Using the data given below, find the height at which a communication satellite can reside. \(\left(G=6.67…
Using the data given below, find the height at which a communication satellite can reside. \(\left(G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}, M=5.98 \times 10^{24}\right.\) \(\mathrm{kg}, R=6.4 \times 10^6 \mathrm{~m}\) )
\(35850 \mathrm{~km}\)
\(3585 \mathrm{~km}\)
\(358.5 \mathrm{~km}\)
\(35.85 \mathrm{~km}\)
Solution
Given, \(G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}\)
\(M=5.98 \times 10^{24} \mathrm{~kg}, R=6.4 \times 10^6 \mathrm{~m}\)
For communication satellite,
\(T=24 \mathrm{~h}=24 \times 60 \times 60 \mathrm{~s}=8.64 \times 10^4 \mathrm{~s}\)
We know that, time period of communication satellite is given as
$\begin{aligned}
T & =\frac{2 \pi(R+h)}{\sqrt{\frac{G M}{R_e+h}}}=2 \pi \sqrt{\frac{\left(R_e+h\right)^3}{G M}} \\
\Rightarrow \quad T^2 & =4 \pi^2 \frac{\left(R_e+h\right)^3}{G M} \Rightarrow R_e+h=\left(\frac{T^2 G M}{4 \pi^2}\right)^{1 / 3} \\
\Rightarrow \quad h & =\left(\frac{T^2 G M}{4 \pi^2}\right)^{1 / 3}-R_e \\
& =\left[\frac{8.64 \times 10^4 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{4 \times(3.14)^2}\right]^{1 / 3}-6.4 \times 10^6 \\
& =42.25 \times 10^6-6.4 \times 10^6=35.85 \times 10^6 \, \text{m} \\
& =35850 \times 10^3 \, \text{m}=35850 \, \text{km}
\end{aligned}$