Question

The time period of polar satellites is about (take $g=9.8{\text{ms}}^{-2}$ and radius of the Earth
$R=6.3\times {10}^6\text{m}$),

[Height of a polar satellite above earth's surface will be around $700\text{km}$]

• A. $100\text{min}$
• B. $6\text{hr}$
• C. $24\text{hr}$
• D. $1000\text{min}$

Answer 1

The correct option is D $1000\text{min}$

Time period of a satellite orbiting around the earth is given by,
$$T=2\pi \sqrt{\frac{(R+h{)}^3}{g{R}^2}}$$Where,
$g=$acceleration due to gravity $=9.8{\text{ms}}^{-2}$
$R=$Radius of the Earth$=6.3\times {10}^6\text{m}$,

Height of the polar satellite above earth surface lies in between $500\text{to}900\text{km}$, so the average value,
$h=700\text{km}$

Substituting the values in the above equation,

$$T=2\pi \sqrt{\frac{(6.3\times {10}^6+700\times {10}^3{)}^3}{9.8\times (6.3\times {10}^6{)}^2}}$$
$\Rightarrow T=5894.29\text{sec}$

$\therefore T=\frac{5894.29}{60}=98.3\approx 100\text{min}$

Thus, time period of polar satellites is about $100\text{minutes}$

Hence, option (b) is the correct answer.