and time consider |h| = |r×v| = rv = r(rd/dt) (12) where v is the component of the velocity vector perpendicular to the position vector. From equation (10), equation (12) becomes (GMp)1/2 = [p/(1+ecos)]2d/dt (GM/p3)1/2t = d/(1+ecos)2 (13) The integral in equation (13) is evaluated using a change of variable. For an elliptic orbit tan(/2) = [(1-e)/(1+e)]1/2tan(/2) (GM/p3)1/2t = ( – esin)/(1-e2)3/2 (14) and for a hyperbolic orbit tanh(/2) = [(e-1)/(e+1)]1/2tan(/2) (GM/p3)1/2t = (esinh – )/(e2-1)3/2 (15) The parabolic case is treated separately. Equation (14) is Kepler’s equation extended by Newton’s Law of Gravity. Kepler derived equation (14) as a proportional equation using his First and Second Laws and a geometric construction. For the elliptic case,  is the true anomaly,  is the eccentric (central) anomaly, and ( – esin) is the mean anomaly with equation (14) expressed in terms of the semi-major axis rather than the semi-latus rectum.

A three-dimensional analysis may be introduced using unit vectors as orthogonal coordinates: h’= h/h e’ = e/e p’ = h’×e’ e’ → x p’ → y h’ → z (16) Then r = [p/(1+ecos)](e’cos + p’sin) v = (GM/p)1/2[p’(e+cos) – e’sin] (17) From equation (17), the energy is E = mv2/2 – GMm/r = (GMm/2p)(e2-1) (18) Equation (10) determines the semi-latus rectum from the specific angular momentum and then equation (18) determines the eccentricity from the energy. Other equations that may be of interest are vr = (GM/p)1/2esin v = (GM/p)1/2(1+ecos) (19) which are the radial and transverse components of the velocity.

References (1) J. B. Marion, Classical Dynamics of Particles and Systems, Academic Press 1970. (2) H. Goldstein, Classical Mechanics, Addison-Wesley, 1980.





References

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