# Orbital Mechanics Questions and Answers – Hyperbolic Orbit – Set 2

This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Hyperbolic Orbit – Set 2”.

1. For a hyperbolic trajectory, what is its semi-major axis if its perigee radius is 9,800 km and apogee radius is -49,800 km?
a) 20,000 km
b) 21,000 km
c) 19,000 km
d) 18,000 km

Explanation: Given,
Perigee radius (rp) = 9,800 km
Perigee radius (ra) = -49,800 km
We know, for hyperbolas,
rp = a(e – 1)
ra = -a(e + 1)
where, a and e are semi-major axis and eccentricity, respectively
Dividing equations mentioned above and substituting the given values, we get,
9800/49,800 = (e – 1)/(e + 1)
0.1968(e + 1) = e – 1
0.8032e = 1.1968
e = 1.49
Substituting back into perigee radius equation, we get,
Semi-major axis (a) = rp/(e – 1)
= 9800/(1.49 – 1)
= 20,000 km

2. What is the specific angular momentum of a hyperbolic trajectory if the speed of the spacecraft is 8.8 km/s, its radius is 14,900 km and the flight path angle is 45°?

a) 100,126 km2/s
b) 89,123 km2/s
c) 92,716 km2/s
d) 15,060 km2/s

Explanation: Given,
Speed of spacecraft (v) = 8.8 km/s
Radius of satellite (r) = 14,900 km
Flight path angle (ν) = 45°
Tangential velocity (vT) = vcos(ν)
= 8.8cos(45°)
= 6.22 km/s
Specific angular momentum (h) = rvT
= 14,900*6.22
= 92,716 km2/s

3. Hyperbolic excess velocity is reached when the distance tends to infinity.
a) True
b) False

Explanation: True. Hyperbolic excess velocity is the relative velocity between helio-centered ellipse and earth’s orbit. It is achieved when a spacecraft accelerates to a speed more than escape velocity. Maintaining a proper hyperbolic excess is important to perform inter-planetary transfers.

4. If the escape speed is more than the spacecraft’s speed, the path is a hyperbola.
a) True
b) False

Explanation: False. At anything more than the escape speed a hyperbolic trajectory starts. If the spacecraft’s speed is less than escape velocity, the orbit could be elliptical or circular. If both spacecraft’s speed and escape speed is the same a parabolic trajectory is formed.

5. What is the aiming radius of a hyperbolic trajectory with eccentricity 1.2 and specific energy 199.3 km2/s2? Gravitational parameter is 398,600 km3/s2
a) 789.97 km
b) 532.90 km
c) 487.54 km
d) 663.32 km

Explanation: Given,
Eccentricity (e) = 1.2
Specific Energy (ε) = 199.3 km2/s2
Gravitational parameter (μ) = 398,600 km3/s2
We know,
Semi-major axis (a) = μ/(2ε)
= 398,600/(2*199.3)
= 1000 km
Aiming radius (Δ) = a*(e2 – 1)1/2
= 1,000*(1.22 – 1)1/2
= 663.32 km

6. What is the semi-minor axis of a hyperbola with specific energy of 199.3 km2/s2 and eccentricity of 1.3? Gravitational parameter is 398,600 km3/s2.
a) 920 km
b) 1200 km
c) 830.66 km
d) 1000 km

Explanation: Given,
Eccentricity (e) = 1.3
Specific Energy (ε) = 199.3 km2/s2
Gravitational parameter (μ) = 398,600 km3/s2
We know,
Semi-major axis (a) = μ/(2ε)
= 398,600/(2*199.3)
= 1000 km
Aiming radius (Δ) = a*(e2 – 1)1/2
= 1,000*(1.32 – 1)1/2
= 830.66 km

7. The specific angular momentum of a hyperbolic trajectory is 94,000 km2/s and the altitude of satellite from earth’s surface is 30,850 km. What is the tangential velocity of the satellite trajectory?
a) 9.562 km/s
b) 2.525 km/s
c) 8.221 km/s
d) 10.24 km/s

Explanation: Given,
Satellite altitude from earth (z) = 30,850 km
Specific angular momentum (h) = 94,000 km2/s
Tangential velocity (vT) = h/r
= 94,000/(30,850 + 6378)
= 2.525 km/s

8. What is the characteristic energy of a satellite with an altitude of 1100 km from earth’s surface? Perigee velocity is 16 km/s and eccentricity is 1.2.
a) 100.39 km2/s2
b) 190.39 km2/s2
c) 149.39 km2/s2
d) 129.39 km2/s2

Explanation: Given,
Satellite Altitude (z) = 1,100 km
Perigee Velocity (vp) = 16 km/s
We know,
Gravitational parameter (μ) = 398,600 km3/s2
Satellite radius (r) = 6378 + z
= 7,478 km
Escape velocity (vesc) = (2μ/r)1/2
= (2*398,600/7,478)1/2
= 10.325 km/s
Hyperbolic excess velocity (v) = (vp2 – vesc2)1/2
= (162 – 10.3252)1/2
= 12.22 km/s
Characteristic Energy (C3) = v2
= 12.222
= 149.39 km2/s2

9. What is the true anomaly of the asymptote of a hyperbolic trajectory with eccentricity 1.2?
a) 45°
b) 146.44°
c) 120.36°
d) 180°

Explanation:
True anomaly of the asymptote (θ) = acos (-1/eccentricity)
= acos (-1/1.2)
= 146.44°

10. For a hyperbolic trajectory, what is its eccentricity if perigee radius and apogee radius are 9,750 km and -49,870 km respectively?
a) 1.811
b) 1.445
c) 1.486
d) 1.256

Explanation: Given,
Perigee radius (rp) = 9,750 km
Perigee radius (ra) = -49,870 km
We know, for hyperbolas,
rp = a(e – 1)
ra = -a(e + 1)
where, a and e are semi-major axis and eccentricity, respectively
Dividing equations mentioned above and substituting the given values, we get,
9,750/49,870 = (e – 1)/(e + 1)
0.1955(e + 1) = e – 1
0.80449e = 1.1955
e = 1.486

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