A space probe is fired from the Earth towards the Sun. How far from the Earth will the Sun’s gravitational pull on the probe balance the Earth’s pull?
(Mass of the Sun, ${\text{m}}_s=2\times {10}^{30}\text{kg}$, mass of the Earth, ${\text{m}}_e=6\times {10}^{24 }\text{kg}$, distance between the Earth and Sun, $\text{r}=1.5\times {10}^{11}\text{m}$)

We need to find the distance of the probe from the Earth. Let the distance of the probe from the Earth be x. Then the distance between the probe and the Sun will be r-x.

${\text{F}}_{\text{e}}=\text{G}.\frac{\text{m}\times {\text{m}}_{\text{e}}}{{\left(x\right)}^2}$

${\text{F}}_{\text{s}}=\text{G}.\frac{\text{m}\times {\text{m}}_{\text{s}}}{{\left(\text{r}-x\right)}^2}$

${\text{F}}_{\text{s}}={\text{F}}_{\text{e}}$                 

$\cancel{G}.\frac{\cancel{m}\times m_e}{x^2}=\cancel{G}.\frac{\cancel{m}\times m_s}{{\left(r-x\right)}^2}$

$\frac{m_e}{x^2}=\frac{m_s}{{\left(r-x\right)}^2}$

$m_e{\left(r-x\right)}^2=m_s\times x^2$

${\cancel{6}}_3\times \cancel{{10}^{24}}\times {(r-x)}^2=\cancel{2}\times \cancel{{10}^{{30}_6}}\times x^2$

$\frac{3}{{10}^6}=\frac{x^2}{{\left(r-x\right)}^2}$

${\left(\frac{\sqrt{3}}{{10}^3}\right)}^{\cancel{2}}={\left(\frac{x}{r-x}\right)}^{\cancel{2}}$

$\frac{x}{r-x}=\frac{\sqrt{3}}{{10}^3}$

${10}^{3}x=\sqrt{3}r-\sqrt{3}x$

$1000x=\sqrt{3}r-\sqrt{3}x$                 

$1000x+\sqrt{3}x=\sqrt{3}r$                

$x\left(1000+\sqrt{3}\right)=\sqrt{3}r$               

$x=\frac{\sqrt{3}r}{1000+\sqrt{3}}$    

$x=\frac{\sqrt{3}\times 1.5\times {10}^{11}}{1000+\sqrt{3}}$                 

$x=\frac{1.732\times 1.5\times {10}^{11}}{1000+1.732}$  

Since 1.732 is very small as compared to 1000, we can ignore it.            

$x=\frac{2.6\times {10}^{11}}{{10}^3}$      

$x=\text{2}.\text{6}\times \text{1}{0}^{\text{8}}$ m