The Physics of DOOM Eternal: blowing holes into Mars

Welcome to the first ever post of fictionphysics! Today I will be diving into the physics behind the following scene from DOOM Eternal-

Although this is clearly impossible in real life, I thought that it should be feasible to calculate the energy it would take to create a hole of that size on Mars and perhaps speculate on how this could be done roughly within the realm of reality.

Methodology

Before I start, this problem needs to be broken down into steps. The obvious first step would be to calculate the size of the hole- we know from contextual information that this hole leads to the mars core, which remains intact. With this and the apparent shape of the hole in mind (Fig. 1), it becomes clear that the hole becomes narrower with increasing depth, suggesting that the shape overall is approximately conical. 

 Fig. 1

Accounting for the curvature, the whole shape is approximately a spherical sector excluding the part that makes up the core. With appropriate pixel scaling, I should be able to calculate the size of the whole, which would bring me to the second step- calculating the energy it would take it make it. For this, I found the Kuz-Ram model, which was developed for the purpose of controlled explosions in quarrying in order to apporximate the size of the debris left after the explosion. THe third and final step would be to speculate on how this could be done.

Step 1: calculating the volume of the hole

We can use pixel scaling to calculate the approximate tangential distance across the crater by comparing the size of the crater to the radius of mars, which according to NASA is 3389.5 km. By measuring the length of the radius in pixels, I can find the metre : pixel ratio as follows-

Fig. 2a- Measurements

The blue line above shows the distance across the crater. However, it is important to specify that this is not the actual distance but the tangential one which doesn't account for the curvature. With thie metre : pixel ratio above, I can calculate the tangential distance to be

Fig. 2b is a cross-sectional view of the crater (excuse the poor drawing) with a few important labels, the curvature here is very exaggerated.  With the assumptions, the crater is a spherical sector with a similar spherical sector removed from it at the end (ie., Of the core).  Thus, the volume of the crater is

r being the radius of the core and R being the radius of Mars. θ is half of the cone angle. a is half of the tangential distance and H is the height of the cone. The sides H, a and R form a right angled triangle, thus, trigonometry can be used to calculate θ-

The final piece of information we need is the radius of the core, which according to NASA is about 1500 km. Now, the volume of the crater can be calculated as follows-

Fig. 2b

This is about

of Mars' volume.

At this stage, a complication arises as Mars has layers: The crust, the mantle and the core (Fig. 2c).  I am only interested in the mantle and crust as the core is left intact.  To make progress with the calculations, I need to know the volumes of these layers in the section of Mars that is destroyed because they have different compositions and thus different energy requirements for destruction.

Fig. 2c

The crust of Mars is approximately 50 kilometres deep.  This information can be used to calculate the volume of the mantle as we can just calculate the volume of the cone left over after 50 km has been reduced from the radius.  This is shown in Fig. 2d.

We can calculate H­2 by subtracting the thickness of the crust

from it's radius-

The other piece of information we need to calculate the volume of the mantle is the length of radius, which is a2.  This can be calculated using the similarity between the larger cone and the smaller cone due to which there is a scale factor between the radii and the heights of the cones, which can be expressed mathematically as

H can be calculated using the curvature of Mars, which is a dip of 2 metres for every 3600 m travelled tangentially.  As we know the tangential distance across the crater, we can calculate the height of the spherical cap assuming that the centre point of the tangent is the maximum height-

By subtracting this value from the radius, we get H to be

With the scale factor above, we can find a2 as folllows:

Now that we have all the measurements we need, we can calculate the volume of the mantle as it is a cone with a spherical sector (that of the core) subtracted from it-

Finally, the volume fo the crust can be calculated by subtracting the volume of the mantle from the overall hole-

Calculating the energy it would take to make the crater

This is the Kuz-Ram model or the Kuznetsov equation and it is used to calculate the average size of each fragment (k5o, in cm) when TNT of mass Q (kg) is used to destroy rock with Volume V (m^3) and a rock factor A.  This equation was made for the purpose of quantifying controlled explosions in mining but can be applied to our situation.  To be specific, the Volume is strictly described as the volume of rock impacted by one borehole and since the 'laser' only fires once we can assume that the equivalent number of boreholes would be one.  Our aim is to find Q, for which we require A and k50 for both the mantle and crust.

Mars’ mantle consists of 73% fayalite, clinopyroxene 12%, garnet 11%, and periclase-wüstite 2%.  Due to a lack of research and testing (as far as I could find) on the other materials, I will assume that fayalite makes up 100% of the volume of the mantle.  Now we are left with finding the rock factor, A, of fayalite.  The equation for rock factor is

HF being the hardness factor, RMD being the rock mass description and RDI being the rock density influence.

The HF of fayalite depends on its Young’s Modulus, if this is less than 50, then we can use it directly to calculate HF as

However, if it is greater than 50, the Unconfined Compressive Strength (UCS) must be used as follows-

Although I couldn’t find the Young’s Modulus (Y) for Fayalite, I could find its Longitudinal Modulus (L), 270, which can be converted into its Young’s Modulus using Poisson’s ratio (σ) for Fayalite.  This is 0.34 and the equation to calculate Young’s Modulus is as follows:

Since this is >50, the UCS must be used to calculate HF-

Now that we have the HF, we can move on to calculating the rock density influence.  The equation for this is

D being the density of the rock in tonne/m^3. Using the density of fayalite, we can calculate the RID to be

We only have the rock mass description left to calculate. The equation for this is

Here, JCF is the joint condition factor, JPS is the vertical joint plane spacing factor, and JPA is the vertical joint plane angle.  The conditions for these are as follows:

1.       JCF- assigned a value depending on the nature of the joints.

Tight joints- 1

Relaxed joints- 1.5

Gouge-filled joints- 2.0

Since we have assumed that the entirety of Mars’ mantle is made up of fayalite, it would be reasonable to assume that the joints are tight to abide by the assumption as closely as possible.

2.      JPS- assigned a value depending on the joint spacing.

joint spacing < 0.1 m, JPS = 10

joint spacing = 0.1–0.3 m, JPS = 20

joint spacing = 0.3 m to 95% of P, JPS = 80

joint spacing > P, 50 (because all blocks will

be intersected).

Likewise to the previous one, as we have assumed that 100% of the mantle is made of fayalite, it would be reasonable to pick the least possible value for joint spacing to abide by the assumption as closely as possible.

3.        JPA- assigned a value depending on the angle at which the joints are facing.

Dip out of face- 40

Strike out of face- 30

Dip into face- 20

Due to their being a lack of research on this topic and this factor not being closely linked to the assumption, I decided to use the dice function on my calculator, assigning the following values to each of the conditions-

Dip out of face- 1,2

Strike out of face- 3,4

Dip into face- 5,6

The calculator performed 999 repetitions and the values with the highest combined frequencies ended up being 3 and 4.  Thus, the value for JPA is 30.

Now that we have all the required values, we can calculate RMD to be

Finally, we can caluclate the rock factor for fayalite-

Mars’ crust primarily consists of picritic basalt, which is mainly made up of a compound called forsterite- we can assume that the crust is completely made of forsterite due to their being a lack of research on the specific components of the composition.  In a similar fashion as before, we need to calculate the rock factor.

The Young’s Modulus of forsterite is 274.4, which is greater than 50, due to which we must use the UCS-

To abide by the assumptions (refer to explanations for fayalite), the JPS is 10 and the JCF is 1.  For the JPA, I decided to use the same parameters as before and ended up with 5,6 being the combination with the highest frequency.  Thus, the JPA is 20. The RMD is

The RDI for Forsterite is

Now, we can calculate the rock factor to be

Due to theri being a lack of debris, It is reasonable to assume that the rocks were broken down to the very unit cell, which measures 10.2 Å (this is our value for k50 for both forsterite and fayalite as they are both types of olivine.

We can rearrange the Kuznetsov equation, making Q the subject-

By substituting the values for both materials and adding the Q values, we get

This can be converted into energy as a kilogramme of TNT is a standardised measurement-

It is very difficult to comprehend how much energy this is.  Due to this, I have quantified this value in terms of a few real life equivalents to somewhat help realise how much energy this is:

Speculation

Now that we have an idea of the energy requirement, it is reasonable to assume that there is no chance of being able to provide the energy required for this kind of weapon in the foreseeable future.  However, if we assume that the supply of energy is sufficient, we could think of possible ways of making this happen.  Firstly, although it may seem to be the case, a laser weapon wouldn’t be able to cause this kind of damage as real-life tests have shown that they tend to cause heating which is unlikely to cause an explosion as the release of energy would be too small.  The first alternative is suggested by the Kuznetsov equation, which is designed for quarrying.  If one were to dig a borehole onto Mars and fill it with 4.56 × 10^37 kg of TNT, the explosion caused upon ignition would be of the same size as the crater in the videogame.  However, the shape won’t necessarily be the same.  From the video, we can see that when the beam hits mars, it causes a circular wavefront to be emitted, similar to an earthquake.  A reason behind this could be that the weapon is emitting a sound wave which is causing the point of impact and the mass below it to vibrate intensely, resulting in wavefronts (in form of tremors, similar to earthquakes) being emitted from the centre, resulting in the destruction of a circular portion of land. A counterargument to this is that sound can’t travel through a vacuum and thus this can’t be the case.  However, it could be argued that the green colour of the beam could be explained (as sound waves don’t have a colour) by a stream of particles being propagated through space, providing a medium as the sound waves travel through the beam.  Although this has further implications regarding how this sort of beam would be created and how it would be directed accurately towards Mars, it is the closest equivalent I can think of.

Final Thoughts

I started working on this problem out of pure curiosity and due to not being able to find a (in my opinion) a sufficiently thorough investigation of it.  I was surprised as working through it I found myself learning materials science to reach the answer.  Other than that, this investigation helped improve my skills of critical thinking and scientific modelling, which I am grateful for. Thank you very much for reading my blog post and I hope that you read the next one, which comes out next month.