What would be the effects of a Chicxulub-sized impact on Earth's magnetic field?

Earth's magnetic field is caused by movements in its core. Its outer core is roughly $2,890,000 \text{ meters}$ below Earth's surface. Therefore, to directly interfere with Earth's magnetic field, this thing is going to have to make a crater $2,890,000 \text{ meters}$ deep.

Issac Newton figured out how to calculate this. The depth $D$ actually doesn't (at least for high-velocity approximations) depend on the velocity of the incoming object, but on the densities of the object ($\rho_o$) and the target ($\rho_t$), and on the length $l$ of the projectile. They are related by $$D=l\frac{\rho_o}{\rho t}$$ Given that the density of an asteroid is about $2 \text{ g/cm}^3$, and the density of Earth is about $5.514 \text{g/cm}^3$, the asteroid would have to have a diameter (assuming a sphere) of $$l=D\frac{\rho_t}{\rho_o}=2,890,000 \frac{2}{5.514}=1,048,240 \text{ meters}$$ By comparison, the object that caused the Chicxulub crater was only about $10,000 \text{ meters}$ in diameter.

Oh, and with a diameter that big, the mass of the object would be $$\frac{4}{3}\pi (1,048,240/2)^3 \times 2,000,000 = \text{something really large}^1$$ Most likely enough to destroy the Earth. So either we're screwed, or we keep our magnetic field. I'd choose the latter.

There are only a few asteroids over $500,000 \text{ meters}$ in diameter, so this thing is probably non-existent in the solar system - unless you count moons and planets. However, they have pretty stable orbits, so I'd say we're okay.

$^1$ I calculated $1.206 \times 10^{24} \text{ meters}$, about $\frac{1}{3}$ the mass of Earth.

posted over 9 years ago

HDE 226868‭

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