<p>Given the mass of the star, <span class="math-container">$M$</span> (or its gravitational parameter), the object's velocity and its position (vector), you can determine if the object's orbit is closed, as @antlersoft wrote in a comment.</p>
<p>Assuming a simple two body problem (with the asteroid having mass 0), you can determine whether the object is in a closed orbit by checking if the velocity is greater than the escape velocity at the given distance from the star as <span class="math-container">$v_{esc}=\sqrt{\dfrac{2GM}{r}}$</span>, where <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$M$</span> is the star's mass, and <span class="math-container">$r$</span> is the distance from the star. If the velocity of the asteroid is less than this value, its orbit is closed. If not, then the asteroid will escape the influence of the star.</p>
<p>If you don't know the velocity of the asteroid, you cannot determine if the asteroid is on an escape trajectory. However, if you <strong>do</strong> know the semi-major axis of the orbit or the eccentricity, you can tell if the orbit is not closed - if the semi-major axis is positive, the orbit is closed. If the semi-major axis is negative, the orbit is hyperbolic (not closed), and if it's <span class="math-container">$\pm \infty$</span>, the orbit is parabolic (also not closed). If the eccentricity is less than 1, the orbit is closed, and if it is greater than or equal to 1, it is not closed. Hopefully this clarifies things.</p>

CC BY-SA 4.0
2022-03-01T15:55:06.837Z
2022-03-01T15:55:06.837Z
WarpPrime
<p>So the sun or any star has a certain gravitational pull. Technically the gravity extends infinitely only it will just become really small. If there is a meteroid around a star how do I calculate when the meteroid is far enough away so that you can say it escaped the system? Or that is ejected from the system? So I know the mass of the star and the meteroid can I calculate with only that information at which radius the meteroid will escape the system. If not, could I calculate it if I also knew the semi major axis and eccentricity of the asteroid?</p>

CC BY-SA 4.0
2022-03-01T15:31:26.713Z
If you know the mass of the star and the position and velocity of the asteroid with respect to the star, it's a trivial two-body problem to show whether the asteroid's orbit is closed.  The tricky part is that asteroid "escapes" (where the asteroid was originally in a closed orbit and then is not) are the result of interactions with third bodies.
antlersoft
CC BY-SA 4.0
2022-03-01T15:39:11.333Z
@antlersoft okay but if you do not know the velocity of the particle can you still calculate if the orbit is closed?
programmingWolf
CC BY-SA 4.0
2022-03-01T21:18:34.190Z
It's just the escape speed.
ProfRob
<orbital-mechanics><planetary-systems><meteoroid>
Meteroids escaping the system
programmingWolf
# Meteroids escaping the system

(Asked by: username_0 on 2022-03-01)

So the sun or any star has a certain gravitational pull. Technically the gravity extends infinitely only it will just become really small. If there is a meteroid around a star how do I calculate when the meteroid is far enough away so that you can say it escaped the system? Or that is ejected from the system? So I know the mass of the star and the meteroid can I calculate with only that information at which radius the meteroid will escape the system. If not, could I calculate it if I also knew the semi major axis and eccentricity of the asteroid?

**Tags:** orbital-mechanics, planetary-systems, meteoroid

> **Comments:**
> **username_1**: If you know the mass of the star and the position and velocity of the asteroid with respect to the star, it's a trivial two-body problem to show whether the asteroid's orbit is closed.  The tricky part is that asteroid "escapes" (where the asteroid was originally in a closed orbit and then is not) are the result of interactions with third bodies.  
> **username_0**: @username_1 okay but if you do not know the velocity of the particle can you still calculate if the orbit is closed?  
> **username_2**: It's just the escape speed.  

---

## 1 answers

### Answer by username_3 on 2022-03-01. Score: 7

Given the mass of the star, $M$ (or its gravitational parameter), the object's velocity and its position (vector), you can determine if the object's orbit is closed, as @username_1 wrote in a comment.

Assuming a simple two body problem (with the asteroid having mass 0), you can determine whether the object is in a closed orbit by checking if the velocity is greater than the escape velocity at the given distance from the star as $v\_{esc}=\sqrt{\dfrac{2GM}{r}}$, where $G$ is the gravitational constant, $M$ is the star's mass, and $r$ is the distance from the star. If the velocity of the asteroid is less than this value, its orbit is closed. If not, then the asteroid will escape the influence of the star.

If you don't know the velocity of the asteroid, you cannot determine if the asteroid is on an escape trajectory. However, if you **do** know the semi-major axis of the orbit or the eccentricity, you can tell if the orbit is not closed - if the semi-major axis is positive, the orbit is closed. If the semi-major axis is negative, the orbit is hyperbolic (not closed), and if it's $\pm \infty$, the orbit is parabolic (also not closed). If the eccentricity is less than 1, the orbit is closed, and if it is greater than or equal to 1, it is not closed. Hopefully this clarifies things.


---

