The energy required to take a satellite to a height h above the...

${U}_{surface} + {E}_{1} = {U}_{h}$

$KE$ of satellite is zero at earth surface and at height $h$

$- \dfrac{G{M}_{e}m}{{R}_{e}} + {E}_{1} = - \dfrac{G{M}_{e}m}{\left({R}_{e} + h\right)}$

${E}_{1} = G{M}_{e}m\left(\dfrac{1}{{R}_{e}} - \dfrac{1}{{R}_{e} + h}\right)$

${E}_{1} = \dfrac{G{M}_{e}m}{\left({R}_{e} + h\right)} \times \dfrac{h}{{R}_{e}}$

Gravitational attraction ${F}_{G} = m{a}_{C} = \dfrac{m{v}^{2}}{\left({R}_{e} + h\right)}$

$⇒ \dfrac{m{v}^{2}}{\left({R}_{e} + h\right)} = \dfrac{G{M}_{e}m}{{\left({R}_{e} + h\right)}^{2}}$

$m{v}^{2} = \dfrac{G{M}_{e}m}{\left({R}_{e} + h\right)}$

${E}_{2} = \dfrac{m{v}^{2}}{2} = \dfrac{G{M}_{e}m}{2\left({R}_{e} + h\right)}$

${E}_{1} = {E}_{2}$

$\dfrac{h}{{R}_{e}} = \dfrac{1}{2} ⇒ h = \dfrac{{R}_{e}}{2} = 3200km$