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Which is closer to Mars, Earth or the Moon?

I'm writing a novel and I'm so grateful for the wealth of information. The only answers I find refer to Phobos and Demos Mars moons.

PearsonArtPhoto
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T. Constantine
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    @Uwe: It is quite easy to see Mars, if you are in an area without a lot of light pollution. At the moment (1/31/18) it's at apparent magnitude 1.2, meaning it's among the brightest stars in the sky. It can be much brighter than that, ranging up to around -3. (Lower magnitudes are brighter.) Which makes it brighter than anything but the sun, moon, and Venus. – jamesqf Feb 01 '18 at 03:52
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    For the record, Luna is not the name of The Moon. The name of The Moon is, quite simply, "The Moon." – corsiKa Feb 02 '18 at 21:55
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    If you want to write a book about spaceflight, you should really play some Kerbal Space program first. especially when you really ask so basic question (that really become non-sensical once you understand what an orbit is). Related xkcd/what-if: https://what-if.xkcd.com/58/ – Polygnome Feb 02 '18 at 23:43
  • Comments are not for extended discussion; this conversation has been moved to chat. – PearsonArtPhoto Feb 03 '18 at 16:23
  • Depends on whether you are sending a radio message or a rocket. If you send a message, the difference between Earth and Moon varies between -1 and +1 seconds, depending on the moon phase (and Earth's and Mars's relative positions in their orbits around the sun, but that changes a bit more slowly.) If you are sending a rocket though, you need a huge rocket on Earth, while a smallish one is enough on the Moon. – Bass Feb 05 '18 at 10:03
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    Something I just thought about... how do you plan to measure "closer" - you could measure in miles or some factor of that, but you could also measure in Delta-V which might be more appropriate for a sci-fi novel! In this case, the Moon is much closer to Mars than the Earth. – corsiKa Feb 05 '18 at 17:12
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    I thank everyone for their answers and I'm happy to close this now asIm starting to get spammy insults from from folks with far too much time on their hands, for the most part I was absolutely astonished by your thoughtful intelligent answers and I've received more than enough to progress with. – T. Constantine Feb 08 '18 at 22:06

10 Answers10

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As you said, it varies.

Imagine I'm in Chicago and you're in London. My little dog is running circles around me. Which is closer to you, me or my dog?

While the correct answer is "it depends on where the dog is in its orbit around me", I'd argue that a better answer is "it doesn't matter" - the distance between you and me is so great that any little extra distance one way or the other is essentially nothing more than a rounding error.

For purposes of your novel, the Earth and the Moon are the same distance from Mars

Dan Pichelman
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Traveling from Mars's surface to Earth's surface requires less energy than traveling from Mars's surface to Luna's surface, but traveling from Luna's surface to Mars's surface requires much less energy than traveling from Earth's surface to Mars's surface.

For the purposes of space-travel, the actual physical distance is much less important than the relative energy needed for the trip. This energy is called Δv ("delta-vee") and is measured in km/s:

\begin{array}{c c|c c} & \Delta\textbf{V} & \textbf{D} \hskip{1em} \textbf{E} \hskip{1em} \textbf{S} & \textbf{T} \hskip{1em} \textbf{I} \hskip{1em} \textbf{N} \hskip{1em} \textbf{A} \hskip{1em} \textbf{T} & \textbf{I} \hskip{1em} \textbf{O} \hskip{1em} \textbf{N} \\ & \textbf{(km/s)} & \textbf{Earth} & \textbf{Luna} & \textbf{Mars} \\ \hline \textbf{S} & \textbf{Earth} & - & 16.1 & 13.5^* \\ \textbf{T} \\ \textbf{A} & \textbf{Luna} & 2.3^* & - & 2.9^* \\ \textbf{R} \\ \textbf{T} & \textbf{Mars} & 6.4^* & 9.3 & - \\ \end{array}

Values are approximate and assume optimal routes.
$^*$ indicates maximum utilisation of aerobraking.
Actual energy expenditure will be more.
Source.

If the characters and/or goods in your novel need to round-trip (or at least there is an approximately equal transfer in each direction), then since Mars-Luna-Mars costs 12.2 km/s and Mars-Earth-Mars costs 19.9 km/s, Luna's surface is significantly closer to Mars's surface than to Earth's surface. Note that in practice, landing on Mars requires retro-propulsion, but that overhead will be approximately equal for travelers arriving from Earth and Luna, so the round-trip comparison stays the same.

Adám
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    So you're saying it takes less energy to get to Mars than to the moon? The optimal Δv for the moon probably doesn't vary more than a few percentage points from the worst case, but the Earth/Mars trips in either direction would have considerably more variance. I don't see how you could escape the Earth/Moon gravity well using less Δv than to get to the moon from earth. – jwdonahue Feb 01 '18 at 01:48
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    @jwdonahue: It takes less Δv to land on Mars (because of aerobraking). If you're willing to just plow straight into the surface, rending your spaceship and everything aboard into their constituent particles, then the trip to Luna would take less Δv. –  Feb 01 '18 at 03:30
  • @Hurkyl and now I'm more than curious to see an actual impact of a several-ton vehicle vs. solid ground at several km/s. – John Dvorak Feb 01 '18 at 04:16
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    'Mars is much closer to Luna than to Earth' - unless I'm misreading something, your table says otherwise. – Rob Feb 01 '18 at 05:11
  • @Hurkyl I think the advantage comes from taking off, it's surely much easier to take off from the moon, this is why a round trip is much 'closer' but a one way trip earth would be – Aequitas Feb 01 '18 at 06:05
  • @Aequitas: Earth is the origin for both of the two trips I was talking about (and, AFAIK, the two trips that "it takes less energy to get to Mars than to the moon" was referring to as well). –  Feb 01 '18 at 07:20
  • Thank you so much for thoughtfully presented answer to my query Adam. Everyone has been incredibly helpful and I'm so pleased to have such a wealth of knowledge at my fingertips. – T. Constantine Feb 01 '18 at 07:48
  • hAlthough this is true, a expedition to mars is pretty much more involved than one to luna. So the craft probably is bigger, so total energetically getting to moon is easier for a expedition. Thus if you need to have some industry to put a lot of stuff to mars then its sane to build the bulk of structures on moon then do personell transfer to moon orbit... or enroute, orbit of moon being abortable. – joojaa Feb 01 '18 at 07:53
  • @BrockAdams Yep. It says mars -> earth 6.4, mars -> luna 9.3. So mars is closer to earth, not luna – Rob Feb 01 '18 at 08:05
  • @Rob: I assign orientation to the relation in the opposite fashion: that those numbers are saying Earth is closer to Mars than Luna is to Mars. For the phrase "Mars is closer to Luna than to Earth" I would look at Earth -> Mars and Luna -> Mars figures. This ambiguity, of course, the problem with taking words that apply to symmetric situations and trying to apply them to an asymmetric situation. However, the answerer did clarify in his paragraph exactly what he means by using the term, which is a third option entirely: he is comparing the cost of a round trip. –  Feb 01 '18 at 09:04
  • @JohnDvorak: Google "airplane vs concrete wall". I think solid ground would be softer, but I'm not sure that matters at the speeds we're talking about! But mainly I wanted to make sure that my comment didn't invoke imagery resembling an ordinary plane crash. –  Feb 01 '18 at 09:13
  • @JohnDvorak Apollo impact sites: http://www.lroc.asu.edu/featured_sites/#ApolloS-IVBImpactSites –  Feb 01 '18 at 09:32
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    I know you tried to keep details and physics complications to a minimum, but is it really OK to call something that is measured in km/s "energy"? Perhaps you could at least add a footnote clarifying that you're doing it for the purposes of simplicity, I don't know... – Pedro A Feb 01 '18 at 10:09
  • @Hurkyl whoa, I didn't know these videos existed. Thanks! It's just one tenth of the speed I was talking about, but still an impressive splash – John Dvorak Feb 01 '18 at 10:29
  • @Hamsterrific The body does say relative energy, which is true: To get the actual momentum needed, just multiply the shipping mass by the change in velocity. – Adám Feb 01 '18 at 10:30
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    @Adám Relative energy is still energy to me. Perhaps this is some jargon of space exploration experts (note that I'm here because of HNQ) so I wouldn't know. Regardless, not only me but I'm sure others will find strange an "energy" or "relative energy" measured in km/s. – Pedro A Feb 01 '18 at 10:33
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    @Hamsterrific Yes, dV (delta V) is a space/physic jargon. It stands for "change in velocity" ("delta" is a common shorthand for "change" or "difference" in physics). So a dV of 10 km/s refers to the energy required to accelerate an object (usually a spacecraft) from 0 km/s to 10 km/s. The actual amount of energy required for that change depends on the mass of the spacecraft. So dV is a way to talk about energy requirements while ignoring the mass of the vehicle. – Kyle A Feb 01 '18 at 17:36
  • So for an Earth observer Mars is closer than Moon? – frarugi87 Feb 05 '18 at 15:22
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    @frarugi87 The savings from completely avoiding using retro-propulsive landing on Mars (though difficult in practice), are greater than the increased energy cost of TMI over a TLI. – Adám Feb 05 '18 at 16:30
  • Mazel tov for the golden badge! – Eric Duminil Feb 05 '18 at 17:51
  • Are these numbers surface to surface or orbit to orbit? It's not clear from the writeup. – Bobson Feb 05 '18 at 17:55
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    @Bobson Surface to surface. – Adám Feb 05 '18 at 17:56
  • @Adám cool! I didn't know you were chilling in space too! – Magic Octopus Urn Jun 19 '18 at 17:54
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Distances to Mars

You can answer the question with Astropy, a Python library for astronomy.

Here's a diagram of distances from Earth to Mars and Moon to Mars, between 2000 and 2030.

enter image description here

You can see that the two curves are so close from each other that they look like one single curve.

Here's a zoom around the last opposition, in May 2016:

enter image description here

Relative difference

Following @gerrit's advice, here's a plot showing the relative difference between the distances Moon-Mars and Earth-Mars. The envelope of the curve oscillates between ±0.089% (during conjunction, when Mars is the furthest) and ±0.69% (during opposition). This difference is positive when Earth is closer to Mars than the Moon.

enter image description here

During this period (2010-2020), Earth is closest 49.6% of the time. Over a longer period of time, this percentage gets very close to 50%.

Code

For reference, here's the code that's been used for the first diagram:

from astropy.time import Time
from astropy.coordinates import solar_system_ephemeris, get_body

import numpy as np
from datetime import datetime, timedelta

import matplotlib.pyplot as plt

start_time = datetime(2000,1,1)
end_time = datetime(2030,1,1)
time_step = timedelta(days=1)

times = np.arange(start_time, end_time, time_step).astype(datetime)
astro_times = Time(times)
with solar_system_ephemeris.set('builtin'):
    mars = get_body('mars', astro_times)
    earth = get_body('earth', astro_times)
    moon = get_body('moon', astro_times)
earth_to_mars = earth.separation_3d(mars).AU
moon_to_mars = moon.separation_3d(mars).AU

plt.plot(times, earth_to_mars, '--', label='From Earth')
plt.plot(times, moon_to_mars, '--', label='From Moon')
plt.legend()

plt.xlabel('Time')
plt.ylabel('Distance (AU)')
plt.title('Distance to Mars')

plt.savefig('earth_moon_mars.png')
plt.show()
Eric Duminil
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  • Nice plots. One additional plot of interest would be the Δ between the distances. – gerrit Feb 01 '18 at 12:15
  • @gerrit: Good idea, the plot is better than just a binary one. Is it what you had in mind? – Eric Duminil Feb 01 '18 at 13:12
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    Excellent plots. It is amazing what is possible using so few lines of code. Without Astropy and Plt, it would be much more difficult. – Uwe Feb 01 '18 at 14:21
  • "During this period (2010-2020), Earth is closest 49.6% of the time." This agrees with my intuition: if you draw a large circle representing their shared orbit around the sun, and a smaller circle representing their mutual orbit, then by a very small margin, the percentage of the smaller circle that is outside of the larger circle should be larger than the percentage inside. Note that both the Earth and the moon orbit around a barycenter which in turn orbits the sun in this larger circle, but the Earth's orbit has a smaller radius than the moon's. – Acccumulation Feb 01 '18 at 15:57
  • @Uwe: I wrote a FORTRAN program to calculate the Moon's position a few years ago. It had almost 300 lines of code. The whole Python packages ecosystem feels like cheating : something like from NASA import brains. – Eric Duminil Feb 01 '18 at 17:20
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    @Acccumulation:My intuition would say "exactly 50%", because the question is basically : "On which side of the Earth-Moon bisector is Mars?". No circle is involved, only a straight line. I'm not sure it's correct though, and the moon has a very complex orbit so there might be a very small bias. For what it's worth, I ran the simulation for 1950-2050, and Earth is closest 50.004 % of the time. – Eric Duminil Feb 01 '18 at 19:06
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    @Acccumulation: And 49.9945% of the time for 2050-2150. – Eric Duminil Feb 01 '18 at 19:13
  • Between 1950-2050, Earth will be closest 50.004% of the time. From 2050-2150 the moon will be, 50.0055% of the time. < answer to the question. I'd assume this is when the moon's epileptic points at Mars. – Mazura Feb 02 '18 at 00:34
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    I wish I had more upvotes to give to this answer. – Erik Feb 02 '18 at 03:22
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    @Eric Duminil: it is decades ago when I wrote my last FORTRAN program. Python is so powerful we could not dream of 40 years ago. We had plotting routines for FORTRAN, but not so powerful as Plt. – Uwe Feb 02 '18 at 15:33
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In terms of distance, the two swap considerably. But perhaps a more interesting question is, which of the two is closer in terms of the energy required to land. For that, let's look at our friend, the delta-v table.

enter image description here

Once one is approaching Earth from Mars, things only become different at the point labeledEarth C3=0 (See $C_3$). From there it is about 2.3 km/s to land on the Moon. To land on Earth requires no rocket propellant, as it can all be lost via the atmosphere, thus it is easier to get from Mars to Earth's Surface than to the Moon.

In the reverse direction, it is far easier to get from the Moon to Mars then it is from Earth to Mars. The energy required to leave the Earth is considerable, while it isn't that much from the Moon, relatively speaking.

PearsonArtPhoto
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    That's fascinating. I think my poor characters may be faced with never returning to earth but they're part of a team that had made the trip back and forth previously. This is all so helpful. Thank you. – T. Constantine Feb 01 '18 at 08:04
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To get a sense of distances in the solar system:

The distance Earth-Moon is 380 thousand km. The distance Earth-Mars varies between 50 and 400 million km, i.e. 3 orders of magnitude more.

Hobbes
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    To improve the sense of distances, I would recommend to walk not a small planet walk (distance Sun to Pluto less than 2 km) but a really large one (distance Sun to Pluto more than 10 km) For long distance walkers, there are also planet walks longer than 30 km. – Uwe Jan 31 '18 at 21:33
  • I love this Hobbes, I think I'll be using this helpful information quite a bit. Thank you for your kindness – T. Constantine Feb 01 '18 at 07:59
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    The Sweden Solar System model has 7.6 km from Sun to Earth, 11.6 km from Sun to Mars, 950 km from Sun to Termination Shock. Earth and Moon are in different rooms within the same museum. – gerrit Feb 01 '18 at 12:14
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I think Adam and PearsonArtPhoto have given you the best answers in terms of the effort of getting from place to place. However, since you are asking from the perspective of writing fiction, I want to give a slightly different angle for consideration. Namely economics, and how they might impact the process.

It sounds like you are trying to decide where your characters departed from on their trip to Mars. My response is that it would depend on how advanced/common space travel is in your setting. So the questions I would ask are: How often are such trips made? What percentage of the lift capacity from Earth is intended for Luna, Mars, or elsewhere? How do those numbers compare with trips from those places back to Earth? What about transshipping between such points without landing on Earth?

As explained in other answers, getting from Earth's surface to orbit is the most energy intensive step of any of the potential trips under discussion. Between atmospheric density and acceleration forces, it is also the phase that has the greatest challenges in terms of engineering. Once we have reached a stable orbit, many considerations that govern the design of a launch vehicle don't apply to one that is intended to remain in space, or land on bodies with lighter gravity and/or a thinner atmosphere. (Compare the Apollo rockets to the lunar lander & ascent module designs for a real example of this.)

The point to all of this is, if your society has reached a point where space travel is frequent enough to make it economical, they will stop trying to build craft that go from the ground on one body to the ground on the other and back. They will instead establish some form of transfer station. A launch vessel would get passengers/cargo from Earth to orbit, where they would transfer to another vessel designed to make the trip to Luna, Mars, or wherever, while the launch vessel returned to Earth for another load.

Whether such a transfer station exists at the other end of any given trip would depend on how much traffic goes there. When the volume reaches a point where it is more cost effective to have dedicated surface to orbit craft, and the personnel to operate them, you can expect one to exist.

In the case of Earth and Luna, they are close enough together that the same station would serve both. For someplace like Mars, there would almost have to be a permanent human presence there, and frequent trips there and back, to make it worth while. Until such a time, the transfer station at Earth would be used.

Nathan Tuggy
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Rozwel
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A couple of other answers have pointed out that the orbit of Mars is outside that of the Earth—that is, it's farther away from the sun.

To be more specific about when the moon will be closer to Mars: a full Moon is a full Moon because we're seeing the side of it that's illuminated by the sun. If we drew it out on a plane as viewed from above, when there was a straight line directly from the center of the sun, going through the center of the Earth, then out to the center of the Moon, that would be a full Moon, and the point at which the Moon is closest to the path of Mars' orbit.

At new Moon, there would be a straight line from the sun, through the Moon, to the Earth. That's the time at which the Moon is farthest away from the path of Mars' orbit.

Note that wording though: "the path of Mars' orbit", not just "closest to Mars" or "farthest from Mars". Depending on exactly when you look, Mars could be on the other side of the sun from the Earth. In this case, a new moon (closer to the sun, farther away from Mars' orbit) is also closer to Mars itself than the Earth is.

The Earth's orbit is about 149.6 million kilometers in diameter and Mars' orbit is about 228 million kilometers in diameter (average, in both cases). At their closest, the Earth and Mars are about 78 million kilometers apart. At their farthest, they're about 189 million kilometers apart.

The Moon's orbit around the earth is about 385 thousand kilometers. So even when the Earth and Mars are at their closest, the distance saved by being on the Moon instead of Earth is so small that in normal calculations, it's pretty much lost to rounding error.

If you're interested in traveling from the Earth to Mars, the actual distance is often nearly irrelevant. When sending rockets over long distances, we often use "slingshot" maneuvers in which the rocket travels (what appears to be) a long ways out of its way. For example, consider this animation of the path Juno followed to get from the earth to Jupiter.

It's launched from Earth, travels outward for a ways, then comes back inward toward the sun, and goes quite a bit closer to the sun than the Earth (or Moon) ever does, before finally spiraling back outward to meet up with Jupiter.

A spacecraft won't normally even try to just follow a straight line from point A to point B, so the closest approach between the Earth and Mars won't necessarily be the time that it's fastest or easiest to travel from one to the other.

Nathan Tuggy
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Jerry Coffin
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    The second and third paragraphs are quite misleading, as full moon/new moon are in no useful relation to the time when the Moon is nearer to Mars (or further away) – as the third paragraph explains. – Paŭlo Ebermann Feb 03 '18 at 00:17
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Earth and Mars both orbit the Sun - because Mars is further out, its orbital path/circle is slightly larger. So think of two circles on a piece of paper - the slightly larger one enveloping the smaller is like Mars' orbit around the sun compared to Earth's orbit around the Sun.

Now for the moon - while Earth is orbiting the Sun in its "circle", the moon is making a bunch of laps around Earth in a small circle.

Now on the paper diagram, you can pick a point on Earth's orbital circle and draw a small circle around that point. This represents the moon's orbit around Earth.

With all this, it's clear now that the moon and Earth take turns being closer to Mars. When the moon is in between the Sun and Earth on its mini-circle, Earth is obviously closer to Mars. Conversly, when the moon is on the other side of its circle (putting it in between Earth's and Mars' big circles) it is obviously closer to Mars than Earth is.

So it really depends on how Earth, Mars, and the moon happen to be positioned in their orbits at any given time. On average though, the Moon and Earth are about an equal distance from Mars.

Best of luck on your novel!

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    "When the moon is in between the Sun and Earth on its mini-circle, Earth is obviously closer to Mars" ... this is not true when Mars is on the other side of the Sun (which will happen around half of the time). (Your end result is the same, though.) – Paŭlo Ebermann Feb 02 '18 at 00:12
  • Right, but then the result is just reversed. When the moon is in-between the Earth and the sun, it is closer to Mars. When the Earth is in between the moon and the sun, Earth is closer to Mars. So in the end Earth and the Moon should take turns being closer, on average. – Inertial Ignorance Feb 04 '18 at 08:30
0

Here's probably the best thing for your story.

During the time period of your story, the Earth is always about the same distance from Mars, whereas the moon spins around Earth.

Every 14 days, it gets 1/2 million miles closer to Mars, then every 14 days, 1/2 million miles further away.

enter image description here

Now just to be clear........

over the course of long periods of time - years - "all of the Earth and Moon" and "all of Mars" of course change distances dramatically.

But the fact is during the time period of your story it will be "about the same" distance from Mars to Earth.

- but -

every 14 days the Moon moves "back and fore" 1/2 million miles.

Whereas during the time period of your story the Earth is always about the same distance.

No matter now far the distance between Mars/Earth - of course, as the years pass that changes dramatically - every 14 days the Moon moves "closer and further away" by 1/2 million miles.

Got it? Good! :)

Fattie
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  • "Now just to be clear........" <-- unfortunately not clear enough. Perhaps your apology needs to come first, before folks read the inaccurate statement. I'll propose an edit. – Phil Feb 02 '18 at 11:40
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    How do you know the time period of the story? The distance from Earth (and Moon) to Mars varies by a factor of 5 to 7 over the span of 26 months, and many stories take longer. – Paŭlo Ebermann Feb 03 '18 at 00:12
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I particularly appreciated the answer @Adám gave.

Wolfram Alpha has several interesting facts about this.

If you take the average distance from Earth to Mars and, the average distance from Luna to Mars, it is the same. The average distance to Mars is 14.1 light minutes.

Willtech
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  • Question to Wolfram Alpha: Average distance from the Earth's Moon to Mars? Answer: Wolfram|Alpha doesn't understand your query.
    Question: Average distance from Luna to Mars? Answer: Check your spelling. Give your input in English.
    – Uwe Feb 02 '18 at 16:44
  • @Uwe Wolfram Alpha has several interesting facts about this on the link I provided. I did not say it knows every detail. – Willtech Feb 02 '18 at 23:49