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.