Spaceflight Mechanics - Basic Orbital Maneuvers Quiz(MCQ)

1 .

Which of these orbits has one-third the Earth's radius?

A)

High Earth Orbit

B)

Low Earth Orbit

C)

Medium Earth Orbit

D)

Geosynchronous Orbit

Correct Answer : Low Earth Orbit

Explanation : Low Earth orbits are the orbits with its center as the Earth’s center. It has an altitude of 2000 km which is approximately one-third the Earth’s Radius, which is 6,371 km.

2 .

Which of these orbits has the lowest ?v requirement for a spacecraft to enter the orbit?

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Equatorial Low Earth Orbit

Correct Answer : Equatorial Low Earth Orbit

Explaination : Equatorial Low Earth Orbit has the lowest ?v requirement because the spacecraft is launched due east which leads to least ?v as Earth rotates from east to west with a surface velocity of 1600 km per hour.

3 .

In which orbit is most of the weather satellites, space stations located?

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Geosynchronous Orbit

Correct Answer : Low Earth Orbit

Explanation : Most of the weather satellites and space stations are located in the circular or elliptical low-earth orbit. Post Apollo mission, there have been no spacecraft orbiting beyond low earth orbit.

4 .

In which orbit range does the Van Allen Belts lie?

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Geosynchronous Orbit

Correct Answer : Medium Earth Orbit

Explanation : Van Allen Belts is the belt of high energy protons that lie within the altitude range of 20,000 to 20,650 km which is the Medium Earth orbit. The orbital period of this orbit range is 12 hours.

5 .

Which of these orbits form a highly inclined figure 8 form?

A)

Polar Orbit

B)

Tundra Orbit

C)

Molniya Orbit

D)

Sun-Synchronous Orbit

Correct Answer : Tundra Orbit

Explanation : Tundra orbit is a type of geosynchronous orbit which has a very high inclination of 63.4 degrees with a low eccentricity of range 0.2-0.3. The orbit is of figure 8 form with one of the smaller loops in either of the hemisphere. Compared to Molniya orbit, this has half the orbital time period.

6 .

What motion does the satellite in a high earth orbit follow?

A)

Prograde motion

B)

Retrograde motion

C)

Apparent Prograde motion

D)

Apparent Retrograde motion

Correct Answer : Apparent Retrograde motion

Explanation : At a high earth orbit of an altitude of approximately 35,000 km, the satellite has a apparent retrograde motion. In this the rotation speed of the earth is higher compared to the orbital velocity which leads to Earth’s surface moving westward.

7 .

Which of these orbits is a high eccentric orbit used for maximizing the viewing time of latitudes?

A)

Polar Orbit

B)

Molniya Orbit

C)

Low-Earth Orbit

D)

Sun-synchronous Orbit

Correct Answer : Molniya Orbit

Explanation : Molniya orbit is a highly eccentric medium earth orbit. It has an eccentricity of 0.722 in order to increase the viewing time of the latitudes as it is an extremely elliptical orbit. The orbit takes 12 hours to complete with Earth rotating underneath it.

8 .

In which orbit is the Communications satellite placed?

A)

High Earth Orbit

B)

Medium Earth Orbit

C)

Geosynchronous Orbit

D)

Equatorial Low Earth Orbit

Correct Answer : Geosynchronous Orbit

Explanation : The Communications satellite is always placed in the Geosynchronous orbit, as the satellite’s position remains same in this orbit and the fixed antennas are able to communicate with them.

9 .

In which of these orbits is the Global Positioning System situated?

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Equatorial Low Earth Orbit

Correct Answer : Medium Earth Orbit

Explanation : GPS is placed at an altitude of 20,200 km which is the medium-earth orbit. The orbital time-period of GPS is about 12 hours. Most of the navigation satellites are placed in the medium- earth orbit.

10 .

Which of these is not a part of in-plane orbit changes?

A)

Change of inclination of the orbits

B)

Hohmann transfer

C)

Adjustment of perigee an apogee height

D)

General coplanar transfer between circular orbits

Correct Answer : Change of inclination of the orbits

Explanation : When the satellite is launches, there’s often error in burnout altitude, speed, flight path angle and the exact orbit that was desired is not achieved. This is corrected by applying a small delta-v in the same plane to adjust the orbital parameters. The in-plane corrections include- Hohmann transfer, adjustment of perigee and apogee height, general coplanar transfer between circular orbits.

11 .

How will themajor axis be affected if the velocity at a point in the orbit is doubled leaving r unchanged?

A)

Major-axis becomes half

B)

Major-axis becomes twice

C)

Major-axis is unaffected

D)

Major-axis becomes one-fourth

Correct Answer : Major-axis becomes twice

12 .

Which of these in-plane transfers is used to transfer the satellite between the circular coplanar orbits using an elliptical transfer orbit?

A)

Bi-elliptical transfer

B)

Phasing maneuver

C)

One-tangent transfer

D)

Hohmann transfer

Correct Answer : Hohmann transfer

Explanation : The transfer between two circular coplanar orbits is carries out using Hohmann transfer. It is a way of moving the satellite to a high-altitude orbit from the parking orbit or the other way round. The transfer takes place using a transfer elliptical orbit in the same plane.

13 .

Which of this parameter in an in-plane maneuver does not change?

A)

Direction of velocity

B)

Magnitude of velocity

C)

Direction of angular momentum

D)

Magnitude of angular momentum

Correct Answer : Magnitude of angular momentum

14 .

Which of these orbital transfers is the most energy efficient?

A)

Bi-elliptic transfer

B)

Hohmann transfer

C)

Non-Hohmann transfer

D)

One tangent-burn transfer

Correct Answer : Hohmann transfer

Explanation : Hohmann transfer is considered at the most energy efficient two-impulse transfer. It does not truly exist. In this, both the initial and final orbital planes are coplanar and the change in velocity magnitude is tangential to both the initial and final orbits.

15 .

How many velocity increments are required to carry out the Hohmann transfer?

A)

1

B)

2

C)

3

D)

4

Correct Answer : 2

Explanation : Two velocity increments in total are required to carry out Hohmann transfer. The first velocity increment is done at the inner circular orbit in the direction of the spacecraft. This places it in a higher energy elliptical trajectory. Once reaching the perigee of the elliptical orbit, the second velocity increment is carries out using thrusters to place it in the outer circular orbit.

16 .

What is the velocity change required to perform a Hohmann transfer to a circular orbit at geosynchronous altitude of 35,768 km if the spacecraft is in an initial circular orbit at an altitude of 200 km?

A)

1.874

B)

2.894

C)

3.933

D)

4.784

Correct Answer : 3.933

Explaination : Given values are: r_A = (6,378 + 200)km = 6,578 km (Where radius of Earth of 6,578 km has to be added)
r_B = 35,786 + 6,328 = 42,164 km

17 .

How many burns are required to successfully carry out Bi-Elliptical Hohmann transfer?

A)

1

B)

2

C)

3

D)

4

Correct Answer : 3

Explanation : In order to carry out Bi-Elliptical Hohmann tranfer 3 burns are required. 1^st burn is to convert the circular orbit at r₁ to elliptical orbit with perigee r_p = r₁ and apogee r_a = r_*. The second burn is carried out at the apogee to raise the perigee of elliptical orbit to r₂. Finally at perigee, the orbit is circulized by lowering the perigee to r₂.

18 .

The radius of the elliptical orbit in bi-elliptical hohmann transfer is a free entity.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : r* is a free parameter and can have any value. Usually as r* tends to infinity, the bi-elliptic tranfer becomes more efficient. As r* tends to infinity, time tends to infinity so there is trade-off between time and efficiency.

19 .

What is the shape of the transfer orbit in the Hohmann transfer?

A)

Circular

B)

Elliptical

C)

Parabolic

D)

Hyperbolic

Correct Answer : Elliptical

Explanation : The Hohmann transfer involves tranfer of the object/satellite between two circular orbits using a transfer elliptical orbit. When transfering from a smaller orbit to a larger one, the change in velocity is applied in the direction of motion. Transfer orbits cannot take the shapes of Parabolic or Hyperbolic as they are usually escapy trajectories.

20 .

What is the time period of the transfer orbit in Hohmann transfer?

A)

Same as the period of orbit

B)

Twice of the period of orbit

C)

One-third of the period of orbit

D)

Half of the period of orbit

Correct Answer : Half of the period of orbit

Explanation : The transfer time is half of the period of orbit.

21 .

The Hohmann transfer is carried out for perpendicular elliptic orbits.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : The Hohmann transfer is carried out between two circular co-planer orbits. The tranfer orbit is the elliptical orbit. The co-planer orbits result in optimising energy.

22 .

In which direction is the delta-v applied to change the orientation of the orbital plane?

A)

Parallel to the plane of orbit

B)

Tangent to the plane of orbit

C)

At an angle to the plane of orbit

D)

Perpendicular to the plane of orbit

Correct Answer : Perpendicular 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.

23 .

On applying a delta-v change, which of the properties of the satellite remain unchanged?

A)

Altitude

B)

Flight path angle

C)

Speed and Flight path angle

D)

Speed of the satellite

Correct Answer : Speed and Flight path angle

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

24 .

Which of these orbital changes is required to change from an inclined orbit to an equatorial orbit?

A)

Inclination change

B)

In-plane orbit change

C)

Out-of-plane orbit change

D)

Three-dimensional orbit change

Correct Answer : Out-of-plane 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.

25 .

For a satellite in a Molniya orbit, large impulse is required for orbital plane change.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

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.

26 .

At which location is the inclination change maneuver the most efficient?

A)

Focus

B)

Apogee

C)

Equator

D)

Perigee

Correct Answer : Apogee

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.

27 .

Which of these parameters is changed while performing an orbital inclination change maneuver?

A)

Eccentricity

B)

Inclination

C)

True anomaly

D)

Right of ascension

Correct Answer : Inclination

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.

28 .

In which direction does the delta-v point with respect to the angular momentum vector at the ascending node?

A)

At an angle i

B)

Perpendicular

C)

In the same direction

D)

In the opposite direction

Correct Answer : In the opposite direction

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 h? when it is applied at the ascending node. At descending node, both the vectors are in the same direction.

29 .

What does impulsive maneuver mean?

A)

Rockets with instantaneous bursts

B)

Rocket fire short bursts for required delta-v

C)

Maneuvers which are analyzed by Lambert�s equation

D)

Rocket fires long bursts for long time for the required delta-v

Correct Answer : Rocket fire short bursts for required delta-v

Explanation : Maneuvers are required by the spacecrafts to move from one orbit to another. Impulsive maneuvers are the ones in which the rockets fire in short bursts for small duration to produce the required velocity change. This changes the magnitude and direction of the velocity vector.

30 .

When the change in velocity represents a change in magnitude, what kind of maneuver is it called?

A)

Pumping maneuver

B)

Cranking maneuver

C)

Orbital raising maneuver

D)

Orbital lowering maneuver

Correct Answer : Pumping maneuver

Explanation : Impulsive maneuver is carried out by firing brief on-board rocket motor. This results in a velocity shift of the spacecraft. If Δv reflects a magnitude change, it is referred to as a pumping maneuver.

31 .

When the change in velocity represents a change in direction, what kind of maneuver is it called?

A)

Pumping maneuver

B)

Orbital raising maneuver

C)

Cranking maneuver

D)

Orbital lowering maneuver

Correct Answer : Cranking maneuver

Explanation : Each impulsive maneuver results in a velocity change of the spacecraft. This velocity change yield both magnitude and direction change of the velocity vector. When Δv represents a change in direction it is referred as a cranking maneuver.

32 .

What does non- impulsive maneuver mean?

A)

Rockets with instantaneous bursts

B)

Rocket fire short bursts for required delta-v

C)

Maneuvers which are analyzed by Lambert�s equation

D)

Rocket fires long bursts for long time for the required delta-v

Correct Answer : Rocket fires long bursts for long time for the required delta-v

Explanation : When the thrust applied by the rocket is done over a longer period to obtain the required delta-v, it is known as the non-impulsive maneuver. Unlike impulsive maneuver it does not involve applying impulse over a short period.

33 .

Which of these is not a type of an impulsive maneuver?

A)

Energy change

B)

Simple plane change

C)

Adjustment of perigee and apogee height

D)

Combined change of apsis altitude and plane orientation

Correct Answer : Energy change

Explanation : There are four types of single impulse maneuver-Adjustment of perigee and apogee height, simple rotation of the line of apsides, simple plane change and combined change of apsis altitude with plane orientation. There is no change in energy involved by applying single impulse.