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So, let's just assume, to no aspect explainable to humans, the sun just magically popped out of existence on January 1st next year (2022), that'll never happen, this is just a scenario to explain my question. If the sun disappeared, regardless of life on Earth, where would the Earth float off to? If it would come close to a gas giant, then a gravity assist would affect the Earth's path through space. Where would it go?

In an answer, I want a feasibly explained way to calculate where Earth would go, not giving much thought to Gravity assists, since I got their calculations from this question, the first answer, and the answer in the link embedded in it.

Also, in the comments section, I've seen comments claiming that my question is a possible duplicate of others. One other addresses what would happen here on Earth, while the other one is about how long it would take for us to stop orbiting the sun after its sudden disappearance. In contrast, I'm wondering about where the Earth would float off to, and where it would go.

Now, this question can seemingly be applied to any celestial object, whether the sun is there or not, I want to figure out trajectory in space. I want to know about trajectory, and how it affects the motion of celestial bodies.

Assuming this can be answered theoretically by the use of equations and mathematics, thanks in advance!

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    $\begingroup$ Earth would fly off in some direction along the ecliptic. The precise destination would depend on the time of year. $\endgroup$ May 14 at 20:57
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    $\begingroup$ New Year's Day is just a few days before perihelion, so the orbital speed is close to its maximum (~30.29 km/s), which means it'll take just under 9900 years to go 1 lightyear (ignoring the galaxy's gravitational field). $\endgroup$
    – PM 2Ring
    May 15 at 0:23
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    $\begingroup$ A partial cross-site duplicate over at World Building: What happens if the sun disappears and then reappears some days later?. $\endgroup$ May 15 at 1:31
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    $\begingroup$ BTW, we don't normally do impossible hypotheticals on this site. But I guess we can just treat your scenario as a "framing device" for a valid computation: the Earth's tangential direction at a specified moment of time. $\endgroup$
    – PM 2Ring
    May 15 at 5:50
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    $\begingroup$ I'm not completely certain of the correct way to calculate this. I've tried 3 different ways, using DE 441 data from Horizons (for 2022-01-01 00:00:00, UT & TDB), and get 3 similar results: near ecliptic longitude 190° (~ RA 12h38), which is roughly in the direction of the binary star Porrima aka Gamma Virginis. $\endgroup$
    – PM 2Ring
    May 15 at 6:02
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On the first of January, the direction of the sun from Earth is towards Sagittarius, with an ecliptic longitude of 280. The velocity of the Earth around the sun is perpendicular that that, and towards Virgo, in particular, it is close to the direction of Gamma Virginis (Porrima), with an ecliptic longitude of 280-90 = 190. The Earth will move in a straight line in that direction. (Conveniently Jan 1st is very close to perihelion, so elliptic corrections are needed even less than normal)

It won't actually get near to Gamma Virginis because that star is 38 light years away and at the Earth's speed of 30km/s it would take 400,000 years to get close and by then the star will have moved away from its current location.

It wouldn't come close to a gas giant (none are in a position in which they would come anywhere near the trajectory of the Earth, even without checking, they are all on the "wrong" side of the sun)

It would be very unlikely for the Earth to come close to a gas giant. If it did then the gravity of the giant planet would change the direction of the Earth, it could go in almost any direction.

For another date, the equation is simple enough: find the ecliptic longitude of the sun on that date and subtract 90. That gives the ecliptic longitude of the Earth's velocity.

You can now use a star map to find the direction that the Earth will travel in. On January 1st the Earth will travel in a direction toward an ecliptic longitude of 190, but remain in the plane of the ecliptic. On the map below, the ecliptic is the line through the middle of the map and you can find where 190 is by looking at the numbers at the top and bottom.

Because in the absence of the sun, there would be nothing to cause the Earth to orbit, it would travel in a straight line towards that point.

enter image description here

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    $\begingroup$ FWIW, the angle between the radial & tangent vectors of an elliptical orbit is only exactly 90° at the apsides, but because the eccentricity of the Earth's orbit is only 0.0167, it's always fairly close to 90°. In the next comment, I'll put a link to a Sage script that shows the angles for any eccentricity, for 30° increments of true anomaly. $\endgroup$
    – PM 2Ring
    May 16 at 1:44
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    $\begingroup$ Ellipse plotter $\endgroup$
    – PM 2Ring
    May 16 at 1:45
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    $\begingroup$ ...yet. "I do not have the ability to code, yet." But my answer doesn't depend on coding. You find out the direction of the sun (roughly 0 on March 21st and increasing by 360/365.35 each day. Then subtract 90 $\endgroup$
    – James K
    May 17 at 16:52
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    $\begingroup$ @Questioner There's a simple calculation for the Sun's ecliptic longitude at en.wikipedia.org/wiki/Position_of_the_Sun#Ecliptic_coordinates If you want a more accurate result than James' $\lambda - 90°$ formula, you can use $$\tan^{-1}\left(\frac{\varepsilon-\cos(\lambda)}{\sin(\lambda)}\right)$$ where $ \varepsilon$ is the eccentricity (~0.1671). But as I said earlier, that correction is small for Earth's orbit: the tangent direction is always between 89° & 91° greater than the Earth's longitude. $\endgroup$
    – PM 2Ring
    May 18 at 10:17
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    $\begingroup$ This plot shows the difference between that previous formula and $\lambda - 90°$. Both axes are in degrees, the horizontal axis is $\lambda$ (the Sun's longitude). $\endgroup$
    – PM 2Ring
    May 18 at 10:58

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