<p>Such forces do exist.</p>

<p>They are utterly negligible in most circumstances. If you imagine a planet "running into" matter whilst travelling in its orbit. Let's say that on average that matter is at rest with respect to the star the planet is orbiting. Then in order to make a very significant change to the linear momentum of the planet, then it must accrete a significant fraction of its own mass.</p>

<p>Such a thing might be important in the very early history of a planetary system when the system is full of gas and debris (e.g. the formation of the Moon), but later on (and bear in mind that the <a href="https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf" rel="noreferrer">IAU definition</a> of a planet includes the specification that it has "cleared the neighbourhood around its orbit") this just isn't an issue.</p>

<p><a href="http://curious.astro.cornell.edu/about-us/75-our-solar-system/comets-meteors-and-asteroids/meteorites/313-how-many-meteorites-hit-earth-each-year-intermediate" rel="noreferrer">This</a> popular science article suggests that 37,000-78,000 tonnes of material hits the Earth every year. Sounds a lot, but in comparison to the mass of the Earth ($6\times 10^{21}$ tonnes) it is very small. If the material was at rest with respect to the Earth's orbit, then an order of magnitude estimate for how long such impacts would take to <em>significantly</em> affect the Earth's orbit would be $\sim 10^{16}$ years.</p>

CC BY-SA 3.0
2017-05-23T12:28:01.200Z
And even that is a hopelessly optimistic number because it assumes that every bit that impacts imparts its momentum in roughly the same direction.
Shadur-don't-feed-the-AI
CC BY-SA 3.0
2017-05-23T12:42:04.433Z
@shadur No it isn't. I assume that the material has no momentum in a heliocentric frame of reference (on average). I suspect that is a reasonable assumption.
ProfRob
CC BY-SA 3.0
2017-05-23T12:43:02.077Z
That or it'd mostly average out, likely.
Shadur-don't-feed-the-AI
CC BY-SA 3.0
2017-05-23T13:05:09.033Z
@shadur I don't understand your comments. Saying that the the material has on average no momentum in the heliocentric frame of reference is the same as saying that the objects orbit the Sun in an isotropic way. That behaviour (on average) is like the Earth hitting a stationary object. I have not assumed that "every bit that impacts imparts its momentum in roughly the same direction". My estimate is not " hopelessly optimistic" (for that reason).
ProfRob
CC BY-SA 3.0
2017-05-23T17:21:01.290Z
I guess that actually most things would orbit the sun in the same direction (as the earth), and therefore have even less effects than a stationary object.
Paŭlo Ebermann
CC BY-SA 3.0
2017-05-23T18:12:19.800Z
@Shadur If the air has no momentum in a geocentric frame, and you're running through it, the effect of the air on your momentum does not average out.  It holds you back.
Cort Ammon
CC BY-SA 3.0
2017-05-23T19:46:30.020Z
The average asteroid has an orbit with the semi-major axis much larger than earth's, and with angular momentum in the same direction - when they cross earth's orbit we would expect them to have a high velocity relative to earth from the heliocentric frame, meaning they would be more likely to increase earth's orbit than decrease it. In simpler terms: an object from the outer solar system colliding with an object from the inner solar system is generally going to drive the inner object outwards.
brandondoge
CC BY-SA 3.0
2017-05-23T08:36:03.150Z
2017-05-24T04:53:03.283Z
2017-05-24T04:53:03.283Z
ProfRob
<p>Actually, the planets do <strong>not</strong> retain their energy or angular momentum. However, this is not (or hardly as explained by Rob) due to collisions with meteroids or any other dust, but due to their mutual gravitational interactions. These are far more important. The gravitational interaction energy of the Earth with Jupiter is about $10^{-4}$ times that with the Sun, $\sim10^{11}$ times more important than the 'dust'.</p>

<p>The gravitational interactions are conservative: energy and angular momentum are not lost, only exchanged, but this does not mean that the Solar system is stable.</p>

<p>Finally, tidal forces result in the dissipation of energy (but not the total angular momentum including spin), for example the Moon's orbit around Earth increases in radius by about 1inch/yr due the tidal interaction with Earth. For the dynamics of the Solar system these tides are the most important non-conservative forces.</p>

CC BY-SA 3.0
2017-05-27T06:56:20.503Z
2017-05-27T06:56:20.503Z
Walter
<p>Obviously, there are objects floating in space. Consider the space a fabric (Here, a Stretched Piece Of Cloth). If you consider the sun as 10Kg metal ball and put it into the center of the Fabric, you will see how deep the fabric is at the center and as you go far away its height increases. Now, consider the earth as a 50 gram ball and put it reasonably away from the metal ball. You will notice that the 'earth' moves towards the 'sun'.</p>

<p>Here's what this implies. The force of attraction is due to gravity. Gravity of a large object <strong>bends</strong> the space (Fabric) around it. This is why nothing escapes a black hole (infinite mass gives it infinite Gravity). Now, gravity causes the <strong>centripetal force</strong> and forces an object to move around it in nearly circular orbits. The direction of movement on any object in a circular path is given by the tangent at is point. Therefore, the earth is always moving away from the sun at the same time  getting pulled towards the sun. </p>

<p>This implies the moving in the orbit, the earth maintains a constant velocity. Since the space is a vacuum, there are no resistance offered and earth doesn't lose its energy. Obviously, if we keep 0.001 gram objects in the path of a 50 gram ball that's under the gravity of a 10kg ball, there will be no effect on the motion of the 'earth'. Hope you got it!</p>

CC BY-SA 3.0
2017-05-26T16:09:41.613Z
This doesn't address the question, which notes that space is not a vacuum, as it contains rarified gas and "debris". It also mixes the "rubbersheet" model of GR with Newtonian mechanics. It contains some errors: Black holes don't have infinite mass and the velocity of the Earth isn't constant.
James K
CC BY-SA 3.0
2017-05-26T08:26:23.107Z
2017-05-26T08:26:23.107Z
Pujan Yadav
<p>I'm watching <a href="https://www.youtube.com/watch?v=MTY1Kje0yLg" rel="noreferrer">this</a> YouTube video, and in it, the lecturer explains a model of planetary orbit. At <a href="http://www.youtube.com/watch?v=MTY1Kje0yLg&amp;t=1m55s" rel="noreferrer">1:55</a> he shows how a marble will naturally orbit the center object, and he briefly claims that planets will not lose energy like the marble does. But I cant see how that is the case. Surely there must be meteors, asteroids, comets, including general space debris, that will over the course of millennia, diminish this energy/momentum.</p>

<p>How then, can planets retain their energy, when there are almost only forces that would slow it down? Slowly increasing the energy wouldn't help either, as it would offset the balance. How can there be stability, especially over the course of millions of years?</p>

CC BY-SA 3.0
2017-05-23T12:29:12.760Z
Short version: Planets are *huge* compared to anything that's likely to impact them. We're talking about "flicking a marble at a speeding truck" orders of magnitude difference.
Shadur-don't-feed-the-AI
CC BY-SA 3.0
2017-05-23T13:09:21.683Z
The video assumes spherical planets in a vacuum. This of course, we all know is not true.
Aron
CC BY-SA 3.0
2017-05-23T15:11:14.133Z
@Shadur the marble breaks the glass, kills the driver, resulting in the truck flying off course (maybe literally if the road is on a mountain).  

I guess that's the bit where the dinosaurs died out though.
UKMonkey
CC BY-SA 3.0
2017-05-23T15:33:28.207Z
@Shadur More like flicking motes of dust at a speeding truck.
Shufflepants
CC BY-SA 3.0
2017-05-23T15:50:29.977Z
@Aron  Yeah, it only works with spherical cows :-)
Carl Witthoft
CC BY-SA 3.0
2017-05-23T18:30:37.543Z
The premise of this question is flawed, because the lecturer did **not** claim that planets will not lose energy. If you listen to it again, you'll hear him say "not noticeably" .. and therein lies the answer to your question.
JBentley
<orbit>
How do planets retain momentum?
Guest_Charlie
# How do planets retain momentum?

(Asked by: username_0 on 2017-05-23)

I'm watching [this](https://www.youtube.com/watch?v=MTY1Kje0yLg) YouTube video, and in it, the lecturer explains a model of planetary orbit. At [1:55](http://www.youtube.com/watch?v=MTY1Kje0yLg&t=1m55s) he shows how a marble will naturally orbit the center object, and he briefly claims that planets will not lose energy like the marble does. But I cant see how that is the case. Surely there must be meteors, asteroids, comets, including general space debris, that will over the course of millennia, diminish this energy/momentum.

How then, can planets retain their energy, when there are almost only forces that would slow it down? Slowly increasing the energy wouldn't help either, as it would offset the balance. How can there be stability, especially over the course of millions of years?

**Tags:** orbit

> **Comments:**
> **username_1**: Short version: Planets are *huge* compared to anything that's likely to impact them. We're talking about "flicking a marble at a speeding truck" orders of magnitude difference.  
> **username_2**: The video assumes spherical planets in a vacuum. This of course, we all know is not true.  
> **username_3**: @Shadur the marble breaks the glass, kills the driver, resulting in the truck flying off course (maybe literally if the road is on a mountain).    I guess that's the bit where the dinosaurs died out though.  
> **username_4**: @Shadur More like flicking motes of dust at a speeding truck.  
> **username_5**: @username_2  Yeah, it only works with spherical cows :-)  
> **username_6**: The premise of this question is flawed, because the lecturer did **not** claim that planets will not lose energy. If you listen to it again, you'll hear him say "not noticeably" .. and therein lies the answer to your question.  

---

## 3 answers

### Answer by username_7 on 2017-05-23. Score: 30

Such forces do exist.

They are utterly negligible in most circumstances. If you imagine a planet "running into" matter whilst travelling in its orbit. Let's say that on average that matter is at rest with respect to the star the planet is orbiting. Then in order to make a very significant change to the linear momentum of the planet, then it must accrete a significant fraction of its own mass.

Such a thing might be important in the very early history of a planetary system when the system is full of gas and debris (e.g. the formation of the Moon), but later on (and bear in mind that the [IAU definition](https://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf) of a planet includes the specification that it has "cleared the neighbourhood around its orbit") this just isn't an issue.

[This](http://curious.astro.cornell.edu/about-us/75-our-solar-system/comets-meteors-and-asteroids/meteorites/313-how-many-meteorites-hit-earth-each-year-intermediate) popular science article suggests that 37,000-78,000 tonnes of material hits the Earth every year. Sounds a lot, but in comparison to the mass of the Earth ($6\times 10^{21}$ tonnes) it is very small. If the material was at rest with respect to the Earth's orbit, then an order of magnitude estimate for how long such impacts would take to *significantly* affect the Earth's orbit would be $\sim 10^{16}$ years.

> **Comments:**
> **username_1**: And even that is a hopelessly optimistic number because it assumes that every bit that impacts imparts its momentum in roughly the same direction.  
> **username_7**: @shadur No it isn't. I assume that the material has no momentum in a heliocentric frame of reference (on average). I suspect that is a reasonable assumption.  
> **username_1**: That or it'd mostly average out, likely.  
> **username_7**: @shadur I don't understand your comments. Saying that the the material has on average no momentum in the heliocentric frame of reference is the same as saying that the objects orbit the Sun in an isotropic way. That behaviour (on average) is like the Earth hitting a stationary object. I have not assumed that "every bit that impacts imparts its momentum in roughly the same direction". My estimate is not " hopelessly optimistic" (for that reason).  
> **username_8**: I guess that actually most things would orbit the sun in the same direction (as the earth), and therefore have even less effects than a stationary object.  
> **username_9**: @Shadur If the air has no momentum in a geocentric frame, and you're running through it, the effect of the air on your momentum does not average out.  It holds you back.  
> **username_10**: The average asteroid has an orbit with the semi-major axis much larger than earth's, and with angular momentum in the same direction - when they cross earth's orbit we would expect them to have a high velocity relative to earth from the heliocentric frame, meaning they would be more likely to increase earth's orbit than decrease it. In simpler terms: an object from the outer solar system colliding with an object from the inner solar system is generally going to drive the inner object outwards.  

---

### Answer by username_11 on 2017-05-27. Score: 1

Actually, the planets do **not** retain their energy or angular momentum. However, this is not (or hardly as explained by Rob) due to collisions with meteroids or any other dust, but due to their mutual gravitational interactions. These are far more important. The gravitational interaction energy of the Earth with Jupiter is about $10^{-4}$ times that with the Sun, $\sim10^{11}$ times more important than the 'dust'.

The gravitational interactions are conservative: energy and angular momentum are not lost, only exchanged, but this does not mean that the Solar system is stable.

Finally, tidal forces result in the dissipation of energy (but not the total angular momentum including spin), for example the Moon's orbit around Earth increases in radius by about 1inch/yr due the tidal interaction with Earth. For the dynamics of the Solar system these tides are the most important non-conservative forces.


---

### Answer by username_12 on 2017-05-26. Score: -2

Obviously, there are objects floating in space. Consider the space a fabric (Here, a Stretched Piece Of Cloth). If you consider the sun as 10Kg metal ball and put it into the center of the Fabric, you will see how deep the fabric is at the center and as you go far away its height increases. Now, consider the earth as a 50 gram ball and put it reasonably away from the metal ball. You will notice that the 'earth' moves towards the 'sun'.

Here's what this implies. The force of attraction is due to gravity. Gravity of a large object **bends** the space (Fabric) around it. This is why nothing escapes a black hole (infinite mass gives it infinite Gravity). Now, gravity causes the **centripetal force** and forces an object to move around it in nearly circular orbits. The direction of movement on any object in a circular path is given by the tangent at is point. Therefore, the earth is always moving away from the sun at the same time getting pulled towards the sun.

This implies the moving in the orbit, the earth maintains a constant velocity. Since the space is a vacuum, there are no resistance offered and earth doesn't lose its energy. Obviously, if we keep 0.001 gram objects in the path of a 50 gram ball that's under the gravity of a 10kg ball, there will be no effect on the motion of the 'earth'. Hope you got it!

> **Comments:**
> **username_13**: This doesn't address the question, which notes that space is not a vacuum, as it contains rarified gas and "debris". It also mixes the "rubbersheet" model of GR with Newtonian mechanics. It contains some errors: Black holes don't have infinite mass and the velocity of the Earth isn't constant.  

---

