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 ${\text{m}}_{\text{s}}=2\times {10}^{30}\text{kg}$ , mass of the earth ${\text{m}}_{\text{e}}=6\times {10}^{24}\text{kg}$ , distance between the earth and sun$\text{r}=1.5\times {10}^{11}\text{m}$ )

let the mass of the space  probe = m

Let us take the gravitational force balance at a distance x from earth.                

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

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

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

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

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

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

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

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

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

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

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

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

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

$x\left(1000+\sqrt{3}\right)=\sqrt{3}r$                $\because \text{r}=1.5\times {10}^{11}$

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

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

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

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

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