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First, I'm sorry if this question is dumb, incoherent or vague - I never really studied any physics and this might be out of my grasp, but I tried to apply some of the math I know and figure these things out myself.

Suppose we're in space, and we ignore things like gravity etc., we have an object $A$, and an object $B$ tied to $A$ with a rope, and $A$ has far larger mass than $B$. Suppose also that the rope has the slack pulled out of it, and object $B$ is moving at tangential speed $10$ right now. I would expect that then object will keep moving at the same speed forever - the speed won't change, but I'm not really sure how to derive this.

I think that in essence I'm asking why momentum will be preserved. How to justify that "infinitesimal" change of direction doesn't change the momentum? Can this be proven/suggested, based on Newton's second law and maybe some other things?

An attempt : denote $p(t)$ the function of the position of object $B$ in time $t$, and suppose $A$ has coordinates $(0,0)$, and $p(0) = (1,0)$. An $\epsilon$ while later, we could say that $p(\epsilon)$ is $(\cos(x),\sin(x))$ for some $x$. We could thus approximate the movement linearly as $g(t) = (1,0) + t ( (\cos(x),\sin(x)) - (1,0))$. Now at time $\epsilon$, the the rope will apply enough force to decelerate the movement of the object with respect to the $(\cos(x),\sin(x))$ direction - so only the perpendicular aspect of the velocity will remain, and it's value will essentially be something like the projection of $(\cos(x),\sin(x)) - (1,0))$ onto this perpendicular direction.

Denoting the function of the magnitude of the velocity by $m(t)$, this gives us an estimate of the magnitude of the velocity at time $\epsilon$ as $m(\epsilon) =m(0) \cos(x)$. Similarly, at the time $2 \epsilon$, $m(2\epsilon) = m(0) \cos^2(x)$, etc.

As we lower $\epsilon$, we see that $m(t)$ keeps going up to $m(0)$, and so the magnitude of the velocity of the actual object $B$ stays constant.

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  • $\begingroup$ Answer: 'A' can't really be immovable. $\endgroup$ – The Photon May 23 at 3:16
  • $\begingroup$ Look I don't know how to phrase this question well. Essentially I have an object B orbiting, and there are no forces/nothing is happening besides the fact that B is held by a rope, and there's a tangential velocity. Another scenario - we have a plane with minimal friction, an object tied to a center with a rope, and an initial tangential velocity. Why will the magnitude of the velocity change minimally? $\endgroup$ – John P May 23 at 3:39
  • $\begingroup$ I'm saying object A must wobble a little bit to maintain conservation of momentum. It can be a very small wobble if the mass of A is much greater than B, but it can't be zero. $\endgroup$ – The Photon May 23 at 3:44
  • $\begingroup$ @ThePhoton I don't understand why you picked this out, when the sentence before I wrote "and we ignore things like gravity etc." - surely you could have made a similar remark like "we can't really ignore gravity, there's always some gravity". I wrote all the assumptions just for simplicity - the point is that $A$ isn't really moving, and $B$ will be orbiting around $A$. Would you please suggest a way I could edit the question so it makes sense? $\endgroup$ – John P May 23 at 3:51
  • $\begingroup$ Because you asked, " I'm asking why momentum will be preserved." and the impossibility of an immovable object is directly related to this question, while ignoring gravity isn't related to the question you asked. $\endgroup$ – The Photon May 23 at 3:56
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Suppose also that the rope has the slack pulled out of it, ...

How to justify that "infinitesimal" change of direction doesn't change the momentum?

Basically, by keeping the rope exactly played out with no slack, you're arranging that the force you apply to object B is always perpendicular to B's velocity. Since the force is perpendicular to the velocity, it can't increase the magnitude of the velocity.

I'm asking why momentum will be preserved

The momentum of B isn't preserved. Momentum is a vector quantity, so if its direction changes, then it is changing.

In order for the system of A and B to have momentum conserved, it's actually not possible for A to be an immovable object. For B to change momentum, A must also change momentum, meaning it must wobble a bit. If the mass of A is much bigger than the mass of B, this wobble could be immeasurably small, but it can't be zero except in the limit that the mass of A goes to infinity.

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  • $\begingroup$ I see. What I should have asked was why the magnitude of velocity doesn't change. I see that this can be deduced from $c' \cdot c'' =0$, as that means $(c' \cdot c')' = 0$. I think what I tried to do instead was to avoid any sort of concept of "force" pulling on the object, and just think of what happens if we approximate a circle as a polygon, and so at each sharp angle the magnitude of velocity gets a bit smaller, but as the approximation gets better, the speed changes less and less. This would instead rely on the "fact" that the velocity changes by the cosine of the angle. $\endgroup$ – John P May 23 at 7:19

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