There was the article An asteroid is about to slip between Earth and the moon — the second near miss in 3 weeks https://www.sott.net/article/340613-An-asteroid-is-about-to-slip-between-Earth-and-the-moon-the-second-near-miss-in-3-weeks
Near misses all the time, how long might it continue before he have a hit? Would it be possible to come up with a rough idea? No, one needs to be an astronomer, an astrophysicist, one needs a huge calculator, a super computer, a...? In any case, below is an approach to the problem using considerations that many will be familiar with:
Let us consider the Earth and the Moon; at one end, point zero, the center of the Earth, at a distance of 6,371 km the surface of the Earth, and then at 384,402 km from the center of the Earth, that is one Lunar distance https://en.wikipedia.org/wiki/Lunar_distance_(astronomy), we would have the center of the Moon.
If we consider this line from the center of the Earth to the center of the Moon as a target line, what would be the chance of missing the area of the Earth Radius?
It should be ( 1-(6,371 km/384,402 km))=0,9834. In other words there is less than a 2 % percent chance of hitting the part of the line, where the Earth radius is located.
Note: The Lunar distance page mentions,
Having considered the target line, could one imagine a target disc, as the ones used for shooting pratices, with the area of the Earth, as seen from the side in the center, compared to the area that could be drawn by a circle with the Lunar distance as the radius?
The chance of missing the area where the Earth is located would then be, using the formular for a circle, pi*r^2: (1-(pi*6,371 km^2)/(pi*384,402 km^2))=1-(6,371/384,402)^2)=0,99972 Apparently only 3 chances in 10000.
If we, instead of seeing the Earth as a circle on a target disc, consider the idea that a danger could come from any direction of space, it may be even more useful to think of the Earth as a globe and the Lunar distance as the radius of a huge ball. Now ,to do this last calculation, I shall include in the model that the meteors that come near the Earth, often begin to loose speed more than 100 km above the surface of the Earth. Therefore, we would be justified to increase the radius of the Earth a bit, lets say from 6371 km to 6500 km.
Using the formular for the volume of a ball: 4/3*pi*r^3, and knowing from our previous attempts that the constants cancel out, we can estimate the chance of missing the Earth by an object travelling anywhere in space between the center of the Earth and the distance to the center of the Moon as:
1-(6,500/384,402)^3=1-(0,0169)^3=1-4.8*10^(-6)=0.9999952 that should give us around 5 hits per million passes.
Expressed differently, what we experience coming into the Earth atmosphere, like Chelyabinsk events, Tunguska events and so on, is just a tiny fraction of what goes on between the Earth and the space extending one lunar distance from its center. It seems based on our calculations, that we will read about many more near passes before getting hit. Or perhaps not, many must have experienced a "shower" of rain with only a drop or two falling, but otherwise no rain, no wet grass, pavement, window screen, just two drops. It is a different story if there comes an intense shower. Then there is nowhere to hide. Besides, I have not included the effect of the gravity field, the shadowing effect of the Moon, or other effects, like human activity, space weather that may change the odds of missing or hitting.
Near misses all the time, how long might it continue before he have a hit? Would it be possible to come up with a rough idea? No, one needs to be an astronomer, an astrophysicist, one needs a huge calculator, a super computer, a...? In any case, below is an approach to the problem using considerations that many will be familiar with:
Let us consider the Earth and the Moon; at one end, point zero, the center of the Earth, at a distance of 6,371 km the surface of the Earth, and then at 384,402 km from the center of the Earth, that is one Lunar distance https://en.wikipedia.org/wiki/Lunar_distance_(astronomy), we would have the center of the Moon.
If we consider this line from the center of the Earth to the center of the Moon as a target line, what would be the chance of missing the area of the Earth Radius?
It should be ( 1-(6,371 km/384,402 km))=0,9834. In other words there is less than a 2 % percent chance of hitting the part of the line, where the Earth radius is located.
Note: The Lunar distance page mentions,
In this initial attempt to calculate, I have not considered this, but worked with the standard Lunar distance.
Having considered the target line, could one imagine a target disc, as the ones used for shooting pratices, with the area of the Earth, as seen from the side in the center, compared to the area that could be drawn by a circle with the Lunar distance as the radius?
The chance of missing the area where the Earth is located would then be, using the formular for a circle, pi*r^2: (1-(pi*6,371 km^2)/(pi*384,402 km^2))=1-(6,371/384,402)^2)=0,99972 Apparently only 3 chances in 10000.
If we, instead of seeing the Earth as a circle on a target disc, consider the idea that a danger could come from any direction of space, it may be even more useful to think of the Earth as a globe and the Lunar distance as the radius of a huge ball. Now ,to do this last calculation, I shall include in the model that the meteors that come near the Earth, often begin to loose speed more than 100 km above the surface of the Earth. Therefore, we would be justified to increase the radius of the Earth a bit, lets say from 6371 km to 6500 km.
Using the formular for the volume of a ball: 4/3*pi*r^3, and knowing from our previous attempts that the constants cancel out, we can estimate the chance of missing the Earth by an object travelling anywhere in space between the center of the Earth and the distance to the center of the Moon as:
1-(6,500/384,402)^3=1-(0,0169)^3=1-4.8*10^(-6)=0.9999952 that should give us around 5 hits per million passes.
Expressed differently, what we experience coming into the Earth atmosphere, like Chelyabinsk events, Tunguska events and so on, is just a tiny fraction of what goes on between the Earth and the space extending one lunar distance from its center. It seems based on our calculations, that we will read about many more near passes before getting hit. Or perhaps not, many must have experienced a "shower" of rain with only a drop or two falling, but otherwise no rain, no wet grass, pavement, window screen, just two drops. It is a different story if there comes an intense shower. Then there is nowhere to hide. Besides, I have not included the effect of the gravity field, the shadowing effect of the Moon, or other effects, like human activity, space weather that may change the odds of missing or hitting.
