# Terminal velocity in space, is there one?

1. Oct 27, 2015

### houlahound

Eg a particle is emitted by a supernova and starts accelerating toward our galaxy, what limits its speed to keep it less than c?

Reasoning: its initial velocity was large, the acceleration due to mass of our galaxy is big allowing for initially it is smaller due to distance.

Initial high velocity + increasing acceleration due to mass of galaxy in the absence of drag = big + increasingly big - 0 drag = big much.

2. Oct 27, 2015

### Staff: Mentor

Relativity itself provides the limit.

3. Oct 27, 2015

### nasu

You forgot the attraction from the supernova itself. I suppose this is the main factor to slow down the expanding gases.
Not that ts has anything to do with the speed of light limit.

4. Oct 27, 2015

### phinds

That's a good point, and that factor is WAYYYY more important than the attraction of a galaxy many light years away. @houlahound, do you get that?

EDIT: Interesting. Now that I think about this, I conclude that houlahound has this all wrong. Let say for the sake of simplicity that the galaxy from in which the supernova occurs has a mass which is identical to that of the Milky Way, and let's say that the particle is ejected from the supernova with velocity V. Now, V is going to be decreasing all the way to the half-way point between the two galaxies and then it is going to be increasing again from that point on. The initial decrease will be roughly the same amount as the later increase, so when it gets to the Milky Way, it's traveling at V again.

All this of course has nothing to do with the limit of c, which as russ pointed out is inherent in relativity.

5. Oct 27, 2015

### Staff: Mentor

Sure, but you could always set up a better scenario, where the particle is ejected toward the supermassive black hole at the center of its own galaxy.

6. Oct 27, 2015

### phinds

Yeah, no argument with that, but the OP specified that it is heading to our galaxy and his scenario is based on that so my response is as well.

7. Oct 27, 2015

### houlahound

The scenario was clumsy, the supernova was just to supply a particle.

Can we go with the supermassive bla k hole that the parts cle is heading for.

8. Oct 27, 2015

### phinds

Your scenario is still really undefined. If the particle is just drifting in space INSIDE a galaxy, and is not attracted to anything along the way, AND is headed in the direction of the center, then it would like end up in the black hole, but not exceeding c along the way. You really need to specify a more precise scenario for it to get meaningful answers.

What is the point, anyway? That is, what are you trying to actually figure out? If it's whether or not something can exceed c, then the answer is no. If it's WHY can't something exceed c, read up on relativity and the energy required for acceleration to near c. If it actually has something to do with the motion of bodies in space, then post a very detailed scenario.

9. Oct 27, 2015

### houlahound

My query is what puts the limit on the acceleration in the absence of drag. Relativity is not a force.

I would be too embarrassed to attempt a properly constructed mathematical problem cos I would only expose my ignorance and lack of technical and mathematical dexterity.

10. Oct 27, 2015

### Staff: Mentor

It works like this:
If you fire an object from a large distance at a large object, the impact speed will be equal to the initial speed plus the escape velocity of the target object (caveat). The OP wants to know if the resulting speed can exceed C.

The answer is no, due to Relativity, and the caveat to the above is that the initial logic only works at non-relativistic speeds.

11. Oct 27, 2015

### Staff: Mentor

Correct, but stating that implies you think it should be to matter here...

12. Oct 27, 2015

### jbriggs444

It would be more correct to say that kinetic energy at impact will be equal to initial kinetic energy plus the escape energy of the target object. That logic works at relativistic speeds as well as non-relativistic.

The relevant point for this formulation is that kinetic energy goes arbitrarily high before you hit light speed.

13. Oct 27, 2015

### phinds

Nothing puts a limit on the force applied to accelerate an object but as the object goes faster, more and more force has to be applied to get just a little more speed, and the object would require an application of infinite force to reach c. Since we can't apply infinite force, we can't reach c. That's why! Does that answer your question? (It's really the same answer Russ has been giving in abbreviated form in several posts).

Last edited: Oct 27, 2015
14. Oct 27, 2015

### houlahound

My confusion is say a light mass particle is ejected at .99c and is heading toward a supermassive black hole it final velocity would be initial velocity + acceleration x time due to mass of black hole.

Then all of a sudden relativity kicks in....seems discontinuous.

15. Oct 27, 2015

### phinds

There is nothing discontinuous about it. The particle, which has an initial velocity of .99c towards the BH gains velocity due to the gravity of the BH and goes a little faster. Since the BH will not apply infinite force, c is not reached. Where do you see the problem with this?

EDIT: and by the way, relativity doesn't "kick in", it's there always.

16. Oct 27, 2015

### houlahound

Prob is there is no relativistic force equation to apply to the particle to calculate its acceleration.......or is there.

17. Oct 27, 2015

### Staff: Mentor

There is, and it is just Newton's second law, the exact same force/acceleration equation you use in classical mechanics: $F=\frac{dp}{dt}$ where $p$ is the momentum.

(You may be more used to seeing the second law written as $F=ma=m\frac{dv}{dt}$ but that form only works when you can write the momentum as $p=mv$. That's a classical approximation that holds when $v$ is small, whereas the relativistic $p=\gamma{m}v$ where $\gamma$ is defined to be $1/\sqrt{1-v^2/c^2}$ works for all values of $v$, whether large or small).

Evaluate the derivative completely and you'll find that $F=\frac{dp}{dt}=\gamma^3ma$; keep the force constant and the acceleration decreases as the speed increases. That's how $c$ can act like a terminal velocity.

Thus, there's no need for relativity to "kick in" - you can always use the relativistic formula, it's just that the difference between it and the classical formula only starts to become noticeable as $v$ increases.

Last edited: Oct 28, 2015
18. Oct 31, 2015

### stevebd1

In this scenario, you can use an equation for velocity from something John Wheeler (Exploring Black Holes) refers to as the 'hail frame'.

$$v_{shell}=\left[\frac{2M}{r}+v_{far}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}$$

where $v_{shell}$ is the shell velocity (i.e. the velocity measured at a specific r by a hovering observer), $v_{far}$ is the velocity of the object hurled inwards from a great distance and $M=Gm/c^2$. You can see that even for an object with an initial velocity of 0.99c, it's shell velocity does not reach c until 2M (the event horizon) where technically there are no hovering observers to measure this (note, the equation only applies to r>2M).

19. Oct 31, 2015

### houlahound

Wow, for some reason those responses appeared as random symbols, today they are equations.

Thank you.