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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.

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    $\begingroup$ +1 A clear, concise question that generates three thoughtful answers deserves to be up voted as well. $\endgroup$ – uhoh Feb 1 '18 at 0:28
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    $\begingroup$ @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. $\endgroup$ – jamesqf Feb 1 '18 at 3:52
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    $\begingroup$ For the record, Luna is not the name of The Moon. The name of The Moon is, quite simply, "The Moon." $\endgroup$ – corsiKa Feb 2 '18 at 21:55
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    $\begingroup$ 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: what-if.xkcd.com/58 $\endgroup$ – Polygnome Feb 2 '18 at 23:43
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    $\begingroup$ Isn't this just a bad question? I'm pretty amazed at how much time people have spent answering it... and it's also standing at 31 votes as I speak. T. Constantine may of course be 5 years old... I'm pretty sure that by the age of 8 (the year people landed on the Moon for the first time, which I watched as an 8-year-old!) I knew a little bit about the planets. Isn't the best answer: look up the Wikipedia page entitled "Solar System" and think about it a bit? $\endgroup$ – mike rodent Feb 4 '18 at 17:44

10 Answers 10

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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

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    $\begingroup$ It doesn't happen often when talking about space, but you overestimated the distance by a lot. 400E6 km / 360000km ⋅ 1m is only about 1km. So you and your dog are at Westminster Abbey, OP is at Trafalgar Square. And that's only at conjunction (~400 million km). At opposition (~55 million km), you'd both be at Trafalgar Square, on opposite corners. $\endgroup$ – Eric Duminil Feb 1 '18 at 10:00
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    $\begingroup$ @EricDuminil This answer does not assert its example is to scale. $\endgroup$ – jpmc26 Feb 1 '18 at 10:29
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    $\begingroup$ @jpmc26 indeed, but since the analogy is used as an argument to ignore the leash as a rounding error, it shouldn't be off by 5 orders of magnitude. With this margin, it could also be "proven" that the OP is within dog's reach. $\endgroup$ – Eric Duminil Feb 1 '18 at 16:41
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    $\begingroup$ @EricDuminil He didn't say the dog was on leash. He just said that it is running in circles around him. Perhaps those circles have a radius of 10 km. :-) Regardless of the details, his basic point is still valid. The distance from the Earth to the Moon is about 0.17% of the distance from the Earth to Mars. (385e3 km vs 225e6 km on average) $\endgroup$ – Jay Feb 1 '18 at 19:43
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    $\begingroup$ Let me rephrase. While it's a great analogy, the distances are off by too many orders of magnitude. Dan should edit it. In short, What @EricDuminil said is, simply, utterly correct. It shouldn't be off by 5 orders of magnitude. It's that simple. The answer should be edited. $\endgroup$ – Fattie Feb 2 '18 at 15:42
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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.

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    $\begingroup$ @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. $\endgroup$ – Hurkyl Feb 1 '18 at 3:30
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    $\begingroup$ 'Mars is much closer to Luna than to Earth' - unless I'm misreading something, your table says otherwise. $\endgroup$ – Rob Feb 1 '18 at 5:11
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    $\begingroup$ 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... $\endgroup$ – Pedro A Feb 1 '18 at 10:09
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    $\begingroup$ @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. $\endgroup$ – Pedro A Feb 1 '18 at 10:33
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    $\begingroup$ @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. $\endgroup$ – Kyle A Feb 1 '18 at 17:36
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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()
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  • $\begingroup$ Nice plots. One additional plot of interest would be the Δ between the distances. $\endgroup$ – gerrit Feb 1 '18 at 12:15
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    $\begingroup$ Excellent plots. It is amazing what is possible using so few lines of code. Without Astropy and Plt, it would be much more difficult. $\endgroup$ – Uwe Feb 1 '18 at 14:21
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    $\begingroup$ @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. $\endgroup$ – Eric Duminil Feb 1 '18 at 19:06
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    $\begingroup$ @Acccumulation: And 49.9945% of the time for 2050-2150. $\endgroup$ – Eric Duminil Feb 1 '18 at 19:13
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    $\begingroup$ I wish I had more upvotes to give to this answer. $\endgroup$ – Erik Feb 2 '18 at 3:22
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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.

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    $\begingroup$ 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. $\endgroup$ – T. Constantine Feb 1 '18 at 8:04
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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.

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    $\begingroup$ 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. $\endgroup$ – Uwe Jan 31 '18 at 21:33
  • $\begingroup$ I love this Hobbes, I think I'll be using this helpful information quite a bit. Thank you for your kindness $\endgroup$ – T. Constantine Feb 1 '18 at 7:59
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    $\begingroup$ 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. $\endgroup$ – gerrit Feb 1 '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.

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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.

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    $\begingroup$ 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. $\endgroup$ – Paŭlo Ebermann Feb 3 '18 at 0:17
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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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    $\begingroup$ "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.) $\endgroup$ – Paŭlo Ebermann Feb 2 '18 at 0:12
  • $\begingroup$ 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. $\endgroup$ – Inertial Ignorance Feb 4 '18 at 8:30
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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! :)

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  • $\begingroup$ "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. $\endgroup$ – Phil Feb 2 '18 at 11:40
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    $\begingroup$ 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. $\endgroup$ – Paŭlo Ebermann Feb 3 '18 at 0:12
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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.

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  • $\begingroup$ 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. $\endgroup$ – Uwe Feb 2 '18 at 16:44
  • $\begingroup$ @Uwe Wolfram Alpha has several interesting facts about this on the link I provided. I did not say it knows every detail. $\endgroup$ – Willtech Feb 2 '18 at 23:49

protected by Hobbes Feb 2 '18 at 8:32

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