The distance of a point from the earth’s centre where the resultant gravitational force due to earth and the moon is zero is___. (Mass of the earth= $6\times {10}^{24}\text{kg}$, Mass of the moon= $7.4\times {10}^{22}$  kg, The distance between the earth and moon= $4\times {10}^5\text{km}$ )

Given:
Mass of earth $=6\times {10}^{24}\text{kg}=\text{M}$ 

Mass of moon $=\text{7}\text{.4}\times {\text{10}}^{22}\text{kg}=\text{m}$

Distance between the earth & the moon, $\text{r}=\text{4}\times {\text{10}}^5\text{km}$

To find:  
A point A along line joining earth & moon where resultant force= 0

Let the distance of A from earth = x                             

At A force due to earth = force due to moon

$\frac{{\text{GM}}_{\text{e}}\text{m}}{x^2}=\frac{\text{G}{\text{.M}}_{\text{m}}\text{m}}{{(\text{r}-x)}^2}$

$\frac{{\text{M}}_e}{x^2}=\frac{{\text{M}}_{\text{m}}}{{(\text{r}-x)}^2}$

$\therefore \frac{\text{M}}{\text{m}}={\left(\frac{x}{\text{r}-x}\right)}^2$

$\sqrt{\frac{6\times {10}^{24}}{7.4\times {10}^{22}}}=\frac{x}{\text{r}-x}$

$\Rightarrow 9=\frac{x}{\text{r}-x}$                                      

$9r-9x=x$                                        $\because r=4\times {10}^5$km

$\therefore x=\frac{9r}{10}=\frac{9\times 4\times {10}^5}{10}=3.6\times {10}^5\text{km}$

We get $x=3.6\times {10}^5\text{km}$