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(Disclaimer: I know the "Hill sphere" is just an approximation of something which isn't genuinely spherical.)

In a two-body system, the approximate formula for the Hill sphere radius of the smaller body is

$$r_{\mathrm{H}} \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}$$

where $a$ is the semi-major axis of the small body's orbit around the larger, $e$ denotes that orbit's eccentricity, $r_H$ is of course the radius of the Hill sphere, $m$ is the mass of the smaller body and $M$ is the mass of the larger one.

If and when the eccentricity is negligible, this is approximated further to

$$r_{\mathrm{H}} \approx a \sqrt[3]{\frac{m}{3M}}$$

This formula is often used to approximate the distance between the smaller body and the L1 and L2 Lagrange points, although these distances are not in fact identical (L2 being further distant than L1.)

The distance $r_1$ between the smaller body and L1 is obtained by solving this equation (which appears also to assume negligible eccentricity):

$$\frac{M}{\left(a-r_{1}\right)^2}=\frac{m}{r_{1}^{2}}+\frac{M}{a^2}-\frac{r_{1}\left(M+m\right)}{a^3}$$

Similarly, the distance $r_2$ between the smaller body and L2 is obtained by solving:

$$\frac{M}{\left(a+r_{2}\right)^2}+\frac{m}{r_{2}^{2}}=\frac{M}{a^2}+\frac{r_{2}\left(M+m\right)}{a^3}$$

As noted above, $r_H$ as obtained using the approximate formula approximates both $r_1$ and $r_2$.

Various online sources use language such as "The Hill sphere lies between the L1 and L2 Lagrangian points" or "the Hill sphere extends somewhere between the L1 and L2 Lagrangian points". I find this language unclear, and can't tell which of the following these sources are saying:

$r_H < min(r_1, r_2)$

$r_H \leq min(r_1, r_2)$

$r_H < max(r_1, r_2)$

$r_H \leq max(r_1, r_2)$

$r_1 < r_H < r_2$

$r_1 < r_H \leq r_2$

$r_1 \leq r_H \leq r_2$

$r_1 \leq r_H < r_2$

or something else.

As I write this, I'm not sure how trivial a task it would be to work through the approximations for the L1 and L2 points to determine if the Hill sphere radius formula overestimates or underestimates each, so I apologise if this is something other users of Astronomy SE would find trivial.

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    $\begingroup$ Related question including derivation of the distance to L1: astronomy.stackexchange.com/q/18844/24157 - L2 (and L3) can be derived in a similar way. Given there are five Lagrangian points, it may not be entirely surprising that the derivation results in quintic equations. $\endgroup$ – antispinwards Dec 28 '19 at 1:29
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    $\begingroup$ Looking at another related question from Space.SE - space.stackexchange.com/a/26037/29381 - @uhoh has used a Python script to do the calculations and plot the results, as well as presenting the L1 and L2 distances and Hill sphere radii for the Sun-Earth and Sun-Mars systems. (I searched for ages last night and never saw that answer!) If uhoh would like to repost that answer here, I'll be delighted to click on the tick and accept it. And to upvote it, of course ;-) $\endgroup$ – Astrid_Redfern Dec 28 '19 at 12:07
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    $\begingroup$ the way @replies is set up the @ will only send a notification if the person has already commented here. If you typed @u and it didn't autocomplete the hoh then it (generally but not always) means that it won't receive a notification. I just happened to see your comment here when reading questions. $\endgroup$ – uhoh Dec 28 '19 at 14:15
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    $\begingroup$ @uhoh I didn't realise that - thanks for telling me! $\endgroup$ – Astrid_Redfern Dec 28 '19 at 16:15
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Since the question is about inequalities and ratios, I took the script from this answer and made it better by normalizing it, now $a=1, M1=1$ and $m \equiv M2/M1$

For physical situations in the spirit of Hill spheres and Lagrange points, I think the answer is always going to be:

$$r_1 < r_H < r_2$$

but I can't prove it, that would need math and it's too late tonight for me for math.


Without normalizing $a=1$, the distances and Hill sphere radii for the Sun-Earth and Sun-Mars systems are as follows:

a_Earth:     149598023  km
Sun-Earth L1:  1491524  km
Sun-Earth L2:  1501504  km
Earth r_Hill:  1496531  km

a_Mars:      227939200  km
Sun-Mars L1:   1082311  km
Sun-Mars L2:   1085748  km
Mars r_Hill:   1084032  km

We now apply the normalizations mentioned above:

Hill sphere and L1/L2 as a function of M2/M1

Python 3 or 2

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import brentq

#unitless   a = 1, M1 = 1, m = M2/M1
# so for the Hill sphere and L1/L2 points of the Earth orbiting the Sun
# m = 5.9724E+24/1.9886E+30  = 4.88987E-07 and
# L1 = 0.0054525 AU, r_Hill = 0.0054625 AU, L2 = 0.0054724 AU

def solve_L1 (r, m):
    return m/r**2 + 1. - r*(1. + m) - (1.-r)**-2

def solve_L2 (r, m):
    return 1 + r*(1 + m) - (1.+r)**-2 - m/r**2

def r_Hill(m):
    return (m / 3.)**(1./3.)

ms = np.logspace(-8, -1, 61)

answers = []
for m in ms:
    r = r_Hill(m)
    L1 = brentq(solve_L1, 0.5*r, 1.5*r, args=(m,))
    L2 = brentq(solve_L2, 0.5*r, 1.5*r, args=(m,))
    answers.append([L2, r, L1])

L2s, rhills, L1s = np.array(list(zip(*answers)))

m_Jup = 1.8982E+27/1.9886E+30
rhill_Jup = r_Hill(m_Jup)
L1_Jup = brentq(solve_L1, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))
L2_Jup = brentq(solve_L2, 0.5*rhill_Jup, 1.5*rhill_Jup, args=(m_Jup,))

if True:
    plt.figure()
    fs = 14
    plt.subplot(2, 1, 1)
    plt.plot(ms, rhills)
    plt.plot([m_Jup], [rhill_Jup], 'ok')
    plt.text(0.001, 0.03, 'Jupiter', fontsize=fs)
    plt.ylabel('r_Hill / a', fontsize=fs)
    plt.xscale('log')
    plt.yscale('log')
    plt.subplot(2, 1, 2)
    plt.plot(ms, L1s/rhills, '-')
    plt.plot(ms, L2s/rhills, '--')
    plt.plot([m_Jup], [L1_Jup/rhill_Jup], 'ok')
    plt.plot([m_Jup], [L2_Jup/rhill_Jup], 'ok')
    plt.xlabel('m = M2/M1', fontsize=fs)
    plt.ylabel('L1 or L2 / r_Hill', fontsize=fs)
    plt.xscale('log')
    plt.show()
|improve this answer|||||
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    $\begingroup$ Thanks - upvoted, but I'm going to edit and paste in the Sun/Earth and Sun/Mars data from the other question's answer before accepting. I think it helps to have the examples in text/numbers. $\endgroup$ – Astrid_Redfern Dec 28 '19 at 16:21
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    $\begingroup$ I've done the edit now, so when you approve it I'll accept the answer. Thanks! $\endgroup$ – Astrid_Redfern Dec 28 '19 at 16:27
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    $\begingroup$ @Astrid_Redfern done, thanks! $\endgroup$ – uhoh Dec 28 '19 at 16:34
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    $\begingroup$ And accepted! Many thanks indeed. $\endgroup$ – Astrid_Redfern Dec 28 '19 at 16:36

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