<p>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 <em>around</em> 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.)</p>

CC BY-SA 4.0
2019-09-23T10:54:43.517Z
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.
TheGreatCabbage
CC BY-SA 4.0
2019-09-23T15:50:34.580Z
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.
Bill K
CC BY-SA 4.0
2019-09-22T22:30:48.177Z
2019-09-22T22:30:48.177Z
Mark Olson
<p>The launch you described is similar to that of the <a href="https://blogs.nasa.gov/parkersolarprobe/2019/09/03/parker-solar-probe-completes-third-close-approach-of-the-sun/" rel="noreferrer">Parker Solar Probe</a> launched August 2018 at 12km/s in a direction opposite Earth's orbital velocity, so it fell toward (rather than <em>into</em>) the Sun, in an elliptical orbit. Its speed at closest approach is expected to be greater than 200km/s</p>

CC BY-SA 4.0
2019-09-25T02:27:08.627Z
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.
NeutronStar
CC BY-SA 4.0
2019-09-23T00:24:19.573Z
2019-09-23T10:08:38.390Z
2019-09-23T10:08:38.390Z
Professor O
<p>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.</p>

CC BY-SA 4.0
2019-09-24T12:00:45.890Z
2019-09-24T12:00:45.890Z
P Sh
<p>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.</p>

<p>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.</p>

<p>What happens next? Well, we can consider the forces acting on the spacecraft:</p>

<ul>
<li>The Earth's gravity causes a force in the direction of Earth.</li>
<li>The Sun's gravity causes a force in the direction of the Sun.</li>
</ul>

<p>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.</p>

<p>To demonstrate the situation, I have created a small simulation which can be viewed in a desktop browser. <a href="https://www.glowscript.org/#/user/TheGreatCabbage/folder/Public/program/StackExchange" rel="noreferrer">Click here to try the simulation.</a> (You can click "View this program" to check the code, and refresh the page to restart the simulation.)</p>

<p>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.</p>

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

<p><a href="https://i.sstatic.net/JBMKZ.png" rel="noreferrer"><img src="https://i.sstatic.net/JBMKZ.png" alt="Screenshot of simulation demonstrating the spacecraft entering an elliptical orbit."></a></p>

<p>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. </p>

<p>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.</p>

CC BY-SA 4.0
2019-09-23T13:57:09.367Z
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 ?
Evargalo
CC BY-SA 4.0
2019-09-23T15:40:36.360Z
@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.
TheGreatCabbage
CC BY-SA 4.0
2019-09-23T17:55:53.453Z
But an accurate simulation depends on the positions of Venus and Mercury, among other things.  As an example, the Galileo Jupiter probe was actually launched towards the Sun, so that it could get gravity boosts from Venus & Earth: https://www.noao.edu/jagi/sepo/atjup/io/launch_to_g29_full.jpg
jamesqf
CC BY-SA 4.0
2019-09-23T20:57:06.560Z
@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.
Mark
CC BY-SA 4.0
2019-09-23T20:57:57.963Z
@TheGreatCabbage thank you for creating that example script! This is exactly what I wanted to know. So depending on how far the object is ejected, Earth may still have enough force to pull the object into a non-colliding path with the Sun.
I found that anything above 1.43*Earths Radius from the Earth at 0m/s results in a collision with the Sun and anything below results in an orbit.
Taku
CC BY-SA 4.0
2019-09-23T22:39:57.367Z
TLDR: hitting the sun is hard.
Draco18s no longer trusts SE
CC BY-SA 4.0
2019-09-24T01:55:38.840Z
Surely there is an acceleration that counteracts it - so if we are moving at V then instantly accelerating it to V+K will counter the Earth's pull and fall into the Sun.
corsiKa
CC BY-SA 4.0
2019-09-24T07:06:53.577Z
@jamesqf That's true, but I think it's beyond the scope of the question and I didn't want to overcomplicate my answer.
TheGreatCabbage
CC BY-SA 4.0
2019-09-24T07:56:38.750Z
@TheGreatCabbage You could just add a simple note to that effect: "The simulation is physically accurate (assuming the effect from Venus, Mercury and all the planets other than Earth is negligible) ...".
Luaan
CC BY-SA 4.0
2019-09-24T09:26:24.780Z
@corsiKa Yes, it's possible to accelerate it to a speed which will cause it to hit the Sun. However, it will never follow a straight line into the Sun; the trajectory will always be slightly curved due to Earth's gravity.
TheGreatCabbage
CC BY-SA 4.0
2019-09-24T12:46:25.363Z
@Draco18s : that's why, if you want to land on the Sun, you should go there by night.
Evargalo
CC BY-SA 4.0
2019-09-25T14:11:39.053Z
Of note is that if your space probe has a perihelion of 3.4 solar radii, it doesn't matter that it _technically_ didn't collide (for comparison, Mercury is about 30x farther), it's still going to melt into a nice diffuse gas of various metals. I do suspect that the effect of the solar wind might have a say in where the metals ultimately end up though.
John Dvorak
CC BY-SA 4.0
2019-09-25T22:46:44.460Z
@TheGreatCabbage: Unless you give it enough of an additional push to cancel out the pull of Earth's gravity.
Vikki
CC BY-SA 4.0
2019-09-23T10:48:26.650Z
2019-09-24T13:40:26.467Z
2019-09-24T13:40:26.467Z
TheGreatCabbage
<h2>Would an object shot from earth fall into the sun?</h2>

<p>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.</p>

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

<p>Could it instead get trapped in the L4 Earth-Sun Lagrange point?</p>

<p><a href="https://i.sstatic.net/LNF1y.png" rel="noreferrer"><img src="https://i.sstatic.net/LNF1y.png" alt="t"></a></p>

CC BY-SA 4.0
2019-09-23T02:59:36.980Z
If you're interested in calculating the details of that trajectory, see https://en.wikipedia.org/wiki/Radial_trajectory
PM 2Ring
CC BY-SA 4.0
2019-09-23T03:08:07.333Z
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.
Greg Miller
CC BY-SA 4.0
2019-09-23T18:10:01.327Z
@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 :-)
Carl Witthoft
CC BY-SA 4.0
2019-09-24T13:40:38.227Z
[See also this question](https://space.stackexchange.com/q/13396/29441).
Draco18s no longer trusts SE
<the-sun><orbital-mechanics>
Would an object shot from earth fall into the sun?
Taku
# Would an object shot from earth fall into the sun?

(Asked by: username_0 on 2019-09-22)

## 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?

[![t](https://i.sstatic.net/LNF1y.png)](https://i.sstatic.net/LNF1y.png)

**Tags:** the-sun, orbital-mechanics

> **Comments:**
> **username_1**: If you're interested in calculating the details of that trajectory, see https://en.wikipedia.org/wiki/Radial_trajectory  
> **username_2**: 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.  
> **username_3**: @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 :-)  
> **username_4**: [See also this question](https://space.stackexchange.com/q/13396/29441).  

---

## 4 answers

### Answer by username_6 on 2019-09-23. Score: 39 | Accepted answer

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.](https://www.glowscript.org/#/user/username_6/folder/Public/program/StackExchange) (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.](https://i.sstatic.net/JBMKZ.png)](https://i.sstatic.net/JBMKZ.png)

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.

> **Comments:**
> **username_11**: 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 ?  
> **username_6**: @username_11 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.  
> **username_12**: But an accurate simulation depends on the positions of Venus and Mercury, among other things.  As an example, the Galileo Jupiter probe was actually launched towards the Sun, so that it could get gravity boosts from Venus & Earth: https://www.noao.edu/jagi/sepo/atjup/io/launch_to_g29_full.jpg  
> **username_13**: @username_6, 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.  
> **username_0**: @username_6 thank you for creating that example script! This is exactly what I wanted to know. So depending on how far the object is ejected, Earth may still have enough force to pull the object into a non-colliding path with the Sun. I found that anything above 1.43*Earths Radius from the Earth at 0m/s results in a collision with the Sun and anything below results in an orbit.  
> **username_4**: TLDR: hitting the sun is hard.  
> **username_14**: Surely there is an acceleration that counteracts it - so if we are moving at V then instantly accelerating it to V+K will counter the Earth's pull and fall into the Sun.  
> **username_6**: @username_12 That's true, but I think it's beyond the scope of the question and I didn't want to overcomplicate my answer.  
> **username_15**: @username_6 You could just add a simple note to that effect: "The simulation is physically accurate (assuming the effect from Venus, Mercury and all the planets other than Earth is negligible) ...".  
> **username_6**: @username_14 Yes, it's possible to accelerate it to a speed which will cause it to hit the Sun. However, it will never follow a straight line into the Sun; the trajectory will always be slightly curved due to Earth's gravity.  
> **username_11**: @Draco18s : that's why, if you want to land on the Sun, you should go there by night.  
> **username_16**: Of note is that if your space probe has a perihelion of 3.4 solar radii, it doesn't matter that it _technically_ didn't collide (for comparison, Mercury is about 30x farther), it's still going to melt into a nice diffuse gas of various metals. I do suspect that the effect of the solar wind might have a say in where the metals ultimately end up though.  
> **username_17**: @username_6: Unless you give it enough of an additional push to cancel out the pull of Earth's gravity.  

---

### Answer by username_8 on 2019-09-23. Score: 17

The launch you described is similar to that of the [Parker Solar Probe](https://blogs.nasa.gov/parkersolarprobe/2019/09/03/parker-solar-probe-completes-third-close-approach-of-the-sun/) 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

> **Comments:**
> **username_9**: 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.  

---

### Answer by username_5 on 2019-09-22. Score: 13

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.)

> **Comments:**
> **username_6**: 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.  
> **username_7**: 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.  

---

### Answer by username_10 on 2019-09-24. Score: 0

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.


---

