Orbits of Planets are Ellipses
Assume a planet with mass M orbiting the Sun with mass S. Conservation of energy can be written as
(1) P.E. + K.E. = E
1/2 Mv2 - GMS/R = E
The velocity v along the orbit is the derivative dS/dt. Hence,
v2 = (dS/dt)2 , which can be written as
(2) v2 = (dS/dt)2 = (dR/dt)2 + R2(dq/dt)2
by noticing from the diagram above that (dS)2 = (dR)2 + R2(dq).
Using conservation of angular momentum L = I w = MR2 dq/dt = constant, we can substitute dq/dt = L/MR2 in equation (2) for v2. Then equation (1) for conservation of energy becomes
E = 1/2 (L2/(MR4)) [ (dR/dq)2 + R2 ] - G (MS/R)
Rearranging terms gives
[ (dR/dq)2 + R2 ] = (2EMR4/L2) + (2GSM2R3/L2)
The derivative dR/dq thus can be written as
(3) dR/dq = R[(2EM/L2) R2 + 2 GSM2/L2 R - 1 ] 1/2
defining combinations of constants p = L2/GSM2 and (e2 - 1)= 2EL2/(G2S2M3) equation (3) can be written as
(4) dR/dq = R [(e2 - 1)/p2 R2 + (2/p) R - 1 ] 1/2
(5) R = p /(1 + e cos (q + c))
This is the equation for an ellipse with the origin at one focus (the sun). The constant of integration c is zero if q = 0 for Rp at perihelion and q = p for Ra at aphelion.
Rp = p/(1+e)
Ra = p/(1-e)
Since the semi-major axis of the ellipse is 2a = Rp + Ra, we have
p = a (1- e2)
Consider four different situations for the ratio of the magnitude of potential energy to kinetic energy and what they imply for the eccentricity e
(1) P.E. > K.E.
The total E is negative and since
e2 = 2EL2/(G2S2M3) + 1 , e2 < 1 and the path of the planet is an ellipse with semimajor axis a and semiminor axis c = a(1- e2)1/2
(2) P.E. = K.E.
When E = 0, the eccentricity e is zero, describing a parabola. This is the condition for escape.
(3) P.E. < K.E.
E > 0 , the eccentricity e is greater than 1 and the path is hyperbolic
(4) When eccentricity e is zero:
e = 2EL2/(G2S2M3) + 1 = 0
or E = - G2S2M3/2L2
This is the minimum value (largest
that E (total energy) can take for any orbital path.
The corresponding K.E. is the minimum value of K.E.
for any possible conic section. The orbit is a circle.