Could a Mars colony use asteroids as weapons against the earth?

Question

If a future Mars colony declared war on the Earth, would asteroid manipulation be a feasible option as an effective weapon against the Earth from a Mars colony? How would this be accomplished? Are there enough asteroids close enough to Mars that current/future rocket technology could steer asteroids towards the Earth - before the Earth could respond with an effective countermeasure?

Answer 1

I've taken the liberty to provide a slightly more general answer. It also answers your question along the way:

Early Stage Colonies (Mars population <50,000) would be incredibly dependent on Earth, and therefore could not afford to risk ending the supply of cargo from Earth. The industrial chains leading to the production of advanced electronics and machinery would likely remain for a long time outside the reach of a colony. Mars would still be under more or less direct Earth control. Asteroid attack chance: N/A, far beyond colonist capabilities. Probably no orbital access.

Late Early Stage (50k < Mars population < 1m) With increasing population, limits on the Earth's Mission Control's capacity to micromanage the affairs of the colonists would likely become apparent. Expect shortages in supply, Martian radicals agitating for independence, extensive mining, the first locally built nuclear reactors, early industrial plants and the first massive infrastructure projects (such as space elevators and giant domes, in the dozen km radius range). At this stage the Martian colonists are most likely to rebel, and most likely to be crushed. Lack of adequate industrial and space infrastructure would likely mean limited orbital capacity, limited manufacturing ability, and limited force-projection capability. Their fragile, easily depressurized habitats are a further handicap. This is likely where the asteroid attack attempt would occur. The asteroid has near zero chance of damaging Earth, of course, since Earth has a larger nuclear arsenal, a larger space fleet, early warning due to distances and superior intelligence networks, and the manufacturing capability to respond effectively to telegraphed threats of this nature (with months to prepare, countermeasures would be developed and implemented)

Mature Colony With a population approaching 10 million, a mature Martian colony would be completely unmanageable from Earth due to the sheer size and complexity of the Martian economy. Since Mars has a permanent labor deficit, the degree of robotization, AI-control and labor-saving technologies in general will be unimaginable by Earth standards, where with 10 billion humans, labor will remain relatively cheap for the foreseeable future. This will have likely given the Martians an immense technological edge in the relevant fields, while the 'frontier' environment will have engendered an openness to daring projects not seen on the Homeworld since the closing of the American West. Martian per capita GDP is similar to that of Switzerland, and growing at a rate of 40% per earth-year, a rate unimaginable on the NIMBY Earth. Slowly, the "Martian Burden" as Earth-politicians will have called the immense investments required to sustain the Martian project in its Early decades, will gradually dwindle, as more and more, the technology flows become reversed, and Martian businessmen and entrepreneurs share of total new Intellectual Property approaches 5%, a 50-fold over-representation of the Martians. Their wealth and influence growing, the Martians achieve through economic strength what their rebellious parents could not by force of arms: self-representation, and a seat in the highest decision-making councils. Asteroid attack success chance: High. Orbital capability is in place, secure communications and counter-intelligence probably in place. However, use is unlikely since in a vast majority of conceivable scenarios sovereignty would have been devolved to the Martians at this point, and Earth still holds a 500:1 industrial advantage

Homeworld With Martian population stable around 40 million, a population density around 0.3 Martians/sq.km, Martians decide by a slim margin to limit any further colonization by the disease-infested short people that swarm the Earth, with the occasional exception made for outstanding scientists working at Earth-branches of the major Martian corporations. Earth's opinion on the matter is irrelevant, since the Martians have a significant technological lead, a near-complete domination in terms of advanced ship technology, and own a majority of the solar space fleets, Asteroid belt mining stations and all three major space elevators on Earth. Earth is engulfed in a dark age due to massive planetary unrest by hyper-conservative Hindutva, Junzi, Quiverful, Mormon, Wahhabi and Hassidic fundamentalists that comprise nearly half of the population, and can spare little attention to distant space concerns. Asteroid attack success chance: Overwhelming. Martians and their Asteroid belt colonies dominate the solar system, with a massive space infrastructure lead on the still-divided Earth nations. Would be considered a low-tech solution, as Martian 'Mining' ships at this stage have mass accelerators that can dispatch 1-10 ton projectiles at a significant fractional $c$. Moreover, while Earth's economy is still over a dozen times larger, Martian GDP per capita is about 30 times larger. An attack would be pointless. More likely to send humanitarian assistance.

Answer 2

The problem is not so much a lack of rocks, as there are plenty all over the place, but their location and the enormous distances involved. This would not be a surprise attack. Presuming the technology to develop a self-sustaining colony on Mars of any significant size to become its own independent polity and engage in warfare with Earth, the technology exists to detect and shoot down (or just redirect) incoming rocks with plenty of warning.

A far better and more likely first strike would be initiating a Kessler syndrome. Using very small rocks, small enough to not be easily spotted (or would be dismissed as no danger), but in very large numbers. This scattershot of rocks timed to go into orbits likely to strike satellites could cause a cascade of debris eventually destroying anything in orbit and denying access to/from the planet. If significant enough, you might then reduce the ability of the Earth to launch sufficient munitions to destroy/deflect the much larger rocks which could actually do damage to the surface. They would probably see the rocks coming for weeks or even months, but couldn't do much about it.

Answer 3

One thing to point out is that Mars has two asteroids right nearby: its moons, Deimos and Phobos. They're both rather small - many times smaller than Earth's Moon - and so are thought to be captured asteroids. Here are some of their characteristics:

• Phobos: Mass: $1.0659 \times 10^{16} \text{ kg}$, semi-major axis: $9,377,000 \text{ m}$, size: $27 \times 21.6 \times 18.8 \text{ km}$
• Deimos: Mass: $1.4762 \times 10^{15} \text{ kg}$, semi-major axis: $23,460,000 \text{ m}$, size: $10 \times 12 \times 16 \text{ km}$

Those are puny! Well, compared to our Moon.

To get one of them out of Mars' orbit, you'd need to move it to escape velocity: $$v= \sqrt{ \frac{2GM}{r}}$$ where $M$ is the mass of Mars and $r$ is the approximate radius of the body's orbit. Let's try it with Phobos: $$v= \sqrt{ \frac{2 \times 6.673 \times 10^{-11} \times 6.4185 \times 10^{23}}{9.377 \times 10^6}}=3022.46 \text{ m/s}$$ So it has a kinetic energy of $$KE=\frac{1}{2}m_{\text{Phobos}}v^2=4.86863 \times 10^{22} \text{ Joules}$$ How are you going to give it that much energy?

The Tsiolkovsky rocket equation of a rocket gives us a good idea of how much oomph needs to be put into this. The formula is $$\Delta v=v_e \ln \frac{m_0}{m_f}$$ Phobos is already moving, at a speed of $$v=\frac{2 \pi r}{t}=\frac{2 \pi 9,377,000}{27552}=2138.41$$ $\Delta v=3022.46-2138.41=884.052$. If all the propellant is used up, $m_f=m_{\text{Phobos}}$ and $m_0=m_{\text{Phobos}}+m_{\text{fuel}}$. So $$884.052=v_e \ln \left( \frac{m_{\text{Phobos}}+M_{\text{fuel}}}{m_{\text{Phobos}}} \right)$$ Let's say that the rocket uses the Variable Specific Impulse Magnetoplasma Rocket (VASIMR), with $v_e=120,000$ (at the best-case scenario). So we solve for $m_{\text{fuel}}$: $$\frac{884.052}{v_e}= \ln \left( 1+\frac{m_{\text{fuel}}}{m_{\text{Phobos}}} \right)$$ $$\exp \left[\frac{884.052}{v_e} \right]=1+\frac{m_{\text{fuel}}}{m_{\text{Phobos}}}$$ $$m_{\text{fuel}}=m_{\text{Phobos}} \left(\exp \left[\frac{884.052}{v_e} \right] -1\right) \approx 1.0659 \times 10^{16} \left(\exp \left[\frac{884.052}{120000} \right] -1 \right) \approx 7.88159 \times 10^{13} \text{ kg}$$ That's not too easy, to say the least. By comparison, the N-1, possibly the most powerful rocket of all time, had a mass of $2.735 \times 10^6$ kilograms - with all its parts! Above, we didn't even count the mass of the rockets needed.

This would not be such a good idea.

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If you're looking for other relatively-easy choices, there are about seven or so Trojan asteroids in Mars' orbit. They're equally far away, though, and they don't have any advantages over choosing Deimos or Phobos.

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PhilFrost brought up an interesting point - going to a Near-Earth asteroid. There are a bunch of issues with that:

• It's a Martian colony. You have to get to the asteroid and back. This is a lot easier when going to Deimos o Phobos than a Near-Earth asteroid. The journey changes from about 16 months - optimistically - to, say, a week. The time for assembling the rockets for the asteroid will take the same in both scenarios - but again, you would have to lug a big honkin' rocket to the asteroid. That's way easier when going to one of the moons.
• You only get one shot with a Near-Earth asteroid. It might go near where you want it again, but not or a while. On the other hand, Deimos and Phobos have really short periods. Earth will have moved in the meantime, but not a significant amount.
• The moons of mars have stable orbits; a Near-Earth asteroid may not. In other words, you'll know the orbital parameters of Deimos or Phobos far better than a Near-Earth asteroid.

So I'll stick with Deimos and Phobos, thank you very much.

Answer 4

From orbital mechanics, we know that the speed of an orbit in elliptical orbit at periapsis (closest approach) and apoapsis (furthest approach) is: $$ v_\text{ap} = \sqrt{\frac{2\mu~r_\text{per}}{r_\text{ap}(r_\text{per}+r_\text{ap})}} \\ v_\text{per} = \sqrt{\frac{2\mu~r_\text{ap}}{r_\text{per}(r_\text{per}+r_\text{ap})}} $$ For an object in circular orbit, where $r_\text{per}=r_\text{ap}=a$, the speed is constant: $$ v = \sqrt{\frac{\mu}{a}} \\ $$

Let's assume that the asteroid is initially in an approximately circular orbit of radius $2.2~\text{AU}$, and we wish it to intersect Earth's orbit at a radius of $1~\text{AU}$. The initial speed of the asteroid is: $$ v = \sqrt{\frac{G~M_\text{Sun}}{2.2~\text{AU}}} = 20.1~\text{km}/\text{s} \\ $$ And the new speed (the apoapsis speed): $$ v_\text{ap} = \sqrt{\frac{2G~M_\text{Sun}~1~\text{AU}}{2.2~\text{AU}(1~\text{AU}+2.2~\text{AU})}} = 15.9~\text{km}/\text{s} \\ $$ Therefore the delta-v required to move the asteroid is: $$ \Delta v=4.2~\text{km}/\text{s} $$ Trying to get enough propellant together to move this asteroid will be difficult, so let's use the asteroid itself as reaction mass with laser ablative propulsion. The specific impulse is around $5000~\text{s}$, corresponding to an effective exhaust velocity of about $49~\text{km}/\text{s}$. Using the rocket equation gives a propellant mass fraction of: $$ \zeta=1-e^{-\Delta v/v_e}=8.2\% $$ Let's assume the mass of your asteroid is about $10^{15}~\text{kg}$, the low-end estimate of the mass of the Chicxulub impactor. According to this xkcd What-if? it takes around $40~\text{MJ}/\text{kg}$ to vaporize rock, so the amount of energy it will take to move our asteroid is $$ 8.2\%\times 40~\text{MJ}/\text{kg}\times 10^{15}~\text{kg}=3.3\cdot 10^{21}~\text{J} $$ This is a lot of energy, it would take a 2000-kilometer reflector a month to decelerate the asteroid. Difficult, but not impossible.

Let's consider a smaller asteroid, say, 100 times the mass of the Chelyabinsk meteor, $1.5\cdot 10^{9}~\text{kg}$. The required energy is now only: $$ 8.2\%\times 40~\text{MJ}/\text{kg}\times 1.5\cdot 10^{9}~\text{kg}=4.9\cdot 10^{15}~\text{J} $$ This would only require a 10 km reflector 2 days! Much more reasonable, and we can send several asteroids in a short amount of time. The solar radiation pressure on the reflector is less than $200~\text{N}$, so we don't need to worry about its orbit changing.

Once you get such a large mass moving, there is basically no way to stop it. Even though we would have about a year until the asteroid reached Earth, that would just give us plenty of time to contemplate our demise.

Answer 5

Earth is around 2.2 AU to 3.2 AU from the asteroid belt. Lets use the minimum distance and assume the Earth is flying around the sun at a similar speed as these asteroids.

2.2 AU is around 329 Million KM. The earth moves around 30KM/Sec...lets use a range of 30KM/s to 60KM/s for these asteroids in the asteroid belt (Ceres, the largest body in the belt moves around 18km per second, so we are picking out the fast movers here).

At 30km per second, it would take an asteroid around 126 days to travel from the asteroid belt to Earth (63 days for a 60km/second moving one). Ceres would take nearly 211 days at it's current speed.

Space is huge and it's hard to remember the timelines when operating at this level...Earth would have a pretty long time to detect that Mars is attempting this, and even if Mars goes through and redirects the asteroid at Earth, your earthlings will see it coming for months before impact. Whether or not they can do anything about it is a different matter.

Answer 6

There is basically one propulsion method efficient enough to be used. Other methods would work but they are so much less efficient nobody would. The method is Nuclear Pulse Propulsion where series of relatively small directed nuclear explosions are used to push a pusher plate attached to whatever you are trying to move. This has both high specific impulse and very high energy density. The technology is relatively well studied and presumed to be already practical by its proponents. Main issues are safety and radiation. During a total war use in a weapon of mass destruction would be quite acceptable.

In fiction nuclear pulse propulsion has been used to kill planetary populations with asteroids. In reality there are some engineering problems associated with building a pusher plate system that can distribute the impulse so that the asteroid would not break up. These are solvable, but the sheer scale would make them take too much time and resources for any kind of surprise to be practical. And unlike the aliens in fiction the defenders of Earth would know what an "orion drive" is and start taking countermeasures right now.

On a positive side, the infrastructure built to make the asteroid not fall apart from nuclear explosions used by the drive would help the asteroid not fall apart from anything else as well. And you'd have time to fill the surface with automated defense systems. So it would be quite difficult to stop.

The response to your enemy building such a weapon would probably be either an all-out-attack or if that is impractical an offer of peace. If no acceptable terms are offered, I'd presume those evil evil EVIL Earthlings would try to destroy so much Martian infrastructure (and population) with less powerful weapons that Mars would be unable to complete the project.

So this really works better as plot device before it is even completed than it ever does as an weapon. You can just use mass drivers to bombard the planet with much better resource efficiency. Moon based mass drivers probably would be what the Earthlings would use to bombard Mars in fact.

Answer 7

There are a number of Mars crossing asteroids that also come close to earth. 433 Eros for example. Occasionally Mars crossing NEAs pass close to the earth, a near miss. Eros came close to the earth in 2012 and we will see another near miss in 2056.

It would take a small amount of delta V to nudge a near miss into an impact.

It's been suggested humans would need to be on the asteroid. Not so. In fact placing a life sustaining hab on the rock would make the mission a lot more difficult. The Keck Report for retrieving an asteroid describes a robotic vehicle driven by ion engines. The proposed Asteroid Redirect Mission is based on this report.

Ion engines have great exhaust velocity. This reduces the amount of reaction mass needed to achieve a given delta V. But they also have very low thrust. This means it will take a long time to execute a burn.

The larger Near Earth Asteroids are too massive to nudge in a short time. Unless the ion rockets are a non neglible fraction of the asteroid's mass, it would take years. The earthlings would probably notice and take counter measures.

It might be plausible to quickly nudge a Tunguska sized asteroid to an earth impact. This would be good for wiping out a city. But if wiping out a city is the goal, the Martians could do it a whole lot more cheaply and efficiently with nuclear bombs.

Answer 8

Mars is at the bottom of a gravity well too, so it's kind of in the same boat as earth.
Kinetic bombardment from Earths moon would be a bigger threat, as would attack from the asteroid belt. The only big rocks that Mars could easily throw would be Phobos and Deimos, but a Belter civilization could send a steady stream of big rocks and overwhelm Earth defenses.