# How big would my telescope have to be if I wanted to see the Mars rover from my backyard?

I imagine that with a big enough telescope, I would be able to zoom in and see the Mars rover in enough detail to make out the details (like the wheels, cameras, etc.). How large would the telescope have to be? (or how can I calculate this value?)

• Link: Why not use Hubble to look at the lunar landers? Answer: Hubble's resolution for objects on the Moon is about 200 meters. – rob Dec 18 '17 at 4:45
• @JEB if it's not on Mars is it still a Mars rover? – njzk2 Dec 18 '17 at 4:56
• Just as a practical matter, the resolution is probably (certainly, in the absense of adaptive optics) going to be limited by atmospheric distortion, more than by the size of the telescope. That's part of the reason Hubble gives better images than Earthbound telescopes, even though they're much larger. – jamesqf Dec 18 '17 at 6:14
• Now I want to see Randall Munroe answer this in an xkcd What If? – Todd Wilcox Dec 18 '17 at 18:30
• I assume we're talking about a frictionless spherical telescope in a vacuum? – Mark Dec 18 '17 at 22:53

Telescope resolution is all about apparent angles. From the sounds of it, the lowest resolution you'd settle for would be something capable of resolving about $1 \operatorname{cm}$ objects, right? Well, the distance between the Earth and Mars varies, depending on the time of year, from around $0.5\operatorname{AU}$ to $2.5\operatorname{AU}$ ($7.5\times 10^{10} \operatorname{m}$ to $3.7\times 10^{11} \operatorname{m}$). At those distances, a $1$ centimeter object subtends an angle of $$\theta = \frac{s}{d},$$ which is $1.5\times 10^{-13}\operatorname{rad}$ to $2.7\times 10^{-14}\operatorname{rad}$.

The resolution of a circular telescope is given by the formula $$\theta = \frac{1.22\lambda}{D}.$$ So, assuming you're using visible light, with $\lambda \approx 500\operatorname{nm}$, to resolve those $1$ centimeter objects it would require telescopes with a diameter of $D=4.6\times 10^6\operatorname{m}$ to $7.4\times 10^7\operatorname{m}$. For reference, the diameter of Earth is about $1.3\times 10^7\operatorname{m}$.

Note that the sheer size is only one of the challenges. In order to achieve this theoretical resolution you would need the surface of the mirror to have the correct shape everywhere to within about a wavelength of light. In other words, this Earth-sized mirror could not have any imperfections larger than about $500\operatorname{nm}$. To see some of the information related to getting ordinary lenses and mirrors correct to this level see the Wikipedia article on optically flat.

• One could add that you don't need a full-sized mirror for that. You only need to capture parts of the wave over such a distance. So in principle, if you had two normal-sized telescopes on opposite sides of the earth, both looking at Mars, and you could do an interference of their captured light, you might also reach that resolution. Then, the problem will probably be the low light intensity (the amount of photons per second coming from the rover and reaching the pair of telescopes will be minuscule). But it would in principle be doable, without making a mirror, the size of the earth. – entrop-x Dec 18 '17 at 5:14
• @entrop-x With two mirrors separated like that you'd only have that resolution in that one direction. To make an image you'd need to be able to get that resolution in multiple directions, so you'd need more than two mirrors, or movable mirrors. Add in the problem of actually coherently interfering the signals, and it's probably intractable. We can only do that sort of thing in radio because of the abundance of photons, and the inter-photon coherence - both are lacking in the optical. – Sean E. Lake Dec 18 '17 at 9:27
• @SeanE.Lake you don't need inter-photon coherence of the signal; if the Mars Rover had a laser which we could use for calibration of a multi-mirror system to capture a wavefront in one scoop, that would be enough. But indeed two mirrors would help little with image resolution (that's only sufficient for localisation of point sources); rather you'd need a whole ensemble or mirrors, all aligned with submicrometre precision... not going to happen soon. – leftaroundabout Dec 18 '17 at 14:13
• @All this, of course, assumes no atmospheric interference. Even for the largest telescopes today, their resolution is limited relative to the theoretical best resolution because of unstable air messing up images from space (though there is technology used to correct for this). A humongous telescope still embedded in Earth's atmosphere will still have very limited resolution. – NeutronStar Dec 18 '17 at 19:12
• @Joshua Given the size described, embedding in Earth's atmosphere is less of a problem than that the rover is embedded in Mars's. – Sean E. Lake Dec 18 '17 at 20:06

# REALLY BIG

There are several parameters that need to be controlled in order to make a properly-functioning telescope, but arguably the two most important parameters are its magnification power and its aperture size.

Magnification is defined fairly straightforward as the ratio between how big the object looks when looking at it through the eyepiece of the telescope to how big the object looks when looking at it with the naked eye. This depends on the properties of the lenses you use in your telescope. For a fairly simple telescope, you will have two lenses: the objective lens at the front of the telescope, and the eyepiece lens, the lens you actually put your eye up to in order to look at the night sky. The magnification can be expressed in terms of each lens' focal length,

$$M = \frac{f_o}{f_e}$$

where $f_o$ is the focal length of the objective lens, and $f_e$ is the focal length of the eyepiece lens. The sum of the two focal lengths will give you a rough estimate on how long the body of the telescope needs to be. This webpage gives a bit more detail on the derivation of this equation and the intuition behind the physics of refraction in lenses.

Once you have the correct lenses for the magnification you desire, the next parameter you need to take into account is the size of the telescope's aperture -- the opening of the telescope that actually captures the light coming from the object you want to look at. If you want to see a brighter, more resolved image, you're going to want a bigger aperture. The equation determining the field of view you can resolve is given on this page in Sean E. Lake's answer. The $D$ in his equation will give you the size of the aperture you need for your telescope.

The above answers have excellent math, but they neglect the most important factor: your hypothetical telescope is in your backyard, on the surface of planet Earth.

No matter how big you make the mirrors or lenses, your telescope cannot see the Mars rover from Earth, because of atmospheric distortion. You know how stars "twinkle"? You know how on a hot day, sometimes the ground shimmers? The air acts like a big blurry wobbly lens, which any ground-based telescope has to look through, limiting the level of detail that is physically possible to see.

When looking 50 million km away (such as Mars at closest approach), an ideal ground-based telescope might see objects that measure many tens of kilometers across. Anything smaller would be blurred out. Adaptive optics might manage to get the resolution into single-digit kilometers, but you're trying to see things 1000x smaller than that. It's not going to happen.

p.s. As @Pere suggests in the comments, you can circumvent this problem by placing the device off-Earth, then sending the image back via radio. Congratulations, your telescope already exists!

• If you look at the numbers the other answers produce, any telescope is going to be tall enough that it sticks out of the atmosphere. – Mark Dec 18 '17 at 22:55
• So what he really needs is an Earth-sized space telescope, with a receiver in his backyard. – Barmar Dec 18 '17 at 23:17
• If the space telescope is wisely placed, it doesn't need to be that big. He example, he can place in it in orbit around Mars or, even better, sitting on Mars a couple of meters from the rover. Anyway, he would need a large receiving antenna that might not fit in his backyard. – Pere Dec 19 '17 at 9:30
• Actually, twinkling or, more accurately, scintillation is not caused by atmospheric distortion/refraction. If it were a result of refraction there would be displacement, which does not occur, and there would not be changes in colour, which there is. Instead, it's the result of interference of the wavefront of plane wave light passing through turbulence. Planets don't scintillate because their light isn't planar relative to the pupil. – Danikov Dec 19 '17 at 13:20
• @Danikov Is Sky & Telescope wrong? – Barmar Dec 19 '17 at 20:28

## protected by Qmechanic♦Dec 18 '17 at 12:27

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).