Question

A satellite is to be placed in equatorial geostationary orbit around earth for communication. The height of such a satellite is $[M_E=6\times {10}^{24}kg,R_E=6400km,T=24h,G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}]$

• 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)

$3.57\times {10}^{7}m$

Time period of satellite, $T=\frac{2\pi (R_E+h)}{\sqrt{\frac{G{M}_E}{(R_E+h)}}}=\frac{2\pi {(R_E+h)}^{3/2}}{\sqrt{G{M}_E}}$
Squaring both sides, we get $T^2=\frac{4\pi {(R_E+h)}^3}{G{M}_E}$
${(R_E+h)}^3=\frac{G{M}_E{T}^2}{4{\pi }^2}$
$(R_E+h)={\left(\frac{G{M}_E{T}^2}{4{\pi }^2}\right)}^{1/3}$
or $h={\left(\frac{G{M}_E{T}^2}{4{\pi }^2}\right)}^{1/3}-R_E$
Here, $M_E=6\times {10}^{24}kg$
$R_E=6400km=6400\times {10}^{3}m=6.4\times {10}^{6}m$
$T=24h=24\times 60\times 60s=86400s$
$G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$
On substituting the given values, we get
$h=\left(\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times {\left(86400\right)}^2}{4\times {\left(3.14\right)}^2}\right)-6.4\times {10}^6$
$=4.21\times {10}^7-6.4\times {10}^6=3.57\times {10}^{7}m$