Yeah, there's something to be said about the simplicity of a kinetic impactor to redirect an asteroid, vs. using a lander. The lander has to slow down, land, stick the landing (poor Philae), and then you have to account for the asteroid's rotation and distribution of mass. And this all assumes the lander remains operational and doesn't have a fault or failure. It's a lot of challenges and risks.
Another method which is popular among the astrophysicists that look into this subject is the gravity tug. Park a massive craft near the asteroid. The asteroid pulls the ship, and the ship pulls on the asteroid. Use thrusters to keep the position relative to the asteroid fixed, and basically you're just towing it with gravity. The benefit of this method is it does not require any interfacing with the asteroid whatsoever. No landing, no surface operations... its as close to the sci-fi tractor beam as you can get.
Then there's the impact method. It is the most simple by far, but we've never done it with the purpose of seeing how well it works in mind. It could also have application toward excavation and mining.
N0B0DY wrote:Source of the post So let me see if I understand that correctly and sorry for any wrong assumptions I may write:
They plan to hit the small one and this will cause a small variation in its orbit around the large one which is marginally measurable. They will then extrapolate the result using computer modelling to simulate a more powerful impact on a larger body? Or the change in the orbit of the small one will also cause the binary orbit (the barycenter) to change slightly and hence affect its solar orbit very, very very slightly (almost zero) and use this to model a larger impact? I am confused..
More the former. The impact will have very little change on the orbit of the binary about the sun. It's mostly a timing difference that it causes in the orbit of the moon about the main asteroid. We can use that to see how effective the transfer of momentum was from the impactor to the asteroid, which then tells us the effectiveness of nudging an asteroid off of an Earth-crossing path.
Let's apply some physics and see how this works:
-The impactor will have a mass of 300kg, travelling 6.25km/s.
-The asteroid's moon is about 170m in diameter.
-Assume the density is about 2g/cm
^{3 }(typical of rocky asteroids). Then its mass is about 5x10
^{9} kg.
Use the principle of conservation of momentum:
[math]m_{asteroid}v_{asteroid,initial} + m_{impactor}v_{impactor} = (m_{asteroid}+m_{impactor})v_{final}Choose a frame of reference where the asteroid's initial velocity is zero, and let the impactor's mass be negligible compared to the asteroid's mass. Then this will simplify to:
[math]m_{impactor}v_{impactor} = m_{asteroid}v_{final}.
Therefore the change in asteroid velocity is
[math]\Delta v = \frac{m_{impactor}}{m_{asteroid}}v_{impactor}Plug in the numbers:
[math]\frac{300kg}{5*10^9 kg}6.25km/s \approx 0.375 mm/sSo this should change the velocity of the asteroid by about 0.4 millimeters per second! This agrees with
wikipedia's numbers.This immediate change in speed by the impact is too small to be measurable. But over time, it will change where the moon is in its orbit about the primary asteroid, and we can measure that. The orbital period is 11.9 hours at a distance of 1.1km. Technically it orbits the barycenter of the asteroid binary, but I'll assume it's a circular orbit with that radius to make it simple. 11.9 hours to make a 1.1km radius orbit is a speed of 161.3 mm/s.
Now the hard part. The impact instantly changed the speed of the moon, but by the orbital mechanics, that ends up changing the orbit, which then changes the speed in a non-intuitive way. If we hit the moon from behind, it speeds the moon up, which boosts it into a larger orbit, which is
slower. If we hit it from the front, we slow it down, which makes it drop to a lower orbit, which is
faster.Suppose we hit the asteroid from behind (how rude). Then the new orbit will have the same periapsis, but a higher apoapsis, which is a slower orbit. Let's use the Vis Viva equation to figure out the size (semimajor axis, "a") of the new orbit:
[math]v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)The mass M of the main asteroid is about 4.3x10
^{11} kg (using the period of a circular orbit, v
^{2} = GM/r).
Then by instantly boosting the velocity from 161.3 to 161.7mm/s, we change the semimajor axis from 1.1km to 1.105km, or a boost of only 5 meters. Again we can't measure this.
But this new, bigger orbit is slower. Using Kepler's Third Law,
[math]P^2 = \frac{4 \pi^2}{GM}a^3the new orbital period is 11.98 hours, or about 290 seconds longer! That's what leads to a visible effect. After about a month, it will have fallen behind by half of its original orbit, so it would be on the opposite side of the main asteroid from where it would be if we had not hit it.
Similarly, a tiny nudge to an asteroid that would hit Earth can shift its orbit in such a way that it would end up missing Earth. The orbit doesn't have to change a whole lot -- it can be more of a timing thing than a positional thing. We can make it so that by the time the asteroid reaches the point in its orbit that originally would have hit Earth, now the Earth isn't in the way... if that makes sense.