Fermi estimate, deflecting an asteroid

Problem

Fermi estimate: deflecting an asteroid. Estimate the change in velocity $\Delta v$ needed to deflect a $10^9\;\text{kg}$ asteroid so that, after a year of travel (distance $\sim 10^9\;\text{km}$), it shifts sideways by $10^4\;\text{km}$ relative to its collision course with Earth. Also estimate the kinetic energy of the $\Delta v$ alone.

Solution

The idea: a small sideways push, applied well in advance, accumulates over the long travel time into an enormous displacement. The sideways displacement is $\Delta v$ times the flight time: $d_\perp = \Delta v \cdot t \quad\Rightarrow\quad \Delta v = \frac{d_\perp}{t}$

Flight time (one year): $t \approx 1\;\text{year} \approx 3{,}15\cdot 10^7\;\mathrm{s}$ Required sideways displacement: $d_\perp = 10^4\;\text{km} = 10^7\;\mathrm{m}$ Hence: $\Delta v = \frac{10^7}{3{,}15\cdot 10^7} \approx 0{,}3\;\mathrm{m/s}$

Kinetic energy associated with the $\Delta v$ alone: $E = \tfrac{1}{2} m\, \Delta v^2 = \tfrac{1}{2}\cdot 10^9 \cdot (0{,}3)^2 \approx 4{,}5\cdot 10^7\;\mathrm{J} \sim 5\cdot 10^7\;\mathrm{J}$ Modest energy (a few tens of MJ, like a small rocket): the real key is not power but acting very early, so that a tiny $\Delta v$ has time to produce an enormous offset. Result: $\ev{\Delta v \approx 0{,}3\;\mathrm{m/s},\quad E \approx 5\cdot 10^7\;\mathrm{J}}$

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Topics: Quantità di moto e urti Concepts: Quantità di moto · Energia cinetica Skills: Stime di Fermi