GATE AE 2007 • Q18 MCQ • 1 Mark Flight Mechanics & Space Dynamics

Flight Mechanics & Space Dynamics

Space Dynamics & Orbital Mechanics

Subtopic: Kepler's Laws & Two-Body Problem

Question Statement:

The Earth's radius is $6.37 \times 10^6\text{ m}$ and the acceleration due to gravity on its surface is $9.81\text{ m/s}^2$. A satellite is in a circular orbit at an altitude of $6.30 \times 10^5\text{ m}$ above the Earth's surface. The minimum additional speed it needs to escape from the Earth's gravitational field is:

Options:

  • (A) $3.66 \times 10^3\text{ m/s}$
  • (B) $3.12 \times 10^3\text{ m/s}$
  • (C) $4.55 \times 10^3\text{ m/s}$
  • (D) $2.84 \times 10^3\text{ m/s}$

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2007 Q.18)** **1. Given Orbital Parameters**: - Earth radius: $R = 6.37 \times 10^6\text{ m}$ - Gravitational parameter: $\mu = g R^2 = 9.81 \times (6.37 \times 10^6)^2 \approx 3.981 \times 10^{14}\text{ m}^3/\text{s}^2$ - Orbital altitude: $h = 6.30 \times 10^5\text{ m}$ - Orbital radius: $r = R + h = 6.37 \times 10^6 + 0.63 \times 10^6 = 7.00 \times 10^6\text{ m}$ **2. Circular Speed and Escape Speed**: - Circular orbital speed: $$v_{\text{circ}} = \sqrt{\frac{\mu}{r}} = \sqrt{\frac{3.981 \times 10^{14}}{7.00 \times 10^6}} = \sqrt{5.687 \times 10^7} \approx 7541\text{ m/s}$$ - Escape speed at distance $r$: $$v_{\text{esc}} = \sqrt{\frac{2\mu}{r}} = \sqrt{2}\,v_{\text{circ}} = 1.4142 \times 7541 \approx 10665\text{ m/s}$$ **3. Minimum Additional Speed Needed**: $$\Delta v = v_{\text{esc}} - v_{\text{circ}} = (\sqrt{2} - 1) v_{\text{circ}} = 0.4142 \times 7541 \approx 3123\text{ m/s} = 3.12 \times 10^3\text{ m/s}$$ **4. Official Master Key**: - **Official Answer Key**: **B** ($3.12 \times 10^3\text{ m/s}$)
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