# How much more likely is it for a colony on Mars to receive a damaging meteorite hit than a similar structure on Earth?

Meteorite damage probability is low both on Mars and Earth, but Earth has a thick atmosphere, which Mars does not. The thinner atmosphere on Mars should yield some increase of impact probability.

Is it possible to estimate the following:

1. How often will meteorites hit some Martian city with an area of several square kilometers?
2. How energetic will they be? Will a once in a hundred year strike be enough to break a solar panel, a greenhouse window, or an actual roof of some pressurized surface building? Many folks will stay underground to avoid radiation but may to up to the surface to enjoy the view once in a while.
• That works for me @uhoh.
– GdD
Jun 15 at 15:51
• Although part of the question is interesting (and uhoh's calculations support it), the threat of a meteorite hitting Earth is more than its kinetic energy. The last meteorite that ended 75% of species on Earth did not achieve that based on its size alone. It is established that the quick climate change it triggered was the actual death blow. Moreover, the geology of the site of impact (Chicxulub) played a critical role according to this study. Without cities, green houses, the Dinosaures were still unlucky. Jun 15 at 18:02
• Hence, more interesting questions: can a meteorite (or other natural phenomena) create a catastrophic climate change on Mars that blocks sunlight in the same way as the Chicxulub meteorite did to Earth, 66 millions years ago? How can we find that there wasn't such a catastrophe in recent Mars history [several millions years]? Jun 15 at 18:18
• @NgPh different but related: Would the one million people on Mars be killed by an impact equivalent to an Extinction Level Event on Earth
– uhoh
Jun 16 at 5:42
• @uhoh, that was a great discussions indeed. Meanwhile, there is this one too. The answer by Boodysaspie linked to a relevant report. With the 200hits/year, each punching a hole of "typically" 4m in diameter, as reported, it is a trivial calculation to compute the prob. that an area is directly hit over a certain time in the future. Jun 16 at 9:11

This is an interesting question! Partial answer!

I don't know about the frequency of Martian meteorites, but for a given one we can estimate its terminal velocity (if it reaches it) and kinetic energy.

If it doesn't reach terminal velocity, then it's going to be going a heck of a lot faster and have a heck of a lot squared more kinetic energy.

A 1 cm radius iron meteorite will have a terminal kinetic energy of 27 Joules on Earth but 540 Joules on Mars!

At 10 cm it's 240 kiloJoules and 5.4 megaJoules!

### But most meteorites are small.

The bigger ones are extremely infrequent.

If you collect all the dust and dirt off of a flat-roofed building and sift through with a strong magnet you can collect iron-containing micrometeorite debris, and if you drag a lot of powerful magnets around in a desert you can collect a lot of larger bits, but these are not likely to do any damage to a colony or be noticeable at all actually.

Their terminal velocity will be faster on Mars due to the lower air density, the lower gravity only does a little bit to mitigate that effect.

From Wikipedia's terminal velocity:

$$v_T = \sqrt{\frac{2mg}{\rho A C_D}}$$

where $$m$$, $$A$$ and $$C_D$$ are the mass, cross-sectional area and drag coefficient of the meteorite, and $$\rho$$ is the density of the atmosphere near the surface.

For the calculation below I've chosen a density of $$\rho$$ of 7.5 g/cm^3 typical for an iron meteorite.

    r (cm)            0.01       0.10        1.00       10.00
------------        ---------  ---------   ---------   ---------
vt_earth  (m/s)      3.88        12.29       38.85      122.85
vt_mars   (m/s)     19.29        60.99      192.87      609.92
KE_earth   (J)       2.37e-07     2.37e-03    2.37e+01    2.37e+05
KE_mars    (J)       5.84e-06     5.84e-02    5.84e+02    5.84e+06


...Cropped section of image of "Egg Rock" from redplanet.asu.edu/?p=21047 showing the spots where Curiosity's ChemCam laser has ablated material.

Python script for plot:

import numpy as np
import matplotlib.pyplot as plt

def vt(r, g, rho_m, rho_a):
Cd = 1.
A = np.pi * r**2
m = rho_m * (4/3) * np.pi * r**3
v_terminal = np.sqrt(2 * m * g / (rho_a * A * Cd))
KE = 0.5 * m * v_terminal**2
return v_terminal, KE

np.set_printoptions(precision=2, suppress=False)
r = np.logspace(-4, -1, 4) # meters
r_cm = 100 * r

rho_meteor = 7500 # kg/m^3

vt_earth, KE_earth = vt(r=r, g=9.81, rho_m=rho_meteor, rho_a=1.3)
vt_mars, KE_mars  = vt(r=r, g=3.72, rho_m=rho_meteor, rho_a=0.02)

print('    r (cm):           ', r_cm)
print('    vt_earth (m/s):   ', vt_earth)
print('    vt_mars (m/s):    ', vt_mars)
print('    KE_earth (J):     ', KE_earth)
print('    KE_mars (J):      ', KE_mars)

fig, axes = plt.subplots(2, 1)
ax1, ax2 = axes
fs = 12
ax1.plot(r_cm, vt_earth, '--g')
ax1.plot(r_cm, vt_mars, '-r')
ax2.plot(r_cm, KE_earth, '--g')
ax2.plot(r_cm, KE_mars, '-r')
ax1.set_ylabel('terminal velocity (m/s)', fontsize=fs)
ax2.set_ylabel('terminal kinetic energy (J)', fontsize=fs)
for ax in axes:
ax.set_xscale('log')
ax.set_yscale('log')
plt.suptitle('meteorite CD=1.0, density = 7.5 g/cm^3')
plt.show()


Uhoh's answer neatly covers the damage a given meteorite would have upon impact.

I'll add to that approach the likelihood that a meteorite even strikes Mars or Earth.

A Martian year is 1.6 times longer than an Earth year (686.96 days versus 365.25). At first approximation, on average, a planet is just as likely to be hit by a meteorite at any point in its orbit (that isn't exactly true and that's why we have yearly meteor showers). In other words, I'm assuming here that the distribution of meteorites comes from a uniform direction in the inner solar system.

Moreover, the radius of Mars is 3397 km compared to Earth's 6378 km. Assuming the planets are spherical, the interception plane (think of it as a "target") of the meteorite traveling in a straight line would be a disk whose radius is the radius of the planet. The area of that disk is what matters here. And the radio of the area of the Earth over the area of Mars is about 3.525 ($$A=r^2\pi$$).

Therefore, Mars is about 5.6 times less likely to be struck by a meteorite, but if it does, the damage is waaaay worse.

• The 1st part of your reasoning seems flawed: If I have a bucket of the same size as my friend's, the 2 of us will collect the same amount of rain, independently of how we run in circles of different sizes under the rain (uniformly distributed in space). Mars being closer to the Main Asteroid belt, it could indicate that your uniform distribution assumption doesn't hold. The conclusion that Mars would be hit more often than Earth (assuming same radius) is still correct though, but due to non-uniformity and not orbit size. Jun 15 at 19:59
• @NgPh I should have clarified that in my mind, when a meteorite is closer to the Sun than the orbit of Earth, then it will never pop out on the other end. In that case, I believe my reasoning is correct. Jun 15 at 21:24
• +1 This is a good answer but it may be necessary to dig a little deeper. The OP specified "several square kilometers" rather than the entire planet, so the size of the planet may not matter, and Mars is closer to the asteroid belt which may or may not have some bearing on this. Meteors from meteor showers are associated with particles from comets which may have a different size distribution than other sources of meteors (e.g. asteroid belt).
– uhoh
Jun 15 at 22:02
• @ChirR, I start to see the subtlety... I think. It's the x1.6 number that confused me. Jun 16 at 14:25

The original OP's question

Could meteorites prevent Mars exploration?

has been transformed into another question that was answered, though partially, into energetic calculations and probabilistic calculations. I will try to answer the original question.

First, how can a meteorite prevent colonization of a planet? You must accept that when people take all the risks of a 9+ month journey in space, and spend \$, it is not because of a breaking window/solar panel, or even a direct deadly hit on an installation that make them think twice. Like the past adventurers, they will rebuild, they will bury their deads and move on. So, to prevent colonization, we must think of a catastrophic event, one that, due to its ramifications in the established ecosystem, can exterminate almost all livings, like the meteorite that hit Earth 66 millions years ago (at Chicxulub).

Now, for the sake of quick-and-dirty estimation, let's say that in 10,000 years the probability of being hit once is 10-4 times the probability being hit once in 100 millions years (I know it is wrong math, but I am on the safe side). Furthermore, assume that for Mars, the probability of being hit once in 100 millions years is 1. At least for Earth, it is much less, from the geological records.

The Chicxulub crater has a radius of 90 km. Probably any living animals in a radius of 2x90 km were killed by its kinetic energy. This is a circle of ~100,000 km2, or about 7% of Mars surface (R=3400 km). Hence, if you put at random a city on Mars, the probability that, over 10,000 years it is erased by a Chicxulub-like event is 7 in one million. In 100 millions years, your city has 93% of being spared, although we assume certainty for the catastrophic event.

As crucial remark, it was not the kinetic energy of Chicxulub that exterminated 75% forms of life on Earth. It was the ramifications of its impact on the atmosphere and the global cooling that resulted from Sun-blocking ashes. Hence, my calculation above is pointless if I cannot predict - which I can't - the impact on the atmosphere of Mars of a big meteorite hit. In fact, my calculation would hold if Mars were completely devoid of atmosphere, since in this scenario it's only the kinetic energy that harms.

Therefore, the answer: with the present knowledge, the likelihood that meteorites constitute a concern for a Martian colonization is non-existent. Even taking into account the facts that Mars is closer to the asteroid belt and its atmosphere is non-protecting.