8
$\begingroup$

An alien race, for reasons of their own, have decided that it's important to make the third and fourth planets in our solar system about the same size and mass. Since they have a strict moratorium on altering planets with clearly visible carbon-based life on them, Mars is the object of their "affections". As a result of their cultural biases, they love iron. Any time they need/want to increase the size of a planet, they use iron to do it. (Yeah, I don't get that either but whatever, aliens, am I right?)

On the day of the drop, they position their ultra-mega cargo freighters full of iron around Mars then let all those gigatons of iron just fall Mars-ward. As you might imagine, the fireworks are spectacular. After the freighters move away, the planet cooling ships move into position but right before they start operations, an urgent call from the Supreme Dear Leader comes in demanding that his entire planet be air-conditioned and the cooling fleet is to report, pronto!

Iron parameters:

  • Initial altitude: 150km (all ingots enter freefall form this height)
  • Iron initial temperature: 250K
  • Speed relative to Mars' surface: 0 m/s
  • Individual Iron Pieces: 100 m^3 ingots

Mars must then cool off on its own. With the cooling fleet gone for an indefinite period of time, how long will the aliens have to wait for Mars to cool down to a comfortable temperature for carbon-based life after dumping all that mass on Mars' surface? Assume that atmospheric insulation/cooling effects can be ignored.

Altering Mars' orbit or the orbit of any of the other planets isn't a concern for these aliens, all they care about is Mars. Besides, they have the capacity to "nudge" planets into stable orbits.

(I realize that this is a fairly fanciful way of asking how long it would take to cool off Mars if you dumped enough iron on it to make it the same size as Earth, but it's more fun to write it this way. I also realize that aliens with these logistic capabilities can do pretty much anything they want.)

$\endgroup$
  • 1
    $\begingroup$ What is the initial temperature of the iron? $\endgroup$ – HDE 226868 Sep 23 '15 at 14:26
  • 2
    $\begingroup$ Thinking about this, I feel like you'd need to know the height the iron was dropped from in order to answer it. I could be wrong though. $\endgroup$ – Dan Smolinske Sep 23 '15 at 14:26
  • $\begingroup$ Dan is right, we need velocity and altitude of the iron before it was dropped. $\endgroup$ – Tim B Sep 23 '15 at 14:29
  • $\begingroup$ @HDE226868 updated parameters. $\endgroup$ – Green Sep 23 '15 at 14:57
  • $\begingroup$ Due to the huge volume of iron you are dropping more or less at the same instance in time, it cannot all be dropped from the same height. At the limit, you are dropping a spherical shell of iron equal to the radius of Mars plus 150 km. Also, the gravity of all the iron dropping must be considered and will in fact dominate the gravity of Mars itself. $\endgroup$ – Gary Walker Sep 23 '15 at 16:49
6
$\begingroup$

Lets start by ignoring that for deep planetary pressures, iron is compressible and that it expands when it is heated.

Basic facts

Mars mass: 6.4171e23 kg 
Earth mass: 5.97237e24 kg
Mars mean radius: 3389.5 km  
Mars surface acceleration: 3.711 m/s^2
Iron density: 7850 kg/m^3
volume of a sphere = 4 / 3 * pi * r^3  
radius of sphere = (volume * 3 / 4 / pi) ^ (1/3)

Mass of iron = m(Earth) - m(Mars) = 5.33066e24 kg
Volume of iron = mass / density = 6.79065e20 m^3 or 6.79065e11 km^3
Gravitational Constant: 6.67385e-11  N m^2 / kg^2

Let us model the required iron as a spherical shell of iron with an inner radius equal to that of mars plus 150 km. So, what is the outer radius? We know that the total volume of the hollow sphere must be the total volume of iron minus the volume of the hollow interior.

vMars = 4 / 3 * pi * 3389.5^3 = 1.63116e11 km^3
vInner = 4 / 3 / * pi * 3539.5^3 = 1.85744e11 km^3
vOuter = vInner + vIron = 1.85744e11 + 6.79065e11 = 8.64809e11 km^3

rOuter = (vOuter*3/4/pi)^(1/3) = 5910.31 km
vIronMars =  8.42181e11 km^3 (calculated below)
rIronMars = 5858.3 km

So, our iron to be dropped consists of a hollow ball of iron with an inner radius of 3539 km and outer radius of 5910 km. At the inner edge of the iron, the downward acceleration would be the acceleration due to Mars alone as the net contribution of the iron mass would be zero for all points inside the shell. At the outer radius, the acceleration would be based on the mass of the entire Earth.

accelInnerInitial = accelMarsSurface * (marsRadius / rInner)^2 = 3.711 * (3539.5/3389.5)^2 = 3.403 m/s^2
accelOuterInitial = accelEarthSurface * (earthRadius / rOuter)^2 = 9.8066 * (6371/5910.3)^2 = 11.395 m/s^2

The differences in acceleration clarify the problem with the iron shell assumption, the iron blocks would crash into each other as they fall. We'll simply ignore this problem by and large.

So, what is the impact velocity of the inner shell? Either we could do calculus since the acceleration increases as the shell falls closer to Mars, or we can take advantage of the formula for gravitational potential:

Gravitational Potential, V(x) = -G * M / x where G is the gravitational constants, M is the mass of the planet and X is the distance to the planets center. 

Note that V(x) is always negative and approaches zero as distance approaches infinity. Since kinetic energy = 1/2 * mV^2 and a falling object converts gravitational energy to kinetic energy we can figure out the impact energy and velocity without having to integrate over radius with variable acceleration.

For example, consider the case of falling from infinity to the surface of Mars.

F(x) = -GM/x thus F(3389500) = - 6.67385e-11 *  6.4171e23 / 3389500 = - 1.2635e7 J/kg. 

Kinetic energy change will have the same magnitude as the gravitational potential energy change due to conservation of energy.

Solving for kinetic energy for velocity, V = sqrt(2*E/m), and using -F(3389500) for E.

V = sqrt(2*1.2635e7/1), V=5.027e3 meters / sec -- this is in perfect agreement with the published value for the escape velocity of Mars, a useful check on our method.

For a mass dropped from 150 km altitude. F(3539500) = -1.20996e7 J/kg. The difference between F(surface) and F(150 km up) is 5.35e5 J/kg, which means the impact velocity is 1035 meters/second if atmospheric drag is ignored. Given the total mass of iron being dropped, this seems like a good assumption.

What about the outermost shell? similar math, but it based on total earth mass as all of the iron lies inside the outermost shell-- Mars radius is now much larger due to all of the rest of the iron already added to Mars. Again ignoring comprehensibility of the iron (and Mars itself), our new Mars planetary volume is the old volume plus the volume of all of the iron:

volumeIronMars = 1.63116e11 + 6.79065e11 = 8.42181e11 km^3. 

Solving for radius yields 5858.3 km so the outer shell will fall a distance of only 52 km, however it does so with the full acceleration due to Earth mass, about 9 times Mars mass.

F(OuterShell) = -GM/x = -6.67385e-11 * 5.97237E+24 / 5910.308044 = -67439276 J/kg

F(IronMarsSurface) =  -6.67385e-11 * 5.97237E+24 / 5858.3.303939 = -68037933 J/kg

Change in outer shell potential from falling = 5.99e5 J/kg`

The difference in energy gain for the inner and outer layers is close enough, that I will just use the geometric mean value of the inner and outer shells as the average energy change (instead of resorting to calculus to compute a more accurate number), i.e., 5.662E5 J/kg

So, finally what is the temperature change? Iron has a specific heat capacity of around 0.45 joules / gram * deg or 450 J/kg*deg so we can finally compute the temperature rise as 5.662e5/450 or 1260 degrees Kelvin, so the final iron temperature is about 1510 Kelvin or 1237 Celsius or 2258 Fahrenheit - this is considered white hot though there is still a yellowish orange appearance -- about the same as a candle flame. Iron melts at 1538C so not molten iron, but it will be much softer / more plastic than iron at Earth surface temperatures.

The incandescent IronMars will be very noticeable from Earth.

Calculation time to cool off is another set of nonlinear problems too. I want to stop here because there are 2 very different solutions.

  1. The iron rests upon the old mars surface or
  2. the iron continues to migrate down towards Mars core due to the impact load and great pressure of the softened iron overburden.

In reality, I think that there would be major penetration of iron. It is a definite possibility that the bulk of the iron will descend further, perhaps even to join with the existing iron core. The extra heat from compressing the crust might be enough to melt all of the iron, in which case it is certainly heading to the core. If this happens (and I think it would) it will take hundreds of millions of years to become Earth-like or Mars-like. Also note that the downward migration of the iron releases additional heat, so if the effect is significant it is also unstoppable, the core size is going to increase greatly.

Note that my basic model was unrealistic from the start (assuming a solid iron shell), in reality dropping from space ships the average drop height would be significantly higher and more energetic.

One major secondary effect, Mars rotation period would become about 10 times as long since the iron has to accelerate up to match the rotational velocity of Mars. This would add another large quantity of kinetic energy to the iron (enough to cause some iron melting near the equator)

So how long to cool? Don't know, and its late and I'm tired so I'm stopping for now.


The assumptions of the Virial theorem do not apply in this artificial case, i.e., we do not start with a stable gravitational bound system of widely dispersed matter. We have a collection designed to collapse upon itself within one day. Even if we allow that the iron will interact and heat up during the infall, nearly all of the radiant energy will terminate on another packet of iron during infall. To reduce this effect, it is necessary to spread out the iron -- but this raises the distance of the drop more than offsetting the increased heat loss. There is also very little time for the heated iron to radiate away its heat before impact. So I don't expect any significant percentage of the heat to radiate away during infall.

What happens to the atmosphere of Mars? It too is heated to incandescence and will remain so as long as the iron remains that hot. I don't expect much of the atmosphere to be lost quickly since the escape velocity for IronMars will be even higher than Earth and the RMS gas velocity is only about 1 km/sec.


I do have a quibble with the problem as posed, do the aliens have antigravity? You can't simply drop iron from orbit, a straight drop required your cargo ships to simply hover in place. If you can do that, why did you have to make such a mess in the first place?

$\endgroup$
  • $\begingroup$ This is the same analysis I was going to do. However I think that the Virial theorem means that only half of the gravitational potential energy becomes thermal energy of the body and the rest is radiated away during the collapse. Obviously the unrealistic solid iron shell is going to trap more heat than in the "hail of ingots" approach, but it might be reasonable to halve the surface temperature to account for the huge blast of radiation during the collapse. $\endgroup$ – Blake Walsh Sep 24 '15 at 8:17
  • $\begingroup$ I figured it would be option 2, that the iron would migrate towards the core. $\endgroup$ – Green Sep 24 '15 at 12:06
  • $\begingroup$ The assumptions of the Virial theorem do not apply in this artificial case, i.e., we do not start with a stable gravitational bound system of widely dispersed matter. We have a collection designed to collapse upon itself within one day. Even if we allow that the iron will interact and heat up during the infall, nearly all of the radiant energy will terminate on another packet of iron during infall. To reduce this effect, it is necessary to spread out the iron -- but this raises the distance of the drop more than offsetting the increased heat loss. $\endgroup$ – Gary Walker Sep 24 '15 at 13:46
2
$\begingroup$

I’ll take a rough historical (and non hard science) stab at an answer.

Infant Earth’s theorized impact with Theia seems like an interesting model for comparison here. We don’t seem to have a particularly clear idea of what Earth looked like at the time, but this event would have introduced quite a bit of material into the Earth’s vicinity (much of it theoretically forming the moon).

At present, it’s thought that the impact event occurred around 4.4-4.45 billion years ago. This would have stirred things up quite a bit and required a cooling period, perhaps not unlike you bombarding the surface with enough iron to double the diameter of Mars.

Determining how long you’d need to wait before the atmosphere reached “pleasant” temperatures is a bigger challenge. It seems that the Earth’s crust cooled and solidified around 4.1 billion years ago, though I’m sure it wouldn’t have been a nice place to be. Earth’s first lifeforms are believed to have developed around 3.8 billion years ago with the beginning of the Archean Period. That puts you at around 600 million years, plus a few hundred million years to get to your preferred temperatures (assuming that the Earth was still a pretty hostile environment even during the Archean Period.)

There are, of course, shortcomings to using this model.

  • The giant impact hypothesis, while a leading theory, may not be fully correct.
  • Earth was most likely still molten when the impact occurred, as opposed to the fully solidified Mars.
  • Further analysis is needed between the start of the Archean period and your target environmental temperatures (though more specification of the target may be useful).
$\endgroup$
1
$\begingroup$

The parameter to the problem where changed midway through my calculation, so I won't finish it. I'll leave this here if someone wants to take over.

Firstly we need to know the thickness of the iron pool.

$\text{Earth's mass} = E_m = 5.972 \cdot 10 ^{24}\text{kg}$

$\text{Mars's mass} = M_m = 6.39 \cdot 10 ^ {23}\text{kg}$

$\text{Difference} = E_m - M_m = 5.333 \cdot 10 ^{24}\text{kg}$

$\text{Iron's density} = D_i = 7850\text{kg}/\text{m}^3$

$\text{Volume of Iron} = V_i = \text{Difference} / D_i = 6.794 \cdot 10^{20}$

$\text{Radius of Mars} = R_m = 3.39 \cdot 10 ^ 6\text{m}$

Volume of a spherical shell

$Vi = 4/3 Pi (R_{mnew}^3 - R_m^3) <=> R_{mnew} = sqrt3(3V_i/(4\pi) + R_m^3) = 5.859 × 10^6\text{m} $

Height of iron is:

$h = R_{mnew} - R_m = 2.469 \cdot 10^6\text{m}$

which makes the radius of Mars slightly lower than Earth's. This is a quite unrealistic result, because compression would play a big role into reducing this value. There are many things that are hard to predict accurately. What would happen to the atmosphere?

Let's assume the aliens choose the lowest temperature at which iron is liquid, it's melting point. Don't waste too much energy and the extra energy from the fall would keep it liquid to spread it evenly. Another inaccuracy, we ignore pressure.

$M_{pi} = 1538ºC$

Due to low amount of oxygen in the Mars atmosphere, I'll assume there's just some red rust on the iron.

$\text{emissivity} = \epsilon = 0.6$

$\text{reflectivity} = Ref = 1 - \epsilon = 0.4 $

Next we want to know the equilibrium temperature of the planet.

$\text{Effective temperature of the Sun} = T_s = 5780K$

$\text{Distance from Mars to the Sun} = D = 2.279*10^{11}\text{m}$

Using planet's emissivity instead of a black body:

$T_eq = T_s * Ref^{2/4} * \sqrt( R_m / (2* D) ) = 9.97\text{K}$

Yes, it will get really really cold eventually. Let's make another assumption and consider the iron shell is empty inside. We've made so many up until now that every error will cancel out and we'll get the right result. Hopefully.

$\endgroup$
  • $\begingroup$ The key thing to notice here, if I'm not mistaken, is that the amount of iron necessary would be huge, you would barely notice the planet that was taking the place. $\endgroup$ – SlySherZ Sep 23 '15 at 15:32
  • 1
    $\begingroup$ Mars mass should be 6e23 not 6e21 $\endgroup$ – Gary Walker Sep 23 '15 at 16:51
  • 1
    $\begingroup$ Uncompressed iron is 7850 kg/m^3. $\endgroup$ – Gary Walker Sep 23 '15 at 16:58
  • $\begingroup$ Er, I don't think the shell is spherical. It's a shell surrounding the entire planet, remember? If it was a sphere, it wouldn't spread around the entire planet. $\endgroup$ – HDE 226868 Sep 23 '15 at 18:06
0
$\begingroup$

Since Mars' atmosphere is only 1% of earth's atmosphere (in mass) so there would hardly be any fireworks visible from earth even with powerful telescopes. There would only be ultra-large clouds of red dust flying around though (iron oxide dust, that constitutes the surface of Mars).

Having an extremely light atmosphere also means that Mars' greenhouse capability is virtually zero, and any heat generated by the event of impact would radiate away into space far quicker than it would on earth. I cannot provide an exact number though, as I am unaware of precise values in this subject.

Also, I think you forgot a very important effect which would result due to this incredible increase in Mars' mass. I am talking about the asteroid belt between Mars and Jupiter. With nearly 3 times increase in Mars' mass, the gravitational balance that keeps the asteroid belt in it's place would be severely disturbed and a lot of asteroids would be deviating from their orbits towards Mars ... and beyond it ... towards ... don't ask me. The very thought horrifies me.


My answer above, is written assuming that the additional iron mass was dumped on the red planet as dust and tiny iron fragments, not as a single planetary sized asteroid. If you are talking about THAT, then it would probably end Mars' existence as a planet as both objects would crash with a force that would crack Mars to the core and pulverize it to dust. Most of the planetoid mass would vaporize from the heat of impact. The phase-2 damage to earth would also be immensely huge. Falling debris and asteroids would end all life on earth (at least all multicellular life) and rip giant fissures on the crust.

$\endgroup$
  • $\begingroup$ I actually doubt that there would be much of a difference in the asteroids' orbits. Mars is still pretty far from the belt, and even a planet with the mass of Earth where Mars is wouldn't have a significant impact. $\endgroup$ – HDE 226868 Sep 23 '15 at 18:12
  • $\begingroup$ Researchers speculate the dinosaur-killing asteroid came from that asteroid reserve. Tells you something about how many others would deviate in times when earth and mars were closest to each other in their orbits ... $\endgroup$ – Youstay Igo Sep 23 '15 at 18:20
  • 1
    $\begingroup$ That assumes that the cause of the perturbation was Mars, which is not necessarily the case. $\endgroup$ – HDE 226868 Sep 23 '15 at 18:23
  • $\begingroup$ It is not necessarily true. I agree. But You cannot completely negate the gravitational effect of making a planet nearly 3 times it's mass. $\endgroup$ – Youstay Igo Sep 23 '15 at 18:27
  • $\begingroup$ Considering the Jupiter is already hundreds of time the mass of Earth, it is safe the say that the asteroid belt survives largely intact. However, asteroid orbits that happen to be near orbital resonance with Mars will be more noticeably affected, so there would be some asteroids moved out of their existing orbits, but more likely to impact Jupiter than Earth. $\endgroup$ – Gary Walker Sep 24 '15 at 5:53
0
$\begingroup$

I suspect the surface will actually be cold from the beginning.

Here is one way to model it. Imagine you take a pillar of iron about 2000km tall, and drop it from 150km. As the pillar strikes the surface of mars, the bottom of the pillar will heat up tremendously through lithobraking and the weight and heat will drive it deep into the surface of mars - I suspect though that being dropped from a mere 150km the 2000km pillar of iron wont embed itself all the way into mars, and the top of it will remain poking out, with the top still being cold, not having experienced any lithobraking and indeed being far away from that heat.

Now take billions of those pillars and drop them all around mars the bottoms will certainly be heated to extreme temperatures in the crunch zone, but the tops will remain cold. (you might want to imagine the pillars being slightly tapered, so when they all press together they form a neat sphere)

In this modelling the boundary between Mars and the iron is super heated to tens of thousands of degrees, but that heat is contained by the mass of iron on top, and that mass of iron on top experiences no heating as such.

And I suspect it doesn't matter whether you use billions of pillars, or trillions of iron ingots, or even a monolithic iron shell. If all the iron is placed 150km above mars, and all dropped simultaneously, that is what you'll get: A super molten layer of iron under a cold crust of iron and under that the "cold" core of mars (maybe heated by being slammed in all directions by 90% earth mass of falling iron). There might be volcanoes of molten iron emerging through fault lines in the collapsing iron, or it might be perfectly contained, nevertheless I am quite certain there will be large areas of cold iron.

Now over time the heat of the super molten layer will migrate upwards and downwards, it certainly seems there is enough heat to heat all the iron to uncomfortable temperatures.

The question is then, how quickly will it radiate out? Will it radiate out quickly enough to super heat the surface? Or will it just radiate out slowly, perhaps gently radiating over hundreds of thousands of years?

I suspect it will be like Earth. There is enough thermal energy in Earth to heat all of Earth to a most uncomfortable temperature, in fact the average temperature of Earth is much hotter than our Iron Mars would be. But the rate of heat transfer from the core to the surface is too slow to do anything more than slightly warm the surface, and even with the wonderful conductive qualities of iron, there would be still be 2000km of it to act as an insulator.

So that is my answer: that the surface of Iron Mars will initially be cold, and over time will be slightly warmed, but not uncomfortably so.

If you want a cataclysmic heating of the surface, then the constraint of releasing all the iron at once should be relaxed, then the iron which is released later will also have a chance to slam into the surface like meteors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.