Question

Imagine a geostationary satellite of earth which is used as an inter continental telecast station. At what height will it have to be established $?$$(\text{Take}G=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$ and mass of earth, $M=6\times {10}^{24}\text{kg})$.

• A. $6.4\times {10}^3\text{m}$
  
• B. $35.94\times {10}^6\text{m}$
  
• C. $\infty$
• D. ${10}^3\text{m}$
  

Answer 1

The correct option is B $35.94\times {10}^6\text{m}$


Time period for a geostationary satellite is $24\text{hour}$, so $T=24\text{hour}$.

We know that time period of a satellite in a circular orbit is given by, $$T^2=\frac{4{\pi }^2(R+h{)}^3}{GM}$$Where,
$M$ is mass of the earth and $R$ is radius of the earth.
$h$ is the height of the satellite above the earth's surface.

Given,
$R=6400\text{km}$
$G=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$ $M=6\times {10}^{24}\text{kg}$.

Substituting the values, we get

$(24\times 60\times 60{)}^2=\frac{4{\pi }^2(64\times {10}^5+h{)}^3}{6.67\times {10}^{-11}\times 6\times {10}^{24}}$

$\Rightarrow h=35786\times {10}^3\text{m}$

$\Rightarrow h=35786\text{km}$

Hence, the above value is approximately equals to that of option (c).

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