Physics - Gravitation Question with Solution | TestHub

The distance of the mass $M$ from the centre of the earth is $r_1=\frac{r}{3}$.

The distance of the mass $M$ from the centre of the moon is $r_2=r-\frac{r}{3}=\frac{2r}{3}$.

The gravitational potential energy of the mass $M$ at this position is given by:

$U=-\frac{G{M}_{1}M}{r_1}-\frac{G{M}_{2}M}{r_2}$

Substituting the values of $r_1$ and $r_2$:

$U=-\frac{3G{M}_{1}M}{r}-\frac{3G{M}_{2}M}{2r}$

To escape to infinity, the total mechanical energy of the mass must be zero. By conservation of energy:

$\frac{1}{2}M{v}^2+U=0$

$\frac{1}{2}M{v}^2=\frac{3G{M}_{1}M}{r}+\frac{3G{M}_{2}M}{2r}$

$\frac{1}{2}{v}^2=\frac{3G}{r}\left(M_1+\frac{M_2}{2}\right)$

$v^2=\frac{6G}{r}\left(M_1+\frac{M_2}{2}\right)$

$v={\left[\frac{6G}{r}\left(M_1+\frac{M_2}{2}\right)\right]}^{1/2}$

Answer: ${\left[\frac{6\mathrm{G}}{\mathrm{r}}\left({\mathrm{M}}_1+\frac{{\mathrm{M}}_2}{2}\right)\right]}^{1/2}$