This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Orbital Elements – Set 2”.
1. Which of these orbital elements determines the size of the conic orbit?
a) Eccentricity
b) Semi-major axis
c) Inclination
d) Argument of periapsis
View Answer
Explanation: Out of the six Keplerian orbital elements, semi-major axis (a) helps in determining the size of the conic orbit. It is defined as the half of the long axis of the ellipse. The orbital period and energy depend on the orbit size.
2. Which of these orbital elements determines the shape of the conic orbit?
a) Eccentricity
b) Semi-major axis
c) Inclination
d) Argument of periapsis
View Answer
Explanation: Eccentricity helps in determining the shape of the conic orbit. It is defined as the ratio of half the foci separation to the semi-major axis. For closed orbit, eccentricity ranges from 0 to 1 and for open orbits it is greater than 1.
3. What is the relation between the argument of periapsis, argument of latitude at epoch and true anomaly at epoch?
a) u0 = ω – v0
b) u0 = ωv0
c) u0 = ω + v0
d) ω = u0 + v0
View Answer
Explanation: Argument of periapsis ω is the angle formed between the ascending node and periapsis point measured in the direction of the object’s motion. Whereas, argument of latitude u0 at epoch is the angle between the ascending node and radius vector to the satellite at some time t and true anomaly at epoch v0 is the angle between periapsis and the position of satellite at a particular time t called epoch.
They are related by:
u0 = ω + v0
4. How many orbits are present based on the angle of inclination?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: The angle between the earth’s equitorial plane and orbital plane is known as inclination. Based on the different inclination angles, there are four types of orbits present.
• Equitorial orbit: Angle of inclination is 0 or 180 deg.
• Polar orbit: Angle of inclination is 90 deg.
• Prograde orbit: Angle of inclincation is between 0 and 90 deg.
• Retrograde orbit: Angle of inclination is between 90 and 180 deg.
5. As the satellite is orbiting closer to the planet, its velocity increase.
a) True
b) False
View Answer
Explanation: Mean motionhepps in understanding how fast the satellite is oriting around the planet (Earth). It is fiven by the relation:
v = \(\frac{GM}{r}\)
Where, v is satellite’s velocity
M is mass of the earth
G is gravitational constant
r is distance between the center of earth and satellite
Since v and r are inversely porportional, as he satellite gets closer to the earth, its velocity also increases.
6. If the semi-major axis and eccentricity of an orbit are 6,900 km and 0.32 respectively, then what is the approximate value of its semi-minor axis?
a) 6,500 km
b) 6,914 km
c) 6,537 km
d) 5,814 km
View Answer
Explanation: Given, a = 6,900 km, e = 0.32
Semi-minor axis is related to the semi-major axis by the relation:
b = \(\sqrt{a^2 – c^2}\)
Where, c = ae
Substituting the values,
c = 6,900 × 0.32 = 2,208 km
b = \(\sqrt{6,900^2 – 2,208^2}\) = 6,537.181 ≅ 6,537 km
7. The Keplerial orbital elemtns change as a function of time.
a) True
b) False
View Answer
Explanation: Keplerian orbits have a fundamental assumption that the only influenece on the orbit is gravitational force of the attracting body and it ha s aspherical potential field. Due to this the orbital elements do not change as a function of time.
Although in real life, potentional field is not precisely sphrecial and there may be few changes in the orbital elements with time which is often neglected.
8. What is the longitude of ascending node for a satellite having angular momentum 15,000\(\hat i\) – 20,000\(\hat j\) + 18,500\(\hat k\)?
a) 36.86 deg
b) 323.14 deg
c) 54.14 deg
d) 21.14 deg
View Answer
Explanation: Given, \(\vec h\) = 15,000\(\hat i\) – 20,000\(\hat j\) + 18,500\(\hat k\)
The magnitude of angular momentum is given by:
h = \(\sqrt{\vec h ∙ \vec h} = \sqrt{(15,000)^2 + (-20,000)^2 + (18,500)^2}\) = 31,100.643 km2/s
\(\vec N = \hat K × \vec h = \begin{bmatrix}\hat I & \hat J & \hat K \\0 & 0 & 1 \\h_x & h_y & h_z \end {bmatrix} = \begin{bmatrix}\hat I & \hat J & \hat K \\0 & 0 & 1 \\15,000 & -20,000 & 18,500 \end {bmatrix}\)
\(\vec N\) = (0 × 18,500 + 20,000) \(\hat I\) – (0 × 18,500 – 15,000) \(\hat J\) + 0\(\hat K\) = 20,000\(\hat I\) + 15,000\(\hat J\)
N = \(\sqrt{\vec N ∙ \vec N} = \sqrt{20,000^2 + 15,000^2}\) = 25,000 km2/s
Thus, the longitude of ascending node is given by:
Ω = cos-1\(\frac{N_x}{N}\) = cos-1\(\frac{20,000}{25,000}\) = 36.86 deg or 323.14 deg
But from the equation of \(\vec N\) we know it lies in the first quadrant; thus, the answer is: Ω = 36.86 deg.
9. What is epoch referred to in orbital mechanics?
a) Time used as reference for mentioning orbital elements
b) Particular position at time t
c) Angle swept by satellite in time t
d) Eclipse
View Answer
Explanation: Epoch is used to refer to a particular time at the moment when the astronomical quantities such as the orbital elements are defined.
10. What is the value of argument of periapsis when the inclination of the orbital plane is 0 deg?
a) 0
b) 90
c) 360
d) Undefined
View Answer
Explanation: Argument of perigee is the angle measured in the direction of the satellite’s motion from the ascending node to the perigee. When inclination is zero, there are no nodes thus its value is undefined.
Sanfoundry Global Education & Learning Series – Orbital Mechanics.
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