Lets start by ignoring that for deep planetary pressures, iron is compressible and that it expands when it is heated.
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.
- The iron rests upon the old mars surface or
- 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?