What time is it on Mars?

We've seen in the last lecture that both the total energy \[ E = \frac{1}{2}\,\mu\dot{r} - \frac{GM\mu}{r} + \frac{L^2}{2\mu r^2} \] and the orbital angular momentum \[ L = \mu r^2 \dot{\varphi} \] are conserved quantities of orbital motion. Further, we also showed that because of the symmetry of the system, all motion must be confined to a single plane. This gives us one constraint and two coupled first-order differential equations, which is enough to uniquely specify the motion of the system.

We can also show that the radial component of motion forms a conic section, eliminating one differential equation in the process. To see this, we could either take the time derivative of the total energy equation, or look at the \(r\)-component of acceleration: \[ \ddot{r} - r\dot{\varphi}^2 = -\frac{GM}{r^2}. \] This might look obscenely complicated because of the mixed time derivatives of \(r\) and \(\varphi\), but notice that we're free to change the independent variable to \(\varphi\), since \[ \frac{dr}{d\varphi} = \frac{dr}{dt} \left(\frac{d\varphi}{dt}\right)^{-1} = \frac{\dot{r}}{\dot{\varphi}}. \] So, dividing through by \(\dot{\varphi}^2\) and changing variables, we have \[ \frac{d^2r}{d\varphi^2} - \frac{2}{r}\left(\frac{dr}{d\varphi}\right)^2 - r = -\frac{GM\mu^2r^2}{L^2} = -\frac{r^2}{r_L} \] where \[ r_L = \frac{L^2}{GM\mu^2} \] is a constant depending on angular momentum.

So far we have a second-order differential equation for \(r\) in terms of \(\varphi\). Our next step is to make this linear (and a lot more familiar-looking) by changing variables again to \(s = 1/r\): \[ \frac{d^2s}{d\varphi^2} + s = \frac{1}{r_L}. \] This is the classic form of Hooke's law, which describes simple harmonic motion. Its general solution has the form \[ s(\varphi) = \frac{1}{r_L} \left[ 1 + e \cos(\varphi - \varphi_0) \right] \] We can set \(\varphi_0 = 0\) if we lay down our coordinate system so that \(\varphi = 0\) corresponds to the \(+x\)-axis. Replacing \(s\) with \(r\), we're left with \[ r(\varphi) = \frac{r_L}{1 + e\cos\varphi}. \] This equation defines a conic section with semi-latus rectum \(r_L\) and eccentricity \(e \geq 0\).

Note, eccentricity is a measure of how far the conic section deviates from a circle. An eccentricity of \(e=0\) is perfectly circular (and has minimum orbital energy); eccentricities \(0\lt e\lt 1\) are elliptical (with energies \(E \lt 0\)); when \(e = 0\) the orbit is parabolic (or marginally bound) with zero total energy; and when \(e \gt 1\) the orbit is hyperbolic with \(E \gt 0\). In the center-of-mass reference frame, the semi-latus rectum \(r_L\) measures half the distance between points where the orbit intersects the \(y\)-axis; for a circular orbit, this is just the radius of the circle.

Under Newtonian gravity, it turns out that any bound two-body orbit is not only elliptical, but also in a kind of harmony with itself: a planetary orbit that starts at either apsis (closest or farthest approach) at \(\varphi = \varphi_0\) will also end at the same apsis with \(\varphi = \varphi_0 + 2\pi\) if you wait exactly one orbital period. In other words, \(r(t)\) and \(\varphi(t)\) have the same intrinsic frequency, and classical orbits have no apsidal precession. You can actually see this from the definition of a conic section. If \[ r(\varphi) = \frac{r_L}{1 + e\cos\varphi} \] and if \(\varphi\rightarrow\varphi+2\pi\) on some periodic timescale \(T\), then \(r(t) = r(t+T)\). Is there a deeper physical reason for this?

Consider the effective radial potential \[ \Phi_L(r) = -\frac{GM\mu}{r} + \frac{L^2}{2\mu r^2}. \] Remember that when we derived this potential we originally had some angular kinetic energy that looked like \(r^2\dot{\varphi}^2/2\), and we lumped this in with potential energy to conveniently eliminate \(\dot{\varphi}\). So what we really have here is a fictitious centrifugal force caused by angular motion in a non-inertial reference frame, which counteracts the inward pull of gravity. Orbital motion then arises from the balance of these two forces.

What are the frequencies that characterize this balance? First, we know from conservation of angular momentum that the angular speed of a circular orbit is \[ \omega_{\varphi} = \dot{\varphi}_{\rm circ} = \frac{L}{\mu r_L^2} = (GM)^2 \left(\frac{\mu}{L}\right)^3. \] To get the characteristic frequency of radial motion, \(\omega_r\), we can look at small deviations away from circular orbits, expanding \(\Phi_L(r)\) to second order around \(r_L\): \begin{align} \Phi_L(r) &= \sum_{n=0}^{\infty} \frac{1}{n!} \frac{d^n\Phi_L}{dr^n}(r_L) \, (r - r_L)^n \\ &= \Phi_L(r_L) + \frac{1}{2}\frac{d^2\Phi_L}{dr^2}(r_L)\,(r-r_L)^2 + \mathcal{O}\left[(r-r_L)^3\right] \end{align} The first-order term vanishes because the inward pull of gravity is exactly balanced by the outward push of the centrifugal acceleration in circular orbits. So, to second order in \(r\), small deviations away from a circular orbit obey a Hooke's law potential with \[ \Phi_L(r) \approx \Phi_L(r_L) + \frac{1}{2}\mu\omega_r^2(r-r_L)^2 \] and the angular frequency can be read off by matching terms: \[ \omega_r = \sqrt{\frac{1}{\mu} \frac{d^2\Phi_L}{dr^2}(r_L)} = (GM)^2 \left(\frac{\mu}{L}\right)^3. \] Hence \(\omega_r = \omega_{\varphi}\), and the total precession angle is exactly zero: \[\delta\varphi = T\left(\omega_{\varphi} - \omega_r\right) = 0. \] In other words, even for non-circular orbits, kinetic and potential energies are always balanced in such a way that both radial and angular motion have the exact same characteristic frequency, so the apses of elliptical orbits cannot precess.

The figure below illustrates this argument for a simulation of the orbit of Mercury around the Sun. As predicted, the evolution of \(r\) returns to the same position after one orbital period, while \(\varphi\) evolves by exactly \(2\pi\) radians (or 360 degrees). In reality the orbit of Mercury actually does precess, in part due to the influence of other heavy planets in the Solar System. However, a precession of just 43 arcseconds (0.01 degrees) per century is unaccounted for by Newtonian gravity, but is significant enough to be measured. We will come back to this mysterious perihelion shift of Mercury in just a couple of weeks when we learn about general relativity.