Question

Q Find Min $U_o$ such that stone will reach other planet.

Answer 1

$\sqrt{\frac{5GM_o}{R}} \frac{2\sqrt{5}-1}{\sqrt{6}}$

Answer 2

To find the minimum initial velocity $U_o$, we use the principle of conservation of energy. The stone is launched from the surface of Planet 1 (mass $25M_o$, radius $2R$) towards Planet 2 (mass $5M_o$, radius $R$). The distance between their centers is $6R$.

The point of zero net gravitational force is at a distance $r$ from Planet 1, where: $\frac{G(25M_o)m_0}{r^2} = \frac{G(5M_o)m_0}{(6R-r)^2}$ Solving this yields $r = \frac{3(5-\sqrt{5})}{2}R$. This is the point of maximum potential energy.

Initial potential energy: $V_i = -\frac{G(25M_o)m_0}{2R} - \frac{G(5M_o)m_0}{4R} = -\frac{55GM_om_0}{4R}$ Potential energy at maximum: $V_{max} = -\frac{5GM_om_0}{3R}(3+\sqrt{5})$

For the stone to reach the other planet, its initial total energy must be at least $V_{max}$. The minimum velocity occurs when the stone has zero kinetic energy at this point. $\frac{1}{2}m_0 U_o^2 + V_i = V_{max}$ $\frac{1}{2}m_0 U_o^2 = V_{max} - V_i = -\frac{5GM_om_0}{3R}(3+\sqrt{5}) + \frac{55GM_om_0}{4R}$ $U_o^2 = \frac{2GM_o}{R} \left( -\frac{5}{3}(3+\sqrt{5}) + \frac{55}{4} \right) = \frac{2GM_o}{R} \left( -5 - \frac{5\sqrt{5}}{3} + \frac{55}{4} \right)$ $U_o^2 = \frac{2GM_o}{R} \left( \frac{-60 - 20\sqrt{5} + 165}{12} \right) = \frac{2GM_o}{R} \left( \frac{105 - 20\sqrt{5}}{12} \right)$ $U_o^2 = \frac{GM_o}{R} \left( \frac{105 - 20\sqrt{5}}{6} \right) = \frac{GM_o}{R} \frac{5(21 - 4\sqrt{5})}{6} = \frac{5GM_o}{6R} (2\sqrt{5}-1)^2$ $U_o = \sqrt{\frac{5GM_o}{R}} \frac{2\sqrt{5}-1}{\sqrt{6}}$