# If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth?

An answer to the question of How well would the Moon protect the Earth from a Meteor? mentions as a possibility that the Moon could get knocked into the Earth.

What is the smallest change to the orbit of the Moon from being impacted by a large meteor that would cause it to eventually impact the Earth (i.e. "circling the drain")? What timeline would that look like (minutes, hours, days, years, etc)?

• Even if some unlikely event changed the Moon's orbit sufficiently for it to strike the Earth, once it got within the Roche limit it would be torn apart by tidal forces. How much of that would form a ring and how much would eventually impact the Earth as a lot of various sized meteorites is another matter, but the Moon itself would never impact the Earth. – BillDOe Jul 30 '19 at 20:04
• @BillDOe I wouldn't be so sure about that; the Roche limit for the Earth-Moon system is extremely close. Even if you assumed only the core is rigid enough to survive the tidal forces, that would still be a 320km diameter ball of iron and nickel. – Luaan Jul 31 '19 at 9:16
• @BillDOe Even if it's torn apart, a lot of it will end up on Earth. – Mast Jul 31 '19 at 13:13
• The longest time is just the free-fall time from the Moon's current orbital distance to the Earth. Depending on the nature of the impact, it could be shorter than that. – zephyr Aug 2 '19 at 14:15

As several people have said, this is incredibly unlikely. Part of the reason why is that the "circling the drain" effect you describe doesn't really happen for solid objects much less dense than black holes. Orbits are not "precarious" in that way.

So, suppose something large enough and fast enough to change its velocity noticeably, but not large enough or fast enough to shatter it, did hit the Moon. The effect would be to shift the Moon from its present almost circular orbit around the Earth, into an elliptical one. Depending on the direction of the impact, it would either get a bit nearer to the Earth than it is now, once per orbit, or a bit further away (it also might swing North and South a bit). What is important though, is that this elliptical track is stable at least for a while. Suppose it gets knocked into an orbit that is 220000 miles from the Earth at its closest and 240000 miles at its furthest, that is where it will stay. It will not "spiral in".

Over a long enough period the gravity of the Sun also comes into play and things may shift a bit, but that is a relatively small effect.

Now, suppose that the impact was really big, or perhaps there were a long series of impacts (starting to look like enemy action..) so that the innermost point of the ellipse was eventually driven down to within a few thousand miles of the Earth, somehow miraculously not smashing the Moon to fragments in the process. At this distance it starts to matter that the near side of the Moon is closer to Earth than the far side, so that Earth's gravity pulls on it more strongly. If it orbited closer than about 3000km to the surface of the Earth for long (the Roche limit) these forces would eventually pull it to pieces, and Earth would probably have a pretty set of rings for a short time before internal collisions between the bits caused them to rain down on Earth and kill everyone.

Finally suppose the impact(s) was(were) so big that they actually put the Moon into an elliptical orbit whose innermost point was so close to Earth that the Earth and Moon touched. This is manifestly impossible without shattering the Moon, but in that case, the Moon would indeed hit the Earth. The time for the impact would be about 1/4 of the Moons current orbital period, which is to say about a week.

• And another scenario would be to kick the moon out of orbit entirely, another highly unlikely scenario that'd almost certainly require enough energy to shatter the moon instead. – jwenting Aug 1 '19 at 10:16

There is no possibility whatsoever of the moon getting knocked out of its orbit by an asteroid impact. Compared to the moon, even a large Chicxulub-type asteroid has a very tiny mass, and the moon has already been struck by several of them, but as you can see, it wasn't knocked out of its orbit. The largest asteroid in the asteroid belt is Ceres, 500 miles in diameter. Its mass is very small compared to the moon, but if a miracle caused it to leap out of its orbit in the asteroid belt, half way to Jupiter, and make a bee line for the moon, an impact at 25 km per sec might just be enough to produce a very slight wobble in the moon's orbit, but nowhere near enough to send it toward the Earth. The moon is actually moving away from us at the rate of several centimetres per year.

• So in essence, if there were an asteroid large enough to push the Moon out of orbit (rogue planet??), it would more likely destroy the Moon than move it? – gilliduck Jul 30 '19 at 19:58
• You're right. I meant 25 km per sec. Why I added three noughts I just don't know. something must have distracted my attention. – Michael Walsby Jul 30 '19 at 21:18
• @gilliduck Yup. It might even destroy the Moon, completely melt it down, only for the Moon to reassemble as it cools down and collides back into a single spherical mass (though part of the mass might be lost or rain down on Earth). In fact, a similar collision formed the Moon in the first place, and didn't much affect the Earth's orbit around the Sun; and despite the mass of the Sun and the Earth, the binding energy is much higher in magnitude for the Earth-Moon system than for the Earth-Sun. – Luaan Jul 31 '19 at 10:16
• Ceres smacking into the Moon at 25 km/s would knock the Moon out of it's orbit if it didn't blow the Moon apart first. That's 3e29 J of energy. 10 times more than it's orbital energy around the Earth and 3 times more than the Moon's gravitational binding energy. – Schwern Aug 1 '19 at 7:04
• @Schwern Agreed. Ceres (9e20kg @ 25km/sec) has about a third the momentum of the moon (7e22kg @ 1km/sec) (wrt earth). That would definitely change the moon's orbit dramatically, assuming everything stayed in more or less one piece. – J... Aug 1 '19 at 19:13

There are two issues at play here, only one of which is real.

It's possible to compute the energy and momentum that an asteroid impact would have to transfer to the Moon, assuming that two solid balls (classic Newtonian billiard balls) hit each other (either a direct impact or a glancing impact). There are certainly cases where the result would be the Moon going into an orbit which hits the Earth.

However long before the impact is big enough to seriously move a solid Moon, both bodies cease acting like solid masses and act more like drops of liquid. They splash, throwing both molten and solid rock into space in all directions at a variety of velocities.

In essence, this would be a smaller version of the events which are theorized to have formed the Moon in the first place, with a Mars-sized protoplanet (named Theia -- h/o/w/ t/h/e/y/ d/i/s/c/o/v/e/r/e/d/ i/t/s/ n/a/m/e/ i/ d/o/n/'/t /k/n/o/w) striking the very young Earth. See the Wikipedia article for a decent short description and pointers to more detail.

There are issues with this hypothesis as an explanation of the Moon's formation, but the broad outlines have been modeled in detail and are well-understood at this point. An impact big enough to seriously move a billiard ball Moon would release a very large amount of energy and throw a very large amount of rock into space in all directions.

Most of the loose rock would form a planetary ring around Earth before being captured by the remnants of the Moon. Enough would hit the Earth to be seriously troublesome. I haven't seen any estimates for a modern-day Lunar strike -- it's really way, way down on the list of things to worry about -- but back-of-the-envelope estimates make me strongly suspect that this would be a very good time to join Elon Musk's Martian colony...

• They know it's name it Theia, because with the Lunar Reconnaissance Orbiter they found the big "Theia was here" written on the back side of the Moon. – Makyen Jul 31 '19 at 6:33
• The disruption of tidal forces by a serious change in the moon's mass (and/or orbit) would probably be far worse than the meteor showers resulting from the impact. Earthquakes, volcanic eruptions, entire families of Godzillas coming up out of the ocean, ancient aliens awakening out of their slumber under the seas and demanding tribute, it'd be ugly. – jwenting Aug 1 '19 at 10:21
• @jwenting Good point. – Mark Olson Aug 1 '19 at 11:40

Here is the math for the fastest scenario, in which the Moon would suddenly stop orbiting and fall straight to Earth:

Moon's mass: $$m_1 = 7.342 \times 10^{22} kg$$

Earth's mass: $$m_2 = 5.9723 \times 10^{24} kg$$

Minimal distance between Moon and Earth: $$r = 356400000 m$$

Gravitational constant: $$G = 6.6743 \times 10^{-11} m^3/(kg \times s^2)$$

Force applied to both Moon and Earth: $$F = G \times m_1 \times m_2 / r^2 = 230.402.044.289.682.584.669 N$$

Initial acceleration of Moon towards Earth: $$a_1 = F / m_1 = 0.00313813735 m/s^2$$

Acceleration of Earth towards Moon: $$a_2 = F / m_2 = 0.00003857844 m/s^2$$ Combined acceleration: $$a = a_1 + a_2 = 0.00317671579 m/s^2$$

Time until impact assuming constant acceleration: $$\sqrt{r/a} = 334949 s = 3.88 days$$

That's not quite 1.5 hours (NoAnswer's answer) but also not a week (Steve Linton's answer). Plus, this is an upper bound (on the lower bound, duh), because the acceleration will increase as Moon closes in on Earth.

The answer to the question is the same as NoAnswer's but for different numbers: Anything between the lower bound (less than 4 days) and infinity (assuming unstable orbits can be achieved by not completely deorbiting Moon in one shot).

• Welcome on the site! We have Latex. :-) The \fracs I let for you. :-) – peterh Aug 1 '19 at 16:11
• Thanks, for improving on readability – Mathius Aug 1 '19 at 16:18
• BTW. This answer assumes no loss or gain in mass. Of course, any impact changing Moons orbit would turn both Moon and the impactor to pieces. Some of the pieces, however might be flung towards Earth, reaching it in under 3 days (for starting speed and increasing acceleration). Coincidentally, 3 days are about what I remember to be the transfer time of Apollo missions. – Mathius Aug 1 '19 at 16:26
• As the Moon nears Earth, also its acceleration will increase. It could be integrated out. – peterh Aug 1 '19 at 16:44

The minimum energy change to the moon's orbit to cause it to impact the earth is 3.2998e28 Joules. After the asteroid hits the moon, it would take 6.2 more days for the moon to hit earth.

We can calculate orbital velocity $$v$$ of the moon at apogee using the Vis-Viva equation: $$v^2=\mu(2/r-1/a)$$ , where $$r=4.046e8$$ meters is the distance between the bodies, $$a=3.844e8$$ meters is the semi-major axis of the orbit, and the gravitational parameter $$\mu=G(m_e+m_m)$$ is calculated using the gravitational constant $$G=6.674e-11N*m^2/kg^2$$ , the earth's mass $$m_e=5.927e24$$ kg and the moon's mass $$m_m=7.342e22$$ kg. I got the values off Wikipedia. So the moon is moving around the earth at $$v=968.4$$ m/s at apogee!

For the moon to actually hit the earth, we can say that perigee of the new orbit needs to be at the sum of the radius of the Earth and Moon or 6.378e6m+1.737e6m = 8.115e6m. So the semi-major axis of the new orbit is one half the perigee plus the apogee or (4.046e8+8.115e6)/2 = 2.0636e8m. Re-evaluate the Vis-Viva equation above with $$a=2.0636e8$$ for a velocity of 197.3m/s at apogee.

So the minimum change to the orbit is if an asteroid hits the moon at apogee and decreases it's orbital velocity from 968.4m/s to 197.3m/s. If we express this as an energy change, recall kinetic energy $$E=m_mv^2/2$$, so the initial energy is 3.4427e28 Joules and the secondary energy is 1.429e27 Joules. So the minimum energy change is the difference between the initial and secondary energies or 3.2998e28 Joules.

How long after the impact would the moon hit the earth? Kepler's 3rd law states that $$a^3/T^2$$ is constant, where $$a$$ is the semi-major axis of the orbit and $$T$$ is the orbital period. This gives us the equation $$a_i^3/T_i^2=a_s^3/T_s^2$$ where $$a_i$$ is the initial semi-major axis, $$T_i=27.3$$ days is the initial period, $$a_s$$ is the secondary semi-major axis and $$T_s$$ is the unknown secondary period. Solve to get $$T_s = \sqrt{a_s^3T_i^2/a_i^3}=12.4$$ days. Kepler's second law states that an orbit sweeps out equal areas in equal time, so the time from apogee (when the asteroid strikes the moon) to perigee (when the moon strikes the earth) is half the orbital period, or 6.2 days.

Notes:

1. A lower energy asteroid impact could still cause fragments of the moon to hit the earth, either in the form of impact ejecta or tidal force fragments. However, I think my answer is in the spirit of the question in that it is the largest lunar orbit that still intersects the earth's surface.

2. I don't explain why an asteroid collision at apogee minimizes the energy required to shift the moon's orbit into intersecting the earth. A straightforward proof falls out nicely from the definition of specific orbital energy (which is strangely a negative value). Also one could run similar calculations for an impact at any other part of the lunar orbit to convince themselves a higher amount of energy is required.

3. Careful readers should note that just a slightly more energetic asteroid collision will bring the moon's orbital velocity to zero, causing it to strike the earth in a direct hit.

4. No known asteroids in our solar system are big enough to shift the moon into an intersection orbit with the earth without blowing up the moon. In our minimum orbital energy loss scenario, the moon looses 771.1m/s orbital velocity. Since conservation of momentum holds, then $$m_mv_m=-m_av_a$$, where $$m_m=7.432e22$$kg and $$v_m=771.1$$m/s are the mass and velocity of the moon, and $$m_a$$ and $$v_a$$ are the mass and velocity of the asteroid. If we put in Ceres mass (the largest known asteroid) of 8.958e20 (about 100th the mass of the moon) and solve for velocity, we get $$v_a = 63200$$ m/s. Then if we calculate the kinetic energy of the impact, $$E = m_mv_m^2/2+m_av_a^2/2$$, we get $$E=1.86e30$$ Joules, which exceeds the moon's gravitational binding energy of 1.2e29 Joules by an order of magnitude! This kinetic energy would cause the moon to disintegrate and never reform. Similar calculations will show that an asteroid on the order of the same mass as the moon could change its orbit without shattering it. These calculations are done for convenience in a zero momentum post impact reference frame.

TL;DR: Anything between 1,5 hours and infinity.

Let's assume the moon would be hit in its perigee by an object of the same mass and speed but opposite direction of movement relative to Earth.

Let's also assume a sizeable chunk of debris left by this colossal impact would remain at the last known position of the moon but with zero orbital velocity. (Maybe the impacting asteroid was made of cheese?) This chunk of debris will be "the moon" for the purpose of this answer.

Next after the incident "the moon" will plummet towards Earth, accelerated by a force of about 1G. This is because gravity doesn't decrease by much for a given distance and 1G is the force exerted by Earth. Actually "the moon" also exerts a force but for simplicity, let's assume it only cancels the effect of distance.

The acceleration of the moon is thus about 9.81 m/s² with a starting distance of the moons perigee (~270.000km if I remember correctly, being too lazy to look it up on wikipedia). If I'm not mistaken "the moon" will take (sqrt(distance/acceleration)=5246,23 seconds) about 1,5 hours to reach Earth. Maybe it'll be a bit less for Earth's radius. It will also arrive with a speed beyond Mach 50 and thus actually "impact" Earth's atmosphere, i.e. experience a resistance equivalent to the sound barrier plus extreme compressive heating, likely to rip it apart.

This is the fastest way for moon to be impacted and then crash into Earth. However the question asked for the slowest way: Well, by decreasing the mass and/or velocity of the asteroid impacting the moon, we can "fine-tune" the effect to take any time between 1,5 hours (complete stop w.r.t. Earth/orbit, see above) and infinity (still having a stable orbit). For moon-crashes later than 1,5 hours after the initial impact, the moon would need to be put in an unstable orbit, e.g. orbiting through low density areas of Earth's atmosphere once in a while.

Also other answers have mentioned ways for the moon to get destroyed or ripped apart in the process of deorbiting it, which definitely apply. I just wanted to focus on the aspect of timeline.

• I felt a specific lack of addressing the timeline aspect when reading the existing answers. – NoAnswer Jul 31 '19 at 15:47
• If you cancelled out the moon's orbital velocity it would take a week to hit the earth. Hint: g isn't 9.8m/s that far out. – Joshua Jul 31 '19 at 21:14
• Your timeline is vague and wrong, since it's making assumptions you really shouldn't. Both bounds of your frame are unsubstantiated without the wrong assumptions. This answer answers the timeline part quite well: roughly a week. – Mast Aug 1 '19 at 13:03

I would argue that the lower limit on impact delay is somewhere around 1.3 seconds.

Any impact that would leave the Moon at rest relative to Earth (see previous answers) would also, ahem, structurally disrupt the Moon. As in, turn it into an expanding cloud of vapor and debris, some of which would strike Earth sooner, some of which would form a ring, some of which would escape into (or out of) the rest of the Solar System. (I'm not sure how well even Elon's Mars colony would fare in the wake of a big collision.)

So, if we're allowing disruption, just hit the Moon with a converging swarm of ultrarelativistic impactors. They essentially turn the Moon into a giant shaped charge. Dial in the energy you want to select the resulting jet's velocity, up to lightspeed-minus-a-small-margin. For a visualization of the resulting impact, look up one of those stop-motion photos of a bullet hitting an apple...