<p>Yes, but only just. The best-case (called &quot;diffraction-limited&quot;) angular resolution of a telescope (i.e. the smallest detail it can make out) is given by
<span class="math-container">$$ \theta = 1.22 \frac{\lambda}{D} $$</span>
Where <span class="math-container">$\lambda$</span> is the wavelength of light you're observing in (say, 500nm for visible light) and <span class="math-container">$D$</span> is the diameter of the telescope. With a 150m diameter, this is <span class="math-container">$4 \times 10^{-9}$</span> radians.</p>
<p>Multiply that by the distance to the moon (400,000km) to get the smallest resolvable detail: <strong>1.6m</strong>, only just big enough to see an astronaut. They would appear as a tiny white dot.</p>
<p>This is an absolute <em>best-case</em> scenario — luckily, real space telescopes tend to get pretty close to the diffraction limit.</p>

CC BY-SA 4.0
2022-09-12T14:59:43.853Z
In fact, this means that it's *easier* to see them from the side. From directly overhead you would need to be able to resolve about half that.
Barmar
CC BY-SA 4.0
2022-09-12T19:38:41.190Z
If the 150m telescope can resolve down to 1.6m, doesn't that mean we still could see the astronaut if s/he was standing near the limb of the moon, as suggested by James K's answer?
William
CC BY-SA 4.0
2022-09-13T00:39:43.740Z
@Barmar, depends on the contrast.  If something is significantly brighter or significantly darker than the surroundings, you can see it even if it's below the diffraction limit.  (See, for example, the stars: every one of them is far below the diffraction limit for your eye.)
Mark
CC BY-SA 4.0
2022-09-13T09:53:37.410Z
So if SpaceX landed a Starship on the limb, that would be relatively easy to spot. Best. Photo. Ever.
DrMcCleod
CC BY-SA 4.0
2022-09-13T15:08:54.773Z
@DrMcCleod: It could be another interesting question : how big should the telescope be in order to see at least a few pixels worth of Starship landing on the moon?
Eric Duminil
CC BY-SA 4.0
2022-09-13T16:51:21.403Z
@EricDuminil: Starship HLS is approximately 50 meters by 9 meters. So if it was right at the limb, you'd need about a 500 mm (20 inch) diameter scope to get an image of roughly 1x6 pixels. Looking at it "head on" you'd need roughly a one meter scope to get a 2x2 pixel image (in both cases, assuming a high contrast between the HLS and the background).
Jerry Coffin
CC BY-SA 4.0
2022-09-13T17:13:45.270Z
@JerryCoffin: Thanks a lot. It would be 6 beautiful pixels! :D
Eric Duminil
CC BY-SA 4.0
2022-09-13T17:31:24.603Z
@EricDuminil: I should add that (much like Steve's answer) that's the theoretical limit. For an earth-based telescope, the atmosphere places a much lower limit on available resolution.
Jerry Coffin
CC BY-SA 4.0
2022-09-12T08:27:52.190Z
2022-09-12T08:27:52.190Z
Steve R
<blockquote>
<p>would it be possible to get a side view rather than a top view of a
person standing on the moon</p>
</blockquote>
<p>It's kind of possible to do both at the same time. Simply wait for the person to be at the <a href="https://en.wikipedia.org/wiki/Terminator_(solar)#Lunar_terminator" rel="noreferrer">lunar terminator</a>. As seen from the moon, the sun will be so low on the horizon that the shadow of the person can get <em>really</em> long:</p>
<p><a href="https://commons.wikimedia.org/wiki/File:Shadow_Person.jpg" rel="noreferrer"><img src="https://i.sstatic.net/T4ttQ.jpg" alt="very long person shadow" /></a></p>
<p>Your powerful telescope will not be able to resolve the head from above, but might resolve a few pixels of the shadow.</p>
<p>This is also the reason why you see different craters on the moon every day : you don't see the ones in the shadow, you don't see much detail at all when they're completely sunlit, but you see very crisp details of the craters that are along the terminator.</p>
<p>This <a href="https://en.wikipedia.org/wiki/Rupes_Recta" rel="noreferrer">&quot;wall&quot; on the moon</a> is only visible during a few hours every month, for example, when the incoming light has just the correct angle:</p>
<p><a href="https://en.wikipedia.org/wiki/Rupes_Recta#/media/File:Ancient_Thebit_(LRO).png" rel="noreferrer"><img src="https://i.sstatic.net/j7zXn.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-09-12T21:40:31.327Z
I love it! I also love how the regolith looks a bit brighter around the edges of the figure's shadow due to [shadow hiding/opposition surge](https://en.wikipedia.org/wiki/Opposition_surge)
uhoh
CC BY-SA 4.0
2022-09-12T11:03:57.680Z
2022-09-12T11:03:57.680Z
Eric Duminil
<p>No, assuming you want to see that the thing you are observing is a human as opposed to a rock.</p>
<p>Steve R beat me to this answer based on the diffraction limit of a telescope. To distinguish between a human and a rock, the telescope would need to be able to have a resolution of 20 centimeters, the size of a human head. Using the Raleigh criterion for the resolution of a telescope,
<span class="math-container">$$\theta \approx 1.22 \frac{\lambda}{D}$$</span>
where <span class="math-container">$\theta$</span> is the angular resolution of the telescope, <span class="math-container">$\lambda$</span> is the wavelength of the electromagnetic radiation being received (I'll use 600 nanometers), and <span class="math-container">$D$</span> is the diameter of the mirror.</p>
<p>Using the small angle approximation, if <span class="math-container">$d_1$</span> is the distance between the telescope and the object of interest, and <span class="math-container">$d_2$</span> is the size of the object, we want
<span class="math-container">$$\theta d_1 &lt; d_2$$</span>
or
<span class="math-container">$$d1 &lt; \frac{d_2 D}{1.22 \lambda}$$</span></p>
<p>Plugging in the numbers (<span class="math-container">$d_2 = 20\,\text{cm}$</span>, <span class="math-container">$D = 150\,\text{m}$</span>, <span class="math-container">$\lambda = 600\,\text{nm}$</span>) yields a distance between the telescope and subject of no more than 41000 km. That's orbiting the Moon rather than the Earth.</p>

CC BY-SA 4.0
2022-09-12T12:59:54.657Z
Actually, you probably wouldn't need anywhere as good resolution as 20 cm to distinguish a person from a _rock_: if she's wearing a bright and/or coloured spacesuit, it will be obvious even in an image that's blurred over 5 m. What you couldn't do is distinuish her from a piece of equipment of the same colour.
leftaroundabout
CC BY-SA 4.0
2022-09-12T14:37:08.230Z
How large would the telescope need to be to allow this? Easier to build a bigger telescope than to move the moon close to earth. ;)
AnoE
CC BY-SA 4.0
2022-09-12T10:19:43.697Z
2022-09-12T10:19:43.697Z
David Hammen
<p>Using an orbit that get close to the Moon regularly can be an option and you can take photos while they are close together, but that still have a huge impact on the orbit in a long run.</p>
<p>But at least once you'll be able to do such picture. I guess you can go quite close to the Moon without any harm to the telescope as it doesn't have real atmosphere, but this will change the telescope's orbit totally.</p>
<p>L2 Lagrange point seems much closer to Moon than the Earth, but I guess it's still not close enough.</p>
<p>    <img src="https://i.sstatic.net/eXWhr.jpg" alt="enter image description here" /></p>
<p>    <sup>The Lagrange points for the Earth-moon system. Credit: David A. Kring, <a href="http://www.lpi.usra.edu/exploration/" rel="nofollow noreferrer">LPI-JSC Center for Lunar Science and Exploration</a></sup></p>

CC BY-SA 4.0
2022-09-14T13:47:29.820Z
This is the only answer that actually comes close to answering the question. All of the discussion of diffraction limits/the limb of the moon/etc is a distraction. It doesn't even need to be a particularly powerful telescope provided it comes close enough, it just needs to be to the side of the person it's imaging...who will of course then appear near the limb of the moon from the vantage point of that telescope. If it's *not* to the side, a more powerful telescope isn't going to help.
Christopher James Huff
CC BY-SA 4.0
2022-09-14T12:21:59.250Z
2022-09-14T12:21:59.250Z
Arpad Horvath
<p>Summary:<br />
To see someone &quot;from the side&quot; you need to view them side on. This occurs, as others have noted, when they are &quot;on a limb&quot; - where limb is the whole circle edge at full moon. The edge on terrain in the images below shows the principle. Other answers indicate how refraction sets a limit to the smallest size of an image which can be viewed.</p>
<hr />
<p>You can see the equivalent now.</p>
<p>Look at good quality high resolution images of the Moon.<br />
Look at the edges of the disk - you'll see they are rough, not smooth. That's the lunar lanscape seen side on. You can see some mountains / large craters quite clearly.</p>
<p><a href="https://cosmicbackground.io/products/the-hunt-for-artemis?variant=42502042681513" rel="nofollow noreferrer">Here</a> you can find A &quot;lower res&quot; version of the highest resolution available. The full resolution version is 11022 x 11022 pixels -
but it costs money.</p>
<p>This is a crop of the top left and bottom left corners of a medium resolution  showing side on detail.</p>
<p><a href="https://i.sstatic.net/GHQXp.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GHQXp.jpg" alt="enter image description here" /></a></p>
<p>.</p>

CC BY-SA 4.0
2022-09-14T07:46:36.973Z
2022-09-15T01:24:30.660Z
2022-09-15T01:24:30.660Z
Russell McMahon
<p>Yes.</p>
<p>The person just has to be standing somewhere near the <em>limb</em> of the moon.  That is near the edge of the apparent disc of the moon.</p>
<p>We see everything near the limb of the moon from the side, at all times.  This is because the moon is (roughly) a sphere. It's not flat.</p>

CC BY-SA 4.0
2022-09-11T20:32:40.640Z
Thank you very much. This perfectly answers my question.
William
CC BY-SA 4.0
2022-09-12T13:04:04.783Z
This would apply to ground-based systems as well.
Carl Witthoft
CC BY-SA 4.0
2022-09-12T14:57:08.497Z
Are Flat-Earthers also Flat-Mooners?
Barmar
CC BY-SA 4.0
2022-09-12T18:25:54.297Z
Is "limb" the same as "horizon" in this context ?
Criggie
CC BY-SA 4.0
2022-09-12T18:32:12.577Z
Probably, depending on what you mean by "horizon", but from the perspective of Earth, "limb" is the correct word.  See https://astronomy.stackexchange.com/questions/49600/name-for-the-boundary-between-near-and-far-sides-relative-to-an-observer/49601#49601
James K
CC BY-SA 4.0
2022-09-12T19:50:15.097Z
@Barmar They are the same.  You never see the moon rotate, you always see the same side from Earth, therefore it must be flat.  If the moon is flat, then the earth also must be.
rtaft
CC BY-SA 4.0
2022-09-13T12:20:23.620Z
@Criggie Since the Moon is tidally locked, it makes sense to talk about a near side and a far side of the Moon. So another way to say the person should stand near the limb, would be to say that the person should stand near or in the "border region" between the near side and the far side.
Jeppe Stig Nielsen
CC BY-SA 4.0
2022-09-14T05:20:48.300Z
@Barmar IIRC, early NASA missions modeled the flight with a flat earth.  Certainly not that usable with Apollo 10 onward.
chux
CC BY-SA 4.0
2022-09-14T12:15:47.443Z
@chux-ReinstateMonica how can one model an orbit using a flat-Earth approximation? Orbiting requires moving full 360°, flat-Earth approximation must break down in this case.
Ruslan
CC BY-SA 4.0
2022-09-14T14:52:28.110Z
@chux-ReinstateMonica I think even artillary calculations at anything other than point-blank range requires taking earth curvature into effect. So I doubt they could really use that simplification in space flight. OTOH, if you're calculating driving distance, the winding of the roads will usually have more significance than earth curvature.
Barmar
CC BY-SA 4.0
2022-09-14T16:47:31.990Z
@Ruslan you could model an orbit as a sinusoidal curve on a 2D map.
Drake P
CC BY-SA 4.0
2022-09-14T18:45:27.377Z
@DrakeP such a model has no predictive power. It might only be useful for a video game, not for flight control.
Ruslan
CC BY-SA 4.0
2022-09-11T20:23:27.110Z
2022-09-11T20:23:27.110Z
James K
<p>If we had a very powerful telescope with say a 150-meter diameter, placed in orbit around earth, would it be possible to get a side view rather than a top view of a person standing on the moon?</p>

CC BY-SA 4.0
2022-09-11T20:11:12.127Z
Size of the telescope has no bearing on whether the view you get is top-down or side on. A larger telescope just allows you to see more detail from the same angle. It's the position of your target on the surface of the moon and the position of your telescope  that determines  the orientation of what's in the image.
notovny
CC BY-SA 4.0
2022-09-11T20:15:11.427Z
@notovny Considering the Earth-moon system, would there be such a configuration at a certain time as to allow a side view?
William
CC BY-SA 4.0
2022-09-12T09:48:54.763Z
Where in orbit about the Earth? LEO? GEO? Further out? Unless it's much further out than GEO, I would say the answer is no.
David Hammen
CC BY-SA 4.0
2022-09-12T19:47:13.070Z
@DavidHammen Thank you David but some of the answers below doesn't seem to agree with you.
William
CC BY-SA 4.0
2022-09-14T07:28:02.183Z
Data point: Look at good quality high resolution images of the Moon. Look at the edges of the disk - you'll see they are rough, not smooth. That's the lunar lanscape seen side on. You can see some mountains / large craters quite clearly.
Russell McMahon
CC BY-SA 4.0
2022-09-14T13:30:51.900Z
@RussellMcMahon that's an issue of location on the lunar surface, not power of the telescope.
Christopher James Huff
CC BY-SA 4.0
2022-09-15T01:22:06.063Z
@ChristopherJamesHuff  To seem someone "from the side" you need to view them side on. This occurs, as others have noted, when they are "on a limb" - where limb is the whole circle edge at full moon. The edge on terraqin shows the principle. The other answers indicate how refraction sets a limit.
Russell McMahon
CC BY-SA 4.0
2022-09-15T13:51:14.290Z
@RussellMcMahon The answers based on *diffraction* are only confusing the issue: if you are viewing them from the side, you don't even need a telescope. There is no limit. If you are *not* viewing them from the side, it doesn't matter how powerful your telescope is.
Christopher James Huff
<telescope>
Can a very powerful telescope show a person standing on the moon from the side rather than from directly overhead?
William
# Can a very powerful telescope show a person standing on the moon from the side rather than from directly overhead?

(Asked by: username_0 on 2022-09-11)

If we had a very powerful telescope with say a 150-meter diameter, placed in orbit around earth, would it be possible to get a side view rather than a top view of a person standing on the moon?

**Tags:** telescope
> **Comments:**
> **username_1**: Size of the telescope has no bearing on whether the view you get is top-down or side on. A larger telescope just allows you to see more detail from the same angle. It's the position of your target on the surface of the moon and the position of your telescope  that determines  the orientation of what's in the image.  
> **username_0**: @notovny Considering the Earth-moon system, would there be such a configuration at a certain time as to allow a side view?  
> **username_2**: Where in orbit about the Earth? LEO? GEO? Further out? Unless it's much further out than GEO, I would say the answer is no.  
> **username_0**: @DavidHammen Thank you David but some of the answers below doesn't seem to agree with you.  
> **username_3**: Data point: Look at good quality high resolution images of the Moon. Look at the edges of the disk - you'll see they are rough, not smooth. That's the lunar lanscape seen side on. You can see some mountains / large craters quite clearly.  
> **username_4**: @RussellMcMahon that's an issue of location on the lunar surface, not power of the telescope.  
> **username_3**: @ChristopherJamesHuff  To seem someone "from the side" you need to view them side on. This occurs, as others have noted, when they are "on a limb" - where limb is the whole circle edge at full moon. The edge on terraqin shows the principle. The other answers indicate how refraction sets a limit.  
> **username_4**: @RussellMcMahon The answers based on *diffraction* are only confusing the issue: if you are viewing them from the side, you don't even need a telescope. There is no limit. If you are *not* viewing them from the side, it doesn't matter how powerful your telescope is.  

---

## 6 answers

### Answer by username_15 on 2022-09-11. Score: 35 | Accepted answer

Yes.

The person just has to be standing somewhere near the *limb* of the moon. That is near the edge of the apparent disc of the moon.

We see everything near the limb of the moon from the side, at all times. This is because the moon is (roughly) a sphere. It's not flat.

> **Comments:**
> **username_0**: Thank you very much. This perfectly answers my question.  
> **username_16**: This would apply to ground-based systems as well.  
> **username_6**: Are Flat-Earthers also Flat-Mooners?  
> **username_17**: Is "limb" the same as "horizon" in this context ?  
> **username_15**: Probably, depending on what you mean by "horizon", but from the perspective of Earth, "limb" is the correct word.  See https://astronomy.stackexchange.com/questions/49600/name-for-the-boundary-between-near-and-far-sides-relative-to-an-observer/49601#49601  
> **username_18**: @Barmar They are the same.  You never see the moon rotate, you always see the same side from Earth, therefore it must be flat.  If the moon is flat, then the earth also must be.  
> **username_19**: @Criggie Since the Moon is tidally locked, it makes sense to talk about a near side and a far side of the Moon. So another way to say the person should stand near the limb, would be to say that the person should stand near or in the "border region" between the near side and the far side.  
> **username_20**: @Barmar IIRC, early NASA missions modeled the flight with a flat earth.  Certainly not that usable with Apollo 10 onward.  
> **username_21**: @chux-ReinstateMonica how can one model an orbit using a flat-Earth approximation? Orbiting requires moving full 360°, flat-Earth approximation must break down in this case.  
> **username_6**: @chux-ReinstateMonica I think even artillary calculations at anything other than point-blank range requires taking earth curvature into effect. So I doubt they could really use that simplification in space flight. OTOH, if you're calculating driving distance, the winding of the roads will usually have more significance than earth curvature.  
> **username_22**: @Ruslan you could model an orbit as a sinusoidal curve on a 2D map.  
> **username_21**: @DrakeP such a model has no predictive power. It might only be useful for a video game, not for flight control.  

---

### Answer by username_5 on 2022-09-12. Score: 28

Yes, but only just. The best-case (called "diffraction-limited") angular resolution of a telescope (i.e. the smallest detail it can make out) is given by
$$ \theta = 1.22 \frac{\lambda}{D} $$
Where $\lambda$ is the wavelength of light you're observing in (say, 500nm for visible light) and $D$ is the diameter of the telescope. With a 150m diameter, this is $4 \times 10^{-9}$ radians.

Multiply that by the distance to the moon (400,000km) to get the smallest resolvable detail: **1.6m**, only just big enough to see an astronaut. They would appear as a tiny white dot.

This is an absolute *best-case* scenario — luckily, real space telescopes tend to get pretty close to the diffraction limit.

> **Comments:**
> **username_6**: In fact, this means that it's *easier* to see them from the side. From directly overhead you would need to be able to resolve about half that.  
> **username_0**: If the 150m telescope can resolve down to 1.6m, doesn't that mean we still could see the astronaut if s/he was standing near the limb of the moon, as suggested by James K's answer?  
> **username_7**: @Barmar, depends on the contrast.  If something is significantly brighter or significantly darker than the surroundings, you can see it even if it's below the diffraction limit.  (See, for example, the stars: every one of them is far below the diffraction limit for your eye.)  
> **username_8**: So if SpaceX landed a Starship on the limb, that would be relatively easy to spot. Best. Photo. Ever.  
> **username_9**: @DrMcCleod: It could be another interesting question : how big should the telescope be in order to see at least a few pixels worth of Starship landing on the moon?  
> **username_10**: @EricDuminil: Starship HLS is approximately 50 meters by 9 meters. So if it was right at the limb, you'd need about a 500 mm (20 inch) diameter scope to get an image of roughly 1x6 pixels. Looking at it "head on" you'd need roughly a one meter scope to get a 2x2 pixel image (in both cases, assuming a high contrast between the HLS and the background).  
> **username_9**: @JerryCoffin: Thanks a lot. It would be 6 beautiful pixels! :D  
> **username_10**: @EricDuminil: I should add that (much like Steve's answer) that's the theoretical limit. For an earth-based telescope, the atmosphere places a much lower limit on available resolution.  

---

### Answer by username_9 on 2022-09-12. Score: 19

> would it be possible to get a side view rather than a top view of a
> person standing on the moon

It's kind of possible to do both at the same time. Simply wait for the person to be at the [lunar terminator](https://en.wikipedia.org/wiki/Terminator_(solar)#Lunar_terminator). As seen from the moon, the sun will be so low on the horizon that the shadow of the person can get *really* long:

[![very long person shadow](https://i.sstatic.net/T4ttQ.jpg)](https://commons.wikimedia.org/wiki/File:Shadow_Person.jpg)

Your powerful telescope will not be able to resolve the head from above, but might resolve a few pixels of the shadow.

This is also the reason why you see different craters on the moon every day : you don't see the ones in the shadow, you don't see much detail at all when they're completely sunlit, but you see very crisp details of the craters that are along the terminator.

This ["wall" on the moon](https://en.wikipedia.org/wiki/Rupes_Recta) is only visible during a few hours every month, for example, when the incoming light has just the correct angle:

[![enter image description here](https://i.sstatic.net/j7zXn.jpg)](https://en.wikipedia.org/wiki/Rupes_Recta#/media/File:Ancient_Thebit_(LRO).png)

> **Comments:**
> **username_11**: I love it! I also love how the regolith looks a bit brighter around the edges of the figure's shadow due to [shadow hiding/opposition surge](https://en.wikipedia.org/wiki/Opposition_surge)  

---

### Answer by username_2 on 2022-09-12. Score: 10

No, assuming you want to see that the thing you are observing is a human as opposed to a rock.

Steve R beat me to this answer based on the diffraction limit of a telescope. To distinguish between a human and a rock, the telescope would need to be able to have a resolution of 20 centimeters, the size of a human head. Using the Raleigh criterion for the resolution of a telescope,
$$\theta \approx 1.22 \frac{\lambda}{D}$$
where $\theta$ is the angular resolution of the telescope, $\lambda$ is the wavelength of the electromagnetic radiation being received (I'll use 600 nanometers), and $D$ is the diameter of the mirror.

Using the small angle approximation, if $d\_1$ is the distance between the telescope and the object of interest, and $d\_2$ is the size of the object, we want
$$\theta d\_1 < d\_2$$
or
$$d1 < \frac{d\_2 D}{1.22 \lambda}$$

Plugging in the numbers ($d\_2 = 20\,\text{cm}$, $D = 150\,\text{m}$, $\lambda = 600\,\text{nm}$) yields a distance between the telescope and subject of no more than 41000 km. That's orbiting the Moon rather than the Earth.

> **Comments:**
> **username_12**: Actually, you probably wouldn't need anywhere as good resolution as 20 cm to distinguish a person from a _rock_: if she's wearing a bright and/or coloured spacesuit, it will be obvious even in an image that's blurred over 5 m. What you couldn't do is distinuish her from a piece of equipment of the same colour.  
> **username_13**: How large would the telescope need to be to allow this? Easier to build a bigger telescope than to move the moon close to earth. ;)  

---

### Answer by username_14 on 2022-09-14. Score: 2

Using an orbit that get close to the Moon regularly can be an option and you can take photos while they are close together, but that still have a huge impact on the orbit in a long run.

But at least once you'll be able to do such picture. I guess you can go quite close to the Moon without any harm to the telescope as it doesn't have real atmosphere, but this will change the telescope's orbit totally.

L2 Lagrange point seems much closer to Moon than the Earth, but I guess it's still not close enough.

![enter image description here](https://i.sstatic.net/eXWhr.jpg)

The Lagrange points for the Earth-moon system. Credit: David A. Kring, [LPI-JSC Center for Lunar Science and Exploration](http://www.lpi.usra.edu/exploration/)

> **Comments:**
> **username_4**: This is the only answer that actually comes close to answering the question. All of the discussion of diffraction limits/the limb of the moon/etc is a distraction. It doesn't even need to be a particularly powerful telescope provided it comes close enough, it just needs to be to the side of the person it's imaging...who will of course then appear near the limb of the moon from the vantage point of that telescope. If it's *not* to the side, a more powerful telescope isn't going to help.  

---

### Answer by username_3 on 2022-09-14. Score: 2

Summary:  
To see someone "from the side" you need to view them side on. This occurs, as others have noted, when they are "on a limb" - where limb is the whole circle edge at full moon. The edge on terrain in the images below shows the principle. Other answers indicate how refraction sets a limit to the smallest size of an image which can be viewed.

---

You can see the equivalent now.

Look at good quality high resolution images of the Moon.  
Look at the edges of the disk - you'll see they are rough, not smooth. That's the lunar lanscape seen side on. You can see some mountains / large craters quite clearly.

[Here](https://cosmicbackground.io/products/the-hunt-for-artemis?variant=42502042681513) you can find A "lower res" version of the highest resolution available. The full resolution version is 11022 x 11022 pixels -
but it costs money.

This is a crop of the top left and bottom left corners of a medium resolution showing side on detail.

[![enter image description here](https://i.sstatic.net/GHQXp.jpg)](https://i.sstatic.net/GHQXp.jpg)

.


---

