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Would an object shot from earth fall into the sun?

If an object is shot at 107,000 km/h via rocket or otherwise, in the opposite direction to our orbit about the sun, it will be traveling at 0 km/h relative to the sun. The moon is not close enough to the object to have a significant force for the purposes of this question.

Will this object start accelerating towards the sun or will it somehow fall into another stable orbit?

Could it instead get trapped in the L4 Earth-Sun Lagrange point?

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    $\begingroup$ If you're interested in calculating the details of that trajectory, see en.wikipedia.org/wiki/Radial_trajectory $\endgroup$
    – PM 2Ring
    Sep 23, 2019 at 2:59
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    $\begingroup$ There is a bit of significant ambiguity left in your question, and that's how far it is from Earth when it reaches 0 velocity relative to the Sun. If it's still anywhere near the Earth (e.g. within the L4 point), it's likely the Earth will have enough effect on it to put it into a very eccentric orbit around the Sun rather than it hitting the Sun. Same with other planets. $\endgroup$
    – Greg Miller
    Sep 23, 2019 at 3:08
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    $\begingroup$ @GregMiller I don't agree: when the rocket is at 0 km/h relative to the sun, the Earth is still moving "very fast" away, so the gravitational field will have died out pretty quickly. But I suppose we should quantify our claims :-) $\endgroup$
    – Carl Witthoft
    Sep 23, 2019 at 18:10
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    $\begingroup$ See also this question. $\endgroup$
    – Draco18s no longer trusts SE
    Sep 24, 2019 at 13:40

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Assume that a spacecraft is instantaneously accelerated at the Earth's surface (disregarding the atmosphere for simplicity). We'll consider this from the Sun's reference frame; in other words, the Sun is stationary and the Earth is moving around it.

The spacecraft is accelerated to a velocity which is precisely equal and opposite to the orbital velocity of the Earth around the Sun, making it completely stationary in the instant after the acceleration.

What happens next? Well, we can consider the forces acting on the spacecraft:

  • The Earth's gravity causes a force in the direction of Earth.
  • The Sun's gravity causes a force in the direction of the Sun.

The stationary spacecraft is therefore going to accelerate towards the Earth and towards the Sun. Since the Earth is moving away quickly on its orbital path, the gravitational force is not enough to pull the spacecraft back into an Earth orbit; however, it will nudge the spacecraft into an elliptical orbit.

To demonstrate the situation, I have created a small simulation which can be viewed in a desktop browser. Click here to try the simulation. (You can click "View this program" to check the code, and refresh the page to restart the simulation.)

The simulation is physically accurate (ignoring the effects of other planets), but the spheres have been enlarged for easy interpretation. The Earth is represented as green, while the Sun is orange and the spacecraft is white. Note that, while the spheres representing the spacecraft and Sun intersect, the distance between the two physical objects is always larger than 3.35 solar radii.

This screenshot shows how the spacecraft has been pulled into an elliptical orbit by the Earth:

Screenshot of simulation demonstrating the spacecraft entering an elliptical orbit.

Finally, we could consider a more realistic scenario where the spacecraft is accelerated until it reaches zero velocity (again, in the Sun's reference frame) at a certain distance from the Earth. At the instant it reaches zero velocity, the engine is stopped.

In this case, the result is essentially the same: there are still forces exerted by the Earth and the Sun, so an elliptical orbit will result. The further the rocket is from the Earth when it reaches zero velocity, the more elliptical the orbit. If the Earth is so far away that its gravity is negligible, the spacecraft will fall directly towards the Sun.

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    $\begingroup$ I suppose this is due to enlarging the spheres, but on your simulation it seems that the object crashes into the Sun. I imagine that it actually passes very close from the Sun, but can you preicse how far ? $\endgroup$
    – Evargalo
    Sep 23, 2019 at 13:57
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    $\begingroup$ @Evargalo Thanks, I've updated the simulation so that it prints the closest approach to the Sun. The simulation will also stop if the spacecraft hits the Sun. In the first orbit the spacecraft travels within 3.4 solar radii of the Sun's centre, but the perihelion seems to get further away in subsequent orbits. $\endgroup$
    – TheGreatCabbage
    Sep 23, 2019 at 15:40
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    $\begingroup$ @TheGreatCabbage, your simulation uses simple Euler integration, which accumulates errors rather quickly (particularly when the body is moving rapidly, such as during perhelion). I'd trust your simulation when it says the object won't collide with the Sun on the first pass, but after that first pass, the simulation's predictions are questionable at best. $\endgroup$
    – Mark
    Sep 23, 2019 at 20:57
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    $\begingroup$ TLDR: hitting the sun is hard. $\endgroup$
    – Draco18s no longer trusts SE
    Sep 23, 2019 at 22:39
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    $\begingroup$ @Draco18s : that's why, if you want to land on the Sun, you should go there by night. $\endgroup$
    – Evargalo
    Sep 24, 2019 at 12:46
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The launch you described is similar to that of the Parker Solar Probe launched August 2018 at 12km/s in a direction opposite Earth's orbital velocity, so it fell toward (rather than into) the Sun, in an elliptical orbit. Its speed at closest approach is expected to be greater than 200km/s

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    $\begingroup$ In order for the Parker Solar Probe to make it as close to the sun as it will eventually go it also has to have some Venus fly-bys that further change its velocity. The 12 km/s from launch by itself isn't sufficient to get close enough to the sun to reach 200 km/s. $\endgroup$
    – NeutronStar
    Sep 25, 2019 at 2:27
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If an object is accelerated away from Earth fast enough that it winds up having no orbital velocity around the Sun, then it will fall radially into the Sun. It's orbital velocity that keeps the object (or the Earth itself) falling around the Sun and not into it. With zero orbital velocity, it simply falls straight down and it can't do anything else (getting trapped at the L4 point requires that it have an orbital motion very nearly the same as Earth's.)

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    $\begingroup$ I believe this answer assumes that the spacecraft's orbital velocity around the Sun is maintained at zero using thrusters, or that the Earth is distant enough for its gravitational effects to be negligible. I've provided an alternative interpretation in my answer. $\endgroup$
    – TheGreatCabbage
    Sep 23, 2019 at 10:54
  • $\begingroup$ I was going to upvote this but I re-read it and the question was: if a rocket was fired with a specific speed... The answer is probably no--there are other variables that would stop that from being the correct speed. If the question had been "If someone fires a rocket backwards at just the right speed near 107,000... to remove all orbital velocity would it fall into the sun", this would be a great answer. $\endgroup$
    – Bill K
    Sep 23, 2019 at 15:50
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The object will be attracted by the Sun's gravitational pull if the Moon and other planets in the Solar System are far enough that they do not significantly alter the object's speed or direction.

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