The minimum energy required to take a satellite to a height ‘h’ above earth's surface (where, radius of earth= 6.4 x 10³km) is E₁ and kinetic energy required for the satellite to be in a circular orbit at this height is E₂. The value of h for which E₁ and E₂ are equals is (in km)

The energy required for taking a satellite up to a height h from earth’s surface is the difference between the energy at h height and energy at surface, then
$\begin{array}{l}\Rightarrow {\mathrm{E}}_1={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}\\ {\mathrm{E}}_1=-\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}+\mathrm{h}}+\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}}\dots .\left(\mathrm{i}\right)\\ {\mathrm{v}}_0=\sqrt{\frac{{\mathrm{GM}}_{\mathrm{e}}}{\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}}\text{ (where, }{\mathrm{M}}_{\mathrm{e}}=\text{ mass of earth) }\end{array}$

So energy required to perform circular motion
$\begin{array}{l}\Rightarrow {\mathrm{E}}_2=\frac{1}{2}{\mathrm{mv}}_0^2=\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{2\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}\\ {\mathrm{E}}_2=\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{2\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}\end{array}$
According to law of conservation of energy,
$\begin{array}{l}\therefore \frac{-{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}+\mathrm{h}}+\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}}=\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{2\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}\\ \Rightarrow 3{\mathrm{R}}_{\mathrm{e}}=2{\mathrm{R}}_{\mathrm{e}}+2\mathrm{h};\mathrm{h}=\frac{{\mathrm{R}}_{\mathrm{e}}}{2}\end{array}$
As radius of earth, ${\mathrm{R}}_{\mathrm{e}}\approx 6.4\times {10}^3\mathrm{km}$
Hence,   $\mathrm{h}=\frac{6.4\times {10}^3}{2}\mathrm{km}$
Or  $=3.2\times {10}^3\mathrm{km}$