solution

We use conservation of energy. The initally

$$E ~=~ K + U ~=~ {1 \over 2}mv^2 - {GM_em\over R}$$ (1.27)

At its maximum height of $2R$ from the center of the earth, the kinetic energy is zero so that all the energy is potential. So

$$E ~=~ U ~=~ - {GM_em\over 2R}$$ (1.28)

Equating initial and final energies, we have,

$${1 \over 2}mv^2 ~=~ {GM_e m\over 2R}$$ (1.29)

So

$$v^2 ~=~ {GM_em\over R}$$ (1.30)

In terms of the escape velocity we have from eqn. 1.23 that

$$v^2 ~=~ v_{esc}^2/2$$ (1.31)

or the velocity is $1/\sqrt{2}$ times the escape velocity.