Can a cheap telescope tell you how many miles/km away an object is that you focus on? (Via some digital readout onscreen, or perhaps via some calculating function that you can access in its computer.) Specifically - for objects within our solar system, and definitely within, say, the orbit of the moon. This would let you know whether a strange object was a planet or just something in near-earth orbit or space junk.

  • $\begingroup$ While most general telescope questions receive good answers in Astronomy SE yours question is interesting and different, as you are asking about measuring distances of objects in the solar system and especially space junk and other objects in Low Earth Orbit, which makes this definitely on-topic here. $\endgroup$
    – uhoh
    Jan 5, 2019 at 8:35
  • $\begingroup$ which cheap telescopes have computers? $\endgroup$
    – user20636
    Jan 5, 2019 at 18:42
  • $\begingroup$ Plenty of consumer telescopes have computer-controlled tracking mounts. Image processing, I don't know. $\endgroup$
    – Hobbes
    Jan 5, 2019 at 18:52

4 Answers 4


If you ask can a distance to a selestial object be calculated from focal distance - the answer is No. Even closest satellites in low Earth orbit will be beyoud hyperfocal distance of a telescope.

But if you want to know "what object did I see in telescope" you can use special software like Stellarium for planets, nebulae, largest asteroids etc. For artifical satellites there are many online tools, my personal choice is www.heavens-above.com

  • 2
    $\begingroup$ It would be great if you could also at least estimate how far the hyperfocal distance of a cheap telescope, or even a fancy telescope really is. "No" + link is nice, but it doesn't really support your answer. How can the OP know your answer is correct? The link just has some ancient equations but no numbers. $\endgroup$
    – uhoh
    Jan 5, 2019 at 10:00

There is an optical formula valid for all spherical lens and mirror telescopes. For simplicty I am ignoring the hyperfocal distance.

1/f = 1/a + 1/b

f is the focal length, a is the distance of the object and b is the distance of the image.

I use a focal length of 1 m for an example. We solve the the formula above for b:

b = (a-1)/a

If the distance of the object a is 100 m, the distance of the image b is 0.99 m. For a = 1000 m we get b = 0.999 m.

If a is 100 km, b is 0.99999 m. For a = 1000 km we get b = 0.999999 m.

The difference between 0.99 m and 0.999 m is 9 mm, this is measurable with good precison. But the difference between 0.99999 m and 0.999999 m is only 9 µm, very close to the wavelength of visible light between 400 and 800 nm. Using a cheap telescope, it is impossible to measure 9 µm precisely. But you can't focus such a telescope so precisely anyway.

But the closest astronomical object, our Moon is about 400,000 km away. No way to distinguish between artifical satellites and natural astronomical objects


Let's try a different challenge; how well could the Hubble Space Telescope, outside of the Earth's atmosphere, tell if a small (unresolved) object was either 100 km away, or at infinity?

Let's assume perfect diffraction-limited optics and instead of a hard-edged Airy diffraction pattern, use a Gaussian beam with a 1/e^2 radius equal to the Hubble's mirror radius of 1.2 meters and a wavelength of 200 nm (it can go lower potentially).

From Wikipedia I get f/24 meaning f = 57.6 meters.

Using Gaussian Beam Optics, $w, z$ = 1.2, 57.6 meters, $\lambda$ = 0.2E-06 meters, I get

$$w_0= \frac{\lambda z}{\pi w} \approx 3 \mu m$$

That's the half-width of a Gaussian beam focussed to a Gaussian spot. The "Rayleigh Range" (a handy and significant parameter) $z_R$ is given by

$$ z_R = \frac{\pi w_0^2}{\lambda} \approx 144 \mu m$$

and the beam width at any point $z$ on either side of best focus is

$$w = w_0 \sqrt{1+\left(\frac{z}{z_R}\right)^2} $$

If the wavefront striking the mirror were not flat, but had a slight curvature because it was a spherical wave from far away, you can (somehow) use

$$R(z) = z \sqrt{1+\left(\frac{z_R}{z}\right)^2} $$

to calculate the shift at the focal plane, but like most things in optics it ends up being the same as the classical behavior shown in @Uwe's answer, the "Lens Equation"

$$\frac{1}{f} = \frac{1}{a} + \frac{1}{b} $$

If $f$ = 57.6 meters and $a$ = 100,000 meters, then b is different from 57.6 meters by 1/100,000, or 10 $u$m.

If I draw that, it looks like the plot below. If you had a lot of time, and both the object of interest and a star (of roughly equal brightness) in the field of view, a very smooth focus mechanism that you could zoom back and forth by a few hundred microns, then eventually, after collecting tons of data, you might be able to notice the 10 micron shift in best focus, by looking at the offsets in slopes at around +/-$z_R$ the Rayleigh distance from best focus.

You can go back and do the Fourier optics for the Airy diffraction and hard circular aperture, and the results will be different but roughly the same. It is theoretically conceivable, but practically impossible, even with the HST above the atmosphere, to detect the difference between an object at infinity, and one at 100 km.

At 10 km, the curve would be shifted ten times farther, and that would be possibly detectable from a practical point of view, but you'd still spend a lot of time measuring very carefully. Below 10 km, for unresolved objects of identical brightness it might be possible routinely, but still a lot of work.

Plot axes are microns:

enter image description here


There won't be a readout, but you can do a quick estimate yourself: when you look at an object through a non-tracking telescope, it'll wander out of view after a while (how long this takes, depends on your field of view).

Close objects will move out of view a lot faster than stars. Satellites in low Earth orbit are difficult to track with a telescope (they'll cross the sky in a few minutes). Observe some of the planets to get an idea of how fast they're moving, that'll give you a base for comparisons.


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