2
$\begingroup$

I do wonder, do interstellar asteroids eventually stop at one point in space after they gradually decelerate (or) even do they decelerate?

Though there is no air like on earth and thus asteroids will not be affected by frictional forces, do they have friction with gravitational forces against their trajectories?

If it is so, they might eventually stop. Are there any known objects like that?

$\endgroup$
  • 2
    $\begingroup$ Stop with respect to what? Remember that motions and velocities are relative. If they would stop w.r.t to the sun, they would fall into it. $\endgroup$ – AtmosphericPrisonEscape Feb 20 at 10:40
  • $\begingroup$ @AtmosphericPrisonEscape, I am thinking like the objects moving in the spaces (not orbiting), like Omuamua. They are travelling through out the spaces. Do those kind of objects eventually stop? $\endgroup$ – Aung Satt Feb 20 at 10:56
  • $\begingroup$ @PM2Ring Yes, I wanted to say on Earth. $\endgroup$ – Aung Satt Feb 21 at 4:27
  • 1
    $\begingroup$ Why would Earth be special? Why not Mars? Or Alpha Centauri? Or some random rock floating out there? See, this is what happens when you think in terms of absolute space. There is no such thing. Space is not a thing. $\endgroup$ – Florin Andrei Feb 21 at 8:05
  • 1
    $\begingroup$ BTW, everyone, I still feel like we've not answering OP's actual question. Perhaps they were thinking in terms of friction with interstellar / intergalactic environment (random atoms, etc). $\endgroup$ – Florin Andrei Feb 21 at 8:14
3
$\begingroup$

An object moving through the interstellar medium will experience a weak drag force. If the drag force is $f(v)$ the velocity will decline as $dv/dt=-f(v)/M$ where $M$ is the mass. In fluids $f(v)\propto v^2$; this appears to work even for low density gas. Note that this differential equation leads to a velocity that declines forever, slowly approaching zero but never reaching it.

We can make a simple ballpark estimate of the timescale of slowdown by estimating how long it takes for it to run into its own mass of interstellar material. If it has radius $r$ it will encounter $\pi r^2 v \rho$ kg of ISM per second. So for mass $M$ the timescale of slowdown is $\tau \sim M/\pi r^2 v \rho$.

If we take a $r=1$ km spherical asteroid of mass $M= 8.8802\cdot 10^{12}$ kg (assuming density 2.12 g/cm$^3$) moving at 26.33 km/s and use $\rho=10^{-20}$ kg/m$^3$ (this varies a lot) then the timescale is about $10^{22}$ seconds, or 340 trillion years. So after maybe a quadrillion years the asteroid would be nearly at rest relative to the local gas if nothing disturbed its trajectory...

However, it is fairly likely for the asteroid to encounter a star during this time. Stellar density is about $\rho_*=0.14$ per cubic parsec, and the timescale for getting within 100 AU from a star is $\sim 1/\pi (100 AU)^2 v \rho_* \approx 10^{11}$years. This is likely to act as a gravity assist giving it some of the star's relative velocity (on the order of kilometers per second). So the true answer is that the asteroid will never settle down as long as there are stars in the galaxy.

$\endgroup$
  • 2
    $\begingroup$ This is a great answer, complete with numeric estimates. $\endgroup$ – Florin Andrei Feb 21 at 18:49
  • $\begingroup$ On the contrary, it is unlikely that an object of these characteristics will encounter another star. Although our galaxy, for example, has a star cluster of about 200 billion stars, it is unlikely that such an object will ever find another star. all this, of course, assuming that the orbit of that object is hyperbolic because otherwise, the object would simply return to the orbiting star. $\endgroup$ – jormansandoval Apr 8 at 0:18
  • $\begingroup$ @jormansandoval - Why do you think it is unlikely? I gave a numerical estimate. Note that I do not claim the object is entering a bound orbit, merely that it gets a velocity change. $\endgroup$ – Anders Sandberg Apr 8 at 5:44
3
$\begingroup$

No. Gravity does not cause "friction". A object will not stop unless it hits something.

Your comment clarifies that you mean "interstellar bodies" and not "asteroids". Newton's first law of motion is "A body will continue in a straight line and a constant speed, unless acted on by an external force". It won't decelerate in deep space where there is no friction.

When an interstellar body happens to approach a star it will start to fall towards it, and speed up. If it doesn't hit the star it will pass by and slow down as it moves up from the star. The velocity it loses as it moves away will be exactly the same as the velocity it gained while falling towards the star.

Gravity is a conservative force, because the total energy (gravitational + Kinetic) remains constant. No energy is lost, and so there is no change in speed.

You should be aware that there is no such thing as "not moving" in the absolute sense. The only words that you can use are "not moving relative to something". When I say "the car is not moving", what I mean is "The car is not moving relative to the ground".

$\endgroup$
  • 2
    $\begingroup$ I would argue this is incorrect: even in deep space there are a few atoms and dust particles here and there, and after a while repeated collisions will slow the object at least to the mean speed of the local environment's particles. $\endgroup$ – Carl Witthoft Feb 20 at 18:12
  • $\begingroup$ @James K, I see, so do you mean, even the object stop relative to it's something, the space around that object itself might be moving, isn't it... $\endgroup$ – Aung Satt Feb 21 at 4:39
  • 1
    $\begingroup$ I would argue that discussions of interstellar dust distract from the the main point,which is that "gravity isn't friction" $\endgroup$ – James K Feb 21 at 6:55
  • 2
    $\begingroup$ @AungSatt "Space" is not a thing. You cannot move relative to it. It cannot move relative to you. It doesn't even exist, it's nothing, just an emptiness that cannot do or be anything. You can only move (or stay motionless) relative to actual things: stars, planets, asteroids, rocks, dust, random molecules. $\endgroup$ – Florin Andrei Feb 21 at 8:02
  • 1
    $\begingroup$ @AungSatt Space is merely the totality of geometric relations (distance, angle, etc) between things. But in a perfect vacuum in an empty universe you could not even say that you're moving or not, because motion cannot be measured relative to space itself. It can only be measured relative to other objects. Relativity only predicts that the character of these geometric relations changes near massive objects. But it's not a "thing" that changes, it's just that when you measure a distance or angle there, it turns out to be different than what you expect in the absence of mass. $\endgroup$ – Florin Andrei Feb 21 at 23:04
-1
$\begingroup$

The asteroids can be accelerated and de-accelerated depending on which side of a much larger rotating object the asteroid passes.

If asteroid passes on the side in the direction of the rotation of the object, the asteroid experiences a gravity boost.

If the asteroid passes on the side opposite to the direction of rotation, the asteroid experiences gravity breaking.

If the breaking slows down the asteroid sufficiently so it can be captured then it collides with object.

An example of gravity assisted flight, the Mars probes used the gravity assist of the Earth to reach Venus, then the gravity assist of Venus to fling it towards Mar. Landing on Mars used gravity breaking.

But for the Mars lander to land on Mars required fuel and a parachute.

Gravity assisted acceleration and de-acceleration is accomplished by exchanging angular momentum with a larger rotating object and is based on conservation of angular momentum.

$\endgroup$
  • $\begingroup$ -1 I never heard of gravity braking, and I'm pretty sure gravity is a conservative force. This isn't a science-based answer. $\endgroup$ – uhoh Feb 25 at 7:14
  • $\begingroup$ Recent Mars landers have used aerodynamic braking. The effect of a gravity assist depends on whether the small body passes on the leading or trailing side of a planet's orbital motion. $\endgroup$ – Mike G Feb 26 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.