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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