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

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Catastrophe

"There never was a good war, or a bad peace."
Thank you for your polite answer. Perhaps you would be good enough to show us the numbers you are using and how you arrive at the result,

Are you using a Barycentre Inertial Coordinate System?

Cat :)
 

Catastrophe

"There never was a good war, or a bad peace."
"I can tell you how big is that "wobbleing", easy maths.
In terms of perpendicular to the system we get a circle area which is 3 times larger than 2D earths area, and in terms of horizontal (ecliptic) view of the system we get an ellipse that is 1,7 times larger that 2D earths area. Also the moon is 5 degreas inclinated to the ecliptic, so the true numbers are biger."

Please explain "we get a circle area which is 3 times larger than 2D Earth's area, and in terms of horizontal (ecliptic) view of the system we get an ellipse that is 1.7 times larger than 2D Earth's area. Also the Moon is 5 degreas [sic] inclinated [sic] to the ecliptic, so the true numbers are biger [sic]."

What do you mean by wobbleing [sic] in mathematical terms? What is this circle, and what is the ellipse? What do they signify, and what is the relevance to the topic?

Your kind response will be much appreciated.

Cat :)
 
Jul 26, 2021
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"I can tell you how big is that "wobbleing", easy maths.
In terms of perpendicular to the system we get a circle area which is 3 times larger than 2D earths area, and in terms of horizontal (ecliptic) view of the system we get an ellipse that is 1,7 times larger that 2D earths area. Also the moon is 5 degreas inclinated to the ecliptic, so the true numbers are biger."

Please explain "we get a circle area which is 3 times larger than 2D Earth's area, and in terms of horizontal (ecliptic) view of the system we get an ellipse that is 1.7 times larger than 2D Earth's area. Also the Moon is 5 degreas [sic] inclinated [sic] to the ecliptic, so the true numbers are biger [sic]."

What do you mean by wobbleing [sic] in mathematical terms? What is this circle, and what is the ellipse? What do they signify, and what is the relevance to the topic?

Your kind response will be much appreciated.

Cat :)
If you make a one month timelaps perpendicular to the Moon-Earth system, the earth will cover (6378km+4671km)(6378km+4671km)π of the picture. If you make a similar timelaps from the field of M-E, the area will be (6378km x 6378km) π + (4671km + 4671km) x (6378km + 6378km). The second, field view area, is not an ellipse, I just aproximated for easyer counting on mind, since I know the formula of the ellipse area (abπ), 1.8 is a beter result than 1.7. Its a stretched circle.

The relevance to the topic is out of question as a matter of course, if you dont see it, please dont coment.
 
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Catastrophe

"There never was a good war, or a bad peace."
Thank you for your kind explanation.

Earth–Moon system, in which the barycenter is located on average 4,671 km (2,902 mi) from Earth's center, 75% of Earth's radius of 6,378 km (3,963 mi).

So you are adding the distance of the barycentre from Earth's centre which is 73% of Earth's radius, to Earth's radius, giving a total of 1.73 Earth radii squared, and then multiplying this by π . I am taking a circle moving in a linear dimension, and you are taking a circle squared.

IIf you wanted to cover the area covered by the Earth's wobble, you would take π x (6378)^2 plus (6378 x 4671) sq km. This is 2 x 0.5 Earth areas plus a rectangle of Earth radius x offset from centre, Thus if offset = 0, the area would default to π x radius ^2. If the offset is 0.73, then the rectangle is 0.73R^2.
The area then becomes π x 1.73R^2".

View: https://imgur.com/a/TBL7JaM


View: https://imgur.com/a/umM00sg


But this is not a still target, it is wobbling. It includes the area where the Earth is not. The area available to be impacted is still π x (6378)^2 sq km..

I cannot imagine why you then want to do this.
I(6378km x 6378km) π + (4671km + 4671km) x (6378km + 6378km).
That is πR^2 + (2 x 0.73)R x 2R = 3.92R^2

compared with my calculation you are using 3.92 vs. 1.73 (R^2)

How can we reconcile this difference?

Cat :)




 
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Jul 26, 2021
17
2
15
Thank you for your kind explanation.

Earth–Moon system, in which the barycenter is located on average 4,671 km (2,902 mi) from Earth's center, 75% of Earth's radius of 6,378 km (3,963 mi).

So you are adding the distance of the barycentre from Earth's centre which is 73% of Earth's radius, to Earth's radius, giving a total of 1.73 Earth radii squared, and then multiplying this by π . I am taking a circle moving in a linear dimension, and you are taking a circle squared.

IIf you wanted to cover the area covered by the Earth's wobble, you would take π x (6378)^2 plus (6378 x 4671) sq km. This is 2 x 0.5 Earth areas plus a rectangle of Earth radius x offset from centre, Thus if offset = 0, the area would default to π x radius ^2. If the offset is 0.73, then the rectangle is 0.73R^2.
The area then becomes π x 1.73R^2".

View: https://imgur.com/a/TBL7JaM


View: https://imgur.com/a/umM00sg


But this is not a still target, it is wobbling. It includes the area where the Earth is not. The area available to be impacted is still π x (6378)^2 sq km..


I cannot imagine why you then want to do this.
I(6378km x 6378km) π + (4671km + 4671km) x (6378km + 6378km).
That is πR^2 + (2 x 0.73)R x 2R = 3.92R^2

compared with my calculation you are using 3.92 vs. 1.73 (R^2)

How can we reconcile this difference?

Cat :)




Easy, becouse you miscalculated cose your pictures are wrong.

Sry for the ugly picture this is the field view for one month.
Beter picture the midle sized circle area is the one month perpendiculary.

However.
In terms of space.
The area of the earth is simplified 4π6378^2. And the area that earth cover in one mount wobbeling around cose of the baricenter aproximate by an elipsoid is more than two times bigger.
Try to calculete the diference in the volume in one month with the"wobbeling", the number will be probably around ...it is 3
 
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Catastrophe

"There never was a good war, or a bad peace."
A=4πr^2 is the surface area of a sphere. The area you need as a basis is the cross-sectional area area which is a circle πr^2.

You are at liberty to believe whatever you wish, and so am I. And so are the good people watching this thread.

In my opinion, involving movement in time simply exaggerates the area which an asteroid might hit. There is an (invalid, imho) argument for involving movement with time, Looking at my second diagram shows that, even if the asteroid is moving very slowly in comparison with the Earth, then the area of Earth available to be struck by the asteroid is very small indeed, and probably less than the area shielded by the Moon when in front of Earth. I could see a possible (inflated) argument of less than 1.7 Earth cross-sectional area, but certainly not 4 times.

You stick with your idea, I will stick with mine.

End of subject as far as I am concerned.
Good Luck. Have a happy life.
Kind regards
Cat :)
 

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