Question

Orbital speed of geo-stationary satellite is (take $G=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$ and mass of earth, $M=6\times {10}^{24}\text{kg}$).
[Height of geostationary satellite from center of earth $=36000\text{km}$]

• A. $3.1\text{km/sec}$ from west to east
• B. $8\text{km/sec}$ from west to east
• C. $11.2\text{km/sec}$ from east to west
• D. Zero

Answer 1

The correct option is A $3.1\text{km/sec}$ from west to east

The orbital velocity of a satellite is given by, $$v=\sqrt{\frac{GM}{r}}$$ Where,
$r=$distance of satellite from the centre of earth.
For a geostationary satellite, $r=36000\text{km}=3.6\times {10}^7\text{m}$

Given,
$G=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$
$M=6\times {10}^{24}\text{kg}$

Substituting, we get

$v=\sqrt{\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{3.6\times {10}^7}}\text{m/s}$

$\Rightarrow v=3.1\times {10}^3\text{m/s}$

$\therefore v=3.1\text{km/s}$

Since, geostationary satellite rotates with earth, so its orbital speed will be $3.1\text{km/s}$ from west to east.

Hence, option(c) is the correct answer.

Why this question: It is important to remember that the distance of a geostationary satellite from the center of the earth is equal to $36000\text{km}$.