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Do we have the capability of moving the Moon just a little bit to time a intercept with a moon size asteroid like the one in the picture?

It just said that if we could intercept an asteroid far out in space predicted to hit the Earth that we can adjust the asteroid just a little bit to miss the Earth. Would it be easier to move the moon just a little bit to intercept the asteroid?

What consequences could come from moving the Moon?enter image description here enter image description here

https://astronomy.stackexchange.com/questions/32844/how-well-would-the-moon-protect-the-earth-from-an-asteroid

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    $\begingroup$ The amount of energy needed to move the moon to intercept the asteroid is going to be millions of times more than the energy needed to move the asteroid to intercept the moon... $\endgroup$ – Arkenstein XII Aug 1 '19 at 21:12
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    $\begingroup$ No... I don't think you understand the scale here. There's not enough uranium on Earth to move the moon more than a metre or so. $\endgroup$ – Arkenstein XII Aug 1 '19 at 21:17
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    $\begingroup$ Any ship that could bring that many nuclear warheads to the Moon could go anywhere in the solar system and back a thousand times. $\endgroup$ – Ryan_L Aug 1 '19 at 21:17
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    $\begingroup$ Not to say "what he said", but it seems worth reinforcing - there is no scenario where moving the moon is less energetically expensive than moving an asteroid, unless the asteroid is a rogue planet, in which case, blocking it with the moon would probably just make things worse. $\endgroup$ – jdunlop Aug 1 '19 at 21:34
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    $\begingroup$ @Muze Then you use that energy to evacuate or throw a big extinction party. You can't block a moon with the moon in any case. $\endgroup$ – stix Aug 2 '19 at 22:15
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Introduction:

This is my first answer on Stackexchange so feel free to correct any mistakes I will undoubtedly make, especially with the math formatting as I am not very good at that either(although I will try my best). I also know that I am a bit late to answer this question, but I hope that this answer will still bring some insight into why moving the Moon is so hard.

I will split this into 2 parts, for the 2 ways I can think of to move the Moon. Part 1 will address the method of hitting the moon with an asteroid to change its speed and part 2 will look at continuous thrust. I will use the same type of setup to Benjamin's answer with an unstoppable asteroid that will hit the Earth in 250 years, but if we can move the Moon in such a way that it is half an orbit early or late we can stop it(The probability of the killer asteroid to be in the same plane as the Moon is very low but I will ignore that small technicality for now).

In both parts I will assume the worst case scenario with the Moon being exactly 180 degrees off where it needs to be. At the end of both parts I will provide a distilled equation for any deadline time other than 250 years, so that you can plug in any year and get out the expected resources that you will need. There will also be a tldr at the very end for those that get horrified by all the math.

Method 1: Guided asteroid impact

So the goal of this part is to find the mass $M_a$ of the asteroid that we send to the Moon. To do that, we first need to figure out how much speed the Moon will lose during the collision. The current speed of the Moon can be figured out using the equation $\sqrt{\frac{\mu}{a_m}}$ where $a_m$ is the radius of the Moons orbit(usually called the semi-major axis, That will be important later when talking about the non-circular orbit of the asteroid) and $\mu$ is the standard gravitational parameter for the Earth and the Moon. Plugging in the values we get:

$a_m=384399000m$

$\mu=4.034713308×10^{14}\frac{m^3}{s^2}$

$v_m=\sqrt{\frac{\mu}{a_m}}=1024.5\frac{m}{s}$

Now when we know the speed of the Moon, we need to know the speed of the asteroid, and That requires knowing the orbit of the asteroid. I will pick an orbit that goes between the asteroid belt and the Earth(a fairly typical path for a near Earth asteroid). So I will pick a perihelion(inner point) of 1 au(Earths orbit), and an aphelion(outer point) at 3 au(in the middle of the asteroid belt). So we have $p_0=1$ and $p_1=3$. Now we need the speed of the asteroid when it is arrives at Earth(when it is at $p_0$), but the simple speed-from-radius equation we used earlier is not going to work this time as the orbit is elliptic instead of circular. So we need an upgraded version called the vis-viva equation:

$v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$

Because I am using units where Earths speed and Earths orbital radius are all set to 1, That $\mu$ term is also 1(note that as both Earth and the asteroid are orbiting around the Sun, That \mu is for the Sun, it is not the same value as earlier). The $r$ and $a$ terms are the current radius and semi-major axis of the orbit respectively. The $r$ we know as the asteroids current position is Earths orbit at $p_0$, the semi-major axis $a$ is simply the average between $p_0$ and $p_1$, ie 2. We also want to multiply the whole thing with the velocity of Earth $v_e=29780 \frac{m}{s}$ to get out of the whole Earth=1 thing and back to normal units. Plugging it all in gives:

$v_a=v_e\sqrt{\frac{2}{1}-\frac{1}{2}}=36472.9\frac{m}{s}$

To get the maximum impact velocity we want the asteroid to hit the Moon when the Moon goes head on to the asteroid from Earths perspective. This gives:

$v_i=\left(v_a-v_e\right)+v_m=7717.4\frac{m}{s}$

So now we know how fast the asteroid will hit the Moon, unfortunately here we need the mass of the asteroid, and we don't have that yet. So from here on I will skip the number crunching, and use only variables.

In any collision, momentum is conserved. I will use this to figure out how much the Moon will slow down when struck by our asteroid. The momentum of an object is $mass\cdot velocity$, so the momentum of the Moon before the collision is $M_mv_m$ where $M_m=7.342\times10^{22}kg$ is the mass of the Moon. The momentum of the asteroid is $M_av_i$. The momentum of the Moon after the collision is $M_mv_m-M_av_i$. To get the new velocity we simply divide by the mass of the Moon:

$v_{m2}=\frac{M_mv_m-M_av_i}{M_m}$

To be able to get the orbital period that we will need later we need to get the new semi-major axis of the Moon after the collision. To do that we use the vis-viva equation again, but in reverse. Things are abit different this time because the current radius $r$ is actually the old semi-major axis $a_m$, and we need to use the $\mu$ this time Because we are orbiting around the Earth. Lets plug things in:

$v_{m2}=\sqrt{\mu\left(\frac{2}{a_m}-\frac{1}{a_{m2}}\right)}$

And solve for $a_{m2}$ to get:

$a_{m2}=\frac{\mu a_m}{2\mu-a_m\left(v_{m2}\right)^2}$

So, remember that I mentioned orbital period? Here is where that comes in. The equation for orbital period is:

$P\left(a\right)=\tau\sqrt{\frac{a^3}{\mu}}$

It takes in the semi-major axis $a$ and spits out the amount of seconds it takes to go one lap around the Earth at that height. We need to use that function in order to construct our final equation for how massive of an asteroid you need to smack the Moon with, so that it can catch up to and be smacked by a much larger asteroid in 250 years. Poor Moon...

$\left(P\left(a_m\right)-P\left(a_{m2}\right)\right)\cdot250\cdot12=\frac{P\left(a_m\right)}{2}$

So here it is. It says that we want the change in time for one orbit times the number of orbits to be equal to half of the time it takes to make one orbit. So when the $M_a$ is just right and this equation is true, the Moon will travel half an orbit early after 250 years and catch the larger asteroid. I will spare you the walk through of solving this, but if you feel brave enough you can try it yourself.

$M_a=\frac{M_m}{v_i}\left(v_m-\sqrt{\frac{u}{a_m}\left(2-\left(1-\frac{1}{2\cdot\left(Y\cdot12\right)}\right)^{-\frac{2}{3}}\right)}\right)$

What a beast! But it does its job perfectly. If i plug in $Y=250$ it spits out $M_a=5.41×10^{17}kg$. That is alot of kilograms. We can also plug that into this formula to get the diameter of the asteroid: $d=2\left(\frac{M_a}{3000}\frac{3}{4\pi}\right)^{\frac{1}{3}}=70km$. 70km is a large asteroid, not many asteroids in the asteroid belt are that large(about 400 or so). Although both the mass and diameter is completely dwarfed by the Moon(Moon is 135567 times more massive and 50 times as wide), and this explains why moving the Moon is hard even when you have 250 years to do it.

Method 2: Continuous Thrust

This part focuses on how much thrust is needed to move the Moon by the same amount as in Part 1, if the thrust is applied continuously over 250 years. Luckily for use the approach is quite similar at the start and we will see many of the same variables from the first part, so I won't explain them twice. So if you see $\mu$ for example, you know that it is the exactly same thing as before.

With that out of the way, lets start with the plan. This time I will use kinetic energy as the conserved quantity instead of momentum. That is because $force\cdot time=energy$, so the decrease in kinetic energy is proportional to time:

$K\left(t\right)=\frac{1}{2}m_mv_m^2-Ft$

As You can see, $K\left(t\right)$ is a function of time, and the only new variable is the thruster force $F$. Yep, that crucial variable that we need to get our hands on at the end of this whole chapter is right there in the very first equation, and that means that I won't do any number crunching until the end. Let's continue in an similar style of "conserved quantity -> $v_{m2}$ -> $a_{m2}$ -> $P(a)$ -> result" that we used in Part 1. But because our conserved quantity is a function of time, so will all of these values be functions of time in this part. Anyways, lets go on to finding $v_{m2}\left(t\right)$. That can be easily solved by reverse engineering the kinetic energy equation to get:

$v_{m2}\left(t\right)=\sqrt{\frac{2K\left(t\right)}{m_m}}$

The equation for velocity to semi-major axis is exactly the same as before, except that it is a function of time, because $v_{m2}\left(t\right)$ is a function of time:

$a_{m2}\left(t\right)=\frac{ua_m}{2u-a_m\left(v_{m2}\left(t\right)\right)^2}$

The period function $P\left(a\right)$ is completely identical, so I won’t bother to show it again. This means that we are already at the point where I show the final equation that puts the constraint that "the difference in each orbit times the amount of orbits should equal half of the original orbit", But hang on a sec... That "difference in each orbit" part is now a function of time, every single second it will be different. Therefore we will need some heavy duty mathematical machinery, namely the (in)famous integral operator from calculus, I know that it sounds scary, but I will both explain it and (painfully) solve it. Anyways, here we go...

$\int_0^T\left(P\left(a_m\right)-P\left(a_{m2}\left(t\right)\right)\right)dt=\frac{P\left(a_m\right)}{2}$

What a monster, isn't it? But if you look closely, You an see that it somewhat resembles the final equation from Part 1, the $P\left(a_m\right)$ and $\frac{P\left(a_m\right)}{2}$ parts are still there, doing the same job as they did before. The $P\left(a_{m2}\right)$ has been upgraded to $P\left(a_{m2}\left(t\right)\right)$ to include time, and the factor of $250\cdot 12$ has been replaced by that scary looking $\int$ integral thing. That integral is there to take care of the fact that during the whole operation from $0$ seconds to $T$ seconds, the value inside the "body" of the integral is constantly changing. I won't go into details as to exactly how the integral manages to do this, rather I will skip the (long) time it took for me to find a solution to this and actually show the solution itself. Oh and by the way, that $T$ is just the amount of seconds there is in 250 years(big number, I know), it is equal to $T=31536000\cdot Y$ where the $Y$ is the same variable for years that we saw earlier. Ok, here is the painfully acquired solution to that scary integral formula thing:

$F=\left(\frac{m_mu}{a_m}\right)\frac{3+\frac{4}{2T-1}-\sqrt{\frac{18T-1}{2T-1}}}{4T}$

That almost looks even scarier than the integral itself, luckily this is just a rather large normal "plug and play" function, where You just plug in all the values as they are and it will spit out the awnser. For the value of $T$ that we need for it to represent 250 years, the formula spits out $F=413266741N$ that is 412 MN(MegaNewtons) or roughly equal to 53 Saturn-V first stage engines, that is 10 copies of the most powerful rocket ever made pushing the Moon for 250 years straight… That approach is also very hard.

Conclusion/tldr:

The Moon is big, very big, and heavy. So pushing it around is not something you can do as if it was any regular old asteroid.

tldr for Method 1: It would take a 70 km asteroid head on to the Moon to be anywhere close to making it in time, that is much larger than the majority of asteroids in the asteroid belt and is about the same size as the asteroid that killed the dinosaurs(although the size of that one is not very well-known).

tldr for Method 2: In order to move the Moon using regular rocket engines,it would take 10 copies of Saturn-V(the rocket that took people to the Moon), blasting away at the Moon for 250 years straight, so that is probably not doable either.

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  • $\begingroup$ I see that you edited my awnser and fixed a bunch of spelling mistakes. Thanks for that. I however think that apogy is not really the right word there as the suffixes for both the apo- and peri- should be the same, and having the -apsis is the general term, although i can change it to -helion for meaning around the sun. I will also remove the !!! at the start becuace that looks abit silly to me... $\endgroup$ – Eriksonn Aug 5 '19 at 23:29
  • $\begingroup$ "The calculation of such a feat is left as an exercise to the reader." -- I see you took this line seriously. Well Done! $\endgroup$ – SurpriseDog Aug 6 '19 at 18:59
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Short answer: No. Also: it'd be a terrible idea to try when you could just move the asteroid instead.

The Moon has a mass of 7.3 * 1022 kg.

The energy change required to move the Moon even one meter would thus be:

U = mgh

(potential energy in a gravitational field)

The gravitational acceleration exerted by the Earth at the distance to the Moon is approximately 0.0025 m/s2

U = 7.3 * 1022 * 0.0025 * 1.0 = 1.86x1020 J

This is about the equivalent of 44 THOUSAND megatons, or about 2.2 MILLION Hiroshima bombs.

Any conceivable way to generate this much energy would be far more than enough to completely vaporize the incoming asteroid instead. It'd be far more feasible to build a giant laser on Earth and point it at the asteroid and burn it up over several days/months/years.

I'd also expect you'd need to move the Moon more than 1 meter to be effective, so your problems get worse from there.

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    $\begingroup$ Your conclusion is spot-on, but your calculation isn't, I believe. You calculated the energy needed to lift the Moon 1 meter on the Earth's surface. But at the Moon's distance, the Earth's gravity is much lower. More to the point, the way to get the Moon out of the way if you have advance notice is to accelerate it, thus changing its orbit slightly and then let the deviations pile up. But the #A1 solution is to move/vaporize/whatever the asteroid which would always be cheaper. $\endgroup$ – Mark Olson Aug 1 '19 at 22:08
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    $\begingroup$ What's being off by even a factor of a thousand matter when you're dealing with numbers on the order of 10^24? :P $\endgroup$ – stix Aug 1 '19 at 22:18
  • $\begingroup$ Regardless of whether you use big bombs to move the Moon or a giant rocket, the energy input required is the same. I'm just using megatons as a reference for the staggering amount of energy required. $\endgroup$ – stix Aug 1 '19 at 22:31
  • $\begingroup$ There should be a unit of measure which is the amount of energy contained in the world's combined nuclear arsenal. $\endgroup$ – Muuski Aug 2 '19 at 21:41
  • $\begingroup$ @Muuski "The SI unit is the armageddon. It is equal to the combined yield of the world's nuclear stockpile in 1980. Its exact value is classified." $\endgroup$ – stix Aug 2 '19 at 22:13
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Normally, I would agree with the other answers that moving the moon is impractical when you can just move the asteroid. However, If your advanced space telescopes can predict the path of a massive asteroid in the Kupier belt many years in advance then it can make some sense.

Let's say we discover an object similar to 90482 Orcus with a mass of 6.4 x 10^20 kg headed on a collision course with earth. This impact would completely devastate the Earth turning it into a ball of liquid magma. We could, of course, wait until it gets close and then try to deflect it, but given the enormous mass this may be impossible. An executive decision is made instead: Move the moon.

Fortunately for us, our telescopes have detected it as it makes a near miss of earth and then it takes another 247 years before it loops back around and smashes into us. Is this enough time to move the Moon as a shield? Let's see:

The moon has a mass of 7.3 x 10^22 kg, quite large but we have time on our hands. We calculate that we need to change the moons position by approximately 1 million km by D-Day (enough to be in the opposite side of the earth). To achieve this we need to increase the moon's velocity by only 0.128 m/s and then wait.

Let's attach 1 million of the biggest ion thrusters made to date: The X3 delivering 5.4 newtons of continuous thrust. Force / Mass = Acceleration so this will give us:

5.4 x 10^6 Newtons / 7.3 x 10^22 KG = 7.4 x 10^-17 m/s^2

Over 247 years that adds up to: 5.7 x 10^-7 m/s.

Not nearly enough. We're doomed!

(unless you can radically supersize your ion thrusters)


Perhaps an alternate strategy would be to find a smaller asteroid with an already high velocity (12.6 km/s) and divert it into hitting the moon and use that to increase the moon's velocity (like interstellar dominoes). The calculation of such a feat is left as an exercise to the reader.

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  • $\begingroup$ Your mass of the moon is off by a factor of 10. However, even with the correct mass, what you suggest requires 6 x 10^21 joules of energy(0.5 * m * v^2, v = 0.128). Spread over 247 years that's 773 gigawatts of continuous power output, every second, for 247 years. Where do you plan to get that power? $\endgroup$ – stix Aug 2 '19 at 22:25
  • $\begingroup$ Fixed. As for the power? The sun! With an unobstructed sky, solar panels on the moon can generate terawatts of power. $\endgroup$ – SurpriseDog Aug 2 '19 at 22:30
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    $\begingroup$ The comments on that article are quite funny and quite accurate. Beaming that much power down to the Earth would itself be a very effective thruster capable of moving the moon. XD $\endgroup$ – stix Aug 2 '19 at 22:37
  • $\begingroup$ If you can predict the course of the asteroid that far in advance it gets even easier to move the asteroid, the further out you apply the thrust the less you need. $\endgroup$ – John Aug 3 '19 at 3:04
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Another problem with moving the Moon to block an asteroid is that changing the Moon's orbit will change the schedule of the tides on Earth. Many lifeforms live full or part time in the tidal zone. Changing the tidal schedule could cause many of those species to go extinct, which could start a chain reaction of extinctions. That would be a very bad disaster, even if it was puny compared to what a large asteroid impact would do.

It might be necessary to move an incoming asteroid by only about one kilometer to deflect it sufficiently months or years before it would otherwise have hit the Earth.

Note that the Moon naturally blocks only a tiny fraction of a percent of the sky due to its size and distance from Earth. An incoming asteroid would be coming from the other half of the Moon's orbit about half the time, so the entire Moon would have to be moved between 90 degrees and 180 degrees about half the time in order to block an incoming asteroid. That would be moving the Moon to a position hundreds of thousands of kilometers or miles from its normal position. Which would require many millions of times as much energy as stix calculated in his answer. And the orbital plane of the Moon would very probably have to be tilted, with more bad results for Earth tides.

The Moon has an equatorial radius of 1,738.1 Kilometers, and thus a diameter of 3,476.2 kilometers. If a dangerous asteroid that needed to be diverted would probably have a diameter of 0.1 to 100 kilometers, the Moon would have a diameter between 34.762 and 34,762 times that of the asteroid. Thus the Moon would probably have a volume between 42,006.282 and42,006,282,000,000 times that of the asteroid. And if the Moon and the asteroid had approximately similar densities, the Moon would have a mass between about 40,000 and 40,000,000,000,000 times that of the asteroid.

So it would take a lot less energy to divert the asteroid than to move the Moon to block it.

[Added 08-03-19. But it might possibly make sense to move some small asteroids and put them in various orbits around the Earth as a last ditch defensive measure to block incoming asteroids detected too late to divert by more usual methods.

All of those guardian asteroids would have to be fitted with powerful rockets to move them into their initial orbits, and those rockets would have to be upgraded to even more powerful ones and kept fueled and maintained and periodically tested to be certain they could move the asteroids fast enough in an emergency.

Some of the guardian asteroids would be equally spaced in orbit around Earth in the ecliptic plane, and others in equatorial orbits around Earth, and others in polar orbits around Earth, and others in orbits at various inclinations.

So if an asteroid was detected heading for Earth too late for normal methods to change its course, one or more of the guardian asteroids would be sent to intercept it as far from Earth as possible. If just one of the guardian asteroids would be large enough to vaporize or shatter the incoming asteroid, another guardian asteroid would be sent on a trajectory to be positioned close to the impact point and directly between the collision of the two other asteroids and Earth. Thus this second guardian asteroid might be hit by fragments from the collision and thus shield a cone of space including Earth from those fragments.

I think such a plan would not be totally impractical for a society with powerful rocket engines - especially as compared to the original suggestion of moving the Moon - and so might possibly be considered as a backup plan to the normal methods of diverting an asteroid from a collision course.]

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There is another big problem to consider when using the moon as a shield. The moon is held in orbit by a balance of it's acceleration and it's mass. If it slows down it will fall into the earth because the effect of the earth's gravitational pull will be stronger than it's orbital momentum. If it speeds up the momentum will be stronger than the gravitational pull and it will leave the earth's orbit and fly off. And that's just the push to get it into place. The crash of the asteroid itself will change the mass, either by knocking material off or more likely by adding itself to the existing mass and probably push it closer into the earth's orbit which will increase the effect of gravity. Also the crash is likely to slow the moon down. All of those factors have the effect of causing the moon to crash into the earth.

However, if you have enough warning to move the moon you probably have enough warning to go push the incoming asteroid enough for it to miss the earth. It doesn't need to be tremendously far away for a small alteration in trajectory to make it miss the earth. I believe that is the method that scientists today have considered most viable if we need to deal with such a situation in real life. We just need to get some drones out in space that are able to fly over and give the asteroid a push.

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    $\begingroup$ Speeding up and slowing down the moon will only result in higher and lower orbits. Simply slowing the moon a little bit would not cause it to crash into the earth, nor would speeding it up necessarily result in it escaping orbit. $\endgroup$ – stix Aug 2 '19 at 19:20
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    $\begingroup$ That's true it depends on the amount it needed to be moved whether that could happen. $\endgroup$ – cyberchis Aug 2 '19 at 19:26

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