Question

Find min $U_0$ such that stone will reach other planet.

Answer 1

$\sqrt{\frac{15 G M_0}{2R}}$

Answer 2

The total energy of the stone is conserved. For the stone to reach the surface of the second planet, its total energy must be at least equal to the potential energy at the surface of the second planet, assuming zero kinetic energy at arrival.

Initial potential energy: $U_{initial} = -\frac{G M_1 m}{R_1} - \frac{G M_2 m}{D-R_1}$ Given $M_1 = 25M_0$, $R_1 = 2R$, $M_2 = 5M_0$, $R_2 = R$, $D = 6R$, and $m = M_0$. $U_{initial} = -\frac{G (25M_0) M_0}{2R} - \frac{G (5M_0) M_0}{6R-2R} = -\frac{25 G M_0^2}{2R} - \frac{5 G M_0^2}{4R} = -\frac{55 G M_0^2}{4R}$

Final potential energy at the surface of the second planet: $U_{final} = -\frac{G M_1 m}{D-R_2} - \frac{G M_2 m}{R_2}$ $U_{final} = -\frac{G (25M_0) M_0}{6R-R} - \frac{G (5M_0) M_0}{R} = -\frac{25 G M_0^2}{5R} - \frac{5 G M_0^2}{R} = -\frac{5 G M_0^2}{R} - \frac{5 G M_0^2}{R} = -\frac{10 G M_0^2}{R}$

By conservation of energy ($E_{initial} = E_{final}$), with $E_{initial} = \frac{1}{2} M_0 U_0^2 + U_{initial}$ and $E_{final} = U_{final}$ (minimum velocity implies zero kinetic energy at destination): $\frac{1}{2} M_0 U_0^2 + U_{initial} = U_{final}$ $\frac{1}{2} M_0 U_0^2 = U_{final} - U_{initial}$ $\frac{1}{2} M_0 U_0^2 = -\frac{10 G M_0^2}{R} - (-\frac{55 G M_0^2}{4R}) = \frac{55 G M_0^2}{4R} - \frac{40 G M_0^2}{4R} = \frac{15 G M_0^2}{4R}$ $U_0^2 = \frac{2 \times 15 G M_0^2}{4R M_0} = \frac{15 G M_0}{2R}$ $U_0 = \sqrt{\frac{15 G M_0}{2R}}$