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Kinetic energy at center= Potential energy at surface-potential energy at center

$\large\frac{1}{2}$$mv^2=-\large\frac{GMm}{R}-\bigg(\frac{-3}{2}\frac{GMm}{R}\bigg)$

When $M$-Mass of earth

$\qquad R$-Radius of earth

$\qquad G$- Gravitational constant

$\large\frac{1}{2}$$mv^2=\large\frac{1}{2}\frac{GMm}{R}$

$v^2=\sqrt {\large\frac{GM}{R}}$$=\sqrt {gR}=\large\frac{v_e}{\sqrt 2}$

Hence d is the correct answer.

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