# Orbital Mechanics Questions and Answers – Orbital Elements

This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Orbital Elements”.

1. If the satellite has an inclination of 180 degrees, then what is that orbit known as?
a) Hohmann
b) Polar

Explanation: Inclination is the measure of angle between the orbital plane of the satellite and the equatorial plane of the orbit it revolves around. If the inclination angle is 0 degrees, it’s called a prograde orbit. When the inclination angle is 90 degrees, it is a polar orbit and when angle is 180 degrees it is known as retrograde orbit. In this particular orbit, the satellite moves in the opposite direction to the rotation of the planet around which it is orbiting.

2. In total how many orbital elements are defined to find the position and orientation of a spacecraft?
a) 3
b) 6
c) 2
d) 4

Explanation: In total, there are six orbital elements defined. Out of these 2 parameters: eccentricity and angular momentum are used to define the orbit. True anomaly is used to find the location of the object in the orbit. Three additional Euler angles are used to describe the orientation of the orbit. These are- inclination, argument of perigee and right ascension of the ascending node.

3. What is the angular momentum of the satellite with the following details given?
$$\vec r$$ = -6000$$\hat I$$ – 3500$$\hat J$$ + 2000$$\hat K$$
$$\vec v$$ = -3$$\hat I$$ + 6$$\hat J$$ + 2.5$$\hat K$$
a) 26,275 km2/s
b) 38005.9 km2/s
c) 51,708.9 km2/s
d) 58,200.4 km2/s

Explanation: The formula to find the angular momentum is given by:
$$\vec h$$ = $$\vec r$$ × $$\vec v$$ = $$\begin{bmatrix}i ̂& j ̂& k ̂\\-6000 & -3500 & 2000 \\-3 & 6 & 2.5 \end {bmatrix}$$
$$\vec h$$ = (-3500 × 2.5 – 2000 × 6)$$\hat i$$ – (-6000 × 2.5 – 2000 × -3)$$\hat j$$ + (-6000 × 6 – (-3500) × – 3)$$\hat k$$ $$\vec h$$ = -20750$$\hat i$$ + 9000$$\hat j$$ – 46500$$\hat k$$
To find the magnitude of the angular momentum:
h = $$\sqrt{\vec h ∙ \vec h}$$
h = $$\sqrt{(-20750)^2 + (9000)^2 + (-46500)^2}$$ = 51,708.9 km2/s.

4. In which quadrant does the true anomaly lie if $$\vec e$$ ∙ $$\vec r$$ < 0 (where $$\vec e$$ is the eccentricity vector and $$\vec r$$ is the position vector)?

Explanation: The true anomaly is computed using the formula:
θ = cos-1$$(\frac{\vec e ∙ \vec r}{er})$$
If the value of $$\vec e$$ ∙ $$\vec r$$ > 0 then the true anomaly will lie in first or fourth quadrant whereas, if the $$\vec e$$ ∙ $$\vec r$$ < 0 then the true anomaly will lie in second or the third quadrant.

5. For which orbit is the true anomaly and argument of perigee undefined?
a) Hyperbolic
b) Circular
c) Elliptical
d) Parabolic

Explanation: For the circular orbits, the eccentricity is zero and there is no defined perigee. Thus, true anomaly and argument of perigee are not defined for the circular orbits.

6. What does the True Anomaly help in determining?
a) Orbit’s plane’s rotation about the Earth
b) Orbit’s shape
c) Satellite’s location in the orbit
d) Orbit’s orientation

Explanation: True anomaly is the angle measured in the direction of satellite motion, from perigee to the satellite’s location. The value varies from 0° to 360° and it is undetermined when the eccentricity is zero (i.e. for circular orbits).

7. What does the Argument of perigee help in determining?
a) Orbit’s plane’s rotation about the Earth
b) Orbit’s shape
c) Satellite’s location in the orbit
d) Orbit’s orientation in the orbital plane

Explanation: Argument of perigee is the angle measured in the direction of satellite motion, from the ascending node to the perigee. The value varies from 0° to 360° and it is undetermined when i = 0° or 180° or when the eccentricity is zero (i.e. for circular orbits).

8. Determine the orbit’s shape if the position and the velocity vectors of the satellite are given as below?
$$\vec r$$ = -6000$$\hat I$$ – 3500$$\hat J$$ + 2000$$\hat K$$
$$\vec v$$ = -3$$\hat I$$ + 6$$\hat J$$ + 2.5$$\hat K$$
a) Circular
b) Elliptical
c) Parabolic
d) Hyperbolic

Explanation: In order to determine the orbit’s shape, we need to compute the magnitude of eccentricity.
$$\vec e = \frac{1}{\mu}$$[(v2 – $$\frac{\mu}{r}) \vec r – (\vec r$$ ∙ $$\vec v)\vec v$$]
r = $$\sqrt{\vec r ∙ \vec r} = \sqrt{(-6000)^2 + (-3500)^2 + (2000)^2}$$ = 7228.42 km
v = $$\sqrt{\vec v ∙ \vec v} = \sqrt{(-3)^2 + (6)^2 + (2.5)^2}$$ = 7.16 km/s
$$\vec e = \frac{1}{398600}$$ [(7.162 – $$\frac{398600}{7228.42}$$)(-6000$$\hat I$$ -3500$$\hat J$$ + 2000$$\hat K$$) – {(-6000$$\hat I$$ – 3500$$\hat J$$ + 2000$$\hat K$$) ⋅ (-3$$\hat I$$ + 6$$\hat J$$ + 2.5$$\hat K$$)} -3$$\hat I$$ + 6$$\hat J$$ + 2.5$$\hat K$$]
$$\vec e = \frac{1}{398600}$$ [(-3.8778)(-6000$$\hat I$$ – 3500$$\hat J$$ + 2000$$\hat K$$) – 2000(-3$$\hat I$$ + 6$$\hat J$$ + 2.5$$\hat K$$)
$$\vec e = \frac{1}{398600}$$ [(23,266.8 $$\hat I$$ + 13,572.3 $$\hat J$$ – 7,755.6 $$\hat K$$) – (-6000$$\hat I$$ – 3500$$\hat J$$ + 2000$$\hat K$$] $$\vec e = \frac{1}{398600}$$ (17,266.8 $$\hat I$$ + 1,572.3 $$\hat J$$ – 12,755.6 $$\hat K$$)
$$\vec e$$ = (0.0433 $$\hat I$$ + 0.0039 $$\hat J$$ – 0.032 $$\hat K$$)
e = $$\sqrt{\vec e ∙ \vec e} = \sqrt{(0.0433)^2 + (0.0039)^2 + (-0.032)^2}$$ = 0.054
Since 0 < e < 1, it is an elliptical orbit.

9. What is the orbital inclination of International Space Station?
a) 0°
b) 52°
c) 98°
d) 100°

Explanation: The orbital inclination of ISS is 52° and it orbits around the Earth from west to east. 98° is the orbital inclination of Mapping and 0° is for Geostationary satellites.

10. What is the eccentricity of the open orbits?
a) 0 ≤ e ≤ 1
b) 1 ≤ e
c) e = 1
d) e ≤ 0

Explanation: Eccentricity is the ratio of half the foci separation to the semi-major axis. For closed orbits, the eccentricity varies from 0 ≤ e ≤ 1 and foe the open orbits 1 ≤ e. Parabolic and Hyperbolic orbits are considered as open orbits as the body flies off to infinity eventually and does not return to the same angular position.

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