Physics - Gravitation Question with Solution | TestHub

Let X be the distance of the point from the centre of earth where gravitational intensity is zero. Therefore, $\frac{{\mathrm{GM}}_{\mathrm{e}}}{{\mathrm{x}}^2}=\frac{{\mathrm{GM}}_{\mathrm{m}}}{{\left(3.85\times 10^8-\mathrm{x}\right)}^2}$ or $\frac{\mathrm{x}}{3.85\times 10^8-\mathrm{x}}=\sqrt{\frac{{\mathrm{M}}_{\mathrm{e}}}{{\mathrm{M}}_{\mathrm{m}}}}=\sqrt{\frac{5.98\times 10^{24}}{7.35\times 10^{22}}}\cong 9$ or $\frac{x}{9}+x=3.85\times 10^8$ or $\mathrm{x}=9\times 3.85\times 10^8/10=3.46\times 10^8\mathrm{m}$ Distance from Moon $=3.85\times 10^8-3.46\times 10^8$ $=3.9\times 10^7\mathrm{m}\cong 3.85\times 10^7\mathrm{m}$