એક ઉપગ્રહ પૃથ્વીની ફરતે વર્તુળાકાર પથ પર એ

As the satellite always remains stationary $w.r.t$ earth surface, thus its time period revolution is equal to time period of rotation of earth i.e $24\, hrs$

Time period of satellite $T=2 \pi \sqrt{\frac{r^{3}}{g R^{2}}}$ where $R=6400 \mathrm{km}=6.4 \times 10^{6} \mathrm{m}$

$\therefore 24 \times 3600=2 \pi \sqrt{\frac{r^{3}}{9.8\left(6.4 \times 10^{6}\right)^{2}}}$

OR $\frac{r^{3}}{401.408 \times 10^{12}}=1.89 \times 10^{8} \quad \Longrightarrow r^{3}=76 \times 10^{21}$

$\Rightarrow r=42400 \mathrm{km}$

Thus height of satellite above earth surface $\quad h=42400-6400=36000 \mathrm{km}$