Is the weapon from GI Joe: Retaliation possible and accurate? [closed]

Question

In the movie G.I. Joe: Retaliation a kinetic bombardment weapon was used to destroy several major world cities. Is that weapon possible and if so, is the destruction accurate?

Note: I asked this question because answers online annoyed me.

Update
I am specifically concerned with the feasibility of launching so much material into space and whether the energies described in the movie are even remotely accurate.

Note that several questions on Physics Stack Exchange have been asked about the accuracy of things that happen in movies (e.g., Questions on the movie Gravity (2013) or Would a giant satellite laser work as a weapon?). Therefore, I do think the physics of this question is on topic and appropriate.

Answer 1

There are so many problems with just this weapon in this movie but let's start with easiest problems.

First, the depiction is that a satellite opens a bay door and the tungsten rod just falls out and hits its target without any guidance. There's some dynamic, cylindrical part on the Earth-facing side that seems to act like a spring-action firing mechanism, but this is still very problematic as I will describe below. Regardless, that they depict the thing being over London and then dropping directly on London is absurd. Even if properly guided, the drop point would need to be very far from London as once it leaves the satellite bay, it starts with a velocity of ~8 km/s transverse to the Earth's surface.

Next, let's look at what type of speeds we are dealing with. Suppose we use the hypothetical parameters of the US Air Force's Project Thor which would have cylindrical tungsten rods with a radius of ~0.3 m and length of ~6.1 m giving a volume of ~1.72 m³. That doesn't sound like much but with a mass density of ~19,300 kg m^-3, this rod would come in at ~33,287 kg.

The Delta IV Heavy rocket's maximum payload to LEO is ~28,790 kg, i.e., about 4000 kg short of just the rod. The satellite holding these rods in the movie is absolutely huge and appears to have at least 6 of these rods, probably more. The strongest launch vehicle available is the Falcon Heavy with a max payload to LEO of ~63,800 kg, i.e., less than 2 rods.

Okay, for the sake of argument, let's assume they brought everything up in pieces over the course of the shuttle program or something and assembled everything in space (still looking at a $100 billion plus price tag but who's counting). Once they are up there, we can approximate the speed of impact if the projectile was dropped from 200 km.

So let's ignore it's orbital velocity to make things simple and just start with an object falling from 200 km that has a drag coefficient, $C_{d}$, of ~0.295 (roughly that of a bullet) and the physical parameters described above. The acceleration can be modeled as: $$ \ddot{r}\left( t \right) = - \frac{ G \ M_{E} }{ \left( R_{E} + r\left( t \right) \right)^{2} } + \frac{ 1 }{ 2 \ m } \rho\left( r \right) \ \dot{r}^{2}\left( t \right) \ A \ C_{d} \tag{0} $$ where $G$ is Newton's gravitational constant, $M_{E}$ ~ 5.9742 x 10²⁴ kg is the mass of Earth, $R_{E}$ ~ 6.6743 x 10⁶ m is the mean equatorial radius of Earth, $m$ is the mass of the falling object (~33287 kg here), $A$ is the cross-sectional area (~0.283 m² here). We model the Earth's atmosphere as a simple exponential given by: $$ \rho\left( r \right) = \rho_{o} e^{ -r\left( t \right)/h_{o} } $$ where $\rho_{o}$ = 1.225 kg m^-3 and $h_{o}$ = 10.4 km.

If we start with $r\left( 0 \right)$ = 200,000 m and zero velocity, i.e., $\dot{r}\left( 0 \right)$ = 0 m/s, then we can numerically solve for the time it takes to reach the ground and speed it reaches when it hits.

Using these numbers/estimates/models, the rod hits Earth after ~207 seconds at a speed of ~1920 m/s or a Mach number of ~5.8. This would carry a kinetic energy of ~6.14 x 10¹⁰ J or ~14.7 kilotons of TNT.

Answers

So is the yield roughly accurate?

Sure, kind of. The fake POTUS states the impact will deliver more damage than a nuclear warhead. The estimate above is in the range of the WWII atomic bombs while many modern nuclear warheads are closer to 500 kilotons of TNT. So perhaps they meant a tactical nuke?
[not sure why I'm pondering... this movie is so absurd]

Will this actually cause the major seismic quakes/ripples shown in the movie?

No, probably not. A magnitude 4.0 earthquake releases ~6.3 x 10¹⁰ J, i.e., slightly more than the total kinetic energy of the impacting rod. A magnitude 4.0 earthquake can be felt, but rarely does significant damage.

Is the weapon possible? Technically, yes. A better question would be, is it feasible?

The answer to that is likely no. Suppose we need to construct our absolutely gigantic satellite in orbit to house six of these tungsten rods, that means we have ~200,000 kg of tungsten to get into space and the spacecraft needs to be more massive than each rod to avoid major orbital trajectory effects during launch, so let's say the spacecraft bus needs to be at least twice one rod so ~66,500 kg. So the total mass to launch into space is now >260,000 kg.

A typical cost per kilogram to LEO is (as of Feburary 5, 2021) ~\$54,500/kg. So just to launch this much mass into space would cost north of $14 billion. That's without assembly or anything else, just launching the parts. Now you need at least a dozen assembly mission launches, each at ~\$1 billion a pop so just to get everything ready would be north of at least \$25 billion.

A quick Google search shows that a tactical nuclear weapon costs around \$30 million a piece and using the weapon probably entails at least another \$1 million, or nearly three orders of magnitude lower. Since money is what governments actually care about, it seems they would prefer nuclear options to \$20+ billion satellites that are so large they are easy targets to see and likely easy to attack from the ground.