Physics - Gravitation Question with Solution | TestHub

Let$\mathrm{P}$be a point at a distance r from the Earth's centre, where gravitational force due to Sun and Earth are equal and opposite and hence, the gravitational force on the rocket is zero.



$\frac{G{M}_{S}m}{{\left(x-r\right)}^2}=\frac{G{M}_{e}m}{r^2}\Rightarrow \frac{G{M}_S}{{\left(x-r\right)}^2}=\frac{G{M}_e}{r^2}$

$\frac{2\times {10}^{30}}{{\left(x-r\right)}^2}=\frac{6\times {10}^{24}}{r^2}$

$\frac{{\left(x-r\right)}^2}{r^2}=\frac{2\times {10}^{30}}{6\times {10}^{24} }=\frac{1}{3}\times {10}^6$

$\Rightarrow \frac{x-r}{r}=\frac{{10}^3}{\sqrt{3}}\Rightarrow r=\left(\frac{3}{1735}\right)x$

$\Rightarrow r=\frac{3}{1375}\times 1.5\times {10}^{11}=\frac{3\times 15\times 100\times {10}^8}{1735}$

$r=2.594\times {10}^8 \mathrm{m}$

$r=2.6\times {10}^8 \mathrm{m}$from Earth