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Let orbital radius of moon be $R_0$ and radius of earth be R. mass of earth M mass of moon m, orbital velocity of moon v and its orbital time period be T

For moon resolving around earth

$\large\frac{GMm}{{R_0}^2}=\frac{mv^2}{R_0}$

Gravitational force =centrepetal force

$GM=gR^2$

$\large\frac{gR^2}{R_0}$$=v^2 =\bigg(\large\frac{2 \pi R_0}{T}\bigg)^2$

$\qquad= \large\frac{4 \pi^2 {R_0}^2}{T^2}$

${R_0}^3=\large\frac{gR^2T^2}{4 \pi ^2}$

$R_0=\bigg[\large\frac{g \pi ^2 T^2}{4 \pi ^2}\bigg]^{\large\frac{1}{3}}$

$R_0=\bigg[\large\frac{9.8 (6.4 \times 10^6)^2 (27.3 \times 86400)^2 }{4(3.14)^2}\bigg]^{\Large\frac{1}{3}}$

$\qquad= 3.83 \times 10^8 \;m$

Hence b is the correct answer.

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