Given r/a, what are the limits on the direction that an orbiting body could be moving (e.g. solid angle vs r/a)?

Question

An orbiting object at distance $r$ and semimajor axis $a$ will move at $\sqrt{2 - \displaystyle \frac{r}{a}}$ times the speed of a circular orbit at $r$, no matter what the eccentricity or which direction it might have!

That comes from the vis-viva equation

$$v = \sqrt{GM \left(\frac{2}{r} - \frac{1}{a} \right)},$$

and if you use AU and years for units, then for orbits only around our Sun it's simply

$$v = 2 \pi \sqrt{\frac{2}{r} - \frac{1}{a}}.$$

If $a$ = 2, it's moving $\sqrt{1.5}$ faster than Earth's $2 \pi$ AU/year, and if it is coming in with $C_3$=0 (heliocentric escape velocity) it's moving $\sqrt{2}$ faster than Earth at 1 AU, which is a handy relationship to remember.

Question: Given $r/a$, what are the limits on the direction that an orbiting body can be going? For example if $r/a = 0.9$ could it be moving in any direction that's say between 80 and 100 degrees with respect to the vector pointing at the Sun?

Possibly an answer could be expressed as solid angle as a function of $r/a$ ranging from 0 to 2, but since I don't know what the answer will look like I won't overly constrain the form.

note: I have not constrained eccentricity, so an answer will (probably?) need to first determine the two limiting eccentricities as a function of $r/a$ and then go from there.

Answer 1

There are no limits on the direction.

The Vis-viva equation will give you a speed. Assuming point-masses and sticking to classical mechanics, the Vis-Viva equation does not care at all about what direction you point your velocity in; It's merely an equation based on how the total Orbital Energy (which is the same for all orbits with the same semimajor axis around the same body) must be distributed between Gravitational Potential Energy and Kinetic Energy.

For Keplerian orbits, the only constraints on $r$ and $a$ are:

• $r$ will be a positive value.
• $a$ must be nonzero.
• If $a$ is positive (meaning an elliptical orbit), $r$ will never exceed $2a$ (If $r$ = $2a$, you're looking at the apoapsis of the linear degenerate ellipse)
• If $a$ is negative (meaning a hyperbolic trajectory), $r$ can be whatever positive value you choose.

To put it another way, by the vis-viva equation, given a radial distance $r$ and a semimajor axis $a$ around a gravitating body defines an orbital speed value $v$. In the ideal two-body Newtonian conditions, regardless of the direction you point that speed $v$, you will always be in a Keplerian Orbit/Trajectory.

Answer 2

Supplemental answer confirming @notovny is correct!

While vis-viva gives you the speed, apparently all directions seem to be still possible!

It seems that I've puzzled myself this time.

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

def deriv(X, t):
    x, v = X.reshape(2, -1)
    acc = -x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

halfpi, pi, twopi   = [f*np.pi for f in (0.5, 1, 2)]
r       = 1.0
answerz = []
titles  = []

for r_over_a in (0.7, 1.4):
    titles.append('r/a = ' + str(round(r_over_a, 2)))
    answers = []
    a     = r / r_over_a
    T     = twopi * np.sqrt(a**3)
    times = np.linspace(0, T, 1001)
    v0    = np.sqrt(2./r - 1./a)

    thetas = np.linspace(0, pi, 8)[:-1] # make the result odd to avoid singularity

    for theta in thetas:
        s, c = [f(theta) for f in (np.sin, np.cos)]
        X0   = np.array([r, 0, s*v0, c*v0])
        answer, info = ODEint(deriv, X0, times, full_output=True)
        answers.append(answer)
    answerz.append(answers)

if True:
    fig = plt.figure()
    for i, (title, answers) in enumerate(zip(titles, answerz)):
        ax  = fig.add_subplot(2, 1, i+1)
        for a in answers:
            x, y = a.T[:2]
            ax.plot(x, y)
        ax.plot([0], [0], 'oy', markersize=12)
        ax.set_aspect('equal')
        ax.set_title(title, fontsize=16)
    plt.show()