The formula for Stokes law when there is an initial velocity The Stokes law states that if we have a spherical object moving in a homogeneous, viscous fluid with laminar flow (in particular, Reynolds number needs to be far less than $1$), then the drag force $F_d$ is given by
$$F_d=6\pi\eta r v$$
Since $v$ is a function of time $t$, we may rewrite this formula as
$$F_d(t)=6\pi\eta r v(t)$$
And in this case, the ratio $\frac{v}{r^2}\to K$, where $K$ is a constant.
There is another assumption that attracts my attention. The spherical object is released in the fluid (my physics textbook says this), which implies that the initial velocity $u$ of the spherical object is $0$. So I'm curious about the formula when the spherical object has an initial velocity $u\neq 0$.
Question: When there is an initial velocity $u$ on the spherical object such that Reynolds number is not too large, what will the terminal velocity of the object be? (a formula or an approximation)
Here is my attempt to derive the equation:
Intuitively, since there is a non-zero initial velocity, it changes the drag force a little bit. So,
$$F_D(t)=F_d(t)+m_S a(t)=6\pi\eta r v(t)+\rho_S V_S a(t)=6\pi\eta r v(t)+\rho_S V_S \dfrac{dv(t)}{dt}$$
where $\eta$ is the coefficient of viscosity of the fluid.
Now, we have another equality involving the buoyancy force when the terminal velocity is reached,
$$6\pi\eta r v_T+\rho_S \dfrac{4}{3}\pi r^3 \dfrac{dv(t)}{dt}+\rho_f\dfrac{4}{3}\pi r^3 g=\rho_S\dfrac{4}{3}\pi r^3g$$
This nearly gives a complete solution, but there is a $\frac{dv}{dt}$ term that remains unkown. I believe that this can be obtained by analyzing the motion of the spherical object, but I'm unable to solve it...
A potential strategy might be looking at Archimedes' principle and establish a differential equation to solve for $v(t)$.
Since the change of the net velocity must be monotonic (otherwise it doesn't makes sense), I guess that the function $v(t)$ is at least bounded.
Thanks for your help!
 A: You always reach terminal velocity, in some way or another.
Let's discuss it in detail, first from the equation you wrote, which we eventually then solve in full.
You have a spherical object of radius $r$ moving in water in the vertical direction you have the following forces acting on it:

*

*Stokes $F_d=-6\pi\eta rv$ (directed against the velocity)

*Gravity minus archimedes $F_a=\Delta\rho {4\over 3}\pi r^3 g$ (directed towards the bottom if $\Delta\rho=\rho_{sphere}-\rho_{water}>0$ i.e. the sphere is heavier than water).

Then, Newton's equation reads, as you pointed out,
$$ m{dv\over dt}=-6\pi\eta rv + \Delta\rho {4\over 3}\pi r^3 g $$
which is valid from the beginning, not only after reaching terminal velocity!
How do we reach $v_f$ with an initial velocity $v_0$?
Now already at this point we can tell some things.
If at any time $m dv/dt=-6\pi\eta rv + \Delta\rho {4\over 3}\pi r^3 g >0$ then the acceleration ($dv/dt$) is positive and the particle will increase its speed.
Let's assume that we have an initial velocity $v_0$ which is lower than the limit speed and such that $-6\pi\eta rv_0 + \Delta\rho {4\over 3}\pi r^3 g >0$. Gravity pushes the particle down, Stokes' law slows it down, but gravity wins. This causes an increase in the speed $v$, which increases Stokes law and at some point $v$ will be big enough for Stokes law to completely compensate gravity and the the acceleration stops and the speed stays constant, which means that the acceleration stays constantly at 0 and nothing more happens: the sphere falls at a limit velocity $v_f$.
So if we start with $v_0<v_f$, even if $v_0\ne0$, then we end up at the limit speed anyways, we just take a shorter time to get there.
If on the other hand $v_0>v_f$ then  $-6\pi\eta rv_0 + \Delta\rho {4\over 3}\pi r^3 g < 0$ which means that the particle decelerates: Stokes' law is stronger than gravity and pushes towards the top. However, the particle already has some speed towards the bottom so it does not rise, it goes towards the bottom but slows down. At some point, again, because of the deceleration, the particles reaches a speed such that $-6\pi\eta rv_0 + \Delta\rho {4\over 3}\pi r^3 g=0$ and the acceleration is now $0$ which means the particles goes again at constant speed. So again, we end up with a limit speed $v_f$ except we got there not by accelerating but by de-celerating.
So even if you have a starting velocity, you always end up at $v_f$ because that's how Stokes law works: it opposes changes in velocity
($dv/dt\sim -v$) so that the acceleration always points in the opposite direction of the instantaneous speed.
Limit velocity is then given, whatever the initial conditions, by
$dv/dt=0$ i.e.
$$-6\pi\eta rv_f + \Delta\rho {4\over 3}\pi r^3 g=0$$
which yields
$$v_f={\Delta\rho {4\over 3}\pi r^3 g \over 6\pi\eta r}$$
semplified into
$$v_f={\Delta\rho 2 r^2 g \over 9\eta}$$
and you see that, if there is no gravity ($g=0$) the particle simply does not move at terminal velocity, because there is nothing to counter the slowing-down due to Stokes': the particle just stops where it arrived thanks to its initial velocity - and if $g=0$ you also have no Archimedes' force to push you back up.
Also, if $\Delta\rho<0$ (i.e. water is denser than the particle), you get a negative limit velocity as you go down, slow down because of both Archimedes and Stokes' and then go back up thanks to Archimedes which starts winning when the speed is small and Stokes vanishes, and now Stokes changes direction (now pointing downwards as the speed is upwards) and it now opposes Archimedes, so that the particle reaches again terminal velocity but on the way up.
Full solution
But we can also solve the full differential equation!
$$ m{dv\over dt}=-6\pi\eta rv + \Delta\rho {4\over 3}\pi r^3 g $$
Let's simplify it a bit as
$${dv\over dt}=-{1\over \tau} v + {\Delta\rho {4\over 3}\pi r^3 g\over m}$$
where I defined $\tau= {m\over 6\pi\eta r}$ (it is a time, dimensionally) and finally we use $m=\rho_s {4\over 3}\pi r^3$ so that
$${dv\over dt}=-{1\over \tau} v + {\Delta\rho \over \rho_s} g$$
which is a simple differential equation with a general solution
$$v(t)={\Delta\rho \over \rho_s} g\tau + c_1 e^{-{t\over \tau}}$$
and to find $c_1$ we have to impose the initial condition $v(0)=-v_0$ (it points downards, so the minus - while assuming $\Delta\rho>0$ then Archimedes points upward and therefore it is positive [NB feel free to re-derive it with your favorite sign convention!) so that
$$-v_0={\Delta\rho \over \rho_s} g\tau + c_1$$ and we get
$$c_1=-v_0-{\Delta\rho \over \rho_s} g\tau$$
which we can rewrite as, using the definition of $\tau$ and of $m$ in terms of the density, as
$$c_1=-v_0-mg{\Delta\rho\over \rho_s 6\pi\eta r}=-v_0-\rho_s{4\over 3}\pi r^3g{\Delta\rho\over \rho_s 6\pi\eta r}$$
and if you remember the definition of limit speed we used above we get
$$c_1=-v_0-v_f$$ so that the equation for the speed reads
$$v(t)=v_f-(v_0+v_f)e^{-{t\over\tau}}$$
which is $-v_0$ if $t=0$ and $v_f=-v_f$ at infinite time, so that you reach terminal velocity whatever the initial velocity and that terminal velocity points downwards (minus sign). You get all other solutions changing the values of $\Delta\rho$ etc. etc.
Of course, if $|v_0|=|v_f|$ (i.e. you start already at terminal velocity) then $v(t)=-v_f$ constantly.
The only thing that changes is that the closer $v_0$ and $v_f$ are, the shortest time it is necessary to reach terminal velocity.
Here's a quick graph of the solution for a limit velocity of $-1$ , $\tau=1$, and a starting velocity of $0$ (black), -0.5 [lower than $v_f$] (orange) and $-2$ higher than $v_f$ (all in invented units)

