An asteroid of mass $M$ explodes into a spherical homogeneous cloud in free space. Due to energy received by the explosion the cloud expands and the expansion is spherically symmetric. At an instant when the radius of cloud is $R_0$, all the particles on the surface are observed receding radially away from the centre with a speed $v_0$. What will the radius of the cloud be when the expansion ceases?

My doubt: In the solution provided, the initial kinetic energy of the cloud has been equated to the change in self gravitational potential energy of the cloud. While finding the K.E., an elemental shell of radius $r$ has been considered and the speed of the particles at that shell has been written as $v=(r/R_0)v_0$. What is the reason behind writing the velocity in that manner?


Since the cloud is given to be homogeneous, its density has to be uniform throughout, at any point in time. Let $\rho_1$ be the density of the outermost shell and $\rho_2$ be the density of an arbitrary shell of radius $R$

$\rho_1 =\rho_2$

$\therefore \frac {V_1}{m_1}=\frac {V_2}{m_2}$

$\therefore \frac 1{m_1} \frac {dV_1}{dt} =\frac 1{m_2} \frac {dV_2}{dt}$

$\therefore \frac {m_2}{m_1}*R_0^2 \frac {dR_0}{dt}=R^2 \frac {dR}{dt}$

$\therefore \frac {V_2}{V_1}*\frac{R_0^2}{R^2}*v_0=v$

$\therefore v=\frac R{R_0} v_0$

  • $\begingroup$ Thank you very much bro $\endgroup$ – Dylan Rodrigues Jul 5 '20 at 17:09

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