<p>I think that @JamesK's <a href="https://astronomy.stackexchange.com/a/28958/7982">answer</a> summarizes the situation nicely. Here I can summarize the way that I see space commonly sliced and diced by people in <a href="https://space.stackexchange.com">Space Exploration SE</a>. I can't defend these as "right", but recollect seeing them typically used this way.</p>

<ol>
<li><p>Earth orbit</p>

<ul>
<li>LEO or low Earth orbit</li>
<li>MEO or medium Earth orbit</li>
<li>GEO or geostationary orbit (though it is sometimes mixed with the broader geosynchronous)</li>
<li>HEO or high Earth orbit</li>
</ul></li>
<li><p>cis-Lunar space</p>

<ul>
<li>out to about 400,000 km from the Earth, still gravitationally bound to Earth (as the Moon is as well) but usually associated with gravitational effects of the Moon as well. Since the Moon is fairly massive (> 1% of the Earth) it has a lot of "pull" in cis-lunar space. Anything a half-million km from the Earth has to deal with the Moon.</li>
<li>includes orbits associated with the Earth-Moon Lagrange points</li>
</ul></li>
<li><p>Heliocentric orbit = "Deep Space"</p>

<ul>
<li>orbits that are bound to the Sun but not strongly bound to the Earth or other bodies.</li>
<li>includes orbits associated with the Sun-Earth Lagrange points (we don't call it "cis-Earth space"). These are heliocentric orbits that are in 1:1 resonance with the Earth's orbit.</li>
<li>exception in <a href="https://space.stackexchange.com/q/23810/12102">some specialized cases</a> where HEO is called "deep space"</li>
</ul></li>
</ol>

<p>So to answer your question, the distance from Earth where we start calling the space "interplanetary" or not associated with the Earth or another planet begins at the <a href="https://en.wikipedia.org/wiki/Hill_sphere" rel="nofollow noreferrer">Hill sphere</a> which is about the same distance as the Sun-Earth Lagrange L1 and L2 points are.</p>

<p>That's about 1.5 million kilometers from Earth, and a little less from Mars, and several million miles for the "big planets"!</p>

<hr>

<p>From <a href="https://space.stackexchange.com/a/26037/12102">this answer</a>:</p>

<p>For L1, find the smallest value of <span class="math-container">$r$</span> such that:</p>

<p><span class="math-container">$$\frac{M_2}{r^2} + \frac{M_1}{R^2} - \frac{r(M_1+M_2)}{R^3} - \frac{M_1}{(R-r)^2} = 0 $$</span></p>

<p>For L2, find the smallest value of <span class="math-container">$r$</span> such that:</p>

<p><span class="math-container">$$\frac{M_1}{R^2} + \frac{r(M_1+M_2)}{R^3}  - \frac{M_1}{(R+r)^2} - \frac{M_2}{R^2}  = 0 $$</span></p>

<p>Even though Mars is 50% farther from the Sun than the Earth, <strong>it's mass is only 11% that of Earth's</strong>, so while the distances to Earth's Lagrange point are about 1% of that to the Sun for Earth, those of Mars are only about 0.5% of the distance to the Sun for Mars.</p>

<p>In either case, a diagram would show two dots very close to each planet. The diagrams on the internet usually exaggerate this <em>greatly</em> to make it easier to see.</p>

<p>The values for the distance from the planets to the Sun and to their Sun-associated L1 and L2 points look like this. </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><a href="https://i.sstatic.net/IXzBf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/IXzBf.png" alt="find Earth and Mars L1 and L2"></a></p>

<p>The Python script based on scipy.optimize's <a href="https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.brentq.html#scipy.optimize.brentq" rel="nofollow noreferrer">Brentq</a>:</p>

<pre><code>def solve_L1 (r, R, M1, M2):
    return M2/r**2 + M1/R**2 - r*(M1 + M2)/R**3 - M1/(R-r)**2

def solve_L2 (r, R, M1, M2):
    return M1/R**2 + r*(M1 + M2)/R**3 - M1/(R+r)**2 - M2/r**2

def r_Hill(R, M1, M2):
    return R * (M2 / (3.*M1))**(1./3.)

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

a_Earth  = 149598023.   # Earth's semi-major axis (km)
a_Mars   = 227939200.   # Mars'   semi-major axis (km)
r_low    =   1000000.   # 1.0  million km  (lower guess)
r_high   =   1600000.   # 1.6  million km  (upper guess)

M_Sun    = 1.9886E+30  # approximate mass (kg)
M_Earth  = 5.9724E+24  # approximate mass (kg)
M_Mars   = 6.4171E+23  # approximate mass (kg)

r_Hill_Earth = r_Hill(a_Earth, M_Sun, M_Earth)
r_Hill_Mars  = r_Hill(a_Mars,  M_Sun, M_Mars)

r = np.linspace(r_low, r_high)
if True:
    plt.figure()
    plt.plot(r, solve_L1(r, a_Earth, M_Sun, M_Earth), '-g')
    plt.plot(r, solve_L1(r, a_Mars,  M_Sun, M_Mars), '-r')

    plt.plot(r, solve_L2(r, a_Earth, M_Sun, M_Earth), '--g')
    plt.plot(r, solve_L2(r, a_Mars,  M_Sun, M_Mars), '--r')

    plt.plot([r_Hill_Earth], [0], 'ok')
    plt.plot([r_Hill_Mars ], [0], 'ok')

    plt.text(1040000, 1.1E+11, 'L1 Mars L2', fontsize=14)
    plt.text(1450000, 3.0E+11, 'L1 Earth L2', fontsize=14)

    plt.plot(r, np.zeros_like(r), '-k')
    plt.ylim(-4E+11, 4E+11)

    plt.show()

# for Mars:
r_L1_Mars = brentq(solve_L1, r_low, r_high, args=(a_Mars, M_Sun, M_Mars))
r_L2_Mars = brentq(solve_L2, r_low, r_high, args=(a_Mars, M_Sun, M_Mars))

# for Earth:
r_L1_Earth = brentq(solve_L1, r_low, r_high, args=(a_Earth, M_Sun, M_Earth))
r_L2_Earth = brentq(solve_L2, r_low, r_high, args=(a_Earth, M_Sun, M_Earth))

print "a_Earth:    ", int(a_Earth), " km"
print "Sun-Earth L1: ", int(r_L1_Earth), " km"
print "Sun-Earth L2: ", int(r_L2_Earth), " km"
print "Earth r_Hill: ", int(r_Hill_Earth), " km"
print ''
print "a_Mars:     ", int(a_Mars), " km"
print "Sun-Mars L1:  ", int(r_L1_Mars), " km"
print "Sun-Mars L2:  ", int(r_L2_Mars), " km"
print "Mars r_Hill:  ", int(r_Hill_Mars), " km"
</code></pre>

CC BY-SA 4.0
2019-01-01T06:19:49.257Z
2019-01-01T06:48:13.450Z
2019-01-01T06:48:13.450Z
uhoh
<p>It it wherever is convenient for the article you are writing. There is no single consenus definintion of "interplanetary space".  If you are in orbit around a planet then you are not in interplanetary space, if you are between planets, then you are.  But there is no substantial change in the matter filling space as you move from planetary orbit to interplanetary trajectory</p>

<p>So you can use any reasonable point that fits the article you happen now to be writing</p>

CC BY-SA 4.0
2019-01-01T03:11:37.433Z
Well I am writing a piece of fiction and for the purpose of my story I was using the Hill Sphere as the definition.  I was genuinely curious if there was one.
Bob516
CC BY-SA 4.0
2019-01-01T03:21:29.380Z
You wrote, "But there is no substantial change in the matter filling space as you move from planetary orbit to interplanetary trajectory."  I was under the impression the solar wind did not penetrate into the magnetosphere.  If that is the case why would there be no difference between the matter in Earth's orbit and interplanetary space.  Wouldn't that also apply to Mercury, Jupiter, Ganymede, Saturn, Uranus, and Neptune?
Bob516
CC BY-SA 4.0
2019-01-01T02:18:53.583Z
2019-01-01T02:18:53.583Z
James K
<p>The Kármán line is often cited as the beginning of outer space.  The Voyager probes have identified the boundary between the heliosphere and interstellar space.  Is there a agreed upon definition of the boundary of interplanetary space?  Is it the Earth's magnetopause?</p>

<solar-system><definition><interplanetary-medium>
How far above the Earth's surface does interplanetary space begin?
Bob516
# How far above the Earth's surface does interplanetary space begin?

(Asked by: username_0 on 2019-01-01)

The Kármán line is often cited as the beginning of outer space. The Voyager probes have identified the boundary between the heliosphere and interstellar space. Is there a agreed upon definition of the boundary of interplanetary space? Is it the Earth's magnetopause?

**Tags:** solar-system, definition, interplanetary-medium


---

## 2 answers

### Answer by username_1 on 2019-01-01. Score: 2 | Accepted answer

I think that @username_2's [answer](https://astronomy.stackexchange.com/a/28958/7982) summarizes the situation nicely. Here I can summarize the way that I see space commonly sliced and diced by people in [Space Exploration SE](https://space.stackexchange.com). I can't defend these as "right", but recollect seeing them typically used this way.

1. Earth orbit

   * LEO or low Earth orbit
   * MEO or medium Earth orbit
   * GEO or geostationary orbit (though it is sometimes mixed with the broader geosynchronous)
   * HEO or high Earth orbit
2. cis-Lunar space

   * out to about 400,000 km from the Earth, still gravitationally bound to Earth (as the Moon is as well) but usually associated with gravitational effects of the Moon as well. Since the Moon is fairly massive (> 1% of the Earth) it has a lot of "pull" in cis-lunar space. Anything a half-million km from the Earth has to deal with the Moon.
   * includes orbits associated with the Earth-Moon Lagrange points
3. Heliocentric orbit = "Deep Space"

   * orbits that are bound to the Sun but not strongly bound to the Earth or other bodies.
   * includes orbits associated with the Sun-Earth Lagrange points (we don't call it "cis-Earth space"). These are heliocentric orbits that are in 1:1 resonance with the Earth's orbit.
   * exception in [some specialized cases](https://space.stackexchange.com/q/23810/12102) where HEO is called "deep space"

So to answer your question, the distance from Earth where we start calling the space "interplanetary" or not associated with the Earth or another planet begins at the [Hill sphere](https://en.wikipedia.org/wiki/Hill_sphere) which is about the same distance as the Sun-Earth Lagrange L1 and L2 points are.

That's about 1.5 million kilometers from Earth, and a little less from Mars, and several million miles for the "big planets"!

---

From [this answer](https://space.stackexchange.com/a/26037/12102):

For L1, find the smallest value of $r$ such that:

$$\frac{M\_2}{r^2} + \frac{M\_1}{R^2} - \frac{r(M\_1+M\_2)}{R^3} - \frac{M\_1}{(R-r)^2} = 0 $$

For L2, find the smallest value of $r$ such that:

$$\frac{M\_1}{R^2} + \frac{r(M\_1+M\_2)}{R^3} - \frac{M\_1}{(R+r)^2} - \frac{M\_2}{R^2} = 0 $$

Even though Mars is 50% farther from the Sun than the Earth, **it's mass is only 11% that of Earth's**, so while the distances to Earth's Lagrange point are about 1% of that to the Sun for Earth, those of Mars are only about 0.5% of the distance to the Sun for Mars.

In either case, a diagram would show two dots very close to each planet. The diagrams on the internet usually exaggerate this *greatly* to make it easier to see.

The values for the distance from the planets to the Sun and to their Sun-associated L1 and L2 points look like this.

```
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

```

[![find Earth and Mars L1 and L2](https://i.sstatic.net/IXzBf.png)](https://i.sstatic.net/IXzBf.png)

The Python script based on scipy.optimize's [Brentq](https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.brentq.html#scipy.optimize.brentq):

```
def solve_L1 (r, R, M1, M2):
    return M2/r**2 + M1/R**2 - r*(M1 + M2)/R**3 - M1/(R-r)**2

def solve_L2 (r, R, M1, M2):
    return M1/R**2 + r*(M1 + M2)/R**3 - M1/(R+r)**2 - M2/r**2

def r_Hill(R, M1, M2):
    return R * (M2 / (3.*M1))**(1./3.)

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

a_Earth  = 149598023.   # Earth's semi-major axis (km)
a_Mars   = 227939200.   # Mars'   semi-major axis (km)
r_low    =   1000000.   # 1.0  million km  (lower guess)
r_high   =   1600000.   # 1.6  million km  (upper guess)

M_Sun    = 1.9886E+30  # approximate mass (kg)
M_Earth  = 5.9724E+24  # approximate mass (kg)
M_Mars   = 6.4171E+23  # approximate mass (kg)

r_Hill_Earth = r_Hill(a_Earth, M_Sun, M_Earth)
r_Hill_Mars  = r_Hill(a_Mars,  M_Sun, M_Mars)

r = np.linspace(r_low, r_high)
if True:
    plt.figure()
    plt.plot(r, solve_L1(r, a_Earth, M_Sun, M_Earth), '-g')
    plt.plot(r, solve_L1(r, a_Mars,  M_Sun, M_Mars), '-r')

    plt.plot(r, solve_L2(r, a_Earth, M_Sun, M_Earth), '--g')
    plt.plot(r, solve_L2(r, a_Mars,  M_Sun, M_Mars), '--r')

    plt.plot([r_Hill_Earth], [0], 'ok')
    plt.plot([r_Hill_Mars ], [0], 'ok')

    plt.text(1040000, 1.1E+11, 'L1 Mars L2', fontsize=14)
    plt.text(1450000, 3.0E+11, 'L1 Earth L2', fontsize=14)

    plt.plot(r, np.zeros_like(r), '-k')
    plt.ylim(-4E+11, 4E+11)

    plt.show()

# for Mars:
r_L1_Mars = brentq(solve_L1, r_low, r_high, args=(a_Mars, M_Sun, M_Mars))
r_L2_Mars = brentq(solve_L2, r_low, r_high, args=(a_Mars, M_Sun, M_Mars))

# for Earth:
r_L1_Earth = brentq(solve_L1, r_low, r_high, args=(a_Earth, M_Sun, M_Earth))
r_L2_Earth = brentq(solve_L2, r_low, r_high, args=(a_Earth, M_Sun, M_Earth))

print "a_Earth:    ", int(a_Earth), " km"
print "Sun-Earth L1: ", int(r_L1_Earth), " km"
print "Sun-Earth L2: ", int(r_L2_Earth), " km"
print "Earth r_Hill: ", int(r_Hill_Earth), " km"
print ''
print "a_Mars:     ", int(a_Mars), " km"
print "Sun-Mars L1:  ", int(r_L1_Mars), " km"
print "Sun-Mars L2:  ", int(r_L2_Mars), " km"
print "Mars r_Hill:  ", int(r_Hill_Mars), " km"

```


---

### Answer by username_2 on 2019-01-01. Score: 2

It it wherever is convenient for the article you are writing. There is no single consenus definintion of "interplanetary space". If you are in orbit around a planet then you are not in interplanetary space, if you are between planets, then you are. But there is no substantial change in the matter filling space as you move from planetary orbit to interplanetary trajectory

So you can use any reasonable point that fits the article you happen now to be writing

> **Comments:**
> **username_0**: Well I am writing a piece of fiction and for the purpose of my story I was using the Hill Sphere as the definition.  I was genuinely curious if there was one.  
> **username_0**: You wrote, "But there is no substantial change in the matter filling space as you move from planetary orbit to interplanetary trajectory."  I was under the impression the solar wind did not penetrate into the magnetosphere.  If that is the case why would there be no difference between the matter in Earth's orbit and interplanetary space.  Wouldn't that also apply to Mercury, Jupiter, Ganymede, Saturn, Uranus, and Neptune?  

---

