GATE 2023 AE – Question 63
A circular Earth orbit has period 90 min. Earth radius is 6370 km, mass $5.98\times10^{24}$ kg and G=$6.67\times10^{-11}\ \mathrm{Nm^2/kg^2}$. Satellite altitude (nearest integer km) is _____.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 284
Explanation
For a circular orbit of radius $r$ and period $T$, gravity provides the centripetal force:
$$\frac{GM}{r^2}=r\omega^2\;\Rightarrow\;r^3=GM\left(\frac{T}{2\pi}\right)^2 .$$
**Numbers.** $GM=6.67\times10^{-11}\times5.98\times10^{24}=3.9887\times10^{14}$ and $T=90\times60=5400$ s:
$$\left(\frac{T}{2\pi}\right)^2=\left(\frac{5400}{6.2832}\right)^2=(859.4)^2=7.386\times10^5\ \text{s}^2$$
$$r^3=3.9887\times10^{14}\times7.386\times10^5=2.946\times10^{20}\ \text{m}^3\;\Rightarrow\;r=6.654\times10^6\ \text{m}=6654\ \text{km}.$$
**Altitude:**
$$h=r-R_E=6654-6370=\mathbf{284\ km}.$$