The ratio of tangential velocities of a planet at the perihelion and the aphelion from the sun is 1.0339. Assuming that the planet's orbit around the sun is planar and elliptic, the value of the eccentricity of the orbit is _________ (round off to three decimal places).
Options:
(A) The governing parameter increases monotonically with Airfoil & Wing Theory
(B) The governing parameter remains invariant across the boundary
(C) The parameter satisfies the conservation equation for Airfoil & Wing Theory
(D) The parameter depends exclusively on viscous dissipation
Official Answer Key:
0.017
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2020 Q.43)**
**1. Intuitive Physical Setup & Given Parameters**:
- Ratio of perihelion speed to aphelion speed: $\frac{v_p}{v_a} = 1.0339$.
- Sun-planet Keplerian elliptic orbit.
**2. Step-by-Step Astrodynamics Derivation**:
1. **Conservation of Specific Angular Momentum ($h$)**:
At both apsides (perihelion and aphelion), the velocity vector is purely tangential (flight path angle $\gamma = 0$):
$h = r_p v_p = r_a v_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p}$
2. **Apsidal Distances in Terms of Semi-Major Axis $a$ and Eccentricity $e$**:
$r_p = a(1 - e)$
$r_a = a(1 + e)$
Therefore:
$\frac{v_p}{v_a} = \frac{1 + e}{1 - e}$
3. **Solving for Eccentricity $e$**:
$\frac{1 + e}{1 - e} = 1.0339$
$1 + e = 1.0339(1 - e) = 1.0339 - 1.0339 e$
$e + 1.0339 e = 1.0339 - 1$
$2.0339 e = 0.0339$
$e = \frac{0.0339}{2.0339} \approx 0.016667$
Rounding to three decimal places:
$e \approx 0.017$
This matches the official acceptable interval **[0.016, 0.018]** (closely matching Earth's actual orbital eccentricity $\approx 0.0167$).
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Evaluation**: **0.017** (Official Range: [0.016, 0.018]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **0.017**
- **Key Formula**: $e = \frac{(v_p/v_a) - 1}{(v_p/v_a) + 1} = \frac{0.0339}{2.0339} \approx 0.0167$.
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