Physical Principle: We analyze the motion of a particle of mass $m$ located on the surface of the cloud. The expansion ceases when the kinetic energy of the surface particles is entirely converted into gravitational potential energy.

1. Initial Energy

Consider a particle of mass $m$ on the surface of the spherical cloud of mass $M$ and radius $R_0$.

• Kinetic Energy ($K_i$): $\frac{1}{2} m v_0^2$
• Potential Energy ($U_i$): Due to the mass $M$ enclosed within the sphere, the potential energy is $- \frac{G M m}{R_0}$.

$$ E_{initial} = \frac{1}{2} m v_0^2 – \frac{G M m}{R_0} $$

2. Final Energy

When the expansion ceases, the velocity of the surface particles momentarily becomes zero. Let the radius at this instant be $R_{max}$.

• Kinetic Energy ($K_f$): $0$
• Potential Energy ($U_f$): $- \frac{G M m}{R_{max}}$

$$ E_{final} = – \frac{G M m}{R_{max}} $$

3. Conservation of Energy

Since the system is in free space, mechanical energy is conserved:

$$ E_{initial} = E_{final} $$ $$ \frac{1}{2} m v_0^2 – \frac{G M m}{R_0} = – \frac{G M m}{R_{max}} $$

Dividing by $m$ and rearranging to solve for $R_{max}$:

$$ \frac{GM}{R_{max}} = \frac{GM}{R_0} – \frac{v_0^2}{2} $$ $$ \frac{GM}{R_{max}} = \frac{2GM – v_0^2 R_0}{2R_0} $$

Inverting both sides:

$$ \frac{R_{max}}{GM} = \frac{2R_0}{2GM – v_0^2 R_0} $$