Bomb versus asteroid

Problem

Estimate, to order of magnitude, the energy needed to deflect a $300\ \text{m}$-diameter asteroid (density $3000\ \text{kg/m}^3$) by an angle sufficient to miss the Earth, acting $30$ days before the predicted impact, with a relative velocity of $20\ \text{km/s}$. Compare the result with the energy of a $1$-megaton H-bomb ($4{,}2\times 10^{15}\ \text{J}$).

Solution

Mass of the asteroid. With radius $r = 150\ \text{m}$, the volume is $V = \tfrac{4}{3}\pi r^3 \approx 1{,}41\times 10^{7}\ \text{m}^3$, so:

$m = \rho\,V = 3000\cdot 1{,}41\times 10^{7} \approx 4{,}2\times 10^{10}\ \text{kg}$

Required lateral displacement. To miss the Earth it is enough to shift the trajectory by roughly one Earth radius, $R_\oplus \approx 6{,}4\times 10^{6}\ \text{m}$, within the remaining time $t = 30\ \text{days} \approx 2{,}6\times 10^{6}\ \text{s}$. The required change in transverse velocity is:

$\Delta v = \frac{R_\oplus}{t} = \frac{6{,}4\times 10^{6}}{2{,}6\times 10^{6}} \approx 2{,}5\ \text{m/s}$

Kinetic energy to be imparted. The energy required to give the asteroid this $\Delta v$ is:

$E = \tfrac{1}{2}m\,\Delta v^2 = \tfrac{1}{2}\cdot 4{,}2\times 10^{10}\cdot (2{,}5)^2 \approx 1{,}3\times 10^{11}\ \text{J} \sim 10^{11}\ \text{J}$

Comparison with the H-bomb. A $1\ \text{Mt}$ bomb releases $4{,}2\times 10^{15}\ \text{J}$: the “useful” energy $\sim 10^{11}\ \text{J}$ is barely $\sim 10^{-4}$ of that bomb’s yield. Note however that the real coupling efficiency between the explosion’s energy and the asteroid’s motion is very low ($\sim 10^{-2}$–$10^{-3}$), because most of the energy is dispersed as radiation and vaporisation. An energy $\sim 1000$ times greater than the theoretical value would therefore be needed, of the order of a few tens of kilotons, i.e. a small nuclear weapon (Hiroshima scale, $\sim 15\ \text{kt}$).

$\ev{\Delta v\approx 2{,}5\ \text{m/s},\quad E\sim 10^{11}\ \text{J}}$

Collegamenti

Argomenti: Fisica nucleare Concetti: Energia cinetica Competenze: Stime di Fermi