GATE AE 2023 • Q63 NAT • 2 Marks Flight Mechanics & Space Dynamics

Flight Mechanics & Space Dynamics

Space Dynamics & Orbital Mechanics

Subtopic: Kepler's Laws & Two-Body Problem

Question Statement:

A satellite is in a circular orbit around Earth with a time period of 90 minutes. The radius of Earth is 6370 km, mass of Earth is $5.98 \times 10^{24}$ kg and the universal gravitational constant is $6.67 \times 10^{-11}$ N m$^2$/kg$^2$. The altitude of the satellite above mean sea level is ____ km. (round off to the nearest integer)

Official Answer Key:

280.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2023 Q.63)** **1. Intuitive Physical Setup & Given Parameters**: - **Orbital Period**: $T = 90\text{ minutes} = 90 \times 60 = 5400\text{ s}$ - **Earth Radius**: $R_E = 6370\text{ km} = 6.370 \times 10^6\text{ m}$ - **Earth Mass**: $M_E = 5.98 \times 10^{24}\text{ kg}$ - **Gravitational Constant**: $G = 6.67 \times 10^{-11}\text{ N m}^2/\text{kg}^2$ - **Gravitational Parameter**: $\mu = G M_E = 6.67 \times 10^{-11} \times 5.98 \times 10^{24} = 3.98866 \times 10^{14}\text{ m}^3/\text{s}^2$ **2. Step-by-Step Mathematical Derivation**: 1. **Kepler's Third Law for a Circular Orbit**: The orbital period $T$ of a satellite at orbital radius $r$ is: T = 2\pi \sqrt{\frac{r^3}{\mu}} \implies T^2 = \frac{4\pi^2 r^3}{\mu} 2. **Solving for Orbital Radius $r$**: r^3 = \frac{\mu T^2}{4\pi^2} Substituting numerical values: r^3 = \frac{(3.98866 \times 10^{14}) \times (5400)^2}{4\pi^2} = \frac{(3.98866 \times 10^{14}) \times (2.916 \times 10^7)}{39.4784} \approx 2.9463 \times 10^{20}\text{ m}^3 Taking the cube root: r = (2.9463 \times 10^{20})^{1/3} \approx 6.654 \times 10^6\text{ m} = 6654\text{ km} 3. **Altitude Above Mean Sea Level ($h$)**: h = r - R_E = 6654\text{ km} - 6370\text{ km} = 284\text{ km} **3. Comprehensive Options & Distractor Analysis**: - **Calculated Altitude**: $h \approx 284\text{ km}$ (rounds cleanly between 270 km and 290 km, accepted official range [270, 290]). - Common pitfalls include forgetting to convert 90 minutes to seconds, or reporting orbital radius $r \approx 6654\text{ km}$ instead of altitude $h = r - R_E$. **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **280.0** (Acceptable Range: **270 to 290**) - **Key Formula**: $r = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}$ and $h = r - R_E$.
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