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As we know, we can find rocks from Mars on Earth, how about Venus and Mercury? Is that not found yet? Or it is impossible to find them because they are closer to the Sun, and debris won't go far away from the Sun?

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2 Answers 2

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You can think of it in terms of Hohmann transfer orbits, which define the minimum $\Delta v$ that needs to be applied to bring something from one orbital radius to another orbital radius when orbiting a massive body. This calculation takes into account that the two objects have Keplerian orbits where the objects begins with at least the orbital speed of the initial orbital radius.

The Hohmann $\Delta v$ is given by $$\Delta v = \sqrt{\frac{GM_{\odot}}{r_1}} \left(\sqrt{\frac{2r_2}{r_1 +r_2}} -1 \right),$$ where $r_1$ and $r_2$ are the initial and final radius from the Sun.

Conceptually what you have to do is give a rock enough energy to escape from the planet and then give it an additional $(m/2)(\Delta v)^2$ of kinetic energy to get it to transfer into the other orbit. If the ejection speed is $v_{ej}$ then $ v_{eg} > \sqrt{(\Delta v)^2 + v_{esc}^2},$ where $v_{esc}$ is the escape velocity.

The numbers for Mercury $\rightarrow$ Earth are $\Delta v = 9.2$ km/s and for Mars $\rightarrow$ Earth $\Delta v = -2.6$ km/s (you have to slow it down to allow it to fall inwards).

The escape velocities for these planets are 4.35 km/s and 5 km/s respectively (so almost the same).

This means you need to give a rock more kinetic energy to get it to Earth from Mercury as from Mars. In the case of Mars, the transfer kinetic energy is almost negligible once the rock can escape Mars' gravity. In the case of Mercury, the rock needs to be given an initial ejection velocity of $> \sqrt{9.2^2 + 4.3^2}= 10.1 $ km/s. This compares with $> \sqrt{2.4^2 + 5^2}= 6.5$ km/s for Mars. At lower ejection speeds most of the ejected objects will be reaccreted by the planet.

Against this, the leading theory to explain migration of rocks between planets is high velocity impacts. Objects falling from much further out will hit Mercury with greater speeds than Mars and impart greater energies to the ejecta.

A paper addressing the possibility of Mecurean meteorites was presented by Gladman & Coffey (2008). They concluded that once ejection speeds are large enough ($\sim 10$ km/s) to produce Earth-crossing ejecta, that significant accretion of meteorites should take place. Several per cent of high speed ejecta should impact the Earth (or its atmosphere at least) within 30 million years. This compares with an efficiency a factor of 2-3 higher for Mars.

There are various reports and speculations that at least one meteorite in existing collections (NWA7325, pictured) may have come from Mercury. See here for example. It appears that the main problem is getting agreement on what the chemical signatures of such meteorites are.

NWA7325

Accretion of material from Venus is a different matter. The required ejection velocities are higher because the escape velocity for Venus 10.4 km/s. But more importantly, drag in the dense Venusian atmosphere would prevent anything emerging from the planet with anything like these speeds.

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  • $\begingroup$ Small point to add / perhaps a different question, but Mars' proximity to the Asteroid belt might increase the number of impacts on Mars' surface, where as the inner solar system and Mercury crossing orbits are a fair bit more rare. That may be a factor that more than compensates for the greater speed of the Mercury impacts. $\endgroup$
    – userLTK
    Mar 16, 2016 at 1:12
  • $\begingroup$ @userLTK Possibly, though I don't think this woild be the source of most "high velocity" impacts. $\endgroup$
    – ProfRob
    Mar 16, 2016 at 6:44
  • $\begingroup$ @userLTK That should be somewhat compensated by Mercury's orbit being smaller and closer to the Sun where sungrazing comets break up. Speaking of high velocity impacts. $\endgroup$
    – LocalFluff
    Mar 16, 2016 at 10:14
  • $\begingroup$ Also, drag in Venus' atmosphere would greatly reduce the impact energy of the incoming object. $\endgroup$
    – TonyK
    Jun 18, 2019 at 22:50
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Possible, but extremely unlikely.

Rocks thrown off Mercury by an impact would not only have to clear Mercury's escape velocity to get off the surface of that planet, but would also have to go faster than the escape velocity from the Sun's gravity well at Mercury's distance from the Sun.

  • The escape velocity from Mercury's gravity well is 4.25 Km per second.
  • The escape velocity from the Sun's gravity well at Mercury's distance is 67.7 Km per second.

That means a rock would have to go faster than 67.7. Kps to even get close to Earth's orbit. Very few rocks will be going that fast, so most rocks knocked off Mercury by an impact will end up orbiting the Sun at the same distance as Mercury. Eventually those rocks will fall back onto Mercury.

The same thinking goes rocks knocked off Venus's surface, but the numbers are a little different than Mercury's.

  • The escape velocity from Venus's gravity well is 10.3 Km per second.
  • The escape velocity from the Sun's gravity well at Venus's distance is 49.5 Km per second.
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  • $\begingroup$ Should meteorites from Phobos and Deimos be expected? Are they maybe too small, or is it unknown how to identify a meteorite as coming from Mars' moons? $\endgroup$
    – LocalFluff
    Mar 15, 2016 at 10:47
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    $\begingroup$ The gist is correct but the numbers aren't. Rocks escaping from Mercury do not have to escape the Sun to reach the Earth. They are also travelling with high speed with respect to the Sun already because they are attached to Mercury! A complete answer also then needs to explain how it works from Mars to Earth. The whole things needs to be discussed in terms of transfer orbits. $\endgroup$
    – ProfRob
    Mar 15, 2016 at 16:35
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    $\begingroup$ So it's -1 from me because the answer is apparently that it is quite likely that there are Mercurean meteorites. $\endgroup$
    – ProfRob
    Mar 15, 2016 at 19:51
  • $\begingroup$ +1 on Rob Jeffries comments. Once a Mercury rock gets into space, gravity assists can toss it well outside of the Sun's escape velocity. That part isn't hard at all. $\endgroup$
    – userLTK
    Mar 16, 2016 at 1:00

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