A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is ${\mathrm{V}}_{\mathrm{e}}$. Its speed with respect to the satellite 

Orbital velocity of the satellite 

$V_0=\sqrt{\frac{GM}{R}}[\text{ Orbit radius }\approx R]$

$V_e=\sqrt{\frac{2GM}{R}}=\sqrt{2}{V}_o$

When the particle was projected from the satellite, with velocity $\overrightarrow{V_r}$ relative to the satellite , then its total velocity $\overrightarrow{V}$ relative to the centre of earth  ${\overrightarrow{V}}_e={\overrightarrow{V}}_0+{\overrightarrow{V}}_r$

$\therefore K.E.\text{ of the particle }=\frac{1}{2}m|\overrightarrow{V}{|}^2$

$\therefore \frac{1}{2}m{\left|{\overrightarrow{V}}_0+{\overrightarrow{V}}_r\right|}^2+\left(-\frac{GMm}{R}\right)=0$

$\therefore \left|{\overrightarrow{V}}_0+{\overrightarrow{V}}_r\right|=\sqrt{\frac{2GM}{R}}=V_e$

Evidently V_r  depends upon direction of projection.