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On October 3, 2007 NASA's Mars Reconnaissance Orbiter HiRISE camera acquired this image of the earth and the moon from Mars enter image description here

The oceans and clouds can be easily distinguished and "the west coast outline of South America at lower right" is also visible.

What magnification would it be considered to render such detail in the image?

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    $\begingroup$ some info on HiRISE in this answer and its resolution on this and this and this and this page and links to how imaging works (but mostly about color) on this page $\endgroup$ – uhoh Mar 19 at 0:33
  • $\begingroup$ I see you've edited your question after I posted an answer. I've been editing it and addressed magnification further, but at some point a question that's primarily about how things "look" to people through telescopes is going to be off-topic in Space Exploration but on-topic in Astronomy SE. $\endgroup$ – uhoh Mar 19 at 2:09
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    $\begingroup$ @DrSheldon say my user name! :-) $\endgroup$ – uhoh Mar 19 at 2:14
  • $\begingroup$ Magnification doesn't matter directly. The pertinent factors are the angular subtense of each pixel and the lens diffraction limit (Airy Disc) relative to pixel size. $\endgroup$ – Carl Witthoft Mar 19 at 13:06
  • $\begingroup$ I didn't know that. Glad I asked here, learned something new. $\endgroup$ – Bob516 Mar 19 at 13:39
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PIA10244: Earth and Moon as Seen from Mars article you link to links to the JPL image catalog page PIA10244: Earth and Moon as Seen from Mars which says:

The High Resolution Imaging Science Experiment (HiRISE) camera would make a great backyard telescope for viewing Mars, and we can also use it at Mars to view other planets. This is an image of Earth and the moon, acquired on October 3, 2007, by the HiRISE camera on NASA's Mars Reconnaissance Orbiter.

At the time the image was taken, Earth was 142 million kilometers (88 million miles) from Mars, giving the HiRISE image a scale of 142 kilometers (88 miles) per pixel, an Earth diameter of about 90 pixels and a moon diameter of 24 pixels. The phase angle is 98 degrees, which means that less than half of the disk of the Earth and the disk of the moon have direct illumination. We could image Earth and moon at full disk illumination only when they are on the opposite side of the sun from Mars, but then the range would be much greater and the image would show less detail.

12756/90 ~ 142 km/pixel  at 142 Mkm is 1.0 uRad
3474/24  ~ 145 km/pixel  at 142 Mkm is 1.0 uRad

Let's call the image scale 143 km/pixel and based on the fact that the transition from white to black at the Moon's terminator is almost complete from one pixel to the next, let's call the resolution 143 km and therefore 1 microradian, which conveniently agrees with Wikipedia's HiRISE design:

HiRISE was designed to be a high resolution camera from the beginning. It consists of a large mirror, as well as a large CCD camera. Because of this, it achieves a resolution of 1 microradian, or 0.3 meter at a height of 300 km. (For comparison purposes, satellite images on Google Mars are available to 1 meter.[21]) It can image in three color bands, 400–600 nm (blue-green or B-G), 550–850 nm (red) and 800–1,000 nm (near infrared or NIR)

Also see https://www.uahirise.org/specs/ for verification of 1.0 uRad

Below I've plotted a quick analysis of the blue channel in the artificially colored TIFF at the image's website linked above. HiRISE's wavelength channels do not really correspond to Human's three visible color channels, so they have worked some magic to make "lifelike" color representations.

But wait! What about magnification?

Mars Reconnaissance Orbiter's High Resolution Imaging Science Experiment (HiRISE) says:

The telescope (Figures 2 and 3) is a 50 cm aperture, f/24 system.

and

2.3. Focal Plane Subsystem (FPS)

The FPS consists of the CCDs, a focal plane substrate of aluminum–graphite composite material, a spectral filter assembly, and CCD processing/memory modules (Figure 4). Each CCD has 2048 12 × 12 um pixels in the cross‐scan direction and 128 TDI elements (stages) in the along‐track direction. The CCDs are staggered to provide full swath coverage without gaps (Figure 5). There are two CCDs for both the BG and NIR bands and 10 CCDs for the RED band. The staggered CCDs overlap by 48±5 pixels with adjacent CCDs in the same color band pass; this provides an effective swath width of ∼20,048 pixels for the RED images and ∼4,048 pixels for the blue‐green and NIR images. The BG and NIR channels overlap with the central two RED CCDs to provide 3‐color coverage.

If 143 kilometers is 12 microns then the magnification is about 8.4E-11, so we can call that the demagnification. Try it another way; the focal length is 0.5*24 = 12 meters. Divide the focal length by the distance of 142 million km and we get almost the same thing: 8.5E-11.

Hey! That's not what I mean by "magnification"!

Okay, what if I were to fly to Mars orbit, visit the MRO spacecraft and put a very nice 12.5 millimeter focal length eyepiece at the focal plane where the CCDs normally sit and scan? M = F_system / F_eyepiece = 960 x

Okay, but what is my Real Question? What if I asked in Astronomy SE?

You can just ask "Ignoring atmospheric effects, what telescope magnification is necessary to resolve 1 microradian by eye?" in Astronomy SE. They might answer

In order to make 1 microradian appear to be 28 arcseconds you need a magnification of at least 135X. For comfort and average eyes better multiply that by 3 and use 400 X.

HiRISE is a red herring here as is the image. It's background/context for why you are interested, but not related to the calculation.

quickie analysis of PIA10244 Earth and Moon as Seen from Mars

import numpy as np
import matplotlib.pyplot as plt
from numpy import fft # didn't use https://docs.scipy.org/doc/numpy/reference/routines.fft.html

earth = plt.imread('PIA10244.png')[535:555, 170:211, :3]
moon  = plt.imread('PIA10244.png')[177:187, 748:763, :3]

if True:
    plt.figure()
    for i, thing in enumerate((earth, moon)):
        plt.subplot(2, 2, i+1)
        plt.imshow(thing, interpolation='none')
        plt.subplot(2, 2, i+2+1)
        for line in thing[..., 2]:
            plt.plot(line)
    plt.show()
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  • $\begingroup$ I've read this over a few times I don't see an answer to the question of the apparent magnification. I readily admit I understand only part of your answer, so I could easily be missing the answer you are offering. In case my question was worded in such a way that I didn't make my goal clear I edited the question. $\endgroup$ – Bob516 Mar 19 at 2:07
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    $\begingroup$ @Bob516 and I've made clear that I feel a question that's primarily about how things "look" to people through telescopes is going to be off-topic in Space Exploration but on-topic in Astronomy SE. I've gone as far as I can go for a Space Exploration SE answer. If you like we can vote to close as off-topic and eventually flag for a moderator request to migrate there. $\endgroup$ – uhoh Mar 19 at 2:11
  • $\begingroup$ I thought it would be ok for Space Exploration because it was based on an image acquired by an orbiter. $\endgroup$ – Bob516 Mar 19 at 2:14
  • $\begingroup$ @Bob516 Imagine the question "Why is the object in this Hubble Space Telescope image (inserted into LEO with an inclination of 28.5° by the Space Shuttle) called the 'Horse Head Nebula'?" $\endgroup$ – uhoh Mar 19 at 2:18

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