Minimum speed of a projectile

Let us assume that there is no rotational motion of the earth or the moon about each other or the sun, and so we are simply modelling a projectile travelling in a straight line between the earth and the moon, both of which are stationary.

If the mass of the earth is $m_1$ , the mass of the moon is $m_2$ , the distance between the earth and the moon is $D$ , then a projectile of mass $m$ will obey the equation of motion $m \ddot{x} \; = \; \frac{Gmm_2}{(D-x)^2} - \frac{Gmm_1}{x^2}$ where $x$ is the distance of the rocket from the centre of the earth. Integrating, we obtain $\tfrac12\dot{x}^2 \; = \; G\left[ \frac{m_2}{D-x} + \frac{m_1}{x} + c\right]$ for some constant $c$ . The minimum value of $\dot{x}^2$ occurs when $\ddot{x}=0$ , which happens when $\frac{m_1}{x^2} = \frac{m_2}{(D-x)^2}$ , or $x \; = \; \frac{D\sqrt{m_1}}{\sqrt{m_1} + \sqrt{m_2}}$ and hence the smallest value of $\dot{x}^2$ is $2G\left[\frac{(\sqrt{m_1} + \sqrt{m_2})^2}{D} + c\right]$ In order to make it all the way to the moon, the speed $\dot{x}$ of the particle must be greater than or equal to $0$ at all times, and hence we must have $c \ge - \frac{(\sqrt{m_1} + \sqrt{m_2})^2}{D}$ If $R$ is the radius of the earth, then the value $v_0$ of $\dot{x}$ at launch that is just sufficient for the projectile to reach the moon is given by $v_0^2 \; = \; 2G\left[\frac{m_2}{D-R} + \frac{m_1}{R} - \frac{(\sqrt{m_1}+\sqrt{m_2})^2}{D}\right]$ With the given values of $m_1,m_2,D,R$ , and using $G = 6.674 \times 10^{-11}$ , we obtain $v_0 = \boxed{11046.2784}\;\mathrm{m}\;\mathrm{s}^{-1}$ .