Question

A satellite is to be placed in equatorial geostationary orbit around earth for communication. The height of such a satellite is.

[$M_{E} = 6 \times 10^{24}kg,R_{E} = 6400km,T = 24hG$

$= 6.67 \times 10^{- 11}Nm^{2}kg^{- 2}$]

• A. $3.57 \times 10^{5}$m
• B. $3.57 \times 10^{6}$m
• C. $3.57 \times 10^{7}$m
• D. $3.57 \times 10^{8}$ m

Answer 1

Answer: (C)

$3.57 \times 10^{7}$m

Answer 2

Time period of satellite.

$T = \frac{2\pi(R_{E} + h)}{\sqrt{\frac{GM_{E}}{(R_{E} + h)}}} = \frac{2\pi(R_{E} + h)^{3/2}}{\sqrt{GM_{E}}}$

Squaring both sides, we get

$T^{2} = \frac{4\pi(R_{E} + h)^{3}}{GM_{E}}$

$(R_{E} + h)^{3} = \frac{GM_{E}T^{2}}{4\pi^{2}}$

$(R_{E} + h) = \left( \frac{GM_{E}T^{2}}{4\pi^{2}} \right)^{1/3}$

Or $h = \left( \frac{GM_{E}T^{2}}{4\pi^{2}} \right)^{1/3} - R_{E}$

Here, $M_{E} = 6 \times 10^{24}kg$

$R_{E} = 6400m = 6400 \times 10^{3}m = 6.4 \times 10^{6}m$

$T = 24h = 24 \times 60 \times 60s = 86400s$

$G = 6.67 \times 10^{- 11}Nm^{2}kg^{- 2}$

On substituting the given values, we get

$h = \left( \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times (86400)^{2}}{4 \times (3.14)^{2}} \right)^{1/3} - 6.4 \times 10^{6} = 4.21 \times 10^{7} - 6.4 \times 10^{6} = 3.57 \times 10^{7}m$