# Orbital Mechanics Questions and Answers – Out-of-Plane Orbit Changes

This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Out-of-Plane Orbit Changes”.

1. In which direction is the delta-v applied to change the orientation of the orbital plane?
a) Perpendicular to the plane of orbit
b) Tangent to the plane of orbit
c) Parallel to the plane of orbit
d) At an angle to the plane of orbit

Explanation: The velocity change delta-v component perpendicular to the plane of orbit changes the orientation of the orbital plane. The velocity can change the size, shape or even rotate the line of apsides.

2. On applying a delta-v change, which of the properties of the satellite remain unchanged?
a) Flight path angle
b) Speed of the satellite
c) Speed and Flight path angle
d) Altitude

Explanation: After applying a finite delta-v, the speed and flight-path angle of the satellite are unchanged, then only the plane of the orbit has been altered. This is called a simple plane change

3. Which of these orbital changes is required to change from an inclined orbit to an equatorial orbit?
a) In-plane orbit change
b) Out-of-plane orbit change
c) Inclination change
d) Three-dimensional orbit change

Explanation: Changing an inclined orbit to an equatorial orbit involves change in plane which is an example of a simple plane change. This is a part of out- of-plane change as it involves change in the orientation of the orbital plane.

4. If the object of the plane change is to equatorialize an orbit, the delta-v must be applied at the polar orbit.
a) True
b) False

Explanation: In order to equatorialize an orbit, the delta-v must be applied at one of the nodes of the orbit. Node is the point where the satellite passes through the equatorial plane. It can be either the ascending or the descending node.

5. In which direction does the delta-v point with respect to the angular momentum vector at the ascending node?
a) In the same direction
b) In the opposite direction
c) Perpendicular
d) At an angle i

Explanation: The orbit normal component of the delta-v that is applied to change the orbital plane is always in the opposite direction to the angular momentum vector $$\vec{h}$$ when it is applied at the ascending node. At descending node, both the vectors are in the same direction.

6. If a satellite orbits in a circular orbit at an altitude of 15,000 km at an inclination of 45 deg, then what is the delta-v impulse required to change the orbital plane at an inclination of 52 deg?
a) 0.411 km/s
b) 0.205 km/s
c) 0.187 km/s
d)0.527 km/s

Explanation: Give, i1 = 45 deg, i2 = 52 deg, R = 15,000 km
The formula to compute the delta-v for a single plane:
Δv = 2v sin $$\frac{\Delta i}{2}$$
Where Δi = 52 – 45deg⁡ = 7 deg
v = $$\sqrt{\frac{\mu}{r}} = \sqrt{\frac{398,600}{15,000 + 6378}}$$ = 4.32 km/s(Where radius of earth of 6378 km has to added to get the value of radius)
Δv = 2 × 4.32 sin $$\frac{7}{2}$$ = 0.527$$\frac{km}{s}$$.

7. Which formula is used to compute the delta-v for changing the orbital inclination of an elliptical orbit?
a) Δv = 2v sin $$\frac{\Delta i}{2}$$
b) Δv = 2v cos $$\frac{\Delta i}{2}$$
c) Δv = 2v cosγ sin $$\frac{\Delta i}{2}$$
d) Δv = 2v sinγ sin $$\frac{\Delta i}{2}$$

Explanation: The delta-v for an elliptical orbit change requires an additional term compared to the one for the circular orbit. We define inclination as the angle measured from the equatorial plane to the projection of the velocity vector onto horizonal plane at the nodal crossing. Thus, we have to add the horizonal velocity component of v cosγ.

8. If a satellite in a Molniya orbit having the orbital elements as: a = 25,000 km, e = 0.54, i = 60 deg, Ω = 45 deg and ω = 88 deg has to undergo an orbital plane change at an inclination of 55 deg then what is the delta-v required?
a) 0.200 km/s
b) 0.414 km/s
c) 0.604 km/s
d) 0.301 km/s

Explanation: Given, a = 25,000 km, e = 0.54, i1 = 60 deg, Ω = 45 deg and ω = 88 deg, i2 = 55 deg
In order to find the delta-v we first need to compute the value of velocity and flight-path angle at the nodal crossing of the elliptical orbit. The satellite crosses the ascending node at a true anomaly θ = 90 deg.
The radius of the trajectory is:
r = $$\frac{p}{1 + e cos\theta} = \frac{a(1 – e^2)}{1 + e cos\theta}$$ = $$\frac{25,000(1 – 0.54^2)}{1 + 0.54 × cos90^°}$$ = $$\frac{17,710}{1}$$ = 17,710 km
The velocity at the ascending node is:
v = $$\sqrt{-\frac{\mu}{a} + \frac{2\mu}{r}}$$ = $$\sqrt{-\frac{398,600}{25,000} + \frac{2 × 398,600}{17,710}}$$ = 5.39 km/s
Flight-path angle at the ascending node is given by:
γ = cos1 ($$\frac{h}{rv}$$) = cos1 ($$\frac{\sqrt{\mu p}}{rv})$$ = cos1 ($$\frac{\sqrt{398,600 × 17,710}}{17,710 × 5.39})$$ = 28.336 deg
Δi = 60 – 55deg⁡ = 5 deg
The impulse is calculated using:
Δv = 2v cosγ sin $$\frac{Δi}{2}$$ = 2 × 5.39cos⁡ 28.336 sin $$\frac{5}{2}$$ = 0.414 km/s

9. For a satellite in a Molniya orbit, large impulse is required for orbital plane change.
a) True
b) False

Explanation: For a satellite orbiting in a Molniya orbit, which is a highly elliptical orbit, the speed is high at the nodal crossing. Due to this, even a small inclination change requires a significantly larger impulse as delta-v is directly proportional to the velocity at the node.

10. Which of these parameters is changed while performing an orbital inclination change maneuver?
a) True anomaly
b) Inclination
c) Eccentricity
d) Right of ascension

Explanation: Orbital inclination change is a maneuver carried out to change the inclination of the orbital plane. This requires a change in the orbital velocity vector by applying a small impulse (delta-v) at the orbital node.

11. At which location is the inclination change maneuver the most efficient?
a) Perigee
b) Apogee
c) Equator
d) Focus

Explanation: The maximum efficiency of inclination is achieved at the apogee where the orbital velocity is the lowest thus the delta-v required is the lowest. This results in high efficiency to change the satellite’s orbital plane.

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