Why don't we get eclipses from other planets between the Earth and our Sun?

 We do, actually!

We even have eclipses from other planets between the Earth and the OTHER STARS!

Only they are known by a different name, so you probably don’t know them as eclipses.

Before I get to them, check out this eclipse of Mercury.

Source Image: Transit of Mercury - Wikipedia

See the black dot I’ve marked in the green circle? That’s Mercury between us and the Sun.

The smudge inside the blue circle is the sunspot. Notice how it’s bigger than the planet of Mercury.

In this image, Mercury is fully intent on blocking the Sun’s light from ever reaching us. But the poor guy is too small to have any meaningful impact.

So, we don’t even notice it.

In fact, Mercury’s obstruction of our view of the Sun is so unremarkable that it’s not even called an eclipse. The same goes for Venus, too.

No effect, but A+ for effort!

When is an Eclipse Not an Eclipse?

During a solar eclipse, the Moon blocks the Sun either fully or partially. Either way, the Moon is capable of blocking a substantial portion of the Sun when viewed from the Earth.

Remember, it’s not the actual size of the Moon that matters here, but its apparent size in the sky.

Since its apparent size in the sky, as seen from Earth, is large enough to block the Sun partially or entirely, its obstruction is known as an eclipse.

Now, would you call it an eclipse if an asteroid just happened to whizz past the Earth between us and the Sun?

You wouldn’t, because it’s simply too small to qualify as an eclipse!

That’s what happens when Mercury or Venus ends up between the Earth and the Sun. They’re just too small for us to experience an eclipse.

So, we call those events Transits!

Transits are Among the Most Important Events in Astronomy

Did you know that planets outside the solar system are too dim to be visible to us?

Several factors, including interstellar dust, brightness of the accompanying star, etc., make it almost impossible for us actually to see these exoplanets even with our most sophisticated telescopes.

So, how do we detect them?

We rely on transits.

When a planet passes before its home star, there’s a small drop in that star’s brightness.

Our sophisticated instruments are capable of detecting and measuring that drop in brightness. Based on this, we ascertain the existence of a planet around a faraway star.

In short, these “eclipses” or “transits,” whatever you wish to call them, play a central role in helping us detect planets outside the solar system.

Are there more planets than we know so far?

 Absolutely ! There are roughly 

${10}^{21}$ ( sextillion ) planets in the visible Universe, and at least 106 known exoplanets only within the surrounding sphere with 100 light years radius.

One of my favorite exo-solar system is the Trappist-1 system:

Trappist-1D:

Trappist-1C:

Proxima Centauri-B:

Furthermore !

There can be bizarre Planets such like our newborn Earth was 4.8 billion years ago:

Moreover:

L4L5 Interferometric Exoplanet Spy System (LIESS)

Basic concept:

Launching two spacecraft to the L4 and L5 points of the Earth's orbit path to study exoplanets within 100 light years.

Expected discoveries and resolutions (without claiming to be complete):

Real-time observation of the planets Proxima Centauri B and C at 13.4

meter/pixel resolution and even real-time monitoring of their meteorology.

Imaging of planets in the Trappist star system at a resolution of 134 meter

per pixel and real-time observation of their meteorology.

Costs and implementation:

The implementation, in contrast to the Proxima Centauri approach planned

for 2069, is much simpler than ion-gun acceleration technology and

would provide continuous observation of at least the surrounding

exoplanets within 100 light years, instead of a one-time journey.

… and another 106 exoplanets, if we only take the ones known so far, and possibly more exoplanets and even exomoons may emerge with a terrain map of unprecedented detail. We can boldly say that with this technology we will see the pimple on the ant's dick on the celestial bodies in the Cuiper belt and the Oort cloud, and we will get global maps comparable to the detail of Google Maps of more than 10 times as many planets as we have known so far. For example, it can easily be found that there are not just 106, but actually 268 exoplanets within the surrounding sphere of 100 light-years radius, and they have 778 exomoons. We can boldly say that with this "penny" and relatively simple tool we can find out what the word "weather" means.

Earth-Mars Orbit Configuration

1. Base Distance:
   Base distance: approximately 1.5 astronomical units (AU) = $2.25\ast {10}^{11}$ meters
2. Observation Wavelength:
   Let's assume an optical wavelength of $500nm$ ($500\ast {10}^{-9}$ meters)
3. Angular Resolution Calculation:
   The formula for angular resolution is: $\theta =\frac{\lambda }{B}$
   Where: $\theta$ is the angular resolution in radians
   $\lambda$ is the observation wavelength
   $B$ is the base distance
   
   Substituting the values: $\theta =\frac{500\ast {10}^{-9}m}{2.25\ast {10}^{11}m}\approx 2.22\ast {10}^{-18}rad$
4. Convert to Distance on Proxima Centauri B:
   Distance to Proxima Centauri B: $4.24$ light-years ($4.02\ast {10}^{16}$ meters)
   $d\cdot \theta =4.02\ast {10}^{16}m\cdot 2.22\ast {10}^{-18}rad\approx 8.92$ meters/pixel

Neptune Configuration

1. Base Distance:
   Base distance: approximately $30.1$ astronomical units (AU) = $4.5\ast {10}^{12}$ meters
2. Observation Wavelength:
   Let's assume an optical wavelength of $500nm$ ($500\ast {10}^{-}9$ meters)
3. Angular Resolution Calculation:
   The formula for angular resolution is: $\theta =\frac{\lambda }{B}$
   Where: $\theta$ is the angular resolution in radians
   $\lambda$ is the observation wavelength
   $B$ is the base distance
   
   Substituting the values: $\theta =\frac{500\ast {10}^{-9}m}{4.5\ast {10}^{12}m}\approx 1.11\ast {10}^{-19}rad$
4. Convert to Distance on Proxima Centauri B:
   Distance to Proxima Centauri B: 4.24 light-years ($4.02\ast {10}^{16}$ meters)$d\cdot \theta =4.02\ast {10}^{16}m\cdot 1.11\ast {10}^{-19}rad\approx 4.46$ meters/pixel

These calculations show the angular resolution and pixel distance for both configurations. The Neptune configuration provides a higher resolution due to the larger base distance between the observation points.

Thus, with the above-mentioned method, we will soon be able to easily observe planetary and lunar systems outside our solar system.