"Little Boy" atomic bomb |

*(*

**Disclaimer:**Since I am not a geologist but only a physicist, I would appreciate if somebody can correct my assumptions and computations. Computations and opinions are solely mine and may not necessarily reflect those of the Department of Physics of Ateneo de Manila University and of the Manila Observatory.)According to Philippine Star, Dr. Renato Solidum of Phivolcs reported that "a magnitude 7 earthquake has an energy equivalent to around 32 Hiroshima atomic bombs". I verified his estimate in Earthquake Energy Calculator and constructed the following table:

Magnitude (R) |
Seismic Moment Energy (J) |
Seismic Radiated Energy (J) |
Hiroshima Bombs |

7.0 | $3.899420\times 10^{19}$ | $1.995262\times 10^{15}$ | 31.8 |

7.1 | $5.508077\times 10^{19}$ | $2.818383\times 10^{15}$ | 45.0 |

7.2 | $7.780366\times 10^{19}$ | $3.981072\times 10^{15}$ | 63.5 |

So a 7.0 quake is equivalent to 32 Hiroshima-type bombs, as Dr. Solidum said. But a 7.2 quake is 64 nuclear bombs!

Little Boy A-Bomb Drop Hiroshima Ruins 11x14 Silver Halide Photo Print |

**the depth of the quake epicenter**. According to USGS, the epicenter's depth is about 20 km and not 0 km (ground level). This means that you make a well 20 km deep, put all the 32 Hiroshima-type bombs there, one on top of each other, cover the well with soil and stones, then detonate the bombs. BOOM! The island of Bohol would then shake and the churches would be leveled to a heap of ruins. Actually, a similar procedure was done by France when they detonated nuclear bombs at the Moruroa Atoll, but the depth of the well is less than 1 km:

France abandoned nuclear testing in the atmosphere in 1974 and moved testing underground in the midst of intense world pressure which was sparked by the New Zealand Government of the time, which sent two frigates, HMNZS Canterbury and Otago, to the atoll in protest for a nuclear free Pacific. Shafts were drilled deep into the volcanic rocks underlying the atolls where nuclear devices were detonated. This practice created much controversy as cracking of the atolls was discovered, resulting in fears that the radioactive material trapped under the atolls would eventually escape and contaminate the surrounding ocean and neighboring atolls. A major accident occurred on 25 July 1979 when a test was conducted at half the usual depth because the nuclear device got stuck halfway down the 800 metre shaft.[2] It was detonated and caused a large submarine landslide on the southwest rim of the atoll, causing a significant chunk of the outer slope of the atoll to break loose and causing a tsunami affecting Mururoa and injuring workers.[2] The blast caused a 2 kilometre long and 40 cm wide crack to appear on the atoll.[2] (Wikipedia: Moruroa)

Richter's Scale: Measure of an Earthquake, Measure of a Man |

**A. Richter Scale and Radiated Energy**

**In order to relate the energy $E$ of an earthquake to the Richter scale $R$, we use the formula**

\begin{equation}

E = 10^{1.5 R + 4.8}

\end{equation}

Let's try for $R=7.0$:

\begin{equation}

E = 10^{1.5(7.0)+4.8} = 10^{15.3} = 10^{0.3}10^{15} = 1.995262\times 10^{15} J,

\end{equation}

where $J$ stands for Joules, a unit of energy. Notice that energy value corresponds not to the seismic moment energy but to the seismic radiated energy. Since Phivolcs says that the magnitude is 7.2 while USGS says it is 7.1, we will just stick to the value of 7.2, which gives

\begin{equation}

E = 10^{1.5(7.2)+4.8} = 10^{15.6}= 3.981\times 10^{15} \approx 4\times 10^{15} J.

\end{equation}

Seismic Wave Propagation and Scattering in the Heterogeneous Earth : Second Edition |

**B. Earthquake Power and Intensity as Function of Distance from Epicenter**

For simplicity, let us assume that the 7.2 magnitude quake happened in 1 second. This means that the power at the source is

\begin{equation}

P = \frac{E}{\Delta t} = \frac{4\times 10^{15} J}{1 s} = 4\times 10^{15} W,

\end{equation}

where the $W$ stands for Watts, a unit of power. If the seismic wave spreads out equally in all directions, then the average intensity at the distance $r$ from the quake epicenter is

\begin{equation}

I_r = \frac{P}{4\pi r^2},

\end{equation}

(c.f. Young and Freedman 2004, p. 567). This simply means that the farther you are from the epicenter, the quake intensity decreases. One way to imagine this is to take a balloon. Blow some air into it, then draw some equally spaced dots on the surface, about 1 cm apart. If you blow more air into the balloon, the balloon expands. The number of dots remain the same, but the number of dots per square centimeter decreases. In the same way, for earthquakes, the power remains the same, but the power per unit area decreases as you go farther from the epicenter.

Intensity of a spherical wave as a function of distance from source (Source: Hyperphysics) |

Since the earthquake epicenter is 20 km deep, we set $r=20\times 10^3 m$ and solve for the intensity $I_r$:

\begin{equation}

I_r = \frac{4\times 10^{15} W}{4\times 3.14\times 400\times 10^6\ m^2} = 7.96\times 10^5 W/m^2 \approx 8\times 10^5 W/m^2.

\end{equation}

Since we assumed that the duration of the quake is $1s$, then the energy $E_r$ absorbed per square meter by the church buildings per square meter is

\begin{equation}

I_r\Delta t = 8\times 10^5 J/m^2 = \frac{E_r}{A},

\end{equation}

where $A= 1 m^2$. Hence,

\begin{equation}

E_r = 8\times 10^5 J.

\end{equation}

Joint Scientific Papers of James Prescott Joule (Cambridge Library Collection - Physical Sciences) (Volume 2) |

**Energy Equivalent of a Falling Mass**

How does one imagine an energy $E_r = 8\times 10^5 J$? Gravitational potential energy is $E = mgy_0$, where $m$ is the mass of the object, $g= 9.8 m/s^2$ is the gravitational acceleration, and $y_0$ is the initial height from the reference (e.g. ground). For simplicity, let us set $y_0 = 1 m$, so that the corresponding weight of the object must be

\begin{equation}

mg = 8\times 10^5 N,

\end{equation}

where $N$ is Newtons, the unit of force. Dividing both sides by the gravitational acceleration $g=9.8 m/s^2$, we get the mass

\begin{equation}

m = 0.8\times 10^5 kg = 80,000 kg.

\end{equation}

ANTIQUE PRINT -ROMAN- BATTERING RAM SYSTEM- 1729 |

**Battering Ram!**

**How does one imagine a mass $m = 80,000 kg$?. Steel has a density of**

\begin{equation}

\rho = \frac{m}{V} = 7,850 kg/m^3.

\end{equation}

For a mass of $80,000 kg$, the corresponding volume of steel is

\begin{equation}

V = \frac{80,000 kg}{7,850 kg/m^3} = 10 m^3.

\end{equation}

If the steel is is a box with base area of $1 m^2$, then the height of the box must be 10 m or about three building floors. So we have a pillar of steel: $1 m\times 1 m\times 10 m$. This is already a battering ram that can destroy the gates of medieval fortresses!

Old Historical Baroque Filipino Church of Loon - 52"W x 35"H - Peel and Stick Wall Decal by Wallmonkeys |

\begin{equation}

mgy_0 = \frac{1}{2}mv^2.

\end{equation}

Solving for the final velocity $v$, we get

\begin{equation}

v = \sqrt{2gy_0} = \sqrt{2\times (9.8 m/s^2)\times (1 m)} = 4.43 m/s.

\end{equation}

This speed is equivalent to

\begin{equation}

v = 4.43 m/s = \frac{4.43 m}{s}\times\frac{3600 s}{hr}\times\frac{1 km}{1000 m} = 15.9 = 16 kph.

\end{equation}

Thus, the earthquake of magnitude 7.2 is equivalent to a battering rams of sizes $1 m\times 1 m \times 10 m$ lifted $1 m$ and then released. The battering rams swing like a pendulum and hit each square meter of the walls at a speed of $4.43 m/s = 16 kph$. Hence, the pile of ruins. (

**Note:**Technically, you still have to compute the moment of inertia of the battering ram as it swings like a pendulum. But for simplicity, we approximated the ram into a ball of the same mass. This is not really important; what is more important is the speed of the ram just before impact, and we set this speed to be 4.43 m/s, the same speed if the ram touches the ground as it fell freely from a height 1 m above the ground.)

Medieval Battering Ram |

**Quake Magnitude and Length of Battering Ram**

**We can repeat the same procedure for 7.0 and 7.1 magnitude quakes. But I think it is best to derive a formula instead that relates the quake magnitude and the length of the battering ram.**

We know that the formula for the quake energy $E$ as function of the Richter magnitude $R$ is

\begin{equation}

E = 10^{1.5 R + 4.8}.

\end{equation}

Dividing this by the time interval $\Delta t$ yields the power

\begin{equation}

P = \frac{E}{\Delta t}.

\end{equation}

The intensity $I_r$ of the quake at a distance $r$ from the epicenter is

\begin{equation}

I_r = \frac{P}{4\pi r^2},

\end{equation}

In terms of energy $E$ and the Richter magnitude $R$, this is

\begin{equation}

I_r =\frac{E}{4\pi r^2\Delta t} = \frac{10^{1.5 R + 4.8}}{4\pi r^2\Delta t}.

\end{equation}

This means that the energy $E_r$ per unit area $A$ at a distance $r$ from the epicenter is

\begin{equation}

\frac{E_r}{A} = \frac{E}{4\pi r^2} = \frac{10^{1.5 R + 4.8}}{4\pi r^2}.

\end{equation}

Replica battering ram at Château des Baux, France |

We define the energy $E_r$ in terms of a mass $m$ falling from a height $h$:

\begin{equation}

E_r = mgy_0.

\end{equation}

For a steel of density $\rho$ which is shaped as solid box of base area $A$ and height $h$, the mass of the steel is

\begin{equation}

m = \rho V = \rho Ah.

\end{equation}

Thus, the energy $E_r$ is

\begin{equation}

E_r = \rho Ahgy_0,

\end{equation}

Using these results, we have

\begin{equation}

\frac{E_r}{A} = \rho hgy_0 = \frac{10^{1.5 R + 4.8}}{4\pi r^2}.

\end{equation}

Solving for the height $h$ of the steel with $1 m^2$ base, we get

\begin{equation}

h = \frac{1}{\rho g y_0} \frac{10^{1.5 R + 4.8}}{4\pi r^2}.

\end{equation}

This height $h$ is the length of the battering ram with $1 m^2$ base area of density $\rho$ raised at a distance $y_0$ from its starting point. This battering ram is the equivalent of the earthquake of Richter magnitude $R$ originating at a depth of $r$ below the surface.

Illustration of Roman Soldiers Forming Testulas and Using Siege Towers to Attack Fortified Wall - 42"W x 37"H - Peel and Stick Wall Decal by Wallmonkeys |

**Steel Ram vs Wooden Ram**

Let us now set the values for a steel ram: $\rho = 7,850 kg/m^3$, $g = 9.8 m/s^2$, $y_0=1m$, $r=20\times 10^3 m$. These give

\begin{equation}

h = h(R) = \frac{10^{1.5 R + 4.8}}{3.867\times 10^{14}}.

\end{equation}

On the other hand, had we set the values for a wooden ram with $1 m^2$ cross-section, we simply change the density to that of wood $\rho=1\times 10^3 kg/m^3$ (e.g. logwood $\approx 0.9\times 10^3 kg/m^3$). This would result to

\begin{equation}

h = h(R) = \frac{10^{1.5 R + 4.8}}{4.926\times 10^{13}}.

\end{equation}

As before, the ramming speed is assumed to be $4.43 m/s$ upon impact.

Let us tabulate our results. Note that this assumes that the structure is immediately above the quake epicenter. Notice that if the quake epicenter is 20 km below the ground, a 6.5 magnitude quake is approximately equivalent to a wrecking ball less slightly less than 1 m in radius; while a 6.9 magnitude quake is equivalent to 28.7 m (94 ft) long wooden ram hitting the building's foundations, such as that used by the Roman army in their siege:

Roman Siege Weapons - The Aries or Battering Ram.The aries, or battering-ram, consisted of a large beam made of the trunk of a tree, frequently one hundred feet in length, to one end of which was fastened a mace of iron or bronze resembling in form the head of a ram; it was often suspended by ropes from a beam fixed transversely over it, so that the soldiers were relieved from supporting its weight, and were able to give it a rapid and forcible swinging motion backward and forward. (Tribunes and Triumphs)

Games Workshop The Siege of Gondor Lord of the Rings Supplement |

The drums rolled louder. Fires leaped up. Great engines crawled across the field; and in the midst was a huge ram, great as a forest-tree a hundred feet in length, swinging on mighty chains. Long had it been forging in the dark smithies of Mordor, and its hideous head founded of black steel, was shaped in the likeness of a ravening wolf; on it spells of ruin lay. Grond they named it, in memory of the Hammer of the Underworld of old. Great beasts drew it, orcs surrounded it, and behind walked mountain-trolls to wield it. (The Siege of Gondor, Lord of the Rings, p. 828)This is only for a 6.9 quake, 20 km deep. How much more for quakes of magnitudes 7.0, 7.1, or 7.2?

Earthquake (R) * |
Steel Ram (m)** |
Wooden Ram (m)** |

6.0 | 0.16 | 1.3 |

6.1 | 0.23 | 1.8 |

6.2 | 0.32 | 2.6 |

6.3 | 0.46 | 3.6 |

6.4 | 0.65 | 5.1 |

6.5 | 0.92 (wrecking ball) | 7.2 |

6.6 | 1.3 | 10.2 |

6.7 | 1.8 | 14.4 |

6.8 | 2.6 | 20.3 |

6.9 | 3.7 | 28.7 (Grond in LotR) |

7.0 | 5.2 | 40.5 |

7.1 | 7.3 | 57.1 |

7.2 | 10.3 | 80.8 |

7.3 | 14.5 | 114.2 |

* Richter magnitude scale

** Cross-sectional area of 1 square meter and moving at a speed of $4.43 m/s$. This speed is equivalent to the speed of an object as it touches the ground after it was released from a height of 1 m above the ground.