Terminal velocity of spherical body.

[PrincePhoenix]

In our textbook, the equation for terminal velocity has been derived from Stokes' law and it comes down as follows,

v_t = (mg)/6(pi)(eta)r

(r is the radius of the spherical body)

then, by putting the value of 'm' from m=(rho)V [where V = 4/3 * (pi)r³]

we get,
v_t = 2(rho)gr² / 9 (eta)

So in one equation, v_t is directly related to r², but in the other it is inversely related. Can someone please explain this to me.
Thak you.

 

[xupid]

The equation v_t = (mg)/6(pi)(eta)r gives the relation between the terminal velocity and the radius for a spherical body of constant mass, that is, you are stretching the body by changing r (mass does not change).

However, in v_t = 2(rho)gr² / 9 (eta) the density is constant and hence as you change the radius, mass also changes.

 

[PrincePhoenix]

So these are two equations for different situations?

 

[xupid]

Exactly.

 

[PrincePhoenix]

Thanks for the help.