<p>Since the question is about inequalities and ratios, I took the script from <a href="https://space.stackexchange.com/a/26037/12102">this answer</a> and made it better by normalizing it, now <span class="math-container">$a=1, M1=1$</span> and <span class="math-container">$m \equiv M2/M1$</span></p>

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

<p><span class="math-container">$$r_1 &lt; r_H &lt; r_2$$</span></p>

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

<hr>

<p>Without normalizing <span class="math-container">$a=1$</span>, the distances and Hill sphere radii for the Sun-Earth and Sun-Mars systems are as follows:</p>

<pre><code>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
</code></pre>

<p>We now apply the normalizations mentioned above:</p>

<p><a href="https://i.sstatic.net/A5r4h.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/A5r4h.png" alt="Hill sphere and L1/L2 as a function of M2/M1"></a></p>

<p><strong>Python 3 or 2</strong></p>

<pre><code>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()
</code></pre>

CC BY-SA 4.0
2019-12-28T16:21:43.597Z
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.
Astrid_Redfern
CC BY-SA 4.0
2019-12-28T16:27:42.920Z
I've done the edit now, so when you approve it I'll accept the answer. Thanks!
Astrid_Redfern
CC BY-SA 4.0
2019-12-28T16:34:59.800Z
@Astrid_Redfern done, thanks!
uhoh
CC BY-SA 4.0
2019-12-28T16:36:19.650Z
And accepted! Many thanks indeed.
Astrid_Redfern
CC BY-SA 4.0
2019-12-28T14:59:16.900Z
2019-12-28T16:34:47.633Z
2019-12-28T16:34:47.633Z
uhoh
<p>(Disclaimer: I know the "Hill sphere" is just an approximation of something which isn't genuinely spherical.)</p>

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

<p><span class="math-container">$$r_{\mathrm{H}} \approx a (1-e) \sqrt[3]{\frac{m}{3 M}}$$</span></p>

<p>where <span class="math-container">$a$</span> is the semi-major axis of the small body's orbit around the larger, <span class="math-container">$e$</span> denotes that orbit's eccentricity, <span class="math-container">$r_H$</span> is of course the radius of the Hill sphere, <span class="math-container">$m$</span> is the mass of the smaller body and <span class="math-container">$M$</span> is the mass of the larger one.</p>

<p>If and when the eccentricity is negligible, this is approximated further to</p>

<p><span class="math-container">$$r_{\mathrm{H}} \approx a \sqrt[3]{\frac{m}{3M}}$$</span></p>

<p>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.)</p>

<p>The distance <span class="math-container">$r_1$</span> between the smaller body and L1 is obtained by solving this equation (which appears also to assume negligible eccentricity):</p>

<p><span class="math-container">$$\frac{M}{\left(a-r_{1}\right)^2}=\frac{m}{r_{1}^{2}}+\frac{M}{a^2}-\frac{r_{1}\left(M+m\right)}{a^3}$$</span></p>

<p>Similarly, the distance <span class="math-container">$r_2$</span> between the smaller body and L2 is obtained by solving:</p>

<p><span class="math-container">$$\frac{M}{\left(a+r_{2}\right)^2}+\frac{m}{r_{2}^{2}}=\frac{M}{a^2}+\frac{r_{2}\left(M+m\right)}{a^3}$$</span></p>

<p>As noted above, <span class="math-container">$r_H$</span> as obtained using the approximate formula approximates both <span class="math-container">$r_1$</span> and <span class="math-container">$r_2$</span>.</p>

<p>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:</p>

<p><span class="math-container">$r_H &lt; min(r_1, r_2)$</span></p>

<p><span class="math-container">$r_H \leq min(r_1, r_2)$</span></p>

<p><span class="math-container">$r_H &lt; max(r_1, r_2)$</span></p>

<p><span class="math-container">$r_H \leq max(r_1, r_2)$</span></p>

<p><span class="math-container">$r_1 &lt; r_H &lt; r_2$</span></p>

<p><span class="math-container">$r_1 &lt; r_H \leq r_2$</span></p>

<p><span class="math-container">$r_1 \leq r_H \leq r_2$</span></p>

<p><span class="math-container">$r_1 \leq r_H &lt; r_2$</span></p>

<p>or something else.</p>

<p>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.</p>

CC BY-SA 4.0
2019-12-28T12:07:52.987Z
Looking at another related question from Space.SE - https://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 ;-)
Astrid_Redfern
CC BY-SA 4.0
2019-12-28T14:15:13.233Z
the way [`@replies`](https://meta.stackexchange.com/q/43019/303080) 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.
uhoh
CC BY-SA 4.0
2019-12-28T16:15:49.850Z
@uhoh I didn't realise that - thanks for telling me!
Astrid_Redfern
<lagrange-point>
Lagrange points and the radius of the Hill sphere
Astrid_Redfern
# Lagrange points and the radius of the Hill sphere

(Asked by: username_0 on 2019-12-27)

(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.

**Tags:** lagrange-point

> **Comments:**
> **username_0**: Looking at another related question from Space.SE - https://space.stackexchange.com/a/26037/29381 - @username_1 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 username_1 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 ;-)  
> **username_1**: the way [`@replies`](https://meta.stackexchange.com/q/43019/303080) 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.  
> **username_0**: @username_1 I didn't realise that - thanks for telling me!  

---

## 1 answers

### Answer by username_1 on 2019-12-28. Score: 2 | Accepted answer

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:

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()

> **Comments:**
> **username_0**: 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.  
> **username_0**: I've done the edit now, so when you approve it I'll accept the answer. Thanks!  
> **username_1**: @username_0 done, thanks!  
> **username_0**: And accepted! Many thanks indeed.  

---

