Question

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

Answer 1

$\frac{5GMm_0}{12R} (21 - 4\sqrt{5})$

Answer 2

The minimum initial kinetic energy $U_0$ is found by setting the total initial energy (kinetic + potential) equal to the potential energy at the neutral point, which is the peak of the gravitational potential barrier.

1. Locate the neutral point: The neutral point is where the gravitational forces from the two planets balance. Let $x_1$ be the distance from Planet 1 (mass $25M$) and $x_2$ be the distance from Planet 2 (mass $5M$). The distance between centers is $6R$. $\frac{G(25M)m_0}{x_1^2} = \frac{G(5M)m_0}{x_2^2} \implies \sqrt{5}x_2 = x_1$ Given $x_1 + x_2 = 6R$, we find $x_1 = \frac{3R(5-\sqrt{5})}{2}$ and $x_2 = \frac{3R(\sqrt{5}-1)}{2}$.

2. Calculate potential energy at the neutral point ($U_{neutral}$): $U_{neutral} = -\frac{G(25M)m_0}{x_1} - \frac{G(5M)m_0}{x_2} = -\frac{5GMm_0}{3R}(3+\sqrt{5})$

3. Calculate initial potential energy ($U_{initial}$): The stone starts at the surface of Planet 1 (distance $2R$ from its center) and is $4R$ from the center of Planet 2. $U_{initial} = -\frac{G(25M)m_0}{2R} - \frac{G(5M)m_0}{4R} = -\frac{55GMm_0}{4R}$

4. Find minimum $U_0$: For the stone to reach the neutral point, $U_0 + U_{initial} \ge U_{neutral}$. The minimum $U_0$ occurs when $U_0 + U_{initial} = U_{neutral}$. $U_0 = U_{neutral} - U_{initial} = -\frac{5GMm_0}{3R}(3+\sqrt{5}) - (-\frac{55GMm_0}{4R})$ $U_0 = \frac{GMm_0}{12R} (165 - 20(3+\sqrt{5})) = \frac{5GMm_0}{12R} (21 - 4\sqrt{5})$