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If there was an asteroid that happened to be made out of something really solid (iron, titanium?) and it got enough velocity (sling shot around the sun?), is it conceivable that it could hit a planet dead on and keep right on going?

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You may create a hole through a planet but the resulting object won't be sufficiently stable. The pressures near the hole will be huge and will drive the material to rupture and rearrange itself so that the hole is filled. That's particularly clear for the Earth whose (outer) core is liquid. Obviously, a hole in a liquid won't stay for too long. Quite generally, planets and celestial bodies want to reduce the gravitational potential energy and they do so by converging towards the perfect spherical/ellipsoidal shape. Every earthquake brings the Earth closer to a perfect sphere, too. –  Luboš Motl Jan 17 '13 at 9:24
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Incidentally, the speed or energy carried by the bullet that would be enough to puncture the Earth would be so insane that it's impossible not just in practice but one could perhaps say that even in principle. Things that are too fast also burn in the atmosphere because of the friction, or they melt and burn while going through the Earth. At any rate, these fast enough bullets would melt the Earth's rock so the resulting liquid would quickly fill the room behind the bullet and there would be no air/vacuum-filled hole left. –  Luboš Motl Jan 17 '13 at 9:27
    
This is a good question. As Luboš says, any hole would be quickly filled in, but the question is whether any projectile could bore a hole or whether it would just blow the planet apart. I suspect that even a small size projectile would just blow the planet apart rather than bore a hole through it, but I don't know of a proof of this. –  John Rennie Jan 17 '13 at 11:15
    
@JohnRennie I think the projectile would have to be hyperdense to bore through the planet. Very roughly speaking, it's mass should be greater than the one of the column of rock that is being displaced. For small impactors with size $D$ boring through the Earth's diameter, their density needs to be about $5\mathrm{\frac{g}{cm^3}}\frac{12000 \mathrm{km}}{D}$. –  mmc Jan 17 '13 at 12:03
    
I understand that the forces would be huge, but what if the resulting doughnut-shaped planed would then be spinning fast enough to keep the doughnut hole empty for a while? –  randunel Jan 17 '13 at 14:34
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1 Answer

Anything that falls from space is going to gain a certain amount of potential energy from falling. In this case, it's

$G \frac{M_E}{r_E} = G \frac{M_E}{r_E^2} * r_E = g * r_E = 9.8 \frac{m}{s^2} * 6.4*10^6 m = 6.3*10^7 J/kg$

of energy. So every kg of (say) iron gains 63 million joules of energy.

At room temperature, Iron has a specific heat of around 500 $\frac{J}{kg K}$. I spent a little time looking for its specific heat curve, but for a back-of-the-envelope calculation, let's assume this is about average from the 3K of space to the melting point of iron, 1800K. So to reach the melting point of iron, we'd need $1800 K * 500 \frac{J}{kg*K} = 900,000 J / kg$, or a little less than 2% of our total KE. If more than about 2% of our energy turns into heat, we're dealing with liquid, rather than solid iron, and too much above that, we're starting to look at gaseous iron.

Now, the earth's core is iron, so in order to take a chunk of it out, we'd need to send a projectile through, and have them go off together. In other words, it would be an inelastic collision. Assume the projectile has mass $m_1$ and speed $v_1$, while the chunk of earth we're carrying away has mass $m_2$ and velocity $v_2 = 0$. In an inelastic collision, $m_1 * v_1 + m_2 * v_2 = (m_1 + m_2) * v_f$, so in this case, $v_f = \frac {m_1 * v_1} {m_1 + m_2}$, initial KE is

$KE_0 = \frac {1}{2} m_1 v_1^2$

and final KE is

$KE_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} (m_1 + m_2) * (\frac{m_1 * v_1} { m_1 + m_2})^2 = \frac{1}{2} \frac{m_1^2 v_1^2}{m_1 + m_2}$

Note that the percentage of KE lost is:

$\frac{KE_0 - KE_f}{KE_0} = \frac{\frac{1}{2}m_1 v_1^2 - \frac{1}{2}\frac{m_1^2 v_1^2}{m_1 + m_2}}{\frac{1}{2}m_1 v_1^2} = 1 - \frac{m_1}{m_1 + m_2}$

which we will approximate as

$\frac {m_2}{m_1}$

So in order to keep less than about 2% of our KE from turning into heat, the stationary target (the part of the earth we're scooping out) should be less than 2% of the mass of the projectile hitting it. In other words, we're looking at a giant iron space rod with a length of at least 50 times the diameter of the earth (probably more like 100 times just to be safe). And since it won't have perfect heat conductivity, yes, a bunch of it will liquefy and/or vaporize anyway.

Also, I cheated. I assumed we just dropped this from a long ways away, but if we just did that, we wouldn't have enough energy to actually carry the giant iron space rod plus the chunk of earth it carries with it away. We'd need to add enough energy to also grab the chunk of earth & carry it out.

I don't want to find the exact ratio of removing some particular geometry of, say, 1/6 of the earth's mass; instead I'm going to pretend I lift this mass from the earth's surface) would require something on the order of 63 million J/kg (like in the first equation above).

If our giant space rod were to have 100 times the mass it wants to "scoop" out, it should have 1/100 of that KE in every kilogram, so it should initially be moving at a speed of at least

$\sqrt{6.3 * 10^5 \frac{m^2}{s^2}} = 800 m/s$

So our giant space rod should be 100 times the diameter of earth, moving at a kilometer per second before it even gets close to the earth, and should hit along the pole in order that the earth not "bend" the rod with its rotation as the rod goes through, and you'll have to aim it at where the earth will be when you eventually hit it. Through the pole. And you'll need to give it some "sideways" velocity so that it moves along with the earth in its orbit around the sun.

The cataclysm would release an extraordinary amount of energy, and the "donut" earth would quickly collapse back into a smaller ball, releasing enough energy to kill all life in the process.

Assuming our donut "hole" is around 1/3 of the diameter of the earth (so that the "hole" is as wide as the rest of it), we're looking at a volume of roughly (approximate with a cylinder)

$\pi (\frac{r_E}{3})^2 * 2r_E $ or about $\frac{2 r_E^3}{3}$

compared with a volume for the earth of around

$\pi \frac{4}{3} r_E^3$, or about $4 r_E^3$

So we're talking about removing 1/6 or so of the earth's mass, or around $3 * 10^{29} kg$. Make the rod smaller if you want to remove less of the earth's mass, and larger if you want to remove more.

Also, I'm just hoping the giant space rod wouldn't bend during the collision or begin collapsing into a sphere itself under its own gravity before it even hit, which it totally would.

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Also, I see I neglected atmospheric friction, but that should only be a minor part of the equation, so we're okay. –  Will Cross Jan 17 '13 at 15:30
    
Oh, and if the rod had enough structural integrity, and maybe some fins on the outside, and were spinning fast enough axially, it might give the earth's remnant enough rotational energy to make something like a ring of saturn, but it would be hard to make a true donut ring; the inside would want to spin faster than the outside, but then friction would make the outside faster & the inside slower & would just make for a bigger ring as the inside dropped to a lower orbit & the outside went into a higher orbit. Also, it would be flat like a washer; not really a donut. –  Will Cross Jan 17 '13 at 15:32
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