Look, if you follow this thread and can not get to this numbers by yourself....like I said, pls dont troll anymore.OK, let's see your calculations please.
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."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
Easy, becouse you miscalculated cose your pictures are wrong.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