# QuestionHow much more likely it would be that a meteorite or an asteroide would hit us if we didn't have a moon

#### RogergR

The moon is about 100 times lighter at a distance of about 400,000 km, so the center of gravity of the system is somewhere 4000 km from the center of the earth towards the moon. I guess the number is between 10 and 100 times more impacts but i would prefer an expert opinion.

barycenter earth moon

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#### sam85geo

Both our Moon and the planet Jupiter act as barriers to comets and asteroids hitting Earth. Recall that in July 1994, the comet Shoemaker-Levy 9 crashed into Jupiter. My opinion is that possible Earth striking "things" are now fewer and far between due to the immense gravity of Jupiter. (It's been 27 years since the Shoemaker-Levy 9 show). To my chagrin, I'm no expert; I suggest that this is an apt question to ask NASA as well as those on this forum. Please post any references and calculations.

Catastrophe

#### RogergR

Both our Moon and the planet Jupiter act as barriers to comets and asteroids hitting Earth. Recall that in July 1994, the comet Shoemaker-Levy 9 crashed into Jupiter. My opinion is that possible Earth striking "things" are now fewer and far between due to the immense gravity of Jupiter. (It's been 27 years since the Shoemaker-Levy 9 show). To my chagrin, I'm no expert; I suggest that this is an apt question to ask NASA as well as those on this forum. Please post any references and calculations.
The bigest target (you call it barrier) is however the sun, but objects evaporize before they hit it. I was hoping to find a link here to some calculations, a solution maybe or something like that as far as just the earth and the moon are concerned.

#### Helio

My hunch is that the Moon adds only a little bit of protection. The Moon will send about as many inbound objects with random trajectories away as it will toward Earth, IMO (non-expert). Of course, those that impact the Moon give us protection, but thus would be a tiny percentage.

#### RogergR

My hunch is that the Moon adds only a little bit of protection. The Moon will send about as many inbound objects with random trajectories away as it will toward Earth, IMO (non-expert). Of course, those that impact the Moon give us protection, but thus would be a tiny percentage.
My point is that objects with random trajectories are targeting the mass center which is not the center of the earth, but the center of the system earth-moon in this case.

#### RogergR

Probably somebody should do a simulation, to get a tangible result.

#### p3orion

The idea is that the moon's gravity changes the trajectory of asteroids whose path would otherwise cause them to strike the Earth. But that works both ways: the moon's gravity is every bit as likely to change the trajectory of other objects which would have missed the Earth, curving them into a collision. It's a wash: if the collection of potential objects is random, then there are just as many nudged into a collision as out of one. All the moon's presence does is complicate the math.

The calculation of the protection offered by the moon, then, is simplified to simple geometry: how big is it? Given the high relative speeds of interplanetary objects, there are very few that could initially miss the face of the moon by such a tiny margin (maybe only a few miles) that they are nevertheless pulled down by its gravity into striking its surface. So the coverage of the moon's "protective shield" is limited to only slightly more than its apparent disc, which is some 3.67 million square miles. The disc of the Earth is about 49.3 million square miles. The moon is catching, therefore, about 7% of the asteroids that would hit the Earth that are coming from a direction that takes them roughly down the moon-Earth axis. Again, assuming a random selection of trajectories, that would be a tiny fraction of all the potential hits.

Fortunately, most asteroids are in roughly the same plane as the Earth-moon system (and the rest of the objects in the solar system.) Still, with the moon covering an angle of only about half a degree of arc, with a useful gravitational influence stretching that out to maybe 5 degrees to either side, that still leaves a whole lot of the sky unblocked.
moon blocking asteroids - Imgur

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#### RogergR

The idea is that the moon's gravity changes the trajectory of asteroids whose path would otherwise cause them to strike the Earth. But that works both ways: the moon's gravity is every bit as likely to change the trajectory of other objects which would have missed the Earth, curving them into a collision. It's a wash: if the collection of potential objects is random, then there are just as many nudged into a collision as out of one. All the moon does is complicate the math.

Given the high relative speeds of interplanetary objects, there are very few that could initially miss the face of the moon by such a tiny margin (maybe only a few miles) that they are nevertheless pulled down by its gravity into striking its surface. The calculation of the protection offered by the moon, then, is simplified to simple geometry: how big is it? So the coverage of the moon's "protective shield" is limited to only a little more than its apparent disc, which is some 3.67 million square miles. The disc of the Earth is about 49.3 million square miles. The moon is catching, therefore, about 7% of the asteroids that would hit the Earth that are coming from a direction that takes them down the moon-Earth axis. Again, assuming a random selection of trajectories, that would be a tiny fraction of all the potential hits.

moon blocking asteroids - Imgur

I disagree with you

Barycenter Earth-Moon

Let's do a simple calculation.
Imagine that the moon does not exist and an object goes towards the earth perpendicular to the ecliptic with an accuracy of lets say 1111 km to hit it in the middle. This is a 2D model, for simplicity, so the target size of 6400x6400x3.14 is about 130,000,000. Now we have a moon, and an object with the same conditions. The area against which this object falls with the same accuracy now measures (6400km + 4700km) x (6400km + 4700km) x3,14 which is about 380,000,000, it is three times larger, which means 3x3 = 10 times less probability of a hit. If we take into account that the moon and the earth revolve around each other, the difference would be even greater. I still think we have 10 to 100 times fewer impacts because of the moon, although I also know that most of them come from the ecliptic plane. As I mentioned, it would be best to program a simulation, I hoped it already existed.

#### p3orion

I disagree with you

Barycenter Earth-Moon

Let's do a simple calculation.
Imagine that the moon does not exist and an object goes towards the earth perpendicular to the ecliptic with an accuracy of lets say 1111 km to hit it in the middle. This is a 2D model, for simplicity, so the target size of 6400x6400x3.14 is about 130,000,000. Now we have a moon, and an object with the same conditions. The area against which this object falls with the same accuracy now measures (6400km + 4700km) x (6400km + 4700km) x3,14 which is about 380,000,000, it is three times larger, which means 3x3 = 10 times less probability of a hit. If we take into account that the moon and the earth revolve around each other, the difference would be even greater. I still think we have 10 to 100 times fewer impacts because of the moon, although I also know that most of them come from the ecliptic plane. As I mentioned, it would be best to program a simulation, I hoped it already existed.
I'm not sure what the phrase is for the logical fallacy you're employing, but it's there. It also took me a bit to figure out where you were finding an extra 4700 kilometers, until I realized you were hung up on thinking that the barycentric center of gravity of the Earth-moon system is in any way relevant. It makes the Earth a moving target, but "moving target" is relative ONLY if one is aiming.

First off, you said yourself you're using a 2-D model, which is appropriate since most asteroids are in roughly the same plane as the rest of the solar system. But your "simple calculations" of a three times larger "target area" only apply from the top or bottom, perpendicular to the plane of the ecliptic. From the perspective of the approaching asteroid it would be just shifting slightly left and right. Even so, it doesn't matter. Even if the Earth swept through a space 100 times larger than its own circular face (instead of about three times as large) it would present the same area no matter where it lies in that "target area," and so the same probability of being intersected by a random trajectory.

It's like shooting a gun at the pendulum of a clock. If you are AIMING at the pendulum, yes, it is harder to hit it if it is moving. However, if you are just shooting randomly, you have exactly the same chance of hitting a moving pendulum as one that is standing still. Asteroids are not aimed, so the fact that the moon makes the Earth shift a bit is just as likely to put it INTO the path of an object as to move it OUT of the way.

#### RogergR

I'm not sure what the phrase is for the logical fallacy you're employing, but it's there. It also took me a bit to figure out where you were finding an extra 4700 kilometers, until I realized you were hung up on thinking that the barycentric center of gravity of the Earth-moon system is in any way relevant. It makes the Earth a moving target, but "moving target" is relative ONLY if one is aiming.

First off, you said yourself you're using a 2-D model, which is appropriate since most asteroids are in roughly the same plane as the rest of the solar system. But your "simple calculations" of a three times larger "target area" only apply from the top or bottom, perpendicular to the plane of the ecliptic. From the perspective of the approaching asteroid it would be just shifting slightly left and right. Even so, it doesn't matter. Even if the Earth swept through a space 100 times larger than its own circular face (instead of about three times as large) it would present the same area no matter where it lies in that "target area," and so the same probability of being intersected by a random trajectory.

It's like shooting a gun at the pendulum of a clock. If you are AIMING at the pendulum, yes, it is harder to hit it if it is moving. However, if you are just shooting randomly, you have exactly the same chance of hitting a moving pendulum as one that is standing still. Asteroids are not aimed, so the fact that the moon makes the Earth shift a bit is just as likely to put it INTO the path of an object as to move it OUT of the way.
Objects are not falling to earth but to the system moon-earth, so we have 2 pendulums at the same time. ..4700 km is the circle of barycenter, thats the target for gravitationaly falling objects, not the center of earth.
Imagine
We have space, gravity, a stacionary moon and earth and an object far away faling perpendicular to them. It will fall exactly to the baricenter (its projection on the sorface), do we agree?
And gravity is however aiming , but the space is not empty and there is always a number of acuracy.

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#### p3orion

It will fall exactly to the baricenter (its projection on the sorface), do we agree?

We do not. An object will fall toward the center of gravity only if it begins at rest. Asteroids do not begin at rest; they are already in motion. That's why they don't fall straight down

#### RogergR

We do not. An object will fall toward the center of gravity only if it begins at rest. Asteroids do not begin at rest; they are already in motion. That's why they don't fall straight down
I said imagine ideal conditions.
If you have a vacuum space and in it only 3 equal 1 cm large stationary ideal balls where the first weighs 10 kg and touches the second which weighs 1 kg and the third at a distance of one meter weighs 1 gram, equidistant from the first two when time starts . Where will the third ball first touch any of the others?

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
Does this help to put things into perspective?
https://www.quora.com/profile/Richard-Mentock
Quote
Richard Mentock
, Studied geodesy and geophysics
Answered March 9, 2015 · Upvoted by
, Member of Planetary Society and Norwegian Astronautical Society since 1990

Imagine you're in a snowstorm, holding a tennis ball. To protect the tennis ball from snowflakes, have someone hold a marble a meter and a half (5 feet) away, and move it around your tennis ball. The snow flakes that hit the marble will not hit your tennis ball, so it is protecting it a bit.
That's approximately the size relation between the earth and moon.
Quote

Seems to apply equally to one snowflake.

Cat

vincenzosassone

#### Lariliss

There are insights made on the subject.

1. There is not much detailed information about the exact number of asteroids that hit the Moon and the Earth during several hundred million years. Firstly, the Earth has erosion, tectonic, etc. processes that hide craters over time. Secondly, the Moon surface is not fully investigated in detail yet.

Still, there are exploration and simulation results that give us some numbers through history.

Reference:
https://www.nature.com/articles/s41467-020-17115-6

2. Current asteroid/meteor activity analysis (specifically yearly showers) is an on-going developing process. Which presumably shows the percentage of ‘snowflakes’ shared by the Moon and the Earth.

3. Except direct physical impact, if any, meteors, reaching the Earth atmosphere with enough amount would change the planet ecosystem.

4. Impact to near-earth spacecrafts endanger factor.

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
Sorry, is this relating to proposed + or - influence of Moon?

Cat

#### Lariliss

No judgement, but my inclination to +
Not to mention moonlight, tides, eclipses, contributing to +

Catastrophe

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
I was inclined to the opinion that the following was pretty convincing:

"To protect the tennis ball from snowflakes, have someone hold a marble a meter and a half (5 feet) away, and move it around your tennis ball. The snow flakes that hit the marble will not hit your tennis ball, so it is protecting it a bit."

I am getting a little lost with "moonlight, tides, eclipses, contributing to + ".

Does that mean that the Sun similarly protects the Earth from impacts?
Vide sunlight, tides, eclipses contributing.

Cat

#### Lariliss

I was inclined to the opinion that the following was pretty convincing:

"To protect the tennis ball from snowflakes, have someone hold a marble a meter and a half (5 feet) away, and move it around your tennis ball. The snow flakes that hit the marble will not hit your tennis ball, so it is protecting it a bit."

I am getting a little lost with "moonlight, tides, eclipses, contributing to + ".

Does that mean that the Sun similarly protects the Earth from impacts?
Vide sunlight, tides, eclipses contributing.

Cat
This would be a different scale

Catastrophe

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
OK, forget the tongue in cheek bit. What do you think about the tennis ball and marble 5 feet away comparison? It certainly shows up the previous diagram when thinking about the asteroid trajectory, particularly bearing in mind the inverse square law.

Cat

The devil is in the detail.

#### Lariliss

OK, forget the tongue in cheek bit. What do you think about the tennis ball and marble 5 feet away comparison? It certainly shows up the previous diagram when thinking about the asteroid trajectory, particularly bearing in mind the inverse square law.

Cat

The devil is in the detail.
Genuinely.
Sorry for any inconsistency or bald approximation in advance.

Input:
NEO Asteroids number:
Apollos: 10,000
Amors: 8,000
Atens: 1,400
Atiras/Apohele: 31

Potentially hazardous asteroids (PHA): 157

Earth disc, sqm: 124,000,000
Moon disc, sqm: 9,000,000

Earth to Moon distance, km: 384,000

Assumption:
• All NEO asteroids presence in the same time;
• According to simulations from asteroid presence from internet, the Moon is roughly covering the Earth from directed asteroids in half day time;
• Calculate 1 year protection impact;
• The asteroid’s position uncertainty region at the time of the potential impact is much larger in both length and breadth than the size of the Earth. Impact could occur anywhere on the forward hemisphere during the hour or so when the Earth crosses the asteroid orbit and sweeps through the uncertainty region (NASA exercise);
• Taking into account inverse square law and Kepler’s 2nd law.

Case 1 (‘snowflake’ approach):
PHA.
1. The Moon takes about 8% of the asteroids: 12 instantaneously.
2. Divide 2 (half of the day), multiply 365: 2190.
Case 2 (‘snowflake’ approach):
Only Apollos and Atens (leave Amors to Mars).
1. The Moon takes about 8% of the asteroids: 912 instantaneously.
2. Divide 2 (half of the day), multiply 365: 166,000.

According to Sentry table, not instantaneous coming of the asteroids, means not oftener that once in 30 years (worst).

In this case, PHA 5 asteroids could come in 1 year. The best help from the Moon would be 1.

Taking into account that PHA does not necessarily hit the Earth, the Moon’s help goes up to about 20%.

The already mentioned study shows 59 hazardous hit the Moon during the past several hundred million years. The investigated number for the Earth for the same period is about 170 asteroids, which is actually 34%.

So, we may conclude valuable impact of the Moon into asteroid hit protection for the Earth.

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
"So, we may conclude valuable impact of the Moon into asteroid hit protection for the Earth."

Good. Thank you. That was a lot of work on your part.

Cat

Lariliss

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
Presumably the same is not going to apply to comets all with greater inclinations to the ecliptic?

Cat

#### Lariliss

Presumably the same is not going to apply to comets all with greater inclinations to the ecliptic?

Cat
Definitely, obviously

NEO comets: 107.

NASA ‘white paper’ sources the most informative:

Comets in the 1-10 kilometer class, 15 of them in short-period orbits that pass inside the Earth's orbit, and an unknown number of long-period comets. Virtually any short-period comet among the 100 or so not currently coming near the Earth could become dangerous after a close passage by Jupiter.

Less predictable are long-period comets, many of which arrive from a region called the Oort Cloud about 100,000 astronomical units (that is, about 100,000 times the distance between Earth and the Sun) from the Sun. These Oort Cloud comets can take as long as 30 million years to complete one trip around the Sun.

What is noticeable, is that HORIZONS gives a warning, that the nearer an object comes, the more predictable it is.

I have checked orbit simulations for about 10 comets as far as 100 years from now. Focusing, if any of them hit the Earth or come near to Jupiter. None of them did.

P.S. NASA simulation confirms there’s no technology able to stop a massive asteroid from hitting.

Catastrophe

#### Catastrophe

##### "Science begets knowledge, opinion ignorance.
Thank you for that information. I would add to "Less predictable are long-period comets, many of which arrive from a region called the Oort Cloud " to say "less, to completely non-predictable".

Cat

#### Lariliss

Thank you for that information. I would add to "Less predictable are long-period comets, many of which arrive from a region called the Oort Cloud " to say "less, to completely non-predictable".

Cat
'Heaven has a logical certainty' - Terry Pratchett

Catastrophe

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