Consider an expanding sphere of instantaneous radius $R$ whose total mass remains constant. The expansion is such that the instantaneous density $\rho$ remains uniform throughout the volume. The rate of fractional change in density $\left(\frac{1}{\rho}\frac{d\rho}{dt}\right)$ is constant. The velocity $v$ of any point on the surface of the expanding sphere is proportional to

Step 1: Express the total mass of the sphere.

The total mass $M$ of a uniform sphere of radius $R$ and density $\rho$ is the product of its volume and density.

$$M = \frac{4}{3}\pi R^{3}\rho$$

Step 2: Differentiate the mass equation with respect to time.

Since the total mass $M$ remains constant, its time derivative is zero.

$$\frac{dM}{dt} = 0$$

Applying the product rule to differentiate the mass expression yields:

$$\frac{4}{3}\pi \left( 3R^{2}\frac{dR}{dt}\rho + R^{3}\frac{d\rho}{dt} \right) = 0$$

Step 3: Simplify the differentiated equation.

Divide the equation by the total mass $M = \frac{4}{3}\pi R^{3}\rho$.

$$\frac{3}{R}\frac{dR}{dt} + \frac{1}{\rho}\frac{d\rho}{dt} = 0$$

Step 4: Substitute the velocity and the fractional change in density.

The velocity of a point on the surface is $v = \frac{dR}{dt}$, and the fractional change in density is a constant $k$.

$$\frac{1}{\rho}\frac{d\rho}{dt} = k$$

Substitute these terms into the simplified equation.

$$\frac{3v}{R} + k = 0$$

Step 5: Determine the proportionality of the velocity.

Solve the equation for the velocity $v$.

$$v = -\frac{k}{3}R$$

Since $k$ is a constant, the velocity is directly proportional to the radius.

$$v \propto R$$

Answer: Proportional to $R$