Question

A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect of earth surface. In such case, its height from the earth surface is:

• A. $32000km$
• B. $36000km$
• C. $6400km$
• D. $4800km$

Answer 1

Answer: (B)

$36000km$

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 $R=6400km=6.4\times {10}^{6}m$

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

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

$⟹$ $r=42400$ $km$

Thus height of satellite above earth surface $h=42400-6400=36000$ $km$