The mass and radius of the earth and moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a…
The mass and radius of the earth and moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\left(\frac{2 d}{3}\right)$ from the centre of $M_1$ so as to escape to $\propto$ is
If the mass $m$ is at a distance $\frac{2}{3} d$ from the centre of $M_1$ then it is located at a distance $\frac{d}{3}$ from the centre of $M_2$.
To calculate the kinetic energy required to make the body escape Earth and moon system:
the change in gravitational potential energy between final and initial location should be equal to the kinetic energy given to the mass $m$.
$\begin{aligned} & \therefore \frac{1}{2} m v_e^2=\left[\frac{G M_1 m}{\left(\frac{2 d}{3}\right)}+\frac{G M_2 m}{\left(\frac{d}{3}\right)}\right] \\ & \Rightarrow \frac{1}{2} m v_e^2=\frac{3 G\left(M_1+2 M_2\right) m}{2 d}\end{aligned}$
Therefore, escape velocity
$v_e=\sqrt{\frac{3 G\left(M_1+2 M_2\right)}{d}}$