Question

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

Answer 1

$\frac{15}{4} \frac{GMm_0}{R}$

Answer 2

The stone needs to be launched from the surface of the larger planet (mass $25M$, radius $2R$) with sufficient initial kinetic energy $U_0$ to reach the surface of the smaller planet (mass $5M$, radius $R$). The distance between the centers of the planets is $6R$.

1. Initial State: The stone is launched from the surface of the larger planet. Its initial distance from the center of the larger planet is $r_i = 2R$. Its initial distance from the center of the smaller planet is $6R - 2R = 4R$. The initial kinetic energy is $U_0$. The initial potential energy $V_i$ due to both planets is: $V_i = -\frac{G(25M)m_0}{2R} - \frac{G(5M)m_0}{4R} = -\frac{55}{4} \frac{GMm_0}{R}$

2. Final State: The stone reaches the surface of the smaller planet. Its final distance from the center of the smaller planet is $r_f = R$. Its final distance from the center of the larger planet is $6R - R = 5R$. For the minimum initial kinetic energy, the stone reaches the destination with zero final kinetic energy ($K_f = 0$). The final potential energy $V_f$ at the surface of the smaller planet is: $V_f = -\frac{G(25M)m_0}{5R} - \frac{G(5M)m_0}{R} = -10 \frac{GMm_0}{R}$

3. Conservation of Energy: $K_i + V_i = K_f + V_f \implies U_0 + V_i = 0 + V_f \implies U_0 = V_f - V_i$ $U_0 = \left( -10 \frac{GMm_0}{R} \right) - \left( -\frac{55}{4} \frac{GMm_0}{R} \right) = \frac{15}{4} \frac{GMm_0}{R}$