Question

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

Answer 1

$\frac{5GMm_0(\sqrt{5}+1)}{6R}$

Answer 2

The minimum initial kinetic energy ($U_o$) required for the stone to escape the gravitational influence of the two planets and reach infinity with zero velocity is equal to the negative of the potential energy at the point of unstable equilibrium. This point is where the net gravitational force on the stone is zero.

Let the larger planet ($M_1 = 25M$) be at $x=0$ and the smaller planet ($M_2 = 5M$) be at $x=6R$. The potential energy of the stone ($m_0$) at position $x$ is $U(x) = -\frac{G M_1 m_0}{x} - \frac{G M_2 m_0}{6R-x}$.

The condition for zero net force is $\frac{G M_1 m_0}{x^2} = \frac{G M_2 m_0}{(6R-x)^2}$. Substituting the masses: $\frac{25M m_0}{x^2} = \frac{5M m_0}{(6R-x)^2}$, which simplifies to $5(6R-x)^2 = x^2$. Solving for $x$ between the planets gives the unstable equilibrium point $x_{unstable} = \frac{3}{2}(5-\sqrt{5})R$.

The potential energy at this point is $U_{unstable} = -\frac{5 G M m_0 (\sqrt{5}+1)}{6R}$. Therefore, the minimum initial kinetic energy required is $U_{o, min} = -U_{unstable} = \frac{5 G M m_0 (\sqrt{5}+1)}{6R}$.