Consider an expanding sphere of instantaneous | vedclass

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(A) The mass $m$ of the sphere is given by $m = \rho \cdot \frac{4}{3} \pi R^3$. Since the total mass $m$ is constant,its derivative with respect to t

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$R$

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(A) The mass $m$ of the sphere is given by $m = \rho \cdot \frac{4}{3} \pi R^3$. Since the total mass $m$ is constant,its derivative with respect to time $t$ is zero: $\frac{dm}{dt} = 0$. Differentiating the mass equation with respect to $t$: $0 = \frac{d\rho}{dt} \cdot \frac{4}{3} \pi R^3 + \rho \cdot 4 \pi R^2 \frac{dR}{dt}$. Dividing by $\rho \cdot \frac{4}{3} \pi R^3$,we get: $0 = \frac{1}{\rho} \frac{d\rho}{dt} + \frac{3}{R} \frac{dR}{dt}$. Rearranging for the velocity of the surface $v = \frac{dR}{dt}$: $\frac{dR}{dt} = -\frac{R}{3} \left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$. Given that the rate of fractional change in density $\left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$ is a constant,let this constant be $k$. Then $v = \frac{dR}{dt} = -\frac{k}{3} R$. Thus,$v \propto R$.

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