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
\(v_e=\sqrt{\frac{3 G\left(M_1+2 M_2\right)}{d}}\)