Proportionality of Surface Expansion Velocity in Expanding Sphere

To find how the expansion velocity $v$ of any point on the surface of the growing sphere depends on its radius $R$, we analyze the relation between its mass, density, and volume.

The total mass $M$ of a sphere with uniform instantaneous density $\rho$ and instantaneous radius $R$ is given by: $M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right)$

Given that the total mass $M$ remains constant over time, its time derivative is zero: $\frac{dM}{dt} = 0$

Taking the natural logarithm of both sides of the mass equation: $\ln M = \ln \rho + \ln \left( \frac{4}{3}\pi \right) + 3 \ln R$

Differentiating this expression with respect to time $t$: $0 = \frac{1}{\rho} \frac{d\rho}{dt} + 0 + \frac{3}{R} \frac{dR}{dt}$

The speed $v$ of any point located on the boundary (surface) of the expanding sphere is the rate of change of its radius, $v = \frac{dR}{dt}$. Substituting $v$ into the equation yields: $\frac{1}{\rho} \frac{d\rho}{dt} + \frac{3}{R} v = 0$

Rearranging the equation to solve for $v$: $v = -\frac{R}{3} \left( \frac{1}{\rho} \frac{d\rho}{dt} \right)$

It is given that the fractional rate of density change, $\left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$, is constant. Therefore, the expansion speed $v$ is directly proportional to the instantaneous radius $R$: $v \propto R$

Hence, the correct option is A.