If the Earth was hit by an asteroid having diameter of about 5 km and moving with a speed close to the speed of light? What would happen?
Would it instantly evaporate? Could it make the Earth evaporate? Could it simply fly through it due to its large momentum?
If the Earth was hit by an asteroid having diameter of about 5 km and moving with the speed close to the speed of light? What would happen?
According to https://what-if.xkcd.com/20/
The momentum would be enough to knock the Earth into a different orbit—but the Earth is no more. The energy deposited is ten thousand times greater than the planet’s gravitational binding energy, and the planet is blown into an expanding cloud of plasma, with a particularly energetic streamer extending away from the far side of the impact site, out into space.
The Sun hiccups and flares as it absorbs waves of dust. The surfaces of Mars and Venus are scoured clean by the waves of incredibly high-energy plasma.
But that is for a trivial 30 meter diameter asteroid. Presumably the results for a 5000 meter diameter asteroid would be fairly inconvenient.
It might be hard to find sources where such collisions have been rigorously modeled, but certainly the amount of kinetic energy available is enough to do some damage.
The amount of kinetic energy is unlimited since $\gamma = 1/\sqrt{1 - v^2/c^2}$ can go to infinity, but let's just use $mc^2$ to define what "close to the speed of light" might mean.
With a density of 2 g/cm^3 a 2500 meter radius sphere will have a mass of 1.3E+14 kg and moving at our "close to the speed of light" a kinetic energy of 1.2E+31 Joules in the Earth's frame.
That a reduced energy of 2E+06 Joules per kilogram of Earth. That is less than the energy necessary to completely disassemble the Earth's mass to infinity, but way more than is necessary to completely destroy it as a solid planet and convert it to a gas or plasma.
But like I said, it might be hard to find citeable sources where someone has done a rigorous simulation including all necessary hydrodynamic transport physics to model such an explosion to see exactly how this would unfold.
and keep going, leaving the earth to fill in a 5 km diameter cylindrical perforation somehow. Due to the high density and relativistic speeds I think there would be so much radiation pressure that the planet would be quickly and completely heated to some mixture of gas and thermal plasma.
But that's just my hunch.
There's a slightly related question in Space SE with some slightly related "target will be vaporized" answers:
The kinetic energy of a relativistic mass is given by $(\gamma -1)mc^2$, where $m$ is the mass of the object and $\gamma$ is the Lorentz factor $(1 - v^2/c^2)^{-1/2}$, where $v$ is the speed.
Asteroids (obviously it wouldn't be an asteroid from our Solar System) have a range of densities from about 1 to 6 g/cm$^3$. Let's leave that as a variable - $\rho$.
The mass of the asteroid is then just its volume multiplied by its density and then the kinetic energy is $$K = (\gamma -1)\frac{\pi d^3}{6}\rho c^2 $$ The momentum of the asteroid is $$ \gamma mv = \gamma \frac{\pi d^3}{6} \rho v$$
How much damage this will do will depend on the size of $\gamma$ and $\rho$ and the question isn't answerable without at least specifying what the former is.
The gravitational binding energy of the Earth is approximately $3GM_E^2/5R_E = 2\times 10^{32}$ J. In an inelastic collision, roughly all of the kinetic energy would be transferred. If we equate the binding energy with $K$, then we find that the value of $\gamma$ that gives enough energy to "unbind" (i.e. explode) the Earth is $$ \gamma_{\rm explode} > 2\times 10^{32}\frac{6}{\pi d^3 \rho c^2} +1 = 12.3 \left(\frac{d}{5{\rm km}}\right)^{-3} \left(\frac{\rho}{3 {\rm g/cm}^3}\right)^{-1}$$
$\gamma =12$ corresponds to a speed of $0.9965c$. This is about 10,000 times faster than the asteroid hypothesised to have killed off the dinosaurs and caused mass extinction.
Note that the asteroid cannot simply "punch through the Earth" because the column of material it would have to displace to do so has a mass that is roughly 2500 times that of the asteroid and is encased in a mass which is many orders of magnitude bigger than that (ignoring glancing blows). It might emerge on the other side, but only after having lost most of its kinetic energy.
But this raises another possibility. Suppose that $\gamma$ was a bit less than this - rather than being destroyed, would the Earth be knocked out of orbit?
Conservation of momentum suggests the change in velocity of the Earth would be $$ \Delta V_E = \frac{\gamma m v}{M_E + m} \simeq \gamma c\left(\frac{m}{M_E}\right) $$ If we say that a "significant orbital perturbation" is a $\Delta V_E > 1$ km/s (the Earth's orbital speed is about 30 km/s), then the $\gamma$ required to achieve this is $$\gamma_{\rm perturb} > 10^5 \left( \frac{\Delta V_E}{1 {\rm km/s}}\right) \left( \frac{d}{5{\rm km}}\right)^{-3}\left( \frac{\rho}{3 {\rm g/cm}^3}\right)^{-1} $$
Thus it seems that the Earth will not get "knocked out of orbit" before it gets totally destroyed by the deposition of kinetic energy. Note, that since $\gamma_{\rm explode}/\gamma_{\rm perturb}$ is independent of asteroid size and density, this conclusion is independent of those factors.
