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(D) Let $M_s = 2 	imes 10^{30} \; kg$ be the mass of the Sun and $M_e = 6 	imes 10^{24} \; kg$ be the mass of the Earth. The distance between them i

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$2.59 	imes 10^{8} \; m$

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(D) Let $M_s = 2 	imes 10^{30} \; kg$ be the mass of the Sun and $M_e = 6 	imes 10^{24} \; kg$ be the mass of the Earth. The distance between them is $r = 1.5 	imes 10^{11} \; m$.Let $x$ be the distance from the Earth's centre where the net gravitational force on the rocket of mass $m$ is zero.At this point,the gravitational pull of the Earth equals the gravitational pull of the Sun:$\frac{G M_e m}{x^2} = \frac{G M_s m}{(r - x)^2}$$\frac{M_e}{x^2} = \frac{M_s}{(r - x)^2}$Taking the square root on both sides:$\frac{\sqrt{M_e}}{x} = \frac{\sqrt{M_s}}{r - x}$$\frac{r - x}{x} = \sqrt{\frac{M_s}{M_e}}$$\frac{r}{x} - 1 = \sqrt{\frac{2 	imes 10^{30}}{6 	imes 10^{24}}} = \sqrt{\frac{1}{3} 	imes 10^6} = \frac{1000}{\sqrt{3}} \approx 577.35$$\frac{r}{x} = 578.35$$x = \frac{1.5 	imes 10^{11}}{578.35} \approx 2.59 	imes 10^8 \; m$.

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